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== 4.11#h(0.6em)Exercises


    These are homework exercises to accompany Libl's "#link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)")[Differential Equations for Engineering]" Textmap. This is a textbook targeted for a one semester first course on differential equations, aimed at engineering students. Prerequisite for the course is the basic calculus sequence.


    
=== 
      #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/4%3A_Fourier_series_and_PDEs/4.01%3A_Boundary_value_problems")[4.1: Boundary value problems]
    


    Hint for the following exercises: Note that when #math.equation(block: false, alt: "λ greater than 0")[$λ > 0$], then #math.equation(block: false, alt: "cos open parenthesis the square root of λ open parenthesis t minus a close parenthesis close parenthesis")[$cos ( sqrt(λ) ( t − a ) )$] and #math.equation(block: false, alt: "sin open parenthesis the square root of λ open parenthesis t minus a close parenthesis close parenthesis")[$sin ( sqrt(λ) ( t − a ) )$] are also solutions of the homogeneous equation.


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Compute all eigenvalues and eigenfunctions of #math.equation(block: false, alt: "x double prime plus λ x equals 0 , x open parenthesis a close parenthesis equals 0 , x open parenthesis b close parenthesis equals 0")[$x^(″) + λ x = 0 , x ( a ) = 0 , x ( b ) = 0$] (assume #math.equation(block: false, alt: "a less than b")[$a < b$]).
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Compute all eigenvalues and eigenfunctions of #math.equation(block: false, alt: "x double prime plus λ x equals 0 , x prime open parenthesis a close parenthesis equals 0 , x prime open parenthesis b close parenthesis equals 0")[$x^(″) + λ x = 0 , x^(′) ( a ) = 0 , x^(′) ( b ) = 0$] (assume #math.equation(block: false, alt: "a less than b")[$a < b$]).
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Compute all eigenvalues and eigenfunctions of #math.equation(block: false, alt: "x double prime plus λ x equals 0 , x prime open parenthesis a close parenthesis equals 0 , x open parenthesis b close parenthesis equals 0")[$x^(″) + λ x = 0 , x^(′) ( a ) = 0 , x ( b ) = 0$] (assume #math.equation(block: false, alt: "a less than b")[$a < b$]).
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Compute all eigenvalues and eigenfunctions of #math.equation(block: false, alt: "x double prime plus λ x equals 0 , x open parenthesis a close parenthesis equals x open parenthesis b close parenthesis , x prime open parenthesis a close parenthesis equals x prime open parenthesis b close parenthesis")[$x^(″) + λ x = 0 , x ( a ) = x ( b ) , x^(′) ( a ) = x^(′) ( b )$] (assume #math.equation(block: false, alt: "a less than b")[$a < b$]).
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
We have skipped the case of #math.equation(block: false, alt: "λ less than 0")[$λ < 0$] for the boundary value problem #math.equation(block: false, alt: "x double prime plus λ x equals 0 , x open parenthesis minus π close parenthesis equals x open parenthesis π close parenthesis , x prime open parenthesis minus π close parenthesis equals x prime open parenthesis π close parenthesis")[$x^(″) + λ x = 0 , x ( − π ) = x ( π ) , x^(′) ( − π ) = x^(′) ( π )$]. Finish the calculation and show that there are no negative eigenvalues.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Consider a spinning string of length 2 and linear density 0.1 and tension 3. Find smallest angular velocity when the string pops out.


      Answer


      #math.equation(block: true, alt: "ω equals π the square root of the fraction 15 over 2")[$ω = π sqrt(frac(15, 2))$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose #math.equation(block: false, alt: "x double prime plus λ x equals 0")[$x^(″) + λ x = 0$] and #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals 1 , x open parenthesis 1 close parenthesis equals 1")[$x ( 0 ) = 1 , x ( 1 ) = 1$]. Find all #math.equation(block: false, alt: "λ")[$λ$] for which there is more than one solution. Also find the corresponding solutions (only for the eigenvalues).


      Answer


      #math.equation(block: false, alt: "λ sub k equals 4 k squared π squared")[$λ_(k) = 4 k^(2) π^(2)$] for #math.equation(block: false, alt: "k equals 1 , 2 , 3 , …")[$k = 1 , #h(0.222em) 2 , #h(0.222em) 3 , …$] #math.equation(block: false, alt: "x sub k equals cos open parenthesis 2 k π t close parenthesis plus B sin open parenthesis 2 k π t close parenthesis")[$x_(k) = cos ( 2 k π t ) + B sin ( 2 k π t )$] (for any #math.equation(block: false, alt: "B")[$B$])
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose #math.equation(block: false, alt: "x double prime plus x equals 0")[$x^(″) + x = 0$] and #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals 0 , x prime open parenthesis π close parenthesis equals 1")[$x ( 0 ) = 0 , x^(′) ( π ) = 1$]. Find all the solution(s) if any exist.


      Answer


      #math.equation(block: true, alt: "x open parenthesis t close parenthesis equals minus sin open parenthesis t close parenthesis")[$x ( t ) = − sin ( t )$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Consider #math.equation(block: false, alt: "x prime plus λ x equals 0")[$x^(′) + λ x = 0$] and #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals 0 , x open parenthesis 1 close parenthesis equals 0")[$x ( 0 ) = 0 , x ( 1 ) = 0$]. Why does it not have any eigenvalues? Why does any first order equation with two endpoint conditions such as above have no eigenvalues?


      Answer


      General solution is #math.equation(block: false, alt: "x equals C e to the power minus λ t")[$x = C e^(− λ t)$]. Since #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals 0")[$x ( 0 ) = 0$] then #math.equation(block: false, alt: "C equals 0")[$C = 0$], and so #math.equation(block: false, alt: "x open parenthesis t close parenthesis equals 0")[$x ( t ) = 0$]. Therefore, the solution is always identically zero. One condition is always enough to guarantee a unique solution for a first order equation.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[(challenging)]


      Suppose #math.equation(block: false, alt: "x to the power ‴ plus λ x equals 0")[$x^(‴) + λ x = 0$] and #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals 0 , x prime open parenthesis 0 close parenthesis equals 0 , x open parenthesis 1 close parenthesis equals 0")[$x ( 0 ) = 0 , x^(′) ( 0 ) = 0 , x ( 1 ) = 0$]. Suppose that #math.equation(block: false, alt: "λ greater than 0")[$λ > 0$]. Find an equation that all such eigenvalues must satisfy. Hint: Note that #math.equation(block: false, alt: "minus the cube root of λ")[$− root(3, λ)$] is a root of #math.equation(block: false, alt: "r cubed plus λ equals 0")[$r^(3) + λ = 0$].


      Answer


      #math.equation(block: true, alt: "the fraction the square root of 3 over 3 e to the power the fraction minus 3 over 2 the cube root of λ minus the fraction the square root of 3 over 3 cos open parenthesis the fraction the square root of 3 the cube root of λ over 2 close parenthesis plus sin open parenthesis the fraction the square root of 3 the cube root of λ over 2 close parenthesis equals 0")[$frac(sqrt(3), 3) e^(frac(− 3, 2) root(3, λ)) − frac(sqrt(3), 3) cos ( frac(sqrt(3) root(3, λ), 2) ) + sin ( frac(sqrt(3) root(3, λ), 2) ) = 0$]
]


    
=== 
      #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/4%3A_Fourier_series_and_PDEs/4.02%3A_The_trigonometric_series")[4.2: The Trigonometric Series]
    


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] is defined on #math.equation(block: false, alt: "[ minus π , π ]")[$[ − π , π ]$] as #math.equation(block: false, alt: "sin open parenthesis 5 t close parenthesis plus cos open parenthesis 3 t close parenthesis")[$sin ( 5 t ) + cos ( 3 t )$]. Extend periodically and compute the Fourier series of #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] is defined on #math.equation(block: false, alt: "[ minus π , π ]")[$[ − π , π ]$] as #math.equation(block: false, alt: "| t |")[$| t |$]. Extend periodically and compute the Fourier series of #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] is defined on #math.equation(block: false, alt: "[ minus π , π ]")[$[ − π , π ]$] as #math.equation(block: false, alt: "| t | cubed")[$| t |^(3)$]. Extend periodically and compute the Fourier series of #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] is defined on #math.equation(block: false, alt: "open parenthesis minus π , π ]")[$( − π , π ]$] as


      #math.equation(block: true, alt: "minus 1 i f minus π less than t less than or equal to 0 ,; 1 i f 0 less than t less than or equal to π .")[$− 1 " " " " " " " " " " italic(i) italic(f) " " − π < t ≤ 0 , \ 1 " " italic(i) italic(f) " " 0 < t ≤ π .$]


      Extend periodically and compute the Fourier series of #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] is defined on #math.equation(block: false, alt: "open parenthesis minus π , π ]")[$( − π , π ]$] as #math.equation(block: false, alt: "t cubed")[$t^(3)$]. Extend periodically and compute the Fourier series of #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] is defined on #math.equation(block: false, alt: "[ minus π , π ]")[$[ − π , π ]$] as #math.equation(block: false, alt: "t squared")[$t^(2)$]. Extend periodically and compute the Fourier series of #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$].
]


    There is another form of the Fourier series using complex exponentials #math.equation(block: false, alt: "e to the power n t")[$e^(n t)$] for #math.equation(block: false, alt: "n equals … , minus 2 , minus 1 , 0 , 1 , 2 , …")[$n = … , − 2 , #h(0.222em) − 1 , #h(0.222em) 0 , #h(0.222em) 1 , #h(0.222em) 2 , …$] instead of #math.equation(block: false, alt: "cos open parenthesis n t close parenthesis")[$cos ( n t )$] and #math.equation(block: false, alt: "sin open parenthesis n t close parenthesis")[$sin ( n t )$] for positive #math.equation(block: false, alt: "n")[$n$]. This form may be easier to work with sometimes. It is certainly more compact to write, and there is only one formula for the coefficients. On the downside, the coefficients are complex numbers.


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let


      #math.equation(block: true, alt: "f open parenthesis t close parenthesis equals the fraction a sub 0 over 2 plus ∑ n equals 1 infinity a sub n cos open parenthesis n t close parenthesis plus b sub n sin open parenthesis n t close parenthesis .")[$f ( t ) = frac(a_(0), 2) + ∑_(n = 1)^(∞) a_(n) cos ( n t ) + b_(n) sin ( n t ) .$]


      Use Euler’s formula #math.equation(block: false, alt: "e to the power i θ equals cos open parenthesis θ close parenthesis plus i sin open parenthesis θ close parenthesis")[$e^(i θ) = cos ( θ ) + i sin ( θ )$] to show that there exist complex numbers #math.equation(block: false, alt: "c sub m")[$c_(m)$] such that


      #math.equation(block: true, alt: "f open parenthesis t close parenthesis equals ∑ m equals minus infinity infinity c sub m e to the power i m t .")[$f ( t ) = ∑_(m = − ∞)^(∞) c_(m) e^(i m t) .$]


      Note that the sum now ranges over all the integers including negative ones. Do not worry about convergence in this calculation. Hint: It may be better to start from the complex exponential form and write the series as


      #math.equation(block: true, alt: "c sub 0 plus ∑ m equals 1 infinity c sub m e to the power i m t plus c sub minus m e to the power minus i m t .")[$c_(0) + ∑_(m = 1)^(∞) c_(m) e^(i m t) + c_(− m) e^(− i m t) .$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] is defined on #math.equation(block: false, alt: "[ minus π , π ]")[$[ − π , π ]$] as #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals sin open parenthesis t close parenthesis")[$f ( t ) = sin ( t )$]. Extend periodically and compute the Fourier series.


      Answer


      #math.equation(block: true, alt: "sin open parenthesis t close parenthesis")[$sin ( t )$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] is defined on #math.equation(block: false, alt: "open parenthesis minus π , π ]")[$( − π , π ]$] as #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals sin open parenthesis π t close parenthesis")[$f ( t ) = sin ( π t )$]. Extend periodically and compute the Fourier series.


      Answer


      #math.equation(block: true, alt: "∑ n equals 1 infinity the fraction open parenthesis π minus n close parenthesis sin open parenthesis π n plus π squared close parenthesis plus open parenthesis π plus n close parenthesis sin open parenthesis π n minus π squared close parenthesis over π n squared minus π cubed sin open parenthesis n t close parenthesis")[$∑_(n = 1)^(∞) frac(( π − n ) sin ( π n + π^(2) ) + ( π + n ) sin ( π n − π^(2) ), π n^(2) − π^(3)) sin ( n t )$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] is defined on #math.equation(block: false, alt: "open parenthesis minus π , π ]")[$( − π , π ]$] as #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals sin squared open parenthesis t close parenthesis")[$f ( t ) = sin^(2) ( t )$]. Extend periodically and compute the Fourier series.


      Answer


      #math.equation(block: true, alt: "the fraction 1 over 2 minus the fraction 1 over 2 cos open parenthesis 2 t close parenthesis")[$frac(1, 2) − frac(1, 2) cos ( 2 t )$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] is defined on #math.equation(block: false, alt: "open parenthesis minus π , π ]")[$( − π , π ]$] as #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals t to the power 4")[$f ( t ) = t^(4)$]. Extend periodically and compute the Fourier series.


      Answer


      #math.equation(block: true, alt: "the fraction π to the power 4 over 5 plus ∑ n equals 1 infinity the fraction open parenthesis minus 1 close parenthesis to the power n open parenthesis 8 π squared n squared minus 48 close parenthesis over n to the power 4 cos open parenthesis n t close parenthesis")[$frac(π^(4), 5) + ∑_(n = 1)^(∞) frac(( − 1 )^(n) ( 8 π^(2) n^(2) − 48 ), n^(4)) cos ( n t )$]
]


    
=== 
      #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/4%3A_Fourier_series_and_PDEs/4.03%3A_More_on_the_Fourier_series")[4.3: More on the Fourier series]
    


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: true, alt: "0 i f minus 1 less than t less than or equal to 0 ,; t i f 0 less than t less than or equal to 1 ,")[$0 italic(i) italic(f) − 1 < italic(t) ≤ 0 , \ t italic(i) italic(f) 0 < italic(t) ≤ 1 ,$] extended periodically.


      + Compute the Fourier series for #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$].
+ Write out the series explicitly up to the #math.equation(block: false, alt: "3 to the power rd")[$3^("rd")$] harmonic.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: true, alt: "minus t i f minus 1 less than t less than or equal to 0 ,; t squared i f 0 less than t less than or equal to 1 ,")[$− t italic(i) italic(f) − 1 < italic(t) ≤ 0 , \ t^(2) italic(i) italic(f) 0 < italic(t) ≤ 1 ,$] extended periodically.


      + Compute the Fourier series for #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$].
+ Write out the series explicitly up to the #math.equation(block: false, alt: "3 to the power rd")[$3^("rd")$] harmonic.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: true, alt: "the fraction minus t over 10 i f minus 10 less than t less than or equal to 0 ,; the fraction t over 10 i f 0 less than t less than or equal to 10 ,")[$frac(− t, 10) italic(i) italic(f) − 10 < italic(t) ≤ 0 , \ frac(t, 10) italic(i) italic(f) 0 < italic(t) ≤ 10 ,$] extended periodically (period is 20).


      + Compute the Fourier series for #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$].
+ Write out the series explicitly up to the #math.equation(block: false, alt: "3 to the power r d")[$3^(italic(r) italic(d))$] harmonic.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals ∑ n equals 1 infinity the fraction 1 over n cubed cos open parenthesis n t close parenthesis")[$f ( t ) = ∑_(n = 1)^(∞) frac(1, n^(3)) cos ( n t )$]. Is #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] continuous and differentiable everywhere? Find the derivative (if it exists everywhere) or justify why #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] is not differentiable everywhere.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals ∑ n equals 1 infinity the fraction open parenthesis minus 1 close parenthesis to the power n over n sin open parenthesis n t close parenthesis")[$f ( t ) = ∑_(n = 1)^(∞) frac(( − 1 )^(n), n) sin ( n t )$]. Is #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] differentiable everywhere? Find the derivative (if it exists everywhere) or justify why #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] is not differentiable everywhere.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: true, alt: "0 i f minus 2 less than t less than or equal to 0 ,; t i f 0 less than t less than or equal to 1 ,; minus t plus 2 i f 1 less than t less than or equal to 2 ,")[$0 " " " " " " italic(i) italic(f) " " − 2 < italic(t) ≤ 0 , \ t italic(i) italic(f) " " 0 < italic(t) ≤ 1 , \ − t + 2 italic(i) italic(f) " " 1 < italic(t) ≤ 2 ,$] extended periodically.


      + Compute the Fourier series for #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$].
+ Write out the series explicitly up to the #math.equation(block: false, alt: "3 to the power rd")[$3^("rd")$] harmonic.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: true, alt: "f open parenthesis t close parenthesis equals e to the power t f o r minus 1 less than t less than or equal to 1")[$f ( t ) = e^(t) " " " " " " " " " " italic(f) italic(o) italic(r) " " − 1 < italic(t) ≤ 1$] extended periodically.


      + Compute the Fourier series for #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$].
+ Write out the series explicitly up to the #math.equation(block: false, alt: "3 to the power rd")[$3^("rd")$] harmonic.
+ What does the series converge to at #math.equation(block: false, alt: "t equals 1")[$t = 1$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: true, alt: "f open parenthesis t close parenthesis equals t squared f o r minus 1 less than t less than or equal to 1")[$f ( t ) = t^(2) " " " " " " " " " " italic(f) italic(o) italic(r) " " − 1 < italic(t) ≤ 1$] extended periodically.


      + Compute the Fourier series for #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$].
+ By plugging in #math.equation(block: false, alt: "t equals 0")[$t = 0$], evaluate #math.equation(block: false, alt: "∑ n equals 1 infinity the fraction open parenthesis minus 1 close parenthesis to the power n over n squared equals 1 minus the fraction 1 over 4 plus the fraction 1 over 9 minus ⋯ .")[$∑_(n = 1)^(∞) frac(( − 1 )^(n), n^(2)) = 1 − frac(1, 4) + frac(1, 9) − ⋯ .$]
+ Now evaluate #math.equation(block: false, alt: "∑ n equals 1 infinity the fraction 1 over n squared equals 1 plus the fraction 1 over 4 plus the fraction 1 over 9 plus ⋯ .")[$∑_(n = 1)^(∞) frac(1, n^(2)) = 1 + frac(1, 4) + frac(1, 9) + ⋯ .$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: true, alt: "0 if minus 3 less than t less than or equal to 0 ,; t if 0 less than t less than or equal to 3 ,")[$0 "if" − 3 < t ≤ 0 , \ t "if" 0 < t ≤ 3 ,$] extended periodically. Suppose #math.equation(block: false, alt: "F open parenthesis t close parenthesis")[$F ( t )$] is the function given by the Fourier series of #math.equation(block: false, alt: "f")[$f$]. Without computing the Fourier series evaluate.


      + #math.equation(block: false, alt: "F open parenthesis 2 close parenthesis")[$F ( 2 )$]
+ #math.equation(block: false, alt: "F open parenthesis minus 2 close parenthesis")[$F ( − 2 )$]
+ #math.equation(block: false, alt: "F open parenthesis 4 close parenthesis")[$F ( 4 )$]
+ #math.equation(block: false, alt: "F open parenthesis minus 4 close parenthesis")[$F ( − 4 )$]
+ #math.equation(block: false, alt: "F open parenthesis 3 close parenthesis")[$F ( 3 )$]
+ #math.equation(block: false, alt: "F open parenthesis minus 9 close parenthesis")[$F ( − 9 )$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: true, alt: "f open parenthesis t close parenthesis equals t squared f o r minus 2 less than t less than or equal to 2")[$f ( t ) = t^(2) " " " " " " " " " " italic(f) italic(o) italic(r) " " − 2 < italic(t) ≤ 2$] extended periodically.


      + Compute the Fourier series for #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$].
+ Write out the series explicitly up to the #math.equation(block: false, alt: "3 to the power rd")[$3^("rd")$] harmonic.


      Answer


      + #math.equation(block: false, alt: "the fraction 8 over 6 plus ∑ n equals 1 infinity the fraction 16 open parenthesis minus 1 close parenthesis to the power n over π squared n squared cos open parenthesis the fraction n π over 2 t close parenthesis")[$frac(8, 6) + ∑_(n = 1)^(∞) frac(16 ( − 1 )^(n), π^(2) n^(2)) cos ( frac(n π, 2) t )$]
+ #math.equation(block: false, alt: "the fraction 8 over 6 minus the fraction 16 over π squared cos open parenthesis the fraction π over 2 t close parenthesis plus the fraction 4 over π squared cos open parenthesis π t close parenthesis minus the fraction 16 over 9 π squared cos open parenthesis the fraction 3 π over 2 t close parenthesis plus ⋯")[$frac(8, 6) − frac(16, π^(2)) cos ( frac(π, 2) t ) + frac(4, π^(2)) cos ( π t ) − frac(16, 9 π^(2)) cos ( frac(3 π, 2) t ) + ⋯$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: true, alt: "f open parenthesis t close parenthesis equals t f o r λ less than t less than or equal to λ open parenthesis f o r s o m e λ close parenthesis")[$f ( t ) = t " " " " " " " " " " italic(f) italic(o) italic(r) " " italic(λ) < italic(t) ≤ italic(λ) " " ( italic(f) italic(o) italic(r) " " italic(s) italic(o) italic(m) italic(e) " " italic(λ) )$] extended periodically.


      + Compute the Fourier series for #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$].
+ Write out the series explicitly up to the #math.equation(block: false, alt: "3 to the power rd")[$3^("rd")$] harmonic.


      Answer


      + #math.equation(block: false, alt: "∑ n equals 1 infinity the fraction open parenthesis minus 1 close parenthesis to the power n plus 1 2 λ over n π sin open parenthesis the fraction n π over λ t close parenthesis")[$∑_(n = 1)^(∞) frac(( − 1 )^(n + 1) 2 λ, n π) sin ( frac(n π, λ) t )$]
+ #math.equation(block: false, alt: "the fraction 2 λ over π sin open parenthesis the fraction π over λ t close parenthesis minus the fraction λ over π sin open parenthesis the fraction 2 π over λ t close parenthesis plus the fraction 2 λ over 3 π sin open parenthesis the fraction 3 π over λ t close parenthesis minus ⋯")[$frac(2 λ, π) sin ( frac(π, λ) t ) − frac(λ, π) sin ( frac(2 π, λ) t ) + frac(2 λ, 3 π) sin ( frac(3 π, λ) t ) − ⋯$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: true, alt: "f open parenthesis t close parenthesis equals the fraction 1 over 2 plus ∑ n equals 1 infinity the fraction 1 over n open parenthesis n squared plus 1 close parenthesis sin open parenthesis n π t close parenthesis .")[$f ( t ) = frac(1, 2) + ∑_(n = 1)^(∞) frac(1, n ( n^(2) + 1 )) sin ( n π t ) .$] Compute #math.equation(block: false, alt: "f prime open parenthesis t close parenthesis")[$f^(′) ( t )$].


      Answer


      #math.equation(block: true, alt: "f prime open parenthesis t close parenthesis equals ∑ n equals 1 infinity the fraction π over n squared plus 1 cos open parenthesis n π t close parenthesis")[$f^(′) ( t ) = ∑_(n = 1)^(∞) frac(π, n^(2) + 1) cos ( n π t )$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: true, alt: "f open parenthesis t close parenthesis equals the fraction 1 over 2 plus ∑ n equals 1 infinity the fraction 1 over n cubed close parenthesis cos open parenthesis n t close parenthesis .")[$f \( t \) = frac(1, 2) + ∑_(n = 1)^(∞) frac(1, n^(3) \)) cos \( n t \) .$]


      + Find the antiderivative.
+ Is the antiderivative periodic?


      Answer


      + #math.equation(block: false, alt: "F open parenthesis t close parenthesis equals the fraction t over 2 plus C plus ∑ n equals 1 infinity the fraction 1 over n to the power 4 sin open parenthesis n t close parenthesis")[$F ( t ) = frac(t, 2) + C + ∑_(n = 1)^(∞) frac(1, n^(4)) sin ( n t )$]
+ no
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: true, alt: "f open parenthesis t close parenthesis equals the fraction t over 2 f o r minus π less than t less than or equal to π")[$f ( t ) = frac(t, 2) " " " " " " " " " " italic(f) italic(o) italic(r) " " − italic(π) < italic(t) ≤ italic(π)$] extended periodically.


      + Compute the Fourier series for #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$].
+ Plug in #math.equation(block: false, alt: "t equals the fraction π over 2")[$t = frac(π, 2)$] to find a series representation for #math.equation(block: false, alt: "the fraction π over 4")[$frac(π, 4)$].
+ Using the first 4 terms of the result from part b) approximate #math.equation(block: false, alt: "the fraction π over 4")[$frac(π, 4)$].


      Answer


      + #math.equation(block: false, alt: "∑ n equals 1 infinity the fraction open parenthesis minus 1 close parenthesis to the power n plus 1 over n sin open parenthesis n t close parenthesis")[$∑_(n = 1)^(∞) frac(( − 1 )^(n + 1), n) sin ( n t )$]
+ #math.equation(block: false, alt: "f")[$f$] is continuous at #math.equation(block: false, alt: "t equals the fraction π over 2")[$t = frac(π, 2)$] so the Fourier series converges to #math.equation(block: false, alt: "f open parenthesis the fraction π over 2 close parenthesis equals the fraction π over 4")[$f ( frac(π, 2) ) = frac(π, 4)$]. Obtain #math.equation(block: false, alt: "the fraction π over 4 equals ∑ n equals 1 infinity the fraction open parenthesis minus 1 close parenthesis to the power n plus 1 over 2 n minus 1 equals 1 minus the fraction 1 over 3 plus the fraction 1 over 5 minus the fraction 1 over 7 plus ⋯")[$frac(π, 4) = ∑_(n = 1)^(∞) frac(( − 1 )^(n + 1), 2 n − 1) = 1 − frac(1, 3) + frac(1, 5) − frac(1, 7) + ⋯$].
+ Using the first #math.equation(block: false, alt: "4")[$4$] terms get #math.equation(block: false, alt: "76 / 105 approximately equals 0.72")[$76 / 105 ≈ 0.72$] (quite a bad approximation, you would have to take about #math.equation(block: false, alt: "50")[$50$] terms to start to get to within #math.equation(block: false, alt: "0.01")[$0.01$] of #math.equation(block: false, alt: "the fraction π over 4")[$frac(π, 4)$]).
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: true, alt: "0 if minus 2 less than t less than or equal to 0 ,; 2 if 0 less than t less than or equal to 2 ,")[$0 "if" − 2 < t ≤ 0 , \ 2 "if" 0 < t ≤ 2 ,$] extended periodically. Suppose #math.equation(block: false, alt: "F open parenthesis t close parenthesis")[$F ( t )$] is the function given by the Fourier series of #math.equation(block: false, alt: "f")[$f$]. Without computing the Fourier series evaluate.


      + #math.equation(block: false, alt: "F open parenthesis 0 close parenthesis")[$F ( 0 )$]
+ #math.equation(block: false, alt: "F open parenthesis minus 1 close parenthesis")[$F ( − 1 )$]
+ #math.equation(block: false, alt: "F open parenthesis 1 close parenthesis")[$F ( 1 )$]
+ #math.equation(block: false, alt: "F open parenthesis minus 2 close parenthesis")[$F ( − 2 )$]
+ #math.equation(block: false, alt: "F open parenthesis 4 close parenthesis")[$F ( 4 )$]
+ #math.equation(block: false, alt: "F open parenthesis minus 8 close parenthesis")[$F ( − 8 )$]


      Answer


      + #math.equation(block: false, alt: "F open parenthesis 0 close parenthesis equals 1")[$F ( 0 ) = 1$]
+ #math.equation(block: false, alt: "F open parenthesis minus 1 close parenthesis equals 0")[$F ( − 1 ) = 0$]
+ #math.equation(block: false, alt: "F open parenthesis 1 close parenthesis equals 2")[$F ( 1 ) = 2$]
+ #math.equation(block: false, alt: "F open parenthesis minus 2 close parenthesis equals 1")[$F ( − 2 ) = 1$]
+ #math.equation(block: false, alt: "F open parenthesis 4 close parenthesis equals 1")[$F ( 4 ) = 1$]
+ #math.equation(block: false, alt: "F open parenthesis minus 9 close parenthesis equals 0")[$F ( − 9 ) = 0$]
]


    
=== 
      #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/4%3A_Fourier_series_and_PDEs/4.04%3A_Sine_and_cosine_series")[4.4: Sine and Cosine Series]
    


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Take #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals open parenthesis t minus 1 close parenthesis squared")[$f ( t ) = ( t − 1 )^(2)$] defined on #math.equation(block: false, alt: "0 less than or equal to t less than or equal to 1")[$0 ≤ t ≤ 1$].


      + Sketch the plot of the even periodic extension of #math.equation(block: false, alt: "f")[$f$].
+ Sketch the plot of the odd periodic extension of #math.equation(block: false, alt: "f")[$f$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the Fourier series of both the odd and even periodic extension of the function #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals open parenthesis t minus 1 close parenthesis squared")[$f ( t ) = ( t − 1 )^(2)$] for #math.equation(block: false, alt: "0 less than or equal to t less than or equal to 1")[$0 ≤ t ≤ 1$]. Can you tell which extension is continuous from the Fourier series coefficients?
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the Fourier series of both the odd and even periodic extension of the function #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals t")[$f ( t ) = t$] for #math.equation(block: false, alt: "0 less than or equal to t less than or equal to π")[$0 ≤ t ≤ π$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the Fourier series of the even periodic extension of the function #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals sin t")[$f ( t ) = sin t$] for #math.equation(block: false, alt: "0 less than or equal to t less than or equal to π")[$0 ≤ t ≤ π$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Consider


      #math.equation(block: true, alt: "x double prime open parenthesis t close parenthesis plus 4 x open parenthesis t close parenthesis equals f open parenthesis t close parenthesis ,")[$x^(″) ( t ) + 4 x ( t ) = f ( t ) ,$]


      where #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals 1")[$f ( t ) = 1$] on #math.equation(block: false, alt: "0 less than t less than 1")[$0 < t < 1$].


      + Solve for the Dirichlet conditions #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals 0")[$x ( 0 ) = 0$], #math.equation(block: false, alt: "x open parenthesis 1 close parenthesis equals 0")[$x ( 1 ) = 0$].
+ Solve for the Neumann conditions #math.equation(block: false, alt: "x prime open parenthesis 0 close parenthesis equals 0")[$x^(′) ( 0 ) = 0$], #math.equation(block: false, alt: "x prime open parenthesis 1 close parenthesis equals 0")[$x^(′) ( 1 ) = 0$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Consider


      #math.equation(block: true, alt: "x double prime open parenthesis t close parenthesis plus 9 x open parenthesis t close parenthesis equals f open parenthesis t close parenthesis ,")[$x^(″) ( t ) + 9 x ( t ) = f ( t ) ,$]


      for #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals sin open parenthesis 2 π t close parenthesis")[$f ( t ) = sin ( 2 π t )$] on #math.equation(block: false, alt: "0 less than t less than 1")[$0 < t < 1$].


      + Solve for the Dirichlet conditions #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals 0 , x open parenthesis 1 close parenthesis equals 0")[$x ( 0 ) = 0 , #h(0.222em) x ( 1 ) = 0$].
+ b) Solve for the Neumann conditions #math.equation(block: false, alt: "x prime open parenthesis 0 close parenthesis equals 0 , x prime open parenthesis 1 close parenthesis equals 0")[$x^(′) ( 0 ) = 0 , #h(0.222em) x^(′) ( 1 ) = 0$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Consider


      #math.equation(block: true, alt: "x double prime open parenthesis t close parenthesis plus 3 x open parenthesis t close parenthesis equals f open parenthesis t close parenthesis , x open parenthesis 0 close parenthesis equals 0 , x open parenthesis 1 close parenthesis equals 0 ,")[$x^(″) ( t ) + 3 x ( t ) = f ( t ) , #h(1em) x ( 0 ) = 0 , #h(1em) x ( 1 ) = 0 ,$]


      where #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals ∑ n equals 1 infinity b sub n sin open parenthesis n π t close parenthesis")[$f ( t ) = ∑_(n = 1)^(∞) b_(n) sin ( n π t )$]. Write the solution #math.equation(block: false, alt: "x open parenthesis t close parenthesis")[$x ( t )$] as a Fourier series, where the coefficients are given in terms of #math.equation(block: false, alt: "b sub n")[$b_(n)$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals t squared open parenthesis 2 minus t close parenthesis")[$f ( t ) = t^(2) ( 2 − t )$] for #math.equation(block: false, alt: "0 less than or equal to t less than or equal to 2")[$0 ≤ t ≤ 2$]. Let #math.equation(block: false, alt: "F open parenthesis t close parenthesis")[$F ( t )$] be the odd periodic extension. Compute #math.equation(block: false, alt: "F open parenthesis 1 close parenthesis , F open parenthesis 2 close parenthesis , F open parenthesis 3 close parenthesis , F open parenthesis minus 1 close parenthesis , F open parenthesis the fraction 9 over 2 close parenthesis , F open parenthesis 101 close parenthesis , F open parenthesis 103 close parenthesis")[$F ( 1 ) , #h(0.222em) F ( 2 ) , #h(0.222em) F ( 3 ) , #h(0.222em) F ( − 1 ) , #h(0.222em) F ( frac(9, 2) ) , #h(0.222em) F ( 101 ) , #h(0.222em) F ( 103 )$]. Note: Do #strong[not]compute using the sine series.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals the fraction t over 3")[$f ( t ) = frac(t, 3)$] on #math.equation(block: false, alt: "0 less than or equal to t less than 3")[$0 ≤ t < 3$].


      + Find the Fourier series of the even periodic extension.
+ Find the Fourier series of the odd periodic extension.


      Answer


      + #math.equation(block: false, alt: "the fraction 1 over 2 plus ∑ n odd n equals 1 infinity the fraction minus 4 over π squared n squared cos open parenthesis the fraction n π over 3 t close parenthesis")[$frac(1, 2) + limits(∑)^(∞)_(limits(n " odd")^(n = 1)) frac(− 4, π^(2) n^(2)) cos ( frac(n π, 3) t )$]
+ #math.equation(block: false, alt: "∑ n equals 1 infinity the fraction 2 open parenthesis minus 1 close parenthesis to the power n plus 1 over π n sin open parenthesis the fraction n π over 3 t close parenthesis")[$limits(∑)^(∞)_(n = 1) frac(2 attach(( − 1 ), t: n + 1), π n) sin ( frac(n π, 3) t )$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals cos open parenthesis 2 t close parenthesis")[$f ( t ) = cos ( 2 t )$] on #math.equation(block: false, alt: "0 less than or equal to t less than π")[$0 ≤ t < π$].


      + Find the Fourier series of the even periodic extension.
+ Find the Fourier series of the odd periodic extension.


      Answer


      + #math.equation(block: false, alt: "cos open parenthesis 2 t close parenthesis")[$cos ( 2 t )$]
+ #math.equation(block: false, alt: "∑ n odd n equals 1 infinity the fraction minus 4 n over π n squared minus 4 π sin open parenthesis n t close parenthesis")[$limits(∑)^(∞)_(limits(n " odd")^(n = 1)) frac(− 4 n, π n^(2) − 4 π) sin ( n t )$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] be defined on #math.equation(block: false, alt: "0 less than or equal to t less than 1")[$0 ≤ t < 1$]. Now take the average of the two extensions #math.equation(block: false, alt: "g open parenthesis t close parenthesis equals the fraction F sub o d d open parenthesis t close parenthesis plus F sub e v e n open parenthesis t close parenthesis over 2")[$g ( t ) = frac(F_(o d d) ( t ) + F_(e v e n) ( t ), 2)$].


      + What is #math.equation(block: false, alt: "g open parenthesis t close parenthesis")[$g ( t )$] if #math.equation(block: false, alt: "0 less than or equal to t less than 1")[$0 ≤ t < 1$] (Justify!)
+ What is #math.equation(block: false, alt: "g open parenthesis t close parenthesis")[$g ( t )$] if #math.equation(block: false, alt: "minus 1 less than t less than 0")[$− 1 < t < 0$] (Justify!)


      Answer


      + #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$]
+ #math.equation(block: false, alt: "0")[$0$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals ∑ n equals 1 infinity the fraction 1 over n squared sin open parenthesis n t close parenthesis")[$f ( t ) = ∑_(n = 1)^(∞) frac(1, n^(2)) sin ( n t )$]. Solve #math.equation(block: false, alt: "x double prime minus x equals f open parenthesis t close parenthesis")[$x^(″) − x = f ( t )$] for the Dirichlet conditions #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals 0")[$x ( 0 ) = 0$] and #math.equation(block: false, alt: "x open parenthesis π close parenthesis equals 0")[$x ( π ) = 0$].


      Answer


      #math.equation(block: true, alt: "∑ n equals 1 infinity the fraction minus 1 over n squared open parenthesis 1 plus n squared close parenthesis sin open parenthesis n t close parenthesis")[$limits(∑)^(∞)_(n = 1) frac(− 1, n^(2) ( 1 + n^(2) )) sin ( n t )$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[(challenging)]


      Let #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals t plus ∑ n equals 1 infinity the fraction 1 over 2 to the power n sin open parenthesis n t close parenthesis")[$f ( t ) = t + ∑_(n = 1)^(∞) frac(1, 2^(n)) sin ( n t )$]. Solve #math.equation(block: false, alt: "x double prime plus π x equals f open parenthesis t close parenthesis")[$x^(″) + π x = f ( t )$] for the Dirichlet conditions #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals 0")[$x ( 0 ) = 0$] and #math.equation(block: false, alt: "x open parenthesis π close parenthesis equals 1")[$x ( π ) = 1$]. Hint: Note that #math.equation(block: false, alt: "the fraction t over π")[$frac(t, π)$] satisfies the given Dirichlet conditions.


      Answer


      #math.equation(block: true, alt: "the fraction t over π plus ∑ n equals 1 infinity the fraction 1 over 2 to the power n open parenthesis π minus n squared close parenthesis sin open parenthesis n t close parenthesis")[$frac(t, π) + limits(∑)^(∞)_(n = 1) frac(1, 2^(n) ( π − n^(2) )) sin ( n t )$]
]


    
=== 
      #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/4%3A_Fourier_series_and_PDEs/4.05%3A_Applications_of_Fourier_series")[4.5: Applications of Fourier series]
    


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "F open parenthesis t close parenthesis equals the fraction 1 over 2 plus ∑ n equals 1 infinity the fraction 1 over n squared cos open parenthesis n π t close parenthesis")[$F ( t ) = frac(1, 2) + ∑_(n = 1)^(∞) frac(1, n^(2)) cos ( n π t )$]. Find the steady periodic solution to #math.equation(block: false, alt: "x double prime plus 2 x equals F open parenthesis t close parenthesis")[$x^(″) + 2 x = F ( t )$]. Express your solution as a Fourier series.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "F open parenthesis t close parenthesis equals ∑ n equals 1 infinity the fraction 1 over n cubed sin open parenthesis n π t close parenthesis")[$F ( t ) = ∑_(n = 1)^(∞) frac(1, n^(3)) sin ( n π t )$]. Find the steady periodic solution to #math.equation(block: false, alt: "x double prime plus x prime plus x equals F open parenthesis t close parenthesis")[$x^(″) + x^(′) + x = F ( t )$]. Express your solution as a Fourier series.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "F open parenthesis t close parenthesis equals ∑ n equals 1 infinity the fraction 1 over n squared cos open parenthesis n π t close parenthesis")[$F ( t ) = ∑_(n = 1)^(∞) frac(1, n^(2)) cos ( n π t )$]. Find the steady periodic solution to #math.equation(block: false, alt: "x double prime plus 4 x equals F open parenthesis t close parenthesis")[$x^(″) + 4 x = F ( t )$]. Express your solution as a Fourier series.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "F open parenthesis t close parenthesis equals t")[$F ( t ) = t$] for #math.equation(block: false, alt: "minus 1 less than t less than 1")[$− 1 < t < 1$] and extended periodically. Find the steady periodic solution to #math.equation(block: false, alt: "x double prime plus x equals F open parenthesis t close parenthesis")[$x^(″) + x = F ( t )$]. Express your solution as a series.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "F open parenthesis t close parenthesis equals t")[$F ( t ) = t$] for #math.equation(block: false, alt: "minus 1 less than t less than 1")[$− 1 < t < 1$] and extended periodically. Find the steady periodic solution to #math.equation(block: false, alt: "x double prime plus π squared x equals F open parenthesis t close parenthesis")[$x^(″) + π^(2) x = F ( t )$]. Express your solution as a series.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "F open parenthesis t close parenthesis equals sin open parenthesis 2 π t close parenthesis plus 0.1 cos open parenthesis 10 π t close parenthesis")[$F ( t ) = sin ( 2 π t ) + 0.1 cos ( 10 π t )$]. Find the steady periodic solution to #math.equation(block: false, alt: "x double prime plus the square root of 2 x equals F open parenthesis t close parenthesis")[$x^(″) + sqrt(2) x = F ( t )$]. Express your solution as a Fourier series.


      Answer


      #math.equation(block: true, alt: "x equals the fraction 1 over the square root of 2 minus 4 π squared sin open parenthesis 2 π t close parenthesis plus the fraction 0.1 over the square root of 2 minus 100 π squared cos open parenthesis 10 π t close parenthesis")[$x = frac(1, sqrt(2) − 4 π^(2)) sin ( 2 π t ) + frac(0.1, sqrt(2) − 100 π^(2)) cos ( 10 π t )$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "F open parenthesis t close parenthesis equals ∑ n equals 1 infinity e to the power minus n cos open parenthesis 2 n t close parenthesis")[$F ( t ) = ∑_(n = 1)^(∞) e^(− n) cos ( 2 n t )$]. Find the steady periodic solution to #math.equation(block: false, alt: "x double prime plus 3 x equals F open parenthesis t close parenthesis")[$x^(″) + 3 x = F ( t )$]. Express your solution as a Fourier series.


      Answer


      #math.equation(block: true, alt: "x equals ∑ n equals 1 infinity the fraction e to the power minus n over 3 minus open parenthesis 2 n close parenthesis squared cos open parenthesis 2 n t close parenthesis")[$x = limits(∑)^(∞)_(n = 1) frac(e^(− n), 3 − ( 2 n )^(2)) cos ( 2 n t )$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "F open parenthesis t close parenthesis equals | t |")[$F ( t ) = | t |$] for #math.equation(block: false, alt: "minus 1 less than or equal to t less than or equal to 1")[$− 1 ≤ t ≤ 1$] extended periodically. Find the steady periodic solution to #math.equation(block: false, alt: "x double prime plus the square root of 3 x equals F open parenthesis t close parenthesis")[$x^(″) + sqrt(3) x = F ( t )$]. Express your solution as a series.


      Answer


      #math.equation(block: true, alt: "x equals the fraction 1 over 2 the square root of 3 plus ∑ n odd n equals 1 infinity the fraction minus 4 over n squared π squared open parenthesis the square root of 3 minus n squared π squared close parenthesis cos open parenthesis n π t close parenthesis")[$x = frac(1, 2 sqrt(3)) + limits(∑)^(∞)_(limits(n " odd")^(n = 1)) frac(− 4, n^(2) π^(2) ( sqrt(3) − n^(2) π^(2) )) cos ( n π t )$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "F open parenthesis t close parenthesis equals | t |")[$F ( t ) = | t |$] for #math.equation(block: false, alt: "minus 1 less than or equal to t less than or equal to 1")[$− 1 ≤ t ≤ 1$] extended periodically. Find the steady periodic solution to #math.equation(block: false, alt: "x double prime plus π squared x equals F open parenthesis t close parenthesis")[$x^(″) + π^(2) x = F ( t )$]. Express your solution as a series.


      Answer


      #math.equation(block: true, alt: "x equals the fraction 1 over 2 the square root of 3 minus the fraction 2 over π cubed t sin open parenthesis π t close parenthesis plus ∑ n odd n equals 3 infinity the fraction minus 4 over n squared π to the power 4 open parenthesis 1 minus n squared close parenthesis cos open parenthesis n π t close parenthesis")[$x = frac(1, 2 sqrt(3)) − frac(2, π^(3)) t sin ( π t ) + limits(∑)^(∞)_(limits(n " odd")^(n = 3)) frac(− 4, n^(2) π^(4) ( 1 − n^(2) )) cos ( n π t )$]
]


    
=== 
      #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/4%3A_Fourier_series_and_PDEs/4.06%3A_PDEs_separation_of_variables_and_the_heat_equation")[4.6: PDEs, Separation of Variables, and the Heat Equation]
    


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Imagine you have a wire of length #math.equation(block: false, alt: "2")[$2$], with #math.equation(block: false, alt: "k equals 0.001")[$k = 0.001$] and an initial temperature distribution of #math.equation(block: false, alt: "u open parenthesis x , 0 close parenthesis equals 50 x")[$u ( x , 0 ) = 50 x$]. Suppose that both the ends are embedded in ice (temperature 0). Find the solution as a series.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find a series solution of


      #math.equation(block: true, alt: "u sub t equals u sub x x ,; u open parenthesis 0 , t close parenthesis equals u open parenthesis 1 , t close parenthesis equals 0 ,; u open parenthesis x , 0 close parenthesis equals 100 f o r 0 less than x less than 1 .")[$u_(t) = u_(x x) , \ u ( 0 , t ) = u ( 1 , t ) = 0 , \ u ( x , 0 ) = 100 " " " " " " " " upright(f) upright(o) upright(r) " " 0 < x < 1 .$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find a series solution of


      #math.equation(block: true, alt: "u sub t equals u sub x x ,; u sub x open parenthesis 0 , t close parenthesis equals u sub x open parenthesis π , t close parenthesis equals 0 ,; u open parenthesis x , 0 close parenthesis equals 3 cos open parenthesis x close parenthesis plus cos open parenthesis 3 x close parenthesis f o r 0 less than x less than π .")[$u_(t) = u_(x x) , \ u_(x) ( 0 , t ) = u_(x) ( π , t ) = 0 , \ u ( x , 0 ) = 3 cos ( x ) + cos ( 3 x ) " " " " " " " " upright(f) upright(o) upright(r) " " 0 < x < π .$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find a series solution of


      #math.equation(block: true, alt: "u sub t equals the fraction 1 over 3 u sub x x ,; u sub x open parenthesis 0 , t close parenthesis equals u sub x open parenthesis π , t close parenthesis equals 0 ,; u open parenthesis x , 0 close parenthesis equals the fraction 10 x over π f o r 0 less than x less than π .")[$u_(t) = frac(1, 3) u_(x x) , \ u_(x) ( 0 , t ) = u_(x) ( π , t ) = 0 , \ u ( x , 0 ) = frac(10 x, π) " " " " " " " " upright(f) upright(o) upright(r) " " 0 < x < π .$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find a series solution of


      #math.equation(block: true, alt: "u sub t equals u sub x x ,; u open parenthesis 0 , t close parenthesis equals 0 , u open parenthesis 1 , t close parenthesis equals 100 ,; u open parenthesis x , 0 close parenthesis equals sin open parenthesis π x close parenthesis f o r 0 less than x less than 1 .")[$u_(t) = u_(x x) , \ u ( 0 , t ) = 0 , " " " " " " u ( 1 , t ) = 100 , \ u ( x , 0 ) = sin ( π x ) " " " " " " " " upright(f) upright(o) upright(r) " " 0 < x < 1 .$]


      Hint: Use the fact that #math.equation(block: false, alt: "u open parenthesis x , t close parenthesis equals 100 x")[$u ( x , t ) = 100 x$] is a solution satisfying #math.equation(block: false, alt: "u sub t equals u sub x x , u open parenthesis 0 , t close parenthesis equals 0 , u open parenthesis 1 , t close parenthesis equals 100")[$u_(t) = u_(x x) , #h(0.222em) u ( 0 , t ) = 0 , #h(0.222em) u ( 1 , t ) = 100$]. Then usesuperposition.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the steady state temperature solution as a function of #math.equation(block: false, alt: "x")[$x$] alone, by letting #math.equation(block: false, alt: "t → infinity")[$t → ∞$] in the solution from exercises #math.equation(block: false, alt: "4")[$4$] and #math.equation(block: false, alt: "5")[$5$]. Verify that it satisfies the equation #math.equation(block: false, alt: "u sub x x equals 0")[$u_(x x) = 0$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Use separation variables to find a nontrivial solution to #math.equation(block: false, alt: "u sub x x plus u sub y y equals 0")[$u_(x x) + u_(y y) = 0$], where #math.equation(block: false, alt: "u open parenthesis x , 0 close parenthesis equals 0")[$u ( x , 0 ) = 0$] and #math.equation(block: false, alt: "u open parenthesis 0 , y close parenthesis equals 0")[$u ( 0 , y ) = 0$]. Hint: Try #math.equation(block: false, alt: "u open parenthesis x , y close parenthesis equals X open parenthesis x close parenthesis Y open parenthesis y close parenthesis")[$u ( x , y ) = X ( x ) Y ( y )$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[(challenging)]


      Suppose that one end of the wire is insulated (say at #math.equation(block: false, alt: "x equals 0")[$x = 0$]) and the other end is kept at zero temperature. That is, find a series solution of


      #math.equation(block: true, alt: "u sub t equals k u sub x x ,; u sub x open parenthesis 0 , t close parenthesis equals u open parenthesis L , t close parenthesis equals 0 ,; u open parenthesis x , 0 close parenthesis equals f open parenthesis x close parenthesis f o r 0 less than x less than L .")[$u_(t) = k u_(x x) , \ u_(x) ( 0 , t ) = u ( L , t ) = 0 , \ u ( x , 0 ) = f ( x ) " " " " " " " " upright(f) upright(o) upright(r) " " 0 < x < L .$]


      Express any coefficients in the series by integrals of #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[(challenging)]


      Suppose that the wire is circular and insulated, so there are no ends. You can think of this as simply connecting the two ends and making sure the solution matches up at the ends. That is, find a series solution of


      #math.equation(block: true, alt: "u sub t equals k u sub x x ,; u open parenthesis 0 , t close parenthesis equals u open parenthesis L , t close parenthesis , u sub x open parenthesis 0 , t close parenthesis equals u sub x open parenthesis L , t close parenthesis; u open parenthesis x , 0 close parenthesis equals f open parenthesis x close parenthesis f o r 0 less than x less than L .")[$u_(t) = k u_(x x) , \ u ( 0 , t ) = u ( L , t ) , " " " " " " " " u_(x) ( 0 , t ) = u_(x) ( L , t ) \ u ( x , 0 ) = f ( x ) " " " " " " " " upright(f) upright(o) upright(r) " " 0 < x < L .$]


      Express any coefficients in the series by integrals of #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Consider a wire insulated on both ends, #math.equation(block: false, alt: "L equals 1")[$L = 1$], #math.equation(block: false, alt: "k equals 1")[$k = 1$], and #math.equation(block: false, alt: "u open parenthesis x , 0 close parenthesis equals cos squared open parenthesis π x close parenthesis")[$u ( x , 0 ) = cos^(2) ( π x )$].


      + Find the solution #math.equation(block: false, alt: "u open parenthesis x , t close parenthesis")[$u ( x , t )$]. Hint: a trig identity.
+ Find the average temperature.
+ Initially the temperature variation is #math.equation(block: false, alt: "1")[$1$] (maximum minus the minimum). Find the time when the variation is #math.equation(block: false, alt: "the fraction 1 over 2")[$frac(1, 2)$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find a series solution of


      #math.equation(block: true, alt: "u sub t equals 3 u sub x x ,; u open parenthesis 0 , t close parenthesis equals u open parenthesis π , t close parenthesis equals 0 ,; u open parenthesis x , 0 close parenthesis equals 5 sin open parenthesis x close parenthesis plus 2 sin open parenthesis 5 x close parenthesis f o r 0 less than x less than π .")[$u_(t) = 3 u_(x x) , \ u ( 0 , t ) = u ( π , t ) = 0 , \ u ( x , 0 ) = 5 sin ( x ) + 2 sin ( 5 x ) " " " " " " " " upright(f) upright(o) upright(r) " " 0 < x < π .$]


      Answer


      #math.equation(block: true, alt: "u open parenthesis x , t close parenthesis equals 5 sin open parenthesis x close parenthesis e to the power 3 t plus 2 sin open parenthesis 5 x close parenthesis e to the power minus 75 t")[$u ( x , t ) = 5 sin ( x ) e^(3 t) + 2 sin ( 5 x ) e^(− 75 t)$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find a series solution of


      #math.equation(block: true, alt: "u sub t equals 0.1 u sub x x ,; u sub x open parenthesis 0 , t close parenthesis equals u sub x open parenthesis π , t close parenthesis equals 0 ,; u open parenthesis x , 0 close parenthesis equals 1 plus 2 cos open parenthesis x close parenthesis f o r 0 less than x less than π .")[$u_(t) = 0.1 u_(x x) , \ u_(x) ( 0 , t ) = u_(x) ( π , t ) = 0 , \ u ( x , 0 ) = 1 + 2 cos ( x ) " " " " " " " " upright(f) upright(o) upright(r) " " 0 < x < π .$]


      Answer


      #math.equation(block: true, alt: "u open parenthesis x , t close parenthesis equals 1 plus 2 cos open parenthesis x close parenthesis e to the power minus 0.1 t")[$u ( x , t ) = 1 + 2 cos ( x ) e^(− 0.1 t)$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Use separation of variables to find a nontrivial solution to #math.equation(block: false, alt: "u sub x t equals u sub x x")[$u_(x t) = u_(x x)$].


      Answer


      #math.equation(block: false, alt: "u open parenthesis x , t close parenthesis equals e to the power λ t e to the power λ x")[$u ( x , t ) = e^(λ t) e^(λ x)$] for some #math.equation(block: false, alt: "λ")[$λ$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Use separation of variables to find a nontrivial solution to #math.equation(block: false, alt: "u sub x plus u sub t equals u")[$u_(x) + u_(t) = u$]. (Hint: try #math.equation(block: false, alt: "u open parenthesis x , t close parenthesis equals X open parenthesis x close parenthesis plus T open parenthesis t close parenthesis")[$u ( x , t ) = X ( x ) + T ( t )$]).


      Answer


      #math.equation(block: true, alt: "u open parenthesis x , t close parenthesis equals A e to the power x plus B e to the power t")[$u ( x , t ) = A e^(x) + B e^(t)$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose that the temperature on the wire is fixed at #math.equation(block: false, alt: "0")[$0$] at the ends, #math.equation(block: false, alt: "L equals 1")[$L = 1$], #math.equation(block: false, alt: "k equals 1")[$k = 1$], and #math.equation(block: false, alt: "u open parenthesis x , 0 close parenthesis equals 100 sin open parenthesis 2 π x close parenthesis")[$u ( x , 0 ) = 100 sin ( 2 π x )$].


      + What is the temperature at #math.equation(block: false, alt: "x equals the fraction 1 over 2")[$x = frac(1, 2)$] at any time.
+ What is the maximum and the minimum temperature on the wire at #math.equation(block: false, alt: "t equals 0")[$t = 0$].
+ At what time is the maximum temperature on the wire exactly one half of the initial maximum at #math.equation(block: false, alt: "t equals 0")[$t = 0$].


      Answer


      + #math.equation(block: false, alt: "0")[$0$]
+ #math.equation(block: false, alt: "minimum minus 100")[$"minimum " − 100$], #math.equation(block: false, alt: "maximum 100")[$"maximum " 100$]
+ #math.equation(block: false, alt: "t equals the fraction ln 2 over 4 π squared")[$t = frac(ln 2, 4 π^(2))$]
]


    
=== 
      #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/4%3A_Fourier_series_and_PDEs/4.07%3A_One_dimensional_wave_equation")[4.7: One dimensional wave equation]
    


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve


      #math.equation(block: true, alt: "y sub t t equals 9 y sub x x ,; y open parenthesis 0 , t close parenthesis equals y open parenthesis 1 , t close parenthesis equals 0 ,; y open parenthesis x , 0 close parenthesis equals sin open parenthesis 3 π x close parenthesis plus the fraction 1 over 4 sin open parenthesis 6 π x close parenthesis f o r 0 less than x less than 1 ,; y sub t open parenthesis x , 0 close parenthesis equals 0 f o r 0 less than x less than 1 .")[$y_(t t) = 9 y_(x x) , \ y ( 0 , t ) = y ( 1 , t ) = 0 , \ y ( x , 0 ) = sin ( 3 π x ) + frac(1, 4) sin ( 6 π x ) upright(f) upright(o) upright(r) " " 0 < x < 1 , \ y_(t) ( x , 0 ) = 0 upright(f) upright(o) upright(r) " " 0 < x < 1 .$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve


      #math.equation(block: true, alt: "y sub t t equals 4 y sub x x ,; y open parenthesis 0 , t close parenthesis equals y open parenthesis 1 , t close parenthesis equals 0 ,; y open parenthesis x , 0 close parenthesis equals sin open parenthesis 3 π x close parenthesis plus the fraction 1 over 4 sin open parenthesis 6 π x close parenthesis f o r 0 less than x less than 1 ,; y sub t open parenthesis x , 0 close parenthesis equals sin open parenthesis 9 π x close parenthesis f o r 0 less than x less than 1 .")[$y_(t t) = 4 y_(x x) , \ y ( 0 , t ) = y ( 1 , t ) = 0 , \ y ( x , 0 ) = sin ( 3 π x ) + frac(1, 4) sin ( 6 π x ) upright(f) upright(o) upright(r) " " 0 < x < 1 , \ y_(t) ( x , 0 ) = sin ( 9 π x ) upright(f) upright(o) upright(r) " " 0 < x < 1 .$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Derive the solution for a general plucked string of length #math.equation(block: false, alt: "L")[$L$], where we raise the string some distance #math.equation(block: false, alt: "b")[$b$] at the midpoint and let go, and for any constant #math.equation(block: false, alt: "a")[$a$] (in the equation #math.equation(block: false, alt: "y sub t t equals a squared y sub x x")[$y_(t t) = a^(2) y_(x x)$]).
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Imagine that a stringed musical instrument falls on the floor. Suppose that the length of the string is 1 and #math.equation(block: false, alt: "a equals 1")[$a = 1$]. When the musical instrument hits the ground the string was in rest position and hence #math.equation(block: false, alt: "y open parenthesis x , 0 close parenthesis equals 0")[$y ( x , 0 ) = 0$]. However, the string was moving at some velocity at impact #math.equation(block: false, alt: "open parenthesis t equals 0")[$\( t = 0$])), say #math.equation(block: false, alt: "y sub t open parenthesis x , 0 close parenthesis equals minus 1")[$y_(t) ( x , 0 ) = − 1$]. Find the solution #math.equation(block: false, alt: "y open parenthesis x , t close parenthesis")[$y ( x , t )$] for the shape of the string at time #math.equation(block: false, alt: "t")[$t$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[(challenging)]


      Suppose that you have a vibrating string and that there is air resistance proportional to the velocity. That is, you have


      #math.equation(block: true, alt: "y sub t t equals a squared y sub x x minus k y sub t ,; y open parenthesis 0 , t close parenthesis equals y open parenthesis 1 , t close parenthesis equals 0 ,; y open parenthesis x , 0 close parenthesis equals f open parenthesis x close parenthesis f o r 0 less than x less than 1 ,; y sub t open parenthesis x , 0 close parenthesis equals 0 f o r 0 less than x less than 1 .")[$y_(t t) = a^(2) y_(x x) − k y_(t) , \ y ( 0 , t ) = y ( 1 , t ) = 0 , \ y ( x , 0 ) = f ( x ) upright(f) upright(o) upright(r) " " 0 < x < 1 , \ y_(t) ( x , 0 ) = 0 upright(f) upright(o) upright(r) " " 0 < x < 1 .$]


      Suppose that #math.equation(block: false, alt: "0 less than k less than 2 π a")[$0 < k < 2 π a$]. Derive a series solution to the problem. Any coefficients in the series should be expressed as integrals of #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose you touch the guitar string exactly in the middle to ensure another condition #math.equation(block: false, alt: "u open parenthesis the fraction L over 2 , t close parenthesis equals 0")[$u ( frac(L, 2) , t ) = 0$] for all time. Which multiples of the fundamental frequency #math.equation(block: false, alt: "the fraction π a over L")[$frac(π a, L)$] show up in the solution?
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve


      #math.equation(block: true, alt: "y sub t t equals y sub x x ,; y open parenthesis 0 , t close parenthesis equals y open parenthesis π , t close parenthesis equals 0 ,; y open parenthesis x , 0 close parenthesis equals sin open parenthesis x close parenthesis f o r 0 less than x less than π ,; y sub t open parenthesis x , 0 close parenthesis equals sin open parenthesis x close parenthesis f o r 0 less than x less than π .")[$y_(t t) = y_(x x) , \ y ( 0 , t ) = y ( π , t ) = 0 , \ y ( x , 0 ) = sin ( x ) upright(f) upright(o) upright(r) " " 0 < x < π , \ y_(t) ( x , 0 ) = sin ( x ) upright(f) upright(o) upright(r) " " 0 < x < π .$]


      Answer


      #math.equation(block: true, alt: "y open parenthesis x , t close parenthesis equals sin open parenthesis x close parenthesis open parenthesis sin open parenthesis t close parenthesis plus cos open parenthesis t close parenthesis close parenthesis")[$y ( x , t ) = sin ( x ) ( sin ( t ) + cos ( t ) )$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve


      #math.equation(block: true, alt: "y sub t t equals 25 y sub x x ,; y open parenthesis 0 , t close parenthesis equals y open parenthesis 2 , t close parenthesis equals 0 ,; y open parenthesis x , 0 close parenthesis equals 0 f o r 0 less than x less than 2 ,; y sub t open parenthesis x , 0 close parenthesis equals sin open parenthesis π x close parenthesis plus 0.1 sin open parenthesis 2 π t close parenthesis f o r 0 less than x less than 2 .")[$y_(t t) = 25 y_(x x) , \ y ( 0 , t ) = y ( 2 , t ) = 0 , \ y ( x , 0 ) = 0 upright(f) upright(o) upright(r) " " 0 < x < 2 , \ y_(t) ( x , 0 ) = sin ( π x ) + 0.1 sin ( 2 π t ) upright(f) upright(o) upright(r) " " 0 < x < 2 .$]


      Answer


      #math.equation(block: true, alt: "y open parenthesis x , t close parenthesis equals the fraction 1 over 5 π sin open parenthesis π x close parenthesis sin open parenthesis 5 π t close parenthesis plus the fraction 1 over 100 π sin open parenthesis 2 π x close parenthesis sin open parenthesis 10 π t close parenthesis")[$y ( x , t ) = frac(1, 5 π) sin ( π x ) sin ( 5 π t ) + frac(1, 100 π) sin ( 2 π x ) sin ( 10 π t )$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve


      #math.equation(block: true, alt: "y sub t t equals 2 y sub x x ,; y open parenthesis 0 , t close parenthesis equals y open parenthesis π , t close parenthesis equals 0 ,; y open parenthesis x , 0 close parenthesis equals x f o r 0 less than x less than π ,; y sub t open parenthesis x , 0 close parenthesis equals 0 f o r 0 less than x less than π .")[$y_(t t) = 2 y_(x x) , \ y ( 0 , t ) = y ( π , t ) = 0 , \ y ( x , 0 ) = x upright(f) upright(o) upright(r) " " 0 < x < π , \ y_(t) ( x , 0 ) = 0 upright(f) upright(o) upright(r) " " 0 < x < π .$]


      Answer


      #math.equation(block: true, alt: "y open parenthesis x , t close parenthesis equals ∑ n equals 1 infinity the fraction 2 open parenthesis minus 1 close parenthesis to the power n plus 1 over n sin open parenthesis n x close parenthesis cos open parenthesis n the square root of 2 t close parenthesis")[$y ( x , t ) = limits(∑)^(∞)_(n = 1) frac(2 ( − 1 )^(n + 1), n) sin ( n x ) cos ( n sqrt(2) t )$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let’s see what happens when #math.equation(block: false, alt: "a equals 0")[$a = 0$]. Find a solution to #math.equation(block: false, alt: "y sub t t equals 0 , y open parenthesis 0 , t close parenthesis equals y open parenthesis π , t close parenthesis equals 0 , y open parenthesis x , 0 close parenthesis equals sin open parenthesis 2 x close parenthesis , y sub t open parenthesis x , 0 close parenthesis equals sin open parenthesis x close parenthesis .")[$y_(t t) = 0 , #h(0.222em) y ( 0 , t ) = y ( π , t ) = 0 , #h(0.222em) y ( x , 0 ) = sin ( 2 x ) , #h(0.222em) y_(t) ( x , 0 ) = sin ( x ) .$]


      Answer


      #math.equation(block: true, alt: "y open parenthesis x , t close parenthesis equals sin open parenthesis 2 x close parenthesis plus t sin open parenthesis x close parenthesis")[$y ( x , t ) = sin ( 2 x ) + t sin ( x )$]
]


    
=== 
      #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/4%3A_Fourier_series_and_PDEs/4.08%3A_DAlembert_solution_of_the_wave_equation")[4.8: D’Alembert solution of the wave equation]
    


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Using the d’Alembert solution solve #math.equation(block: false, alt: "y sub t t equals 4 y sub x x")[$y_(t t) = 4 y_(x x)$], #math.equation(block: false, alt: "0 less than x less than π , t greater than 0 , y open parenthesis 0 , t close parenthesis equals y open parenthesis π , t close parenthesis equals 0 , y open parenthesis x , 0 close parenthesis equals sin x")[$0 < x < π , t > 0 , y ( 0 , t ) = y ( π , t ) = 0 , y ( x , 0 ) = sin x$], and #math.equation(block: false, alt: "y sub t open parenthesis x , 0 close parenthesis equals sin x")[$y_(t) ( x , 0 ) = sin x$]. Hint: Note that #math.equation(block: false, alt: "sin x")[$sin x$] is the odd extension of #math.equation(block: false, alt: "y open parenthesis x , 0 close parenthesis")[$y ( x , 0 )$] and #math.equation(block: false, alt: "y sub t open parenthesis x , 0 close parenthesis")[$y_(t) ( x , 0 )$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Using the d’Alembert solution solve #math.equation(block: false, alt: "y sub t t equals 2 y sub x x")[$y_(t t) = 2 y_(x x)$], #math.equation(block: false, alt: "0 less than x less than 1 , t greater than 0 , y open parenthesis 0 , t close parenthesis equals y open parenthesis 1 , t close parenthesis equals 0 , y open parenthesis x , 0 close parenthesis equals sin to the power 5 open parenthesis π x close parenthesis")[$0 < x < 1 , t > 0 , y ( 0 , t ) = y ( 1 , t ) = 0 , y ( x , 0 ) = sin^(5) ( π x )$], and #math.equation(block: false, alt: "y sub t open parenthesis x , 0 close parenthesis equals sin cubed open parenthesis π x close parenthesis")[$y_(t) ( x , 0 ) = sin^(3) ( π x )$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Take #math.equation(block: false, alt: "y sub t t equals 4 y sub x x")[$y_(t t) = 4 y_(x x)$], #math.equation(block: false, alt: "0 less than x less than π , t greater than 0 , y open parenthesis 0 , t close parenthesis equals y open parenthesis π , t close parenthesis equals 0 , y open parenthesis x , 0 close parenthesis equals x open parenthesis π minus x close parenthesis")[$0 < x < π , t > 0 , y ( 0 , t ) = y ( π , t ) = 0 , y ( x , 0 ) = x ( π − x )$], and #math.equation(block: false, alt: "y sub t open parenthesis x , 0 close parenthesis equals 0")[$y_(t) ( x , 0 ) = 0$].


      + Solve using the d’Alembert formula. Hint: You can use the sine series for #math.equation(block: false, alt: "y open parenthesis x , 0 close parenthesis")[$y ( x , 0 )$].
+ Find the solution as a function of #math.equation(block: false, alt: "x")[$x$] for a fixed #math.equation(block: false, alt: "t equals 0.5 , t equals 1 ,")[$t = 0.5 , #h(0.222em) t = 1 ,$] and #math.equation(block: false, alt: "t equals 2")[$t = 2$]. Do not use the sine series here.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Derive the d’Alembert solution for #math.equation(block: false, alt: "y sub t t equals a squared y sub x x")[$y_(t t) = a^(2) y_(x x)$], #math.equation(block: false, alt: "0 less than x less than π , t greater than 0 , y open parenthesis 0 , t close parenthesis equals y open parenthesis π , t close parenthesis equals 0 , y open parenthesis x , 0 close parenthesis equals f open parenthesis x close parenthesis")[$0 < x < π , t > 0 , y ( 0 , t ) = y ( π , t ) = 0 , y ( x , 0 ) = f ( x )$], and #math.equation(block: false, alt: "y sub t open parenthesis x , 0 close parenthesis equals 0")[$y_(t) ( x , 0 ) = 0$], using the Fourier series solution of the wave equation, by applying an appropriate trigonometric identity. Hint: Do it first for a single term of the Fourier series solution, in particular do it when #math.equation(block: false, alt: "y")[$y$] is #math.equation(block: false, alt: "sin open parenthesis the fraction n π over L x close parenthesis sin open parenthesis the fraction n π a over L t close parenthesis")[$sin ( frac(n π, L) x ) sin ( frac(n π a, L) t )$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
The d’Alembert solution still works if there are no boundary conditions and the initial condition is defined on the whole real line. Suppose that #math.equation(block: false, alt: "y sub t t equals y sub x x")[$y_(t t) = y_(x x)$] (for all #math.equation(block: false, alt: "x")[$x$] on the real line and #math.equation(block: false, alt: "t greater than or equal to 0")[$t ≥ 0$]), #math.equation(block: false, alt: "y open parenthesis x , 0 close parenthesis equals f open parenthesis x close parenthesis")[$y ( x , 0 ) = f ( x )$], and #math.equation(block: false, alt: "y sub t open parenthesis x , 0 close parenthesis")[$y_(t) ( x , 0 )$], where


      #math.equation(block: true, alt: "0 i f x less than minus 1 ,; x plus 1 i f minus 1 less than or equal to x less than 0 ,; minus x plus 1 i f 0 less than or equal to x less than 1; 0 i f x greater than 1 .")[$0 upright(i) upright(f) x < − 1 , \ x + 1 upright(i) upright(f) − 1 ≤ x < 0 , \ − x + 1 upright(i) upright(f) 0 ≤ x < 1 \ 0 upright(i) upright(f) x > 1 .$]


      Solve using the d’Alembert solution. That is, write down a piecewise definition for the solution. Then sketch the solution for #math.equation(block: false, alt: "t equals 0 , t equals 1 / 2 , t equals 1")[$t = 0 , t = 1 / 2 , t = 1$], and #math.equation(block: false, alt: "t equals 2")[$t = 2$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Using the d’Alembert solution solve #math.equation(block: false, alt: "y sub t t equals 9 y sub x x")[$y_(t t) = 9 y_(x x)$], #math.equation(block: false, alt: "0 less than x less than 1 , t greater than 0 , y open parenthesis 0 , t close parenthesis equals y open parenthesis 1 , t close parenthesis equals 0 , y open parenthesis x , 0 close parenthesis equals sin open parenthesis 2 π x close parenthesis")[$0 < x < 1 , t > 0 , y ( 0 , t ) = y ( 1 , t ) = 0 , y ( x , 0 ) = sin ( 2 π x )$], and #math.equation(block: false, alt: "y sub t open parenthesis x , 0 close parenthesis equals sin open parenthesis 3 π x close parenthesis")[$y_(t) ( x , 0 ) = sin ( 3 π x )$].


      Answer


      #math.equation(block: true, alt: "y open parenthesis x , t close parenthesis equals the fraction sin open parenthesis 2 π open parenthesis x minus 3 t close parenthesis close parenthesis plus sin open parenthesis 2 π open parenthesis 3 t plus x close parenthesis close parenthesis over 2 plus the fraction cos open parenthesis 3 π open parenthesis x minus 3 t close parenthesis close parenthesis minus cos open parenthesis 3 π open parenthesis 3 t plus x close parenthesis close parenthesis over 18 π")[$y ( x , t ) = frac(sin ( 2 π ( x − 3 t ) ) + sin ( 2 π ( 3 t + x ) ), 2) + frac(cos ( 3 π ( x − 3 t ) ) − cos ( 3 π ( 3 t + x ) ), 18 π)$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Take #math.equation(block: false, alt: "y sub t t equals 4 y sub x x")[$y_(t t) = 4 y_(x x)$], #math.equation(block: false, alt: "0 less than x less than 1 , t greater than 0 , y open parenthesis 0 , t close parenthesis equals y open parenthesis 1 , t close parenthesis equals 0 , y open parenthesis x , 0 close parenthesis equals x minus x squared")[$0 < x < 1 , t > 0 , #h(0.222em) y ( 0 , t ) = y ( 1 , t ) = 0 , #h(0.222em) y ( x , 0 ) = x − x^(2)$], and #math.equation(block: false, alt: "y sub t open parenthesis x , 0 close parenthesis equals 0")[$y_(t) ( x , 0 ) = 0$]. Using the D’Alembert solution find the solution at


      + #math.equation(block: false, alt: "t equals 0.1")[$t = 0.1$],
+ #math.equation(block: false, alt: "t equals 1 / 2")[$t = 1 / 2$],
+ #math.equation(block: false, alt: "t equals 1")[$t = 1$].


      You may have to split your answer up by cases.


      Answer


      + #math.equation(block: false, alt: "x minus x squared minus 0.04 i f 0.2 less than or equal to x less than or equal to 0.8; 0.6 x i f x less than or equal to 0.2; 0.6 minus 0.6 x i f x greater than or equal to 0.8")[$x − x^(2) − 0.04 upright(i) upright(f) 0.2 ≤ x ≤ 0.8 \ 0.6 x upright(i) upright(f) x ≤ 0.2 \ 0.6 − 0.6 x upright(i) upright(f) x ≥ 0.8$]
+ #math.equation(block: false, alt: "y open parenthesis x , the fraction 1 over 2 close parenthesis equals minus x plus x squared")[$y ( x , frac(1, 2) ) = − x + x^(2)$]
+ #math.equation(block: false, alt: "y open parenthesis x , 1 close parenthesis equals x minus x squared")[$y ( x , 1 ) = x − x^(2)$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Take #math.equation(block: false, alt: "y sub t t equals 100 y sub x x")[$y_(t t) = 100 y_(x x)$], #math.equation(block: false, alt: "0 less than x less than 4 , t greater than 0 , y open parenthesis 0 , t close parenthesis equals y open parenthesis 4 , t close parenthesis equals 0 , y open parenthesis x , 0 close parenthesis equals F open parenthesis x close parenthesis")[$0 < x < 4 , t > 0 , #h(0.222em) y ( 0 , t ) = y ( 4 , t ) = 0 , #h(0.222em) y ( x , 0 ) = F ( x )$], and #math.equation(block: false, alt: "y sub t open parenthesis x , 0 close parenthesis equals 0")[$y_(t) ( x , 0 ) = 0$]. Suppose that #math.equation(block: false, alt: "F open parenthesis 0 close parenthesis equals 0 , F open parenthesis 1 close parenthesis equals 2 , F open parenthesis 2 close parenthesis equals 3 , F open parenthesis 3 close parenthesis equals 1")[$F ( 0 ) = 0 , #h(0.222em) F ( 1 ) = 2 , #h(0.222em) F ( 2 ) = 3 , #h(0.222em) F ( 3 ) = 1$]. Using the D’Alembert solution find


      + #math.equation(block: false, alt: "y open parenthesis 1 , 1 close parenthesis")[$y ( 1 , 1 )$],
+ #math.equation(block: false, alt: "y open parenthesis 4 , 3 close parenthesis")[$y ( 4 , 3 )$],
+ #math.equation(block: false, alt: "y open parenthesis 3 , 9 close parenthesis")[$y ( 3 , 9 )$].


      Answer


      + #math.equation(block: false, alt: "y open parenthesis 1 , 1 close parenthesis equals minus the fraction 1 over 2")[$y ( 1 , 1 ) = − frac(1, 2)$]
+ #math.equation(block: false, alt: "y open parenthesis 4 , 3 close parenthesis equals 0")[$y ( 4 , 3 ) = 0$]
+ #math.equation(block: false, alt: "y open parenthesis 3 , 9 close parenthesis equals the fraction 1 over 2")[$y ( 3 , 9 ) = frac(1, 2)$]
]


    
=== 
      #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/4%3A_Fourier_series_and_PDEs/4.09%3A_Steady_state_temperature_and_the_Laplacian")[4.9: Steady state temperature and the Laplacian]
    


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "R")[$R$] be the region described by #math.equation(block: false, alt: "0 less than x less than π")[$0 < x < π$] and #math.equation(block: false, alt: "0 less than y less than π")[$0 < y < π$]. Solve the problem


      #math.equation(block: true, alt: "Δ u equals 0 , u open parenthesis x , 0 close parenthesis equals sin x , u open parenthesis x , π close parenthesis equals 0 , u open parenthesis 0 , y close parenthesis equals 0 , u open parenthesis π , y close parenthesis equals 0 .")[$Δ u = 0 , " " " " " " " " u ( x , 0 ) = sin x , " " " " " " " " u ( x , π ) = 0 , " " " " " " " " u ( 0 , y ) = 0 , " " " " " " " " u ( π , y ) = 0 .$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "R")[$R$] be the region described by #math.equation(block: false, alt: "0 less than x less than 1")[$0 < x < 1$] and #math.equation(block: false, alt: "0 less than y less than 1")[$0 < y < 1$]. Solve the problem


      #math.equation(block: true, alt: "u sub x x plus u sub y y equals 0 ,; u open parenthesis x , 0 close parenthesis equals sin open parenthesis π x close parenthesis minus sin open parenthesis 2 π x close parenthesis , u open parenthesis x , 1 close parenthesis equals 0 ,; u open parenthesis 0 , y close parenthesis equals 0 , u open parenthesis 1 , y close parenthesis equals 0 .")[$u_(x x) + u_(y y) = 0 , \ u ( x , 0 ) = sin ( π x ) − sin ( 2 π x ) , " " " " " " u ( x , 1 ) = 0 , \ u ( 0 , y ) = 0 , " " " " " " u ( 1 , y ) = 0 .$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "R")[$R$] be the region described by #math.equation(block: false, alt: "0 less than x less than 1")[$0 < x < 1$] and #math.equation(block: false, alt: "0 less than y less than 1")[$0 < y < 1$]. Solve the problem


      #math.equation(block: true, alt: "u sub x x plus u sub y y equals 0 ,; u open parenthesis x , 0 close parenthesis equals u open parenthesis x , 1 close parenthesis equals u open parenthesis 0 , y close parenthesis equals u open parenthesis 1 , y close parenthesis equals C .")[$u_(x x) + u_(y y) = 0 , \ u ( x , 0 ) = u ( x , 1 ) = u ( 0 , y ) = u ( 1 , y ) = C .$]


      for some constant #math.equation(block: false, alt: "C")[$C$]. Hint: Guess, then check your intuition.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "R")[$R$] be the region described by #math.equation(block: false, alt: "0 less than x less than π")[$0 < x < π$] and #math.equation(block: false, alt: "0 less than y less than π")[$0 < y < π$]. Solve


      #math.equation(block: true, alt: "Δ u equals 0 , u open parenthesis x , 0 close parenthesis equals 0 , u open parenthesis x , π close parenthesis equals π , u open parenthesis 0 , y close parenthesis equals y , u open parenthesis π , y close parenthesis equals y .")[$Δ u = 0 , " " " " " " u ( x , 0 ) = 0 , " " " " " " u ( x , π ) = π , " " " " " " u ( 0 , y ) = y , " " " " " " u ( π , y ) = y .$]


      Hint: Try a solution of the form #math.equation(block: false, alt: "u open parenthesis x , y close parenthesis equals X open parenthesis x close parenthesis plus Y open parenthesis y close parenthesis")[$u ( x , y ) = X ( x ) + Y ( y )$] (different separation of variables).
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Use the solution of Exercise #math.equation(block: false, alt: "4")[$4$] to solve


      #math.equation(block: true, alt: "Δ u equals 0 , u open parenthesis x , 0 close parenthesis equals sin x , u open parenthesis x , π close parenthesis equals π , u open parenthesis 0 , y close parenthesis equals y , u open parenthesis π , y close parenthesis equals y .")[$Δ u = 0 , " " " " " " u ( x , 0 ) = sin x , " " " " " " u ( x , π ) = π , " " " " " " u ( 0 , y ) = y , " " " " " " u ( π , y ) = y .$]


      Hint: Use superposition.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "R")[$R$] be the region described by #math.equation(block: false, alt: "0 less than x less than w")[$0 < x < w$] and #math.equation(block: false, alt: "0 less than y less than h")[$0 < y < h$]. Solve the problem


      #math.equation(block: true, alt: "u sub x x plus u sub y y equals 0 ,; u open parenthesis x , 0 close parenthesis equals 0 , u open parenthesis x , h close parenthesis equals f open parenthesis x close parenthesis ,; u open parenthesis 0 , y close parenthesis equals 0 , u open parenthesis w , y close parenthesis equals 0 .")[$u_(x x) + u_(y y) = 0 , \ u ( x , 0 ) = 0 , u ( x , h ) = f ( x ) , \ u ( 0 , y ) = 0 , u ( w , y ) = 0 .$]


      The solution should be in series form using the Fourier series coefficients of #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "R")[$R$] be the region described by #math.equation(block: false, alt: "0 less than x less than w")[$0 < x < w$] and #math.equation(block: false, alt: "0 less than y less than h")[$0 < y < h$]. Solve the problem


      #math.equation(block: true, alt: "u sub x x plus u sub y y equals 0 ,; u open parenthesis x , 0 close parenthesis equals 0 , u open parenthesis x , h close parenthesis equals 0 ,; u open parenthesis 0 , y close parenthesis equals f open parenthesis y close parenthesis , u open parenthesis w , y close parenthesis equals 0 .")[$u_(x x) + u_(y y) = 0 , \ u ( x , 0 ) = 0 , u ( x , h ) = 0 , \ u ( 0 , y ) = f ( y ) , u ( w , y ) = 0 .$]


      The solution should be in series form using the Fourier series coefficients of #math.equation(block: false, alt: "f open parenthesis y close parenthesis")[$f ( y )$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "R")[$R$] be the region described by #math.equation(block: false, alt: "0 less than x less than w")[$0 < x < w$] and #math.equation(block: false, alt: "0 less than y less than h")[$0 < y < h$]. Solve the problem


      #math.equation(block: true, alt: "u sub x x plus u sub y y equals 0 ,; u open parenthesis x , 0 close parenthesis equals 0 , u open parenthesis x , h close parenthesis equals 0 ,; u open parenthesis 0 , y close parenthesis equals 0 , u open parenthesis w , y close parenthesis equals f open parenthesis y close parenthesis .")[$u_(x x) + u_(y y) = 0 , \ u ( x , 0 ) = 0 , u ( x , h ) = 0 , \ u ( 0 , y ) = 0 , u ( w , y ) = f ( y ) .$]


      The solution should be in series form using the Fourier series coefficients of #math.equation(block: false, alt: "f open parenthesis y close parenthesis")[$f ( y )$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "R")[$R$] be the region described by #math.equation(block: false, alt: "0 less than x less than 1")[$0 < x < 1$] and #math.equation(block: false, alt: "0 less than y less than 1")[$0 < y < 1$]. Solve the problem


      #math.equation(block: true, alt: "u sub x x plus u sub y y equals 0 ,; u open parenthesis x , 0 close parenthesis equals sin open parenthesis 9 π x close parenthesis , u open parenthesis x , 1 close parenthesis equals sin open parenthesis 2 π x close parenthesis ,; u open parenthesis 0 , y close parenthesis equals 0 , u open parenthesis 1 , y close parenthesis equals 0 .")[$u_(x x) + u_(y y) = 0 , \ u ( x , 0 ) = sin ( 9 π x ) , u ( x , 1 ) = sin ( 2 π x ) , \ u ( 0 , y ) = 0 , u ( 1 , y ) = 0 .$]


      Hint: Use superposition.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "R")[$R$] be the region described by #math.equation(block: false, alt: "0 less than x less than 1")[$0 < x < 1$] and #math.equation(block: false, alt: "0 less than y less than 1")[$0 < y < 1$]. Solve the problem


      #math.equation(block: true, alt: "u sub x x plus u sub y y equals 0 ,; u open parenthesis x , 0 close parenthesis equals sin open parenthesis π x close parenthesis , u open parenthesis x , 1 close parenthesis equals sin open parenthesis π x close parenthesis ,; u open parenthesis 0 , y close parenthesis equals sin open parenthesis π y close parenthesis , u open parenthesis 1 , y close parenthesis equals sin open parenthesis π y close parenthesis .")[$u_(x x) + u_(y y) = 0 , \ u ( x , 0 ) = sin ( π x ) , u ( x , 1 ) = sin ( π x ) , \ u ( 0 , y ) = sin ( π y ) , u ( 1 , y ) = sin ( π y ) .$]


      Hint: Use superposition.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[(challenging)]


      Using only your intuition find #math.equation(block: false, alt: "u open parenthesis 1 / 2 , 1 / 2 close parenthesis")[$u ( 1 / 2 , 1 / 2 )$], for the problem #math.equation(block: false, alt: "Δ u equals 0")[$Δ u = 0$], where #math.equation(block: false, alt: "u open parenthesis 0 , y close parenthesis equals u open parenthesis 1 , y close parenthesis equals 100")[$u ( 0 , y ) = u ( 1 , y ) = 100$] for #math.equation(block: false, alt: "0 less than y less than 1")[$0 < y < 1$], and #math.equation(block: false, alt: "u open parenthesis x , 0 close parenthesis equals u open parenthesis x , 1 close parenthesis equals 0")[$u ( x , 0 ) = u ( x , 1 ) = 0$] for #math.equation(block: false, alt: "0 less than x less than 1")[$0 < x < 1$]. Explain.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "R")[$R$] be the region described by #math.equation(block: false, alt: "0 less than x less than 1")[$0 < x < 1$] and #math.equation(block: false, alt: "0 less than y less than 1")[$0 < y < 1$]. Solve the problem


      #math.equation(block: true, alt: "Δ u equals 0 , u open parenthesis x , 0 close parenthesis equals ∑ n equals 1 infinity sin open parenthesis n π x close parenthesis , u open parenthesis x , 1 close parenthesis equals 0 , u open parenthesis 0 , y close parenthesis equals 0 , u open parenthesis 1 , y close parenthesis equals 0 .")[$Δ u = 0 , " " " " " " u ( x , 0 ) = ∑_(n = 1)^(∞) sin ( n π x ) , " " " " " " u ( x , 1 ) = 0 , " " " " " " u ( 0 , y ) = 0 , " " " " " " u ( 1 , y ) = 0 .$]


      Answer


      #math.equation(block: true, alt: "u open parenthesis x , y close parenthesis equals ∑ n equals 1 infinity the fraction 1 over n squared sin open parenthesis n π x close parenthesis open parenthesis the fraction sinh open parenthesis n π open parenthesis 1 minus y close parenthesis close parenthesis over sinh open parenthesis n π close parenthesis close parenthesis")[$u ( x , y ) = limits(∑)^(∞)_(n = 1) frac(1, n^(2)) sin ( n π x ) ( frac(sinh ( n π ( 1 − y ) ), sinh ( n π )) )$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "R")[$R$] be the region described by #math.equation(block: false, alt: "0 less than x less than 1")[$0 < x < 1$] and #math.equation(block: false, alt: "0 less than y less than 2")[$0 < y < 2$]. Solve the problem


      #math.equation(block: true, alt: "Δ u equals 0 , u open parenthesis x , 0 close parenthesis equals 0.1 sin open parenthesis π x close parenthesis , u open parenthesis x , 2 close parenthesis equals 0 , u open parenthesis 0 , y close parenthesis equals 0 , u open parenthesis 1 , y close parenthesis equals 0 .")[$Δ u = 0 , " " " " " " u ( x , 0 ) = 0.1 sin ( π x ) , " " " " " " u ( x , 2 ) = 0 , " " " " " " u ( 0 , y ) = 0 , " " " " " " u ( 1 , y ) = 0 .$]


      Answer


      #math.equation(block: true, alt: "u open parenthesis x , y close parenthesis equals 0.1 sin open parenthesis π x close parenthesis open parenthesis the fraction sinh open parenthesis π open parenthesis 2 minus y close parenthesis close parenthesis over sinh open parenthesis 2 π close parenthesis close parenthesis")[$u ( x , y ) = 0.1 sin ( π x ) ( frac(sinh ( π ( 2 − y ) ), sinh ( 2 π )) )$]
]


    
=== 
      #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/4%3A_Fourier_series_and_PDEs/4.10%3A_Dirichlet_Problem_in_the_Circle_and_the_Poisson_Kernel")[4.10: Dirichlet problem in the circle and the Poisson kernel]
    


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Using series solve #math.equation(block: false, alt: "Δ u equals 0 , u open parenthesis 1 , θ close parenthesis equals | θ |")[$Δ u = 0 , u ( 1 , θ ) = | θ |$] for #math.equation(block: false, alt: "minus π less than θ less than or equal to π")[$− π < θ ≤ π$].
]


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Using series solve #math.equation(block: false, alt: "Δ u equals 0 , u open parenthesis 1 , θ close parenthesis equals g open parenthesis θ close parenthesis")[$Δ u = 0 , u ( 1 , θ ) = g ( θ )$] for the following data. Hint: trig identities.


      + #math.equation(block: false, alt: "g open parenthesis θ close parenthesis equals 1 / 2 plus 3 sin open parenthesis θ close parenthesis plus cos open parenthesis 3 θ close parenthesis")[$g ( θ ) = 1 / 2 + 3 sin ( θ ) + cos ( 3 θ )$]
+ #math.equation(block: false, alt: "g open parenthesis θ close parenthesis equals cos open parenthesis 3 θ close parenthesis plus 3 sin open parenthesis 3 θ close parenthesis plus sin open parenthesis 9 θ close parenthesis")[$g ( θ ) = cos ( 3 θ ) + 3 sin ( 3 θ ) + sin ( 9 θ )$]
+ #math.equation(block: false, alt: "g open parenthesis θ close parenthesis equals 2 cos open parenthesis θ plus 1 close parenthesis")[$g ( θ ) = 2 cos ( θ + 1 )$]
+ #math.equation(block: false, alt: "g open parenthesis θ close parenthesis equals sin squared open parenthesis θ close parenthesis")[$g ( θ ) = sin^(2) ( θ )$]
]


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Using the Poisson kernel, give the solution to #math.equation(block: false, alt: "Δ u equals 0")[$Δ u = 0$], where #math.equation(block: false, alt: "u open parenthesis 1 , θ close parenthesis")[$u ( 1 , θ )$] is zero for #math.equation(block: false, alt: "θ")[$θ$] outside the interval #math.equation(block: false, alt: "[ minus π / 4 , π / 4 ]")[$[ − π / 4 , π / 4 ]$] and #math.equation(block: false, alt: "u open parenthesis 1 , θ close parenthesis")[$u ( 1 , θ )$] is #math.equation(block: false, alt: "1")[$1$] for #math.equation(block: false, alt: "θ")[$θ$] on the interval #math.equation(block: false, alt: "[ minus π / 4 , π / 4 ]")[$[ − π / 4 , π / 4 ]$].
]


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+ Draw a graph for the Poisson kernel as a function of #math.equation(block: false, alt: "α")[$α$] when #math.equation(block: false, alt: "r equals 1 / 2")[$r = 1 / 2$] and #math.equation(block: false, alt: "θ equals 0")[$θ = 0$].
+ Describe what happens to the graph when you make #math.equation(block: false, alt: "r")[$r$] bigger (as it approaches 1).
+ Knowing that the solution #math.equation(block: false, alt: "u open parenthesis r , θ close parenthesis")[$u ( r , θ )$] is the weighted average of #math.equation(block: false, alt: "g open parenthesis θ close parenthesis")[$g ( θ )$] with Poisson kernel as the weight, explain what your answer to part b means.
]


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Take the function #math.equation(block: false, alt: "g open parenthesis θ close parenthesis")[$g ( θ )$] to be the function #math.equation(block: false, alt: "x y equals cos open parenthesis θ close parenthesis sin open parenthesis θ close parenthesis")[$x y = cos ( θ ) sin ( θ )$] on the boundary. Use the series solution to find a solution to the Dirichlet problem #math.equation(block: false, alt: "Δ u equals 0 , u open parenthesis 1 , θ close parenthesis equals g open parenthesis θ close parenthesis")[$Δ u = 0 , u ( 1 , θ ) = g ( θ )$]. Now convert the solution to Cartesian coordinates #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$]. Is this solution surprising? Hint: use your trig identities.
]


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Carry out the computation we needed in the separation of variables and solve #math.equation(block: false, alt: "r squared R double prime plus r R prime minus n squared R equals 0")[$r^(2) R^(″) + r R^(′) − n^(2) R = 0$], for #math.equation(block: false, alt: "n equals 0 , 1 , 2 , 3 , . . .")[$n = 0 , 1 , 2 , 3 , . . .$].
]


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#emph[(challenging)]


      Derive the series solution to the Dirichlet problem if the region is a circle of radius #math.equation(block: false, alt: "ρ")[$ρ$] rather than #math.equation(block: false, alt: "1")[$1$]. That is, solve #math.equation(block: false, alt: "Δ u equals 0 , u open parenthesis ρ , θ close parenthesis equals g open parenthesis θ close parenthesis")[$Δ u = 0 , u ( ρ , θ ) = g ( θ )$].
]


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#emph[(challenging)]


      + Find the solution for #math.equation(block: false, alt: "Δ u equals 0")[$Δ u = 0$], #math.equation(block: false, alt: "u open parenthesis 1 , θ close parenthesis equals x squared y cubed plus 5 x squared")[$u ( 1 , θ ) = x^(2) y^(3) + 5 x^(2)$]. Write the answer in Cartesian coordinates.
+ Now solve #math.equation(block: false, alt: "Δ u equals 0")[$Δ u = 0$], #math.equation(block: false, alt: "u open parenthesis 1 , θ close parenthesis equals x to the power k y to the power l")[$u ( 1 , θ ) = x^(k) y^(l)$]. Write the solution in Cartesian coordinates.
+ Suppose you have a polynomial #math.equation(block: false, alt: "P open parenthesis x , y close parenthesis equals ∑ j equals 0 m ∑ k equals 0 n c sub j , k x to the power j y to the power k")[$P ( x , y ) = ∑_(j = 0)^(m) ∑_(k = 0)^(n) c_(j , k) x^(j) y^(k)$], solve #math.equation(block: false, alt: "Δ u equals 0")[$Δ u = 0$], #math.equation(block: false, alt: "u open parenthesis 1 , θ close parenthesis equals P open parenthesis x , y close parenthesis")[$u ( 1 , θ ) = P ( x , y )$] (that is, write down the formula for the answer). Write the answer in Cartesian coordinates.


      Notice the answer is again a polynomial in #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$]. See also Exercise #math.equation(block: false, alt: "5")[$5$].
]


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Using series solve #math.equation(block: false, alt: "Δ u equals 0 , u open parenthesis 1 , θ close parenthesis equals 1 plus ∑ n equals 1 infinity the fraction 1 over n squared sin open parenthesis n θ close parenthesis")[$Δ u = 0 , #h(0.222em) u ( 1 , θ ) = 1 + ∑_(n = 1)^(∞) frac(1, n^(2)) sin ( n θ )$].


      Answer


      #math.equation(block: true, alt: "u equals 1 ∑ n equals 1 infinity the fraction 1 over n squared r to the power n sin open parenthesis n θ close parenthesis")[$u = 1 limits(∑)^(∞)_(n = 1) frac(1, n^(2)) r^(n) sin ( n θ )$]
]


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Using the series solution find the solution to #math.equation(block: false, alt: "Δ u equals 0 , u open parenthesis 1 , θ close parenthesis equals 1 minus cos open parenthesis θ close parenthesis")[$Δ u = 0 , #h(0.222em) u ( 1 , θ ) = 1 − cos ( θ )$]. Express the solution in Cartesian coordinates (that is, using #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$]).


      Answer


      #math.equation(block: true, alt: "u equals 1 minus x")[$u = 1 − x$]
]


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+ Try and guess a solution to #math.equation(block: false, alt: "Δ u equals minus 1 , u open parenthesis 1 , θ close parenthesis equals 0")[$Δ u = − 1 , #h(0.222em) u ( 1 , θ ) = 0$]. Hint: try a solution that only depends on #math.equation(block: false, alt: "r")[$r$]. Also first, don’t worry about the boundary condition.
+ Now solve #math.equation(block: false, alt: "Δ u equals minus 1 , u open parenthesis 1 , θ close parenthesis equals sin open parenthesis 2 θ close parenthesis")[$Δ u = − 1 , #h(0.222em) u ( 1 , θ ) = sin ( 2 θ )$] using superposition.


      Answer


      + #math.equation(block: false, alt: "u equals the fraction minus 1 over 4 r squared plus the fraction 1 over 4")[$u = frac(− 1, 4) r^(2) + frac(1, 4)$]
+ #math.equation(block: false, alt: "u equals the fraction minus 1 over 4 r squared plus the fraction 1 over 4 plus r squared sin open parenthesis 2 θ close parenthesis")[$u = frac(− 1, 4) r^(2) + frac(1, 4) + r^(2) sin ( 2 θ )$]
]


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#emph[(challenging)]


      Derive the Poisson kernel solution if the region is a circle of radius #math.equation(block: false, alt: "ρ")[$ρ$] rather than #math.equation(block: false, alt: "1")[$1$]. That is, solve #math.equation(block: false, alt: "Δ u equals 0 , u open parenthesis ρ , θ close parenthesis equals g open parenthesis θ close parenthesis")[$Δ u = 0 , #h(0.222em) u ( ρ , θ ) = g ( θ )$].


      Answer


      #math.equation(block: true, alt: "u open parenthesis r , θ close parenthesis equals the fraction 1 over 2 π ∫ minus π π the fraction ρ squared minus r squared over ρ minus 2 r ρ cos open parenthesis θ minus α close parenthesis plus r squared g open parenthesis α close parenthesis d α")[$u ( r , θ ) = frac(1, 2 π) ∫_(− π)^(π) frac(ρ^(2) − r^(2), ρ − 2 r ρ cos ( θ − α ) + r^(2)) g ( α ) d α$]
]