5.4 Exercises
These are homework exercises to accompany Libl's "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.
Your Turn
Find eigenvalues and eigenfunctions of
y
″
+
λ
y
=
0
,
y
(
0
)
−
y
′
(
0
)
=
0
,
y
(
1
)
=
0
.
y''+ \lambda y=0,~~~y(0)-y'(0)=0,~~~y(1)=0.
Your Turn
Expand the function f ( x ) = x f(x)=x 0 ≤ x ≤ 1 0 \leq x \leq 1
y
″
+
λ
y
=
0
,
y
′
(
0
)
=
0
,
y
(
1
)
=
0
.
y''+ \lambda y=0,~~~y'(0)=0,~~~y(1)=0.
Your Turn
Suppose that you had a Sturm-Liouville problem on the interval [ 0 , 1 ] [0,1] y n ( x ) = sin ( γ n x ) y_n(x)=\sin(\gamma nx) γ > 0 \gamma >0 f ( x ) = x , 0 < x < 1 f(x)=x, 0<x<1
Your Turn
Find eigenvalues and eigenfunctions of
y
′
(
4
)
+
λ
y
=
0
,
y
(
0
)
=
0
,
y
′
(
0
)
=
0
,
y
(
1
)
=
0
y
′
(
1
)
=
0
.
y'^{(4)}+ \lambda y=0,~~~y(0)=0,~~~y'(0)=0,~~~y(1)=0~~~y'(1)=0.
This problem is not a Sturm-Liouville problem, but the idea is the same.
Your Turn
(more challenging)
Find eigenvalues and eigenfunctions for
d
d
x
(
e
x
y
′
)
+
λ
e
x
y
=
0
,
y
(
0
)
=
0
,
y
(
1
)
=
0
.
\frac{d}{dx}(e^xy')+ \lambda e^xy=0,~~~y(0)=0,~~~y(1)=0.
Hint: First write the system as a constant coefficient system to find general solutions. Do note that Theorem 5.1.1 guarantees λ ≥ 0 \lambda \geq 0
Your Turn
Find eigenvalues and eigenfunctions of
y
″
+
λ
y
=
0
,
y
(
−
1
)
=
0
,
y
(
1
)
=
0
.
y''+ \lambda y=0,~~~y(-1)=0,~~~y(1)=0.
Answer
λ n = ( 2 n − 1 ) π 2 , n = 1 , 2 , 3 , ⋯ , \lambda_{n}=\frac{(2n-1)\pi}{2},\:n=1,\: 2,\: 3,\cdots, y n = cos ( ( 2 n − 1 ) π 2 x ) y_{n}=\cos\left(\frac{(2n-1)\pi}{2}x\right)
Your Turn
Put the following problems into the standard form for Sturm-Liouville problems, that is, find p ( x ) , q ( x ) , r ( x ) , α 1 , α , β 1 , β 1 , p(x),q(x), r(x), \alpha_1,\alpha_,\beta_1,\beta_1,
x y ″ + λ y = 0 xy''+\lambda y=0 0 < x < 1 , y ( 0 ) = 0 , y ( 1 ) = 0 , 0<x<1,\: y(0)=0,\: y(1)=0, ( 1 + x 2 ) y ″ + 2 x y ′ + ( λ − x 2 ) y = 0 (1+x^2)y''+2xy'+(\lambda -x^2)y=0 − 1 < x < 1 , y ( − 1 ) = 0 , y ( 1 ) + y ′ ( 1 ) = 0 -1<x<1,\: y(-1)=0,\: y(1)+y'(1)=0 Answer
p ( x ) = 1 , q ( x ) = 0 , r ( x ) = 1 x , α 1 = 1 , α 2 = 0 , β 1 = 1 , β 2 = 0 p(x)=1,\: q(x)=0,\: r(x)=\frac{1}{x},\:\alpha_{1}=1,\:\alpha_{2}=0,\:\beta_{1}=1,\:\beta_{2}=0 p ( x ) = 1 + x 2 , q ( x ) = x 2 , r ( x ) = 1 , α 1 = 1 , α 2 = 0 , β 1 = 1 , β 2 = 1 p(x)=1+x^{2},\: q(x)=x^{2},\: r(x)=1,\:\alpha_{1}=1,\:\alpha_{2}=0,\:\beta_{1}=1,\:\beta_{2}=1 Your Turn
Suppose you have a beam of length 5 5 y y x x ( 0 < x < 5 ) (0<x<5) y t t + 4 y x x x x = 0 y_{tt}+4y_{xxxx}=0 x ( 5 − x ) x(5-x) 2 2 x x y y
Your Turn
Suppose you have a beam of length 5 5 x = 5 x=5 u u x x ( 0 < x < 5 ) (0<x<5) u t t = 4 u x x u_{tt}=4u_{xx} x − 5 50 \frac{x-5}{50} − ( x − 5 ) 100 \frac{-(x-5)}{100} u u
Your Turn
Suppose the beam is L L
Your Turn
Suppose you have
a
4
y
x
x
x
x
+
y
t
t
=
0
(
0
<
x
<
1
,
t
>
0
)
,
y
(
0
,
t
)
=
y
x
x
(
0
,
t
)
=
0
,
y
(
1
,
t
)
=
y
x
x
(
1
,
t
)
=
0
,
y
(
x
,
0
)
=
f
(
x
)
,
y
t
(
x
,
0
)
=
g
(
x
)
.
\begin{align}\begin{aligned} & a^4 y_{xxxx} + y_{tt} = 0 \quad (0 < x < 1, t > 0), \\ & y(0,t) = y_{xx}(0,t) = 0,\\ & y(1,t) = y_{xx}(1,t) = 0,\\ & y(x,0) = f(x), \quad y_{t}(x,0) = g(x). \end{aligned}\end{align}
That is, you have also an initial velocity. Find a series solution. Hint: Use the same idea as we did for the wave equation.
Your Turn
Suppose you have a beam of length 1 1 y y x x 0 < x < 1 0<x<1 y t t + 4 y x x x x = 0 y_{tt}+4y_{xxxx}=0 sin ( π x ) \sin(\pi x) 0 0 y y
Answer
y
(
x
,
t
)
=
sin
(
π
x
)
cos
(
2
π
2
t
)
y(x,t)=\sin (\pi x)\cos (2\pi^{2}t)
Your Turn
Suppose you have a beam of length 10 10 y y x x 0 < x < 10 0<x<10 y t t + 9 y x x x x = 0 y_{tt}+9y_{xxxx}=0 sin ( π x ) \sin(\pi x) x ( 10 − x ) x(10-x)
Answer
9 y x x x x + y t t = 0 9y_{xxxx}+y_{tt}=0 ( 0 < x < 10 , t > 0 ) (0<x<10,\: t>0) y ( 0 , t ) = y x ( 0 , t ) = 0 \quad y(0,t)=y_{x}(0,t)=0 y ( 10 , t ) = y x ( 10 , t ) = 0 \quad y(10,t)=y_{x}(10,t)=0 y ( x , 0 ) = sin ( π x ) \quad y(x,0)=\sin (\pi x) y t ( x , 0 ) = x ( 10 − x ) \quad y_{t}(x,0)=x(10-x)
Your Turn
Suppose that the forcing function for the vibrating string is F 0 sin ( ω t ) F_0 \sin(\omega t) y p y_p
Your Turn
Take the forced vibrating string. Suppose that L = 1 , a = 1 L=1,a=1 1 1 0 < x < 1 0<x<1 − 1 -1 − 1 < x < 0 -1<x<0 5.3 .1 \PageIndex{5.3.1}
Your Turn
The units are cgs (centimeters-grams-seconds). For k = 0.005 , ω = 1.991 × 10 − 7 , A 0 = 20 k=0.005, \omega =1.991 \times 10^{-7},A_0=20 ± 10 \pm 10
Your Turn
Take the forced vibrating string. Suppose that L = 1 , a = 1 L=1,a=1 | x | − 1 2 |x|-\frac{1}{2} − 1 < x < 1 -1<x<1
Answer
y
p
(
x
,
t
)
=
∑
n
odd
n
=
1
∞
−
4
n
4
π
4
(
cos
(
n
π
x
)
−
cos
(
n
π
)
−
1
sin
(
n
π
)
sin
(
n
π
x
)
−
1
)
cos
(
n
π
t
)
.
y_{p}(x,t)=\sum\limits_{\overset{n=1}{n\text{ odd}}}^\infty \frac{-4}{n^{4}\pi^{4}}\left(\cos (n\pi x)-\frac{\cos (n\pi )-1}{\sin (n\pi )}\sin (n\pi x)-1\right)\cos (n\pi t).
Your Turn
The units are cgs (centimeters-grams-seconds). For k = 0.01 , ω = 1.991 × 10 − 7 , A 0 = 25 k=0.01, \omega =1.991 \times 10^{-7},A_0=25
Answer
Approximately 1991 centimeters
Adapted from Differential Equations for Engineers by Jiří Lebl (https://www.jirka.org/diffyqs/), © Jiří Lebl, licensed under CC BY-SA 4.0. Changes were made. License: CC-BY-SA-4.0 .
