#set document(title: "6.1 The Laplace Transform", author: "Jiří Lebl")
#set page(width: 8.5in, height: auto, margin: 1in)
#import "@preview/cetz:0.5.2"
#set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en")
#show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math"))
#set par(justify: true, leading: 0.62em, spacing: 0.9em)
#set enum(spacing: 1.1em)   // room between list items so tall inline fractions don't collide
#set list(spacing: 1.1em)
#set table(stroke: 0.5pt + rgb("#c7ccd3"))

#let BLUE = rgb("#183B6F")     // brand navy — section bars + example/solution labels (white on navy 11.09:1)
#let ORANGE = rgb("#A94509")   // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text)
#let RED = rgb("#DC2626")      // brand error-600
#let GREEN = rgb("#059669")    // brand success-600 (decoration only; small green text uses green-text #007942)

#show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt,
  fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt,
  text(fill: white, weight: "bold", size: 19pt, it.body))
#show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt,
  fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt,
  text(fill: white, weight: "bold", size: 12pt, it.body))
#show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body)
#show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body)

#let examplebox(label, title, body) = block(width: 100%, breakable: true,
  fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt,
  above: 12pt, below: 12pt)[
  #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt,
    text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]]
  #body]

// rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade
#let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint,
    stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt),
    above: 11pt, below: 11pt)[
    #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)]
    #linebreak() #body]

#let solutionbox(body) = block(above: 4pt, below: 8pt)[
  #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body]

#let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"),
  stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[
  #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))]

// Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a
// consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the
// print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a
// degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero.
#let _STD_W = 3.5
#let _WIDE_W = 5.6
#let _MAX_H = 3.4
#let _ASPECT_WIDE = 2.2
#let _UPSCALE_MAX = 1.15
#let stdfig(body) = context {
  let m = measure(body)
  let w = calc.max(m.width / 1in, 0.01)
  let h = calc.max(m.height / 1in, 0.01)
  let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W }
  let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX)
  align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body)))
}

#show figure: set block(breakable: false)
#set figure(gap: 8pt)
#show figure.caption: set text(size: 8.5pt, fill: rgb("#555"))


    
== 6.1#h(0.6em)The Laplace Transform


    
=== Transform


    In this chapter we will discuss the Laplace transform#math.equation(block: false, alt: "to the power 1")[$1$]. The Laplace transform turns out to be a very efficient method to solve certain ODE problems. In particular, the transform can take a differential equation and turn it into an algebraic equation. If the algebraic equation can be solved, applying the inverse transform gives us our desired solution. The Laplace transform also has applications in the analysis of electrical circuits, NMR spectroscopy, signal processing, and elsewhere. Finally, understanding the Laplace transform will also help with understanding the related Fourier transform, which, however, requires more understanding of complex numbers.


    The Laplace transform also gives a lot of insight into the nature of the equations we are dealing with. It can be seen as converting between the time and the frequency domain. For example, take the standard equation


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


    We can think of #math.equation(block: false, alt: "t")[$t$] as time and #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] as incoming signal. The Laplace transform will convert the equation from a differential equation in time to an algebraic (no derivatives) equation, where the new independent variable #math.equation(block: false, alt: "s")[$s$] is the frequency.


    We can think of the #emph[Laplace transform] as a black box that eats functions and spits out functions in a new variable. We write #math.equation(block: false, alt: "ℒ { f open parenthesis t close parenthesis } equals F open parenthesis s close parenthesis")[$ℒ \{ f ( t ) \} = F ( s )$] for the Laplace transform of #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$]. It is common to write lower case letters for functions in the time domain and upper case letters for functions in the frequency domain. We use the same letter to denote that one function is the Laplace transform of the other. For example #math.equation(block: false, alt: "F open parenthesis s close parenthesis")[$F ( s )$] is the Laplace transform of #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$]. Let us define the transform.


    #math.equation(block: true, alt: "ℒ { f open parenthesis t close parenthesis } equals F open parenthesis s close parenthesis equals d e f ∫ 0 infinity e to the power minus s t f open parenthesis t close parenthesis d t .")[$ℒ \{ f ( t ) \} = F ( s ) limits(=)^(upright(d) upright(e) upright(f)) ∫_(0)^(∞) e^(− s t) f ( t ) d t .$]


    We note that we are only considering #math.equation(block: false, alt: "t greater than or equal to 0")[$t ≥ 0$] in the transform. Of course, if we think of #math.equation(block: false, alt: "t")[$t$] as time there is no problem, we are generally interested in finding out what will happen in the future (Laplace transform is one place where it is safe to ignore the past). Let us compute some simple transforms.


    #examplebox("Example 1")[][
Suppose #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals 1")[$f ( t ) = 1$], then


        #math.equation(block: true, alt: "ℒ { 1 } equals ∫ 0 infinity e to the power minus s t d t equals [ the fraction e to the power minus s t over minus s ] t equals 0 infinity equals lim sub h → infinity [ the fraction e to the power minus s t over minus s ] t equals 0 h equals lim sub h → infinity open parenthesis the fraction e to the power minus s h over minus s minus the fraction 1 over minus s close parenthesis equals the fraction 1 over s .")[$ℒ \{ 1 \} = ∫_(0)^(∞) e^(− s t) d t = attach([ frac(e^(− s t), − s) ], t: ∞, b: t = 0) = lim_(h → ∞) attach([ frac(e^(− s t), − s) ], t: h, b: t = 0) = lim_(h → ∞) ( frac(e^(− s h), − s) − frac(1, − s) ) = frac(1, s) .$]


        The limit (the improper integral) only exists if #math.equation(block: false, alt: "s greater than 0")[$s > 0$]. So #math.equation(block: false, alt: "ℒ { 1 }")[$ℒ \{ 1 \}$] is only defined for #math.equation(block: false, alt: "s greater than 0")[$s > 0$].
]


    #examplebox("Example 2")[][
Suppose #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals e to the power minus a t")[$f ( t ) = e^(− a t)$], then


        #math.equation(block: true, alt: "ℒ { e to the power minus a t } equals ∫ 0 infinity e to the power minus s t e to the power minus a t d t equals ∫ 0 infinity e to the power minus open parenthesis s plus a close parenthesis t d t equals [ the fraction e to the power minus open parenthesis s plus a close parenthesis t over minus s open parenthesis s plus a close parenthesis ] t equals 0 infinity equals the fraction 1 over s plus a .")[$ℒ \{ e^(− a t) \} = ∫_(0)^(∞) e^(− s t) e^(− a t) d t = ∫_(0)^(∞) e^(− ( s + a ) t) d t = attach([ frac(e^(− ( s + a ) t), − s ( s + a )) ], t: ∞, b: t = 0) = frac(1, s + a) .$]


        The limit only exists if #math.equation(block: false, alt: "s plus a greater than 0")[$s + a > 0$]. So #math.equation(block: false, alt: "ℒ { e to the power minus a t }")[$ℒ \{ e^(− a t) \}$] is only defined for #math.equation(block: false, alt: "s plus a greater than 0")[$s + a > 0$].
]


    #examplebox("Example 3")[][
Suppose #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals t")[$f ( t ) = t$], then using integration by parts


        #math.equation(block: true, alt: "ℒ { t } equals ∫ 0 infinity e to the power minus s t t d t; equals [ the fraction minus t e to the power minus s t over s ] t equals 0 infinity plus the fraction 1 over s ∫ 0 infinity e to the power minus s t d t; equals 0 plus the fraction 1 over s [ the fraction e to the power minus s t over minus s ] t equals 0 infinity; equals the fraction 1 over s squared .")[$ℒ \{ t \} = ∫_(0)^(∞) e^(− s t) t d t \ = attach([ frac(− t e^(− s t), s) ], t: ∞, b: t = 0) + frac(1, s) ∫_(0)^(∞) e^(− s t) d t \ = 0 + frac(1, s) attach([ frac(e^(− s t), − s) ], t: ∞, b: t = 0) \ = frac(1, s^(2)) .$]


        Again, the limit only exists if #math.equation(block: false, alt: "s greater than 0")[$s > 0$].
]


    #examplebox("Example 4")[][
A common function is the unit step function, which is sometimes called the #emph[Heaviside]#emph[function]#math.equation(block: false, alt: "squared")[$2$]. This function is generally given as


        #math.equation(block: true, alt: "0 i f t less than 0 ,; 1 i f t greater than or equal to 0 .")[$0 upright(i) upright(f) " " t < 0 , \ 1 upright(i) upright(f) " " t ≥ 0 .$]


        Let us find the Laplace transform of #math.equation(block: false, alt: "u open parenthesis t minus a close parenthesis")[$u ( t − a )$], where #math.equation(block: false, alt: "a greater than or equal to 0")[$a ≥ 0$] is some constant. That is, the function that is 0 for #math.equation(block: false, alt: "t less than a")[$t < a$] and 1 for #math.equation(block: false, alt: "t greater than or equal to a")[$t ≥ a$].


        #math.equation(block: true, alt: "ℒ { u open parenthesis t minus a close parenthesis } equals ∫ 0 infinity e to the power minus s t u open parenthesis t minus a close parenthesis d t equals ∫ a infinity e to the power minus s t d t equals [ the fraction e to the power minus s t over minus s ] t equals a e to the power minus a s equals the fraction e to the power minus a s over s ,")[$ℒ \{ u ( t − a ) \} = ∫_(0)^(∞) e^(− s t) u ( t − a ) d t = ∫_(a)^(∞) e^(− s t) d t = attach([ frac(e^(− s t), − s) ], t: e^(− a s), b: t = a) = frac(e^(− a s), s) ,$]


        where of course #math.equation(block: false, alt: "s greater than 0")[$s > 0$] (and #math.equation(block: false, alt: "a greater than or equal to 0")[$a ≥ 0$] as we said before).
]


    By applying similar procedures we can compute the transforms of many elementary functions. Many basic transforms are listed in Table #math.equation(block: false, alt: "1")[$1$].


    #figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#strong[#math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$]]], [#strong[#math.equation(block: false, alt: "{ f open parenthesis t close parenthesis }")[$\{ f ( t ) \}$]]]),
  [#math.equation(block: false, alt: "C")[$C$]], [#math.equation(block: false, alt: "the fraction C over s")[$frac(C, s)$]],
  [#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "the fraction 1 over s squared")[$frac(1, s^(2))$]],
  [#math.equation(block: false, alt: "t squared")[$t^(2)$]], [#math.equation(block: false, alt: "the fraction 2 over s cubed")[$frac(2, s^(3))$]],
  [#math.equation(block: false, alt: "t cubed")[$t^(3)$]], [#math.equation(block: false, alt: "the fraction 6 over s to the power 4")[$frac(6, s^(4))$]],
  [#math.equation(block: false, alt: "t to the power n")[$t^(n)$]], [#math.equation(block: false, alt: "the fraction n ! over s to the power s plus 1")[$frac(n !, s^(s + 1))$]],
  [#math.equation(block: false, alt: "e to the power minus a t")[$e^(− a t)$]], [#math.equation(block: false, alt: "the fraction 1 over s plus a")[$frac(1, s + a)$]],
  [#math.equation(block: false, alt: "sin open parenthesis ω t close parenthesis")[$sin ( ω t )$]], [#math.equation(block: false, alt: "the fraction ω over s squared plus ω squared")[$frac(ω, s^(2) + ω^(2))$]],
  [#math.equation(block: false, alt: "cos open parenthesis ω t close parenthesis")[$cos ( ω t )$]], [#math.equation(block: false, alt: "the fraction s over s squared plus ω squared")[$frac(s, s^(2) + ω^(2))$]],
  [#math.equation(block: false, alt: "sinh open parenthesis ω t close parenthesis")[$sinh ( ω t )$]], [#math.equation(block: false, alt: "the fraction ω over s squared minus ω squared")[$frac(ω, s^(2) − ω^(2))$]],
  [#math.equation(block: false, alt: "cosh open parenthesis ω t close parenthesis")[$cosh ( ω t )$]], [#math.equation(block: false, alt: "the fraction s over s squared minus ω squared")[$frac(s, s^(2) − ω^(2))$]],
  [#math.equation(block: false, alt: "u open parenthesis t minus a close parenthesis")[$u ( t − a )$]], [#math.equation(block: false, alt: "the fraction e to the power minus a s over s")[$frac(e^(− a s), s)$]],
))


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Verify all the entries in Table #math.equation(block: false, alt: "1")[$1$].
]


    Since the transform is defined by an integral. We can use the linearity properties of the integral. For example, suppose #math.equation(block: false, alt: "C")[$C$] is a constant, then


    #math.equation(block: true, alt: "ℒ { f open parenthesis t close parenthesis } equals ∫ o infinity e to the power minus s t C f open parenthesis t close parenthesis d t equals C ∫ 0 infinity e to the power minus s t f open parenthesis t close parenthesis d t equals C ℒ { f open parenthesis t close parenthesis } .")[$ℒ \{ f ( t ) \} = ∫_(o)^(∞) e^(− s t) C f ( t ) d t = C ∫_(0)^(∞) e^(− s t) f ( t ) d t = C ℒ \{ f ( t ) \} .$]


    So we can “pull out” a constant out of the transform. Similarly we have linearity. Since linearity is very important we state it as a theorem.


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#strong[Linearity of the Laplace Transform]


      Suppose that #math.equation(block: false, alt: "A")[$A$], #math.equation(block: false, alt: "B")[$B$], and #math.equation(block: false, alt: "C")[$C$] are constants, then


      #math.equation(block: true, alt: "ℒ { A f open parenthesis t close parenthesis plus B g open parenthesis t close parenthesis } equals A ℒ { f open parenthesis t close parenthesis } plus B ℒ { g open parenthesis t close parenthesis }")[$ℒ \{ A f ( t ) + B g ( t ) \} = A ℒ \{ f ( t ) \} + B ℒ \{ g ( t ) \}$]


      and in particular


      #math.equation(block: true, alt: "ℒ { C f open parenthesis t close parenthesis } equals C ℒ { f open parenthesis t close parenthesis } .")[$ℒ \{ C f ( t ) \} = C ℒ \{ f ( t ) \} .$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Verify theorem #math.equation(block: false, alt: "1")[$1$]. That is, show that #math.equation(block: false, alt: "ℒ { A f open parenthesis t close parenthesis plus B g open parenthesis t close parenthesis } equals A ℒ { f open parenthesis t close parenthesis } plus B ℒ { g open parenthesis t close parenthesis }")[$ℒ \{ A f ( t ) + B g ( t ) \} = A ℒ \{ f ( t ) \} + B ℒ \{ g ( t ) \}$].
]


    These rules together with Table #math.equation(block: false, alt: "1")[$1$] make it easy to find the Laplace transform of a whole lot of functions already. But be careful. It is a common mistake to think that the Laplace transform of a product is the product of the transforms. In general


    #math.equation(block: true, alt: "ℒ { f open parenthesis t close parenthesis g open parenthesis t close parenthesis } not equal to ℒ { f open parenthesis t close parenthesis } ℒ { g open parenthesis t close parenthesis } .")[$ℒ \{ f ( t ) g ( t ) \} ≠ ℒ \{ f ( t ) \} ℒ \{ g ( t ) \} .$]


    It must also be noted that not all functions have a Laplace transform. For example, the function #math.equation(block: false, alt: "1 / t")[$1 / t$] does not have a Laplace transform as the integral diverges for all #math.equation(block: false, alt: "s")[$s$]. Similarly, #math.equation(block: false, alt: "tan t")[$tan t$] or #math.equation(block: false, alt: "e to the power t squared")[$e^(t^(2))$]do not have Laplace transforms.


    
=== Existence and Uniqueness


    Let us consider when does the Laplace transform exist in more detail. First let us consider functions of exponential order. The function #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] is of #emph[exponential order] as #math.equation(block: false, alt: "t")[$t$] goes to infinity if


    #math.equation(block: true, alt: "| f open parenthesis t close parenthesis | less than or equal to M e to the power c t ,")[$| f ( t ) | ≤ M e^(c t) ,$]


    for some constants #math.equation(block: false, alt: "M")[$M$] and #math.equation(block: false, alt: "c")[$c$], for sufficiently large #math.equation(block: false, alt: "t")[$t$] (say for all #math.equation(block: false, alt: "t greater than t sub o")[$t > t_(o)$] for some #math.equation(block: false, alt: "t sub o")[$t_(o)$]). The simplest way to check this condition is to try and compute


    #math.equation(block: true, alt: "lim sub t → infinity the fraction f open parenthesis t close parenthesis over e to the power c t")[$lim_(t → ∞) frac(f ( t ), e^(c t))$]


    If the limit exists and is finite (usually zero), then #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] is of exponential order.


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Use L'Hopital's rule from calculus to show that a polynomial is of exponential order. Hint: Note that a sum of two exponential order functions is also of exponential order. Then show that #math.equation(block: false, alt: "t to the power n")[$t^(n)$] is of exponential order for any #math.equation(block: false, alt: "n")[$n$].
]


    For an exponential order function we have existence and uniqueness of the Laplace transform.


    
      The existence theorem's own integral, drawn. For f(t) = e^t — of exponential order with c = 1 — the transform integrand e^(−st) f(t) = e^((1−s)t) is the solid curve, with f itself dashed behind it, and only t ≥ 0 is plotted because the definition integrates from 0 to ∞. Drag s down through the threshold s = c = 1: above it the integrand decays and the improper integral converges, so F(s) exists; at s = 1 it flattens to the constant 1 and the area is already infinite; below it the integrand grows and there is no transform at all.
      {"functions":\[{"color":"\#1f4e79","domain":\[0,20\],"expression":"exp((1-s)\*t)","expression2":"","id":"f-integrand","inequality":"lt","lineDash":\[\],"lineWidth":2.5,"mode":"cartesian","tMax":10,"tMin":-10,"variable":"t","visible":true},{"color":"\#6b3fa0","domain":\[0,20\],"expression":"exp(t)","expression2":"","id":"f-original","inequality":"lt","lineDash":\[6,4\],"lineWidth":1.5,"mode":"cartesian","tMax":10,"tMin":-10,"variable":"t","visible":true}\],"grid":{"axisColor":"\#333333","gridColor":"\#cccccc","majorSpacing":1,"minorGridColor":"\#eeeeee","showAxes":true,"showLabels":true},"parameters":\[{"id":"p-s","label":"Transform variable s","max":3,"min":0,"name":"s","step":0.05,"unit":"","value":1.5}\],"title":"Transform integrand and the s \> c threshold","version":1,"viewport":{"centerX":3.9,"centerY":2.2,"scale":85}}
    
    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#strong[Existence]


      Let #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] be continuous and of exponential order for a certain constant #math.equation(block: false, alt: "c")[$c$]. Then #math.equation(block: false, alt: "F open parenthesis s close parenthesis equals ℒ { f open parenthesis t close parenthesis }")[$F ( s ) = ℒ \{ f ( t ) \}$] is defined for all #math.equation(block: false, alt: "s greater than c")[$s > c$].
]


    The existence is not difficult to see. Let #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] be of exponential order, that is #math.equation(block: false, alt: "| f open parenthesis t close parenthesis | less than or equal to M e to the power c t")[$| f ( t ) | ≤ M e^(c t)$] for all #math.equation(block: false, alt: "t greater than 0")[$t > 0$] (for simplicity #math.equation(block: false, alt: "t sub 0 equals 0")[$t_(0) = 0$]). Let #math.equation(block: false, alt: "s greater than c")[$s > c$], or in other words #math.equation(block: false, alt: "open parenthesis c minus s close parenthesis less than 0")[$( c − s ) < 0$]. By the comparison theorem from calculus, the improper integral defining #math.equation(block: false, alt: "ℒ { f open parenthesis t close parenthesis }")[$ℒ \{ f ( t ) \}$] exists if the following integral exists


    #math.equation(block: true, alt: "∫ 0 infinity e to the power minus s t open parenthesis M e to the power c t close parenthesis d t equals M ∫ 0 infinity e to the power open parenthesis c minus s close parenthesis t d t equals M [ the fraction e to the power open parenthesis c minus s close parenthesis t over c minus s ] t equals 0 infinity equals the fraction M over c minus s .")[$∫_(0)^(∞) e^(− s t) ( M e^(c t) ) d t = M ∫_(0)^(∞) e^(( c − s ) t) d t = M attach([ frac(e^(( c − s ) t), c − s) ], t: ∞, b: t = 0) = frac(M, c − s) .$]


    The transform also exists for some other functions that are not of exponential order, but that will not be relevant to us. Before dealing with uniqueness, let us note that for exponential order functions we obtain that their Laplace transform decays at infinity:


    #math.equation(block: true, alt: "lim sub s → infinity F open parenthesis s close parenthesis equals 0")[$lim_(s → ∞) F ( s ) = 0$]


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#strong[Uniqueness]


      Let #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] and #math.equation(block: false, alt: "g open parenthesis t close parenthesis")[$g ( t )$] be continuous and of exponential order. Suppose that there exists a constant #math.equation(block: false, alt: "C")[$C$], such that #math.equation(block: false, alt: "F open parenthesis s close parenthesis equals G open parenthesis s close parenthesis")[$F ( s ) = G ( s )$] for all #math.equation(block: false, alt: "s greater than C")[$s > C$]. Then #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals g open parenthesis t close parenthesis")[$f ( t ) = g ( t )$] for all #math.equation(block: false, alt: "t greater than or equal to 0")[$t ≥ 0$].
]


    Both theorems hold for piecewise continuous functions as well. Recall that piecewise continuous means that the function is continuous except perhaps at a discrete set of points where it has jump discontinuities like the Heaviside function. Uniqueness however does not “see” values at the discontinuities. So we can only conclude that #math.equation(block: false, alt: "F open parenthesis s close parenthesis equals G open parenthesis s close parenthesis")[$F ( s ) = G ( s )$] outside of discontinuities. For example, the unit step function is sometimes defined using #math.equation(block: false, alt: "u open parenthesis 0 close parenthesis equals 1 / 2")[$u ( 0 ) = 1 / 2$]. This new step function, however, has the exact same Laplace transform as the one we defined earlier where #math.equation(block: false, alt: "u open parenthesis 0 close parenthesis equals 1")[$u ( 0 ) = 1$].


    
=== 6.1.3Inverse Transform


    As we said, the Laplace transform will allow us to convert a differential equation into an algebraic equation. Once we solve the algebraic equation in the frequency domain we will want to get back to the time domain, as that is what we are interested in. If we have a function #math.equation(block: false, alt: "F open parenthesis s close parenthesis")[$F ( s )$], to be able to find #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] such that #math.equation(block: false, alt: "ℒ { f open parenthesis t close parenthesis } equals F open parenthesis s close parenthesis")[$ℒ \{ f ( t ) \} = F ( s )$], we need to first know if such a function is unique. It turns out we are in luck by Theorem #math.equation(block: false, alt: "3")[$3$]. So we can without fear make the following definition.


    If #math.equation(block: false, alt: "F open parenthesis s close parenthesis equals ℒ { f open parenthesis t close parenthesis }")[$F ( s ) = ℒ \{ f ( t ) \}$] for some function #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$]. We define the #emph[inverse Laplace transform] as


    #math.equation(block: true, alt: "ℒ to the power minus 1 { F open parenthesis s close parenthesis } equals d e f f open parenthesis t close parenthesis")[$ℒ^(− 1) \{ F ( s ) \} limits(=)^(upright(d) upright(e) upright(f)) f ( t )$]


    There is an integral formula for the inverse, but it is not as simple as the transform itself—it requires complex numbers and path integrals. For us it will suffice to compute the inverse using Table #math.equation(block: false, alt: "1")[$1$].


    #examplebox("Example 5")[][
Find the inverse Laplace transform of #math.equation(block: false, alt: "F open parenthesis s close parenthesis equals the fraction 1 over s plus 1")[$F ( s ) = frac(1, s + 1)$]


      
      #solutionbox[
We look at the table to find


        #math.equation(block: true, alt: "ℒ to the power minus 1 { the fraction 1 over s plus 1 } equals e to the power minus t")[$ℒ^(− 1) \{ frac(1, s + 1) \} = e^(− t)$]
]
]


    As the Laplace transform is linear, the inverse Laplace transform is also linear. That is,


    #math.equation(block: true, alt: "ℒ to the power minus 1 { A F open parenthesis s close parenthesis plus B G open parenthesis s close parenthesis } equals A ℒ to the power minus 1 { F open parenthesis s close parenthesis } plus B ℒ to the power minus 1 { G open parenthesis s close parenthesis }")[$ℒ^(− 1) \{ A F ( s ) + B G ( s ) \} = A ℒ^(− 1) \{ F ( s ) \} + B ℒ^(− 1) \{ G ( s ) \}$]


    Of course, we also have #math.equation(block: false, alt: "ℒ to the power minus 1 { A F open parenthesis s close parenthesis equals A ℒ to the power minus 1 { F open parenthesis s close parenthesis } .")[$ℒ^(− 1) \{ A F ( s ) = A ℒ^(− 1) \{ F ( s ) \} .$] Let us demonstrate how linearity can be used.


    #examplebox("Example 6")[][
Find the inverse Laplace transform of


        #math.equation(block: true, alt: "F open parenthesis s close parenthesis equals the fraction s squared plus s plus 1 over s cubed plus s .")[$F ( s ) = frac(s^(2) + s + 1, s^(3) + s) .$]


      
      #solutionbox[
First we use the #emph[method of partial fractions]to write #math.equation(block: false, alt: "F")[$F$] in a form where we can use Table #math.equation(block: false, alt: "1")[$1$]. We factor the denominator as #math.equation(block: false, alt: "s open parenthesis s squared plus 1 close parenthesis")[$s ( s^(2) + 1 )$] and write


        #math.equation(block: true, alt: "the fraction s squared plus s plus 1 over s cubed plus s equals the fraction A over s plus the fraction B s plus C over s squared plus 1")[$frac(s^(2) + s + 1, s^(3) + s) = frac(A, s) + frac(B s + C, s^(2) + 1)$]


        Putting the right hand side over a common denominator and equating the numerators we get #math.equation(block: false, alt: "A open parenthesis s squared plus 1 close parenthesis plus s open parenthesis B s plus C close parenthesis equals s squared plus s plus 1")[$A ( s^(2) + 1 ) + s ( B s + C ) = s^(2) + s + 1$]. Expanding and equating coefficients we obtain #math.equation(block: false, alt: "A plus B equals 1")[$A + B = 1$], #math.equation(block: false, alt: "C equals 1")[$C = 1$], #math.equation(block: false, alt: "A equals 1")[$A = 1$] and thus #math.equation(block: false, alt: "B equals 0")[$B = 0$]. In other words,


        #math.equation(block: true, alt: "F open parenthesis s close parenthesis equals the fraction s squared plus s plus 1 over s cubed plus s equals the fraction 1 over s plus the fraction 1 over s squared plus 1")[$F ( s ) = frac(s^(2) + s + 1, s^(3) + s) = frac(1, s) + frac(1, s^(2) + 1)$]


        By linearity of the inverse Laplace transform we get


        #math.equation(block: true, alt: "ℒ to the power minus 1 { the fraction s squared plus s plus 1 over s cubed plus s } equals ℒ to the power minus 1 { the fraction 1 over s } plus ℒ to the power minus 1 { the fraction 1 over s squared plus 1 } equals 1 plus sin t .")[$ℒ^(− 1) \{ frac(s^(2) + s + 1, s^(3) + s) \} = ℒ^(− 1) \{ frac(1, s) \} + ℒ^(− 1) \{ frac(1, s^(2) + 1) \} = 1 + sin t .$]
]
]


    
=== Shifting Property of Laplace Transforms


    Another useful property is the so-called #emph[shifting property] or the#emph[first shifting property]


    #math.equation(block: true, alt: "ℒ { e to the power minus a t f open parenthesis t close parenthesis } equals F open parenthesis s plus a close parenthesis")[$ℒ \{ e^(− a t) f ( t ) \} = F ( s + a )$]


    where #math.equation(block: false, alt: "F open parenthesis s close parenthesis")[$F ( s )$] is the Laplace transform of #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] and #math.equation(block: false, alt: "a")[$a$] is a constant.


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Derive the first shifting property from the definition of the Laplace transform.
]


    The shifting property can be used, for example, when the denominator is a more complicated quadratic that may come up in the method of partial fractions. We complete the square and write such quadratics as #math.equation(block: false, alt: "open parenthesis s plus a close parenthesis squared plus b")[$( s + a )^(2) + b$] and then use the shifting property.


    #examplebox("Example 7")[][
Find #math.equation(block: true, alt: "ℒ to the power minus 1 { the fraction 1 over s squared plus 4 s plus 8 } .")[$ℒ^(− 1) \{ frac(1, s^(2) + 4 s + 8) \} .$]


      
      #solutionbox[
First we complete the square to make the denominator #math.equation(block: false, alt: "open parenthesis s plus 2 close parenthesis squared plus 4")[$( s + 2 )^(2) + 4$]. Next we find


        #math.equation(block: true, alt: "ℒ { the fraction 1 over s squared plus 4 } equals the fraction 1 over 2 sin open parenthesis 2 t close parenthesis .")[$ℒ \{ frac(1, s^(2) + 4) \} = frac(1, 2) sin ( 2 t ) .$]


        Putting it all together with the shifting property, we find


        #math.equation(block: true, alt: "ℒ { the fraction 1 over s squared plus 4 s plus 8 } equals ℒ { the fraction 1 over open parenthesis s plus 2 close parenthesis squared plus 4 } equals the fraction 1 over 2 e to the power minus 2 t sin open parenthesis 2 t close parenthesis .")[$ℒ \{ frac(1, s^(2) + 4 s + 8) \} = ℒ \{ frac(1, ( s + 2 )^(2) + 4) \} = frac(1, 2) e^(− 2 t) sin ( 2 t ) .$]
]
]


    In general, we want to be able to apply the Laplace transform to #emph[rational functions], that is functions of the form


    #math.equation(block: true, alt: "the fraction F open parenthesis s close parenthesis over G open parenthesis s close parenthesis")[$frac(F ( s ), G ( s ))$]


    where #math.equation(block: false, alt: "F open parenthesis s close parenthesis")[$F ( s )$] and #math.equation(block: false, alt: "G open parenthesis s close parenthesis")[$G ( s )$] are polynomials. Since normally, for the functions that we are considering, the Laplace transform goes to zero as #math.equation(block: false, alt: "s → infinity")[$s → ∞$], it is not hard to see that the degree of #math.equation(block: false, alt: "F open parenthesis s close parenthesis")[$F ( s )$] must be smaller than that of #math.equation(block: false, alt: "G open parenthesis s close parenthesis")[$G ( s )$]. Such rational functions are called #emph[proper rational functions]and we can always apply the method of partial fractions. Of course this means we need to be able to factor the denominator into linear and quadratic terms, which involves finding the roots of the denominator,


    
=== Footnotes


    \[1\] Just like the Laplace equation and the Laplacian, the Laplace transform is also named after Pierre-Simon, marquis de Laplace (1749 – 1827).


    \[2\] The function is named after the English mathematician, engineer, and physicist Oliver Heaviside (1850–1925). Only by coincidence is the function “heavy” on “one side.”