KindiakMath

More Laplace Transforms

Problem 1. Prove that for any real constant a > 0,

\displaystyle \mathcal L\{ \sin (at) \} = \frac{a}{s^2+a^2},\quad \mathcal L\{ \cos (at) \} = \frac{s}{s^2+a^2}.

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Solution. By Euler’s formula, e^{i(at)} = \cos(at) + i \sin (at). Taking Laplace transforms on both sides and applying linearity,

\mathcal L\{ e^{i(at)} \} = \mathcal L\{ \cos (at) \} + i \cdot \mathcal L\{ \sin (at) \}.

On the left-hand side,

\displaystyle \begin{aligned} \mathcal L\{ e^{(ia)t} \} &= \frac{1}{s - ia} \\ &= \frac{1}{s-ia} \cdot \frac{s+ia}{s+ia} \\ &= \frac{s+ia}{s^2 + a^2} \\ &= \frac{s}{s^2 + a^2} + i \cdot \frac{a}{s^2 + a^2}.\end{aligned}

Combining these results,

\displaystyle \mathcal L\{ \cos (at) \} + i \cdot \mathcal L\{ \sin (at) \} = \frac{s}{s^2 + a^2} + i \cdot \frac{a}{s^2 + a^2}.

Problem 2. Prove that for any n, \mathcal L\{t^n\} = n!/s^{n+1}.

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Solution. By induction, we can prove that for any reasonably-defined function f,

\mathcal L\{ f^{(n)} (t) \} = s^n \mathcal L\{ f(t)\},

if f^{(k)}(0) = 0 for k = 0, 1, \dots, n-1.

Defining f(t) = t^n, we have f^{(k)}(0) = 0 for k = 0, 1, \dots, n-1 and f^{(n)}(t) = n!. Applying these results into the derivative result,

\mathcal L\{ n! \} = s^n \mathcal L\{ t^n \}.

Hence,

\displaystyle \begin{aligned} \mathcal L\{ t^n \} &= \frac 1{s^n} \cdot \mathcal L\{ n! \} \\ &= \frac 1{s^n} \cdot n! \cdot \mathcal L\{ 1 \} \\ &= \frac 1{s^n} \cdot n! \cdot \frac 1s = \frac{n!}{s^{n+1}}. \end{aligned}

Problem 3. Suppose there exists a function \delta such that U' = \delta. Prove that for any real constant a > 0, \mathcal L\{\delta(t - a)\} = e^{-as}.

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Solution. Assuming there exists such a function,

\begin{aligned} \mathcal L\{\delta(t-a)\} &= \mathcal L\{U'(t-a) \cdot (t-a)'\} \\ &= \mathcal L\{(U(t-a))'\} \\ &= s \cdot \mathcal L\{U(t-a)\} - U(0 - a) \\ &= s \cdot \frac{e^{-as}}{s} - 0 = e^{-as}.\end{aligned}

This notion, defined properly, is known as the Dirac delta distribution.

—Joel Kindiak, 13 Feb 25, 0152H

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