STEP III 2009 4

4

Write \mathcal L \{f\} for the Laplace transform of f.

(i)

We have:

\begin{align*}\mathcal L \left\{e^{-bt} f(t)\right\} & = \int_0^\infty e^{-st} e^{-bt} f(t) \mathrm dt \\ & = \int_0^\infty e^{-(s + b)t } f(t) \mathrm dt \\ & = \int_0^\infty e^{-ut} f(t) \mathrm dt\end{align*}

where we write u = s + b for clarity. This is just the Laplace transform of f evaluated at u, so:

\begin{align*}\mathcal L \left\{e^{-bt} f(t)\right\} & = \int_0^\infty e^{-ut} f(t) \mathrm dt \\ & = F(u) \\ & = F(s+b)\end{align*}

(ii)

The Laplace transform of f(at) is given by:

\displaystyle \mathcal L \{f(at)\} = \int_0^\infty e^{-st} f(at) \mathrm dt

Substituting u = at, we have:

\displaystyle \int_0^\infty e^{-st} f(at) \mathrm dt = \frac 1 a\int_0^\infty e^{-u \times \frac s a} f(t) \mathrm dt

The integral is just the Laplace transform of f evaluated at \dfrac s a, so:

\displaystyle \mathcal L \{f(at)\} = a^{-1} F \left(\frac s a\right)

(iii)

The Laplace transform of f'(t) is given by:

\displaystyle \mathcal L\{f'(t)\} = \int_0^\infty e^{-st} f'(t) \mathrm dt

Using IBP with the notation:

\displaystyle \int uv' = uv - \int u'v

we set u = e^{-st}, v' = f'(t), with a view to write this integral in terms of the Laplace transform of f, which gives:

\begin{align*}\int_0^\infty e^{-st} f'(t) \mathrm dt & = \left[e^{-st} f(t)\right]_0^\infty - \int_0^\infty \left(-se^{-st} f(t)\right) \mathrm dt \\ & = \lim_{t \to \infty} e^{-st} f(t) - e^0 f(0) + s \int_0^\infty e^{-st} f(t) \mathrm dt \\ & = -f(0) + s F(s)\end{align*}

(iv)

The Laplace transform of \sin t is given by:

\displaystyle \mathcal L \{\sin t\} = \int_0^\infty e^{-st} \sin t \mathrm dt

We want to use IBP twice, first with u = e^{-st}, v' = \sin t: (u' = -se^{-st}, v = -\cos t)

\begin{align*} \int_0^\infty e^{-st} \sin t \mathrm dt &= \left[-e^{-st} \cos t\right]_0^\infty - \int_0^\infty se^{-st} \cos t \mathrm dt \\ & = \lim_{t \to \infty} \left(-e^{-st} \cos t\right) + e^0 \cos 0 - s \int_0^\infty e^{-st} \cos t \mathrm dt \\ & = 1 - s \int_0^\infty e^{-st} \cos t \mathrm dt\end{align*}

We now want to do IBP on:

\displaystyle \int_0^\infty e^{-st} \cos t \mathrm dt

with u = e^{-st}, v' = \cos t (u' = -se^{-st}, v = \sin t). We have:

\begin{align*}\int_0^\infty e^{-st} \cos t \mathrm dt & = \left[e^{-st} \sin t\right]_0^\infty - \int_0^\infty \left(-se^{-st} \sin t \right) \mathrm dt \\ & = \lim_{t \to \infty} e^{-st} \sin t - e^0 \sin 0 + s \int_0^\infty e^{-st} \sin t \mathrm dt \\ & = s \int_0^\infty e^{-st} \sin t \mathrm dt \end{align*}

Substituting back we have:

\displaystyle \int_0^\infty e^{-st} \sin t \mathrm dt = 1 - s^2 \int_0^\infty e^{-st} \sin t \mathrm dt

so:

\displaystyle \left(1 + s^2\right) \int_0^\infty e^{-st} \sin t \mathrm dt = 1

giving:

\displaystyle \int_0^\infty e^{-st} \sin t \mathrm dt = \frac 1 {s^2 + 1}

(iv)

Using part (iii) and part (iv) we know that the Laplace transform of \cos t is:

\displaystyle \frac s {s^2 + 1} - \sin 0 = \frac s {s^2 + 1}

noting that (\sin t)' = \cos t.

Using part (ii) we know that the Laplace transform of \cos qt is:

\begin{align*}\frac 1 a \times \frac {\frac s a} {\left(\frac s a\right)^2 + 1} & = \frac s {a^2 \left(\frac {s^2} {a^2} + 1\right)} \\ & = \frac s {s^2 + a^2}\end{align*}

Using part (i), the Laplace transform of e^{-pt} \cos qt is then:

\displaystyle \frac {s + p} {(s + p)^2 + a^2}

More on Laplace transforms (including more precise formulations) can be found on https://proofwiki.org/wiki/Category:Laplace_Transforms and https://en.wikipedia.org/wiki/Laplace_transform.

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