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Mathematics LibreTexts

1.7: Applications to Laplace transforms

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    17328
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    The Laplace transform of a function \(f\) locally integrable on \([0,\infty)\) is

    \[F(s)=\int_{0}^{\infty}e^{-sx}f(x)\,dx\]

    for all \(s\) such that integral converges. Laplace transforms are widely applied in mathematics, particularly in solving differential equations.

    We leave it to you to prove the following theorem (Exercise [exer:26]).

    [theorem:12] Suppose \(f\) is locally integrable on \([0,\infty)\) and \(|f(x)|\le M e^{s_{0}x}\) for sufficiently large \(x\). Then the Laplace transform of \(F\) converges uniformly on \([s_{1},\infty)\) if \(s_{1}>s_{0}\).

    [theorem:13] If \(f\) is continuous on \([0,\infty)\) and \(H(x)=\int_{0}^{\infty}e^{-s_{0}u}f(u)\,du\) is bounded on \([0,\infty),\) then the Laplace transform of \(f\) converges uniformly on \([s_{1},\infty)\) if \(s_{1}>s_{0}.\)

    If \(0\le r\le r_{1}\),

    \[\int_{r}^{r_{1}}e^{-sx}f(x)\,dx =\int_{r}^{r_{1}}e^{-(s-s_{0})x}e^{-s_{0}x}f(x)\,dt =\int_{r}^{r_{1}}e^{-(s-s_{0})t}H'(x)\,dt.\]

    Integration by parts yields

    \[\int_{r}^{r_{1}}e^{-sx}f(x)\,dt=e^{-(s-s_{0})x}H(x)\biggr|_{r}^{r_{1}} +(s-s_{0})\int_{r}^{r_{1}}e^{-(s-s_{0})x} H(x)\,dx.\]

    Therefore, if \(|H(x)|\le M\), then

    \[\begin{aligned} \left|\int_{r}^{r_{1}}e^{-sx}f(x)\,dx\right|&\le& M\left|e^{-(s-s_{0})r_{1}} +e^{-(s-s_{0})r} +(s-s_{0})\int_{r}^{r_{1}}e^{-(s-s_{0})x}\,dx\right|\\ &\le &3Me^{-(s-s_{0})r}\le 3Me^{-(s_{1}-s_{0})r},\quad s\ge s_{1}.\end{aligned}\]

    Now Theorem [theorem:4] implies that \(F(s)\) converges uniformly on \([s_{1},\infty)\).

    The following theorem draws a considerably stonger conclusion from the same assumptions.

    [theorem:14] If \(f\) is continuous on \([0,\infty)\) and

    \[H(x)=\int_{0}^{x}e^{-s_{0}u}f(u)\,du\]

    is bounded on \([0,\infty),\) then the Laplace transform of \(f\) is infinitely differentiable on \((s_{0},\infty),\) with

    \[\label{eq:30} F^{(n)}(s)=(-1)^{n}\int_{0}^{\infty} e^{-sx} x^{n}f(x)\,dx;\]

    that is, the \(n\)-th derivative of the Laplace transform of \(f(x)\) is the Laplace transform of \((-1)^{n}x^{n}f(x)\).

    First we will show that the integrals

    \[I_{n}(s)=\int_{0}^{\infty}e^{-sx}x^{n}f(x)\,dx,\quad n=0,1,2, \dots\]

    all converge uniformly on \([s_{1},\infty)\) if \(s_{1}>s_{0}\). If \(0<r<r_{1}\), then

    \[\int_{r}^{r_{1}}e^{-sx}x^{n}f(x)\,dx= \int_{r}^{r_{1}}e^{-(s-s_{0})x}e^{-s_{0}x}x^{n}f(x)\,dx =\int_{r}^{r_{1}}e^{-(s-s_{0})x}x^{n}H'(x)\,dx.\]

    Integrating by parts yields

    \[\begin{aligned} \int_{r}^{r_{1}}e^{-sx}x^{n}f(x)\,dx &=&r_{1}^{n}e^{-(s-s_{0})r_{1}}H(r)-r^{n}e^{-(s-s_{0})r}H(r)\\ &&-\int_{r}^{r_{1}}H(x)\left(e^{-(s-s_{0})x}x^{n}\right)'\,dx,\end{aligned}\]

    where \('\) indicates differentiation with respect to \(x\). Therefore, if \(|H(x)|\le M\le \infty\) on \([0,\infty)\), then

    \[\left|\int_{r}^{r_{1}}e^{-sx}x^{n}f(x)\,dx\right|\le M\left(e^{-(s-s_{0})r}r^{n}+e^{-(s-s_{0})r}r^{n} +\int_{r}^{\infty}|(e^{-(s-s_{0})x})x^{n})'|\,dx\right).\]

    Therefore, since \(e^{-(s-s_{0})r}r^{n}\) decreases monotonically on \((n,\infty)\) if \(s>s_{0}\) (check!),

    \[\left|\int_{r}^{r_{1}}e^{-sx}x^{n}f(x)\,dx\right|<3Me^{-(s-s_{0})r}r^{n},\quad n<r<r_{1},\]

    so Theorem [theorem:4] implies that \(I_{n}(s)\) converges uniformly \([s_{1},\infty)\) if \(s_{1}>s_{0}\). Now Theorem [theorem:11] implies that \(F_{n+1}=-F_{n}'\), and an easy induction proof yields [eq:30] (Exercise [exer:25]).

    [example:13] Here we apply Theorem [theorem:12] with \(f(x)=\cos ax\) (\(a\ne0\)) and \(s_{0}=0\). Since

    \[\int_{0}^{x}\cos au\,du=\frac{\sin ax}{a}\]

    is bounded on \((0,\infty)\), Theorem [theorem:12] implies that

    \[F(s)=\int_{0}^{\infty}e^{-sx}\cos ax\,dx\]

    converges and

    \[\label{eq:31} F^{(n)}(s)=(-1)^{n}\int_{0}^{\infty}e^{-sx}x^{n}\cos ax\,dx, \quad s>0.\]

    (Note that this is also true if \(a=0\).) Elementary integration yields

    \[F(s)=\frac{s}{s^{2}+a^{2}}.\]

    Hence, from [eq:31],

    \[\int_{0}^{\infty}e^{-sx}x^{n}\cos ax=(-1)^{n}\frac{d^n}{ds^n} \frac{s}{s^{2}+a^{2}}, \quad n=0,1, \dots.\]