Differential equations – Laplace transform – Extra

The gamma function

For \(x>0\) the gamma function is defined as \(\displaystyle\Gamma(x)=\int_0^{\infty}e^{-t}t^{x-1}\,dt\). It can be shown that this integral converges for \(x>0\).

Note that \(\Gamma(1)=\displaystyle\int_0^{\infty}e^{-t}\,dt=-e^{-t}\bigg|_0^{\infty}=1\).

Using integration by parts we obtain for \(x>0\)

\[\Gamma(x+1)=\int_0^{\infty}e^{-t}t^x\,dt=-\int_0^{\infty}t^x\,de^{-t}=-e^{-t}t^x\bigg|_0^{\infty}+\int_0^{\infty}e^{-t}\,dt^x =0+x\int_0^{\infty}e^{-t}t^{x-1}\,dt=x\Gamma(x).\]

Note that this implies that \(\Gamma(n+1)=n!\) for \(n=0,1,2,\ldots\). So, the gamma function can be seen as a generalization of the factorial.

Furthermore, using the substitution \(t=x^2\) we have

\[\Gamma(\tfrac{1}{2})=\int_0^{\infty}e^{-t}t^{-\frac{1}{2}}\,dt=\int_0^{\infty}e^{-x^2}x^{-1}\cdot2x\,dx=2\int_0^{\infty}e^{-x^2}\,dx.\]

It can be shown that \(\displaystyle\int_0^{\infty}e^{-x^2}\,dx=\tfrac{1}{2}\sqrt{\pi}\). This implies that \(\Gamma(\frac{1}{2})=\sqrt{\pi}\).

This can be used to find the Laplace transform of \(t^{\alpha}\) for real \(\alpha>-1\). Using the substitution \(st=x\) or \(t=\dfrac{x}{s}\) we obtain:

\[\mathcal{L}\{t^{\alpha}\}(s)=\int_0^{\infty}e^{-st}t^{\alpha}\,dt=\frac{1}{s^{\alpha+1}}\int_0^{\infty}e^{-u}u^x\,du =\frac{\Gamma(\alpha+1)}{s^{\alpha+1}},\quad\alpha>-1,\quad s>0.\]

For instance, this implies that

\[\mathcal{L}\left\{\frac{1}{\sqrt{t}}\right\}(s)=\frac{\Gamma(\frac{1}{2})}{s^{\frac{1}{2}}}=\sqrt{\frac{\pi}{s}}\quad\text{and}\quad \mathcal{L}\left\{\sqrt{t}\right\}(s)=\frac{\Gamma(\frac{3}{2})}{s^{\frac{3}{2}}}=\frac{\frac{1}{2}\Gamma(\frac{1}{2})}{s^{\frac{3}{2}}}=\frac{\sqrt{\pi}}{2s\sqrt{s}}.\]

Taylor series

Since \(\mathcal{L}\{t^n\}(s)=\dfrac{n!}{s^{n+1}}\) for \(n=0,1,2,\ldots\) the Taylor series of a function around zero can be used to find its Laplace transform. Some examples:

1) Using \(e^{at}=\displaystyle\sum_{n=0}^{\infty}\frac{(at)^n}{n!}\) we obatin \(\displaystyle\mathcal{L}\{e^{at}\}(s) =\sum_{n=0}^{\infty}\frac{a^n}{n!}\cdot\frac{n!}{s^{n+1}}=\sum_{n=0}^{\infty}\frac{a^n}{s^{n+1}} =\frac{1}{s}\cdot\frac{1}{1-\frac{a}{s}}=\frac{1}{s-a}\) for \(s>a\).

2) Using \(\sin(at)=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}(at)^{2n+1}\) we obtain

\[\mathcal{L}\{\sin(at)\}(s)=\sum_{n=0}^{\infty}\frac{(-1)^na^{2n+1}}{(2n+1)!}\cdot\frac{(2n+1)!}{s^{2n+2}} =\frac{a}{s^2}\sum_{n=0}^{\infty}\left(-\frac{a^2}{s^2}\right)^n=\frac{a}{s^2}\cdot\frac{1}{1+\frac{a^2}{s^2}} =\frac{a}{s^2+a^2},\quad s>a\geq 0.\]

3) Using \(\cos(at)=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}(at)^{2n}\) we obtain

\[\mathcal{L}\{\cos(at)\}(s)=\sum_{n=0}^{\infty}\frac{(-1)^na^{2n}}{(2n)!}\cdot\frac{(2n)!}{s^{2n+1}} =\frac{1}{s}\sum_{n=0}^{\infty}\left(-\frac{a^2}{s^2}\right)^n=\frac{1}{s}\cdot\frac{1}{1+\frac{a^2}{s^2}} =\frac{s}{s^2+a^2},\quad s>a\geq 0.\]

This principle can be used to find Laplace transforms of other functions as well. For instance: since \(\displaystyle\frac{\sin(at)}{t}=\sum_{n=0}^{\infty}\frac{(-1)^na^{2n+1}}{(2n+1)!}t^{2n}\) we obtain

\[\mathcal{L}\left\{\frac{\sin(at)}{t}\right\}(s)=\sum_{n=0}^{\infty}\frac{(-1)^na^{2n+1}}{(2n+1)!}\cdot\frac{(2n)!}{s^{2n+1}} =\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\left(\frac{a}{s}\right)^{2n+1}=\arctan\left(\frac{a}{s}\right),\quad s>a\geq0.\]

Bessel functions

The Bessel function \(J_0(t)\) of the first kind of order \(0\) is defined as: \(J_0(t)=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^nt^{2n}}{2^{2n}(n!)^2}\). Now we use

\[\binom{-\frac{1}{2}}{n}=\frac{(-\frac{1}{2})(-\frac{3}{2})\cdots(-\frac{1}{2}-n+1)}{n!}=\frac{(-1)^n1\cdot3\cdots(2n-1)}{2^n\,n!} =\frac{(-1)^n(2n)!}{2^{2n}(n!)^2}=\frac{(-1)^n}{2^{2n}}\binom{2n}{n},\quad n=0,1,2,\ldots\]

to find that

\begin{align*} \mathcal{L}\{J_0(t)\}(s)&=\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{2n}(n!)^2}\cdot\frac{(2n)!}{s^{2n+1}} =\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{2n}}\binom{2n}{n}\frac{1}{s^{2n+1}}\\[2.5mm] &=\frac{1}{s}\sum_{n=0}^{\infty}\binom{-\frac{1}{2}}{n}\left(\frac{1}{s^2}\right)^n =\frac{1}{s}\left(1+\frac{1}{s^2}\right)^{-\frac{1}{2}}=\frac{1}{s\sqrt{1+\frac{1}{s^2}}}=\frac{1}{\sqrt{s^2+1}},\quad s>1. \end{align*}

This Bessel function \(J_0(t)\) satisfies the differential equation \(ty''(t)+y'(t)+ty(t)=0\) for \(t>0\). Using \(\displaystyle\mathcal{L}\{tf(t)\}(s)=-\frac{d}{ds}\mathcal{L}\{f(t)\}(s)\) we obtain for \(Y(s)=\mathcal{L}\{y(t)\}(s)\):

\[-\frac{d}{ds}\left(s^2Y(s)-sy(0)-y'(0)\right)+sY(s)-y(0)-\frac{d}{ds}Y(s)=0\quad\Longleftrightarrow\quad -s^2Y'(s)-2sY(s)+y(0)+sY(s)-y(0)-Y'(s)=0.\]

This implies that \((s^2+1)Y'(s)+sY(s)=0\). This is a separable differential equation with general solution \(Y(s)=\dfrac{c}{\sqrt{s^2+1}}\) with \(c\in\mathbb{R}\). Note that this implies that \(y(t)=cJ_0(t)\).

The Bessel differential equation of order \(\nu\geq0\) is: \(t^2y''(t)+ty'(t)+(t^2-\nu^2)y(t)=0\) for \(t>0\). Let \(Y(s)=\mathcal{L}\{y(t)\}(s)\), then we have:

\begin{align*} &\frac{d^2}{ds^2}(s^2Y(s)-sy(0)-y'(0))-\frac{d}{ds}(sY(s)-y(0))+\frac{d^2}{ds^2}Y(s)-\nu^2Y(s)=0\\[2.5mm] &{}\quad\Longleftrightarrow\quad\frac{d}{ds}(s^2Y'(s)+2sY(s)-y(0))-sY'(s)-Y(s)+Y''(s)-\nu^2Y(s)=0\\[2.5mm] &{}\quad\Longleftrightarrow\quad s^2Y''(s)+4sY'(s)+2Y(s)-sY'(s)-Y(s)+Y''(s)-\nu^2Y(s)=0\\[2.5mm] &{}\quad\Longleftrightarrow\quad (s^2+1)Y''(s)+3sY'(s)+(1-\nu^2)Y(s)=0. \end{align*}

Let \(Y(s)=\displaystyle\frac{u(s)}{\sqrt{s^2+1}}\), then: \(Y'(s)=\displaystyle\frac{u'(s)}{\sqrt{s^2+1}}-\frac{su(s)}{(s^2+1)^{3/2}}\) and \(Y''(s)=\displaystyle\frac{u''(s)}{\sqrt{s^2+1}}-\frac{2su'(s)}{(s^2+1)^{3/2}}+\frac{3s^2u(s)}{(s^2+1)^{5/2}}-\frac{u(s)}{(s^2+1)^{3/2}}\). Substitution then leads to

\begin{align*} &\sqrt{s^2+1}\,u''(s)-\frac{2su'(s)}{\sqrt{s^2+1}}+\frac{3s^2u(s)}{(s^2+1)^{3/2}}-\frac{u(s)}{\sqrt{s^2+1}}+\frac{3su'(s)}{\sqrt{s^2+1}} -\frac{3s^2u(s)}{(s^2+1)^{3/2}}+(1-\nu^2)\frac{u(s)}{\sqrt{s^2+1}}=0\\[2.5mm] &{}\quad\Longleftrightarrow\quad\frac{d}{ds}\left(\sqrt{s^2+1}\,u'(s)\right)=\nu^2\frac{u(s)}{\sqrt{s^2+1}} \quad\Longleftrightarrow\quad\sqrt{s^2+1}\,u'(s)\cdot\frac{d}{ds}\left(\sqrt{s^2+1}\,u'(s)\right)=\nu^2u(s)\cdot u'(s)\\[2.5mm] &{}\quad\Longleftrightarrow\quad\left(\sqrt{s^2+1}\,u'(s)\right)^2=\left(\nu\,u(s)\right)^2 \quad\Longleftrightarrow\quad\sqrt{s^2+1}\,u'(s)=\pm\nu\,u(s). \end{align*}

This is a separable differential equation with implicit solution \(\ln|u(s)|=\mp\nu\ln(\sqrt{s^2+1}-s)+K\). This implies that

\[u(s)=c_1(\sqrt{s^2+1}-s)^{\nu}+c_2(\sqrt{s^2+1}-s)^{-\nu}\quad\Longrightarrow\quad Y(s)=c_1\frac{(\sqrt{s^2+1}-s)^{\nu}}{\sqrt{s^2+1}}+c_2\frac{(\sqrt{s^2+1}-s)^{-\nu}}{\sqrt{s^2+1}}.\]

This is the Laplace transform of \(y(t)=c_1J_{\nu}(t)+c_2J_{-\nu}(t)\), where \(J_{\nu}(t)=\displaystyle\sum_{k=0}^{\infty}\frac{(-1)^k}{\Gamma(\nu+k+1)k!}\left(\frac{t}{2}\right)^{\nu+2k}\).

---

Last modified on May 5, 2021