How to find inverse Laplace transform of the following function?
$$ U(x,s)=\exp\left(P-\frac{\sqrt{P^2+4Ps}}{2}x\right) $$
I obtained this equation by solving a PDE using Laplace Transform, now I have to take the inverse Laplace of this to obtain the actual solution to the PDE.
I am looking for any hint or calculations, kindly suggest any efficient method with some steps.
Best Answer
To compute this ILT of $U$, we need to use the definition:
$$u(x,t) = \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, B e^{-a \sqrt{P+4 s}} e^{s t} $$
where $B = e^{P}$ and $a = \sqrt{P} x/2$.
To evaluate this integral, we use Cauchy's theorem and define the following contour integral:
$$\oint_C dz \, B e^{-a \sqrt{P+4 z}} e^{z t} $$
where $C$ is the following contour:
where the large arc has radius $R$ and the small arc has radius $\epsilon$. Along this contour, the above integral is equal to
$$\int_{c-i \sqrt{R^2-c^2}}^{c+i \sqrt{R^2-c^2}} ds \, B e^{-a \sqrt{P+4 s}} e^{s t} + i R \int_{\pi/2-\arcsin{(c/R)}}^{\pi} d\theta \,e^{i \theta}\, B e^{-a \sqrt{P+4 R e^{i \theta}}} e^{R t e^{i \theta}}\\ + e^{i \pi} \int_R^{P/4+\epsilon} dx \, B e^{-i a \sqrt{4 x-P}} e^{-x t} + i \epsilon \int_{\pi}^{-\pi} d\phi \, e^{i \phi} \, B e^{-a \sqrt{P+4 \epsilon e^{i \phi}}} e^{\epsilon t e^{i \phi}}\\ + e^{-i \pi} \int_{P/4+\epsilon}^R dx \, B e^{+i a \sqrt{4 x-P}} e^{-x t}+ i R \int_{\pi}^{3 \pi/2+\arcsin{(c/R)}} d\theta \,e^{i \theta}\, B e^{-a \sqrt{P+4 R e^{i \theta}}} e^{R t e^{i \theta}}$$
By Cauchy's theorem, this contour integral is zero. Further, we consider the limits as $R \to \infty$ and $\epsilon \to 0$; in this limit, the second, fourth, and sixth integrals vanish. This means that we have, by Cauchy's theorem:
$$\int_{c-i \infty}^{c+i \infty} ds \, B e^{-a \sqrt{P+4 s}} e^{s t} = -\int_{\infty}^{P/4} dx \, B e^{-i a \sqrt{4 x-P}} e^{-x t} + \int_{P/4}^{\infty} dx \, B e^{+i a \sqrt{4 x-P}} e^{-x t} $$
or, the ILT is
$$\begin{align}\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, B e^{-a \sqrt{P+4 s}} e^{s t} &= \frac{B}{\pi} \int_{P/4}^{\infty} dx \, \sin{\left (a \sqrt{4 x-P} \right )} e^{-x t} \\ &= \frac{B}{\pi} e^{-P t/4} \int_0^{\infty} dy \,\sin{\left ( 2 a \sqrt{y} \right )} e^{-y t} \\ &= \frac{2 B}{\pi} e^{-P t/4} \int_0^{\infty} du \, u \, \sin{(2 a u)} e^{-t u^2} \\ &= \frac{2 B}{\pi} e^{-P t/4} \int_0^{\infty} du \, u^2 \frac{\sin{(2 a u)}}{u} e^{-t u^2} \\ &= -\frac{2 B}{\pi} e^{-P t/4} \frac{\partial}{\partial t} \int_0^{\infty} du \, \frac{\sin{(2 a u)}}{u} e^{-t u^2} \\ &= -\frac{B}{\pi} e^{-P t/4} \frac{\partial}{\partial t} \int_{-\infty}^{\infty} du \, \frac{\sin{v}}{v} e^{-(t/(4 a^2)) v^2}\end{align}$$
Now, to do the integral on the RHS we use Parseval's theorem for Fourier transforms.
$$\begin{align} \int_{-\infty}^{\infty} du \, \frac{\sin{v}}{v} e^{-(t/(4 a^2)) v^2} &= \frac1{2 \pi} 2 a \sqrt{\frac{\pi}{t}} \pi \int_{-1}^1 dk \, e^{-a^2 k^2/t} \\ &= 2 \sqrt{\pi} \int_0^{a/\sqrt{t}} dw \, e^{-w^2} \\ &= \pi \, \operatorname{erf}{\left (\frac{a}{\sqrt{t}} \right )}\end{align} $$
Hence,
$$\begin{align} \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, B e^{-a \sqrt{P+4 s}} e^{s t} &= -\frac{B}{\pi} e^{-P t/4} \frac{\partial}{\partial t} \left [ \pi \, \operatorname{erf}{\left (\frac{a}{\sqrt{t}} \right )}\right ] \\ &= \frac{a B}{\sqrt{\pi t^3}} e^{-P t/4} e^{-a^2/t} \end{align}$$
and as $a = \sqrt{P} x/2$ and $B=e^P$, we have