## How to find triple numerical inverse Laplace trans...

I have the following fuction in Laplace domain,

restart:with(plots):with(inttrans):

u:=Pi^4*s3^(alpha-1)/((s1^2+Pi^2)*(s2^2+Pi^2)*(-s1^2+Pi^2*s3^alpha-s2^2))-Pi*s1*s2^(alpha-1)/(s3*(1+s2^alpha)*(-s1^2+Pi^2*s3^alpha-s2^2))-Pi*s1*s2^(alpha-1)/(-s1^2+Pi^2*s3^alpha-s2^2);

Where, s1, s2, s3 are the Laplace variables.

x1:=invlaplace(u, s1, x);

This worked. But the next two doesn't work.

y1:=invlaplace(x1, s2, y);

uu:=invlaplace(y1, s3,t);

Even, I tried to plot the unevaluated invlaplace but no luck.

alpha:=1:t:=1:
plot3d(uu,x =1..2, y=1..2);

Am I missing something?

## Symbolic Laplace Transform...

Is Symbolic Laplace and Inverse Laplace transform possible on Maple? if Yes, how do I find the inverse laplace of this function

Thanks.

## how to evaluate numerical inverse Laplace transfor...

I would like to apply inverse Laplace transform to U(x,p), which is defined by

For simplicity with my calculations, I assumed p:=i*beta^2. That is why I have the following equation after applying Laplace transform

(beta=0 is not a pole, that is why I removed the last term in my calculations later. Because there is no contribution) where

Here p and beta are complex values, we can write Re(p)=-2*Re(beta)*Im(beta), Im(p)=(Re(beta))^2-(Im(beta))^2 due to p:=i*beta^2. I numerically compute the roots of h(beta), you can find the numerical values of beta (I assumed digits are 50 due to accuracy ) betap.mw

Finally, I would like to plot U(x,t) with the values t=0.8, lambda=1, L=10, k=1. For checking the figure give t=0 and observe that U(x,0)=0.

I am expecting the plot is more or less like the following figure

PS: I already tried to solve and plot the problem, but I could not find where I make a mistake. I  share the worksheet below. Thank you!

complexplot.mw

## How to find invlaplace of a certain function?...

Here is the function I want to find the inverse of

U(x):=-(s^2/(s^2+1)+1/(s^2+1)-1)*exp(sqrt(s)*x/sqrt(phi))/(s^2+1)+exp(-sqrt(s)*x/sqrt(phi))/(s^2+1);

where phi>0

Now taking the inverselaplace of U(x)

u:=invlaplace(U(x),s,t);

which give me an unevaluated integral.

Any idea, how to get the complete inverse, maybe interms of erf?