MHB Plotting an infinite domain PDE

[Dustinsfl]

Is anyone familiar with plotting an infinite domain PDE where the solution is an integral.

Take the solution
\[
T(x,t) = \frac{100}{\pi}\int_0^{\infty}\int_{-\infty}^{\infty} \frac{\sinh(u(10-y)}{\sinh(10u)} \cos(u(\xi-x))d\xi du
\]
How could I plot this in Matlab, Mathematica, or Python?

As a note, Mathematica is slow because I have code that does it, but when I increase u, the time to complete goes up exponentially. I am talking hours.

Everything I tried in Matlab failed.

 

[Ackbach]

I'm curious: can you post your Mathematica code?

 

[Dustinsfl]

I'm curious: can you post your Mathematica code?

f[x_, y_, u_, \[Xi]_] =  Sinh[u*(5 - y)]/Sinh[u*5]*100/\[Pi]*Cos[u*(\[Xi] - x)];
 
g[x_?NumericQ, y_?NumericQ] := NIntegrate[f[x, y, u, \[Xi]], {u, 1, 10}, {\[Xi], -5, 5}];

DensityPlot[g[x, y], {x, -5, 5}, {y, 0, 5}, PlotPoints -> 5]

This code only integrate u from 1 to 10 and \(\xi\) from -5 to 5. However, I would like to increase the integration to have a more accurate plot.

Here is Python code that takes a long time but can handle a bigger integration bounds in the same time Mathematica can but the code isn't producing an accurately plot either so there must be something wrong as well.

import numpy as np
import pylab as pl 
from scipy.integrate import dblquad  

b = 50.0  
x = np.linspace(-10, 10, 1000) 
y = np.linspace(0, 10, 1000)  def f(xx, u, xi, yj):     
    return ((np.exp(u * (b - yj)) - np.exp(-u * (b - yj))) / 
              (np.exp(u * b) - np.exp(-u * b)) * np.cos(u * (xx - xi)))  T = np.zeros([len(x), len(y)])

 
for i, xi in enumerate(x):     
    for j, yj in enumerate(y):         
        T[i, j] += dblquad(             
                   f, -10, 10, lambda x: 0.1, lambda x: 10, args=(xi, yj))[0]fig = pl.figure()
ax = fig.add_subplot(111)
pax = ax.pcolormesh(y, x, T)
fig.colorbar(pax)
pl.xlim((-X, X))
pl.ylim((0, b))
pl.grid(True)

pl.show()

 

[Ackbach]

You might save some computational time if you pull everything out of the inner integral that you can:
$$T(x,t)= \frac{100}{ \pi} \int_{0}^{ \infty} \frac{ \sinh(u(10-y))}{ \sinh(10u)}
\underbrace{ \int_{- \infty}^{ \infty} \cos(u( \xi-x)) \, d \xi}_{ \text{define this as }f(u,x)} \, du.$$
This way, the ratio on sinh's isn't evaluated for every inner loop. I should point out, however, that the inner integral, which is $f(u,x)$, doesn't converge. The Riemann-Lebesgue Lemma might come in handy here, depending on where you want to look at things.