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.

 

[alphy]

I am familiar with plotting infinite domain PDEs where the solution is an integral. In this case, the solution provided is a function of two variables, x and t, and involves an integral over an infinite domain. This type of solution is common in many physical systems, such as heat transfer and diffusion processes.

To plot this solution in a computer program like Matlab, Mathematica, or Python, you will need to first define the function T(x,t) with the given integral expression. Then, you can use the built-in plotting functions in these programs to plot the function over a desired range of x and t values.

However, it is worth noting that as u increases, the time to complete the computation will also increase exponentially. This is because the integral becomes more complex and requires more computational resources to solve. Therefore, it is expected that the computation time will be longer for larger values of u.

In order to improve the computational efficiency, you can try using numerical methods such as finite difference or finite element methods to approximate the solution. This will allow you to plot the solution for larger values of u without significantly increasing the computation time.

In conclusion, plotting an infinite domain PDE with an integral solution is possible in Matlab, Mathematica, or Python, but the computation time may increase significantly for larger values of u. Using numerical methods can help improve the efficiency of the computation.