Python Solving the heat equation numerically using Python

[Wrichik Basu]

I want to solve the heat equation numerically. The equation is: $$k \dfrac{\partial^2 T}{\partial x^2} = \dfrac{\partial T}{\partial t}.$$ This is a parabolic PDE.

Following this pdf (specifically, equation 7 given on page 3), I wrote the following Python function to implement the explicit algorithm:

import numpy as np

def heat_equation_explicit(t0, t_end, dt, dx, k, initial_profile):
    """
    Solves the heat equation using explicit algorithm.

    Parameters
    ----------
    t0 : float
        The start of time.
    t_end : float
        The end of time.
    dt : float
        Time step.
    dx : float
        Step size on x-axis.
    k : float
        Diffusivity.
    initial_profile : numpy.ndarray
        The initial heat profile for all values of x over which the solution has to be calculated,
        (i.e. T(x, 0) ∀ x ∈ np.arange(x0, x_end, dx)). The first and the  last elements should be the
        boundary values, i.e. T(x0, t) and T(x_end, t), which will remain constant with time.

    Returns
    -------
    numpy.ndarray
        The heat profile at t = t_end.

    """

    # Number of steps:
    num_t = round((t_end - t0) / dt)
    num_x = np.size(initial_profile)

    old_profile = initial_profile

    global finalProfile_numerical

    lamb = (k * dt) / (dx ** 2)

    for j in range(0, num_t):

        # Initialise the variable:
        finalProfile_numerical = np.zeros(num_x)

        # Keep the boundary values at the two ends:
        finalProfile_numerical[0] = initial_profile[0]
        finalProfile_numerical[-1] = initial_profile[-1]

        # Compute the new profile:
        for i in range(1, num_x - 1):

            u_old_iPlus1 = old_profile[i + 1]
            u_old_iMinus1 = old_profile[i - 1]

            finalProfile_numerical[i] = old_profile[i] + lamb * (u_old_iPlus1 - 2 * old_profile[i] + u_old_iMinus1)

        old_profile = finalProfile_numerical

    return finalProfile_numerical

Our college Professor gave us a sample problem to test our programs. The given values are as follows:

• $x \in [0, \pi]$
• $T(0, t) = T(\pi,t) = 0$
• $T(x,0) = \sin x$
• $k = 0.05$
• Analytical solution: $T(x,t) = \exp(-t) \sin x$
• I have to compute the final profile at $t = 0.5s.$

Well and good. I called my function with the necessary values:

import math
import numpy as np
import matplotlib.pyplot as plt

x0 = 0
xEnd = math.pi
dx = math.pi / 100

t0 = 0
tEnd = 0.5
dt = 0.01

k = 0.05

xVal = np.arange(x0, xEnd, dx)

# Values of T(x, 0) at all x. Includes T(x0, t) at the beginning, and T(x_end, t) at the end.
initialProfile = np.sin(xVal)

finalProfile_numerical = heat_equation_explicit(t0, tEnd, dt, dx, k, initialProfile)
finalProfile_analytic = math.exp(-0.5) * np.sin(xVal)

# Plot the numerical solution:
plt.plot(xVal, finalProfile_numerical, '-o', label="Numerical", markevery=2)

# Plot the analytical solution:
plt.plot(xVal, finalProfile_analytic, '-o', label='Analytic', markevery=2)

plt.xlabel(r'x$\rightarrow$')
plt.ylabel(r'$T(x)\rightarrow$')
plt.title("Temperature profile at t = " + f'{tEnd:.2f}' + "s")
plt.legend(loc='best')
plt.show()

And I get this graph:

Obviously, the numerical solution doesn't match the analytic solution.

I double-checked my function, and there seems to be no error. But there is an error somewhere, right?

If I print final_profile_numerical / final_profile_analytic, I get:

[       nan 1.60800745 1.60800745 1.60800745 1.60800745 1.60800745
 1.60800745 1.60800745 1.60800745 1.60800745 1.60800745 1.60800745
 1.60800745 1.60800745 1.60800745 1.60800745 1.60800745 1.60800745
 1.60800745 1.60800745 1.60800745 1.60800745 1.60800745 1.60800745
 1.60800745 1.60800745 1.60800745 1.60800745 1.60800745 1.60800745
 1.60800745 1.60800745 1.60800745 1.60800745 1.60800745 1.60800745
 1.60800745 1.60800745 1.60800745 1.60800745 1.60800745 1.60800745
 1.60800745 1.60800745 1.60800745 1.60800745 1.60800745 1.60800745
 1.60800745 1.60800745 1.60800745 1.60800745 1.60800745 1.60800745
 1.60800745 1.60800745 1.60800745 1.60800745 1.60800745 1.60800745
 1.60800745 1.60800745 1.60800745 1.60800745 1.60800745 1.60800745
 1.60800745 1.60800745 1.60800745 1.60800745 1.60800746 1.60800746
 1.60800747 1.60800748 1.60800752 1.60800757 1.60800769 1.60800788
 1.60800825 1.60800886 1.60800999 1.60801179 1.60801496 1.60801993
 1.6080283  1.60804126 1.60806232 1.60809452 1.60814542 1.60822265
 1.60834263 1.60852485 1.60880718 1.60924297 1.60993333 1.61105178
 1.61296191 1.61650076 1.6242027  1.64872127]

Ignore the nan in the front. Interestingly, note that $1 / \exp(-0.5) = 1.648$.

Any idea where the problem is occurring?

 

[willem2]

Any idea where the problem is occurring?

The analytical solution doesn't contain a k. It's only valid if k=1. With k = 0.05 the temperatures will vary much more slowly than with k=1.

If you want to compute k=1 with your python program you need to set dt much smaller. For larger values of k*dt, the solution will become unstable and will produce large heatwaves with wavelength dt.

 

[pbuk]

I don't think line 58 of your code does what you think it does:

        old_profile = finalProfile_numerical

Do you want to do this instead?
Edit:

        old_profile = finalProfile_numerical.copy()

(Before the edit this showed the following, which is not as good a recommendation).

        old_profile = numpy.copy(finalProfile_numerical)

You need to develop some techniques for debugging numerical methods; one common technique is to run for a range of step sizes and see if the solution converges to the analytical solution at a rate you would expect.

 

[Wrichik Basu]

I don't think line 58 of your code does what you think it does:

        old_profile = finalProfile_numerical

Do you want to do this instead?

        old_profile = numpy.copy(finalProfile_numerical)

Unfortunately, that doesn't change the graph.

The analytical solution doesn't contain a k. It's only valid if k=1. With k = 0.05 the temperatures will vary much more slowly than with k=1.

That does it. I changed k = 1, tEnd = 1e-9, dt = 1e-10, and updated the analytic solution accordingly: finalProfile_analytic = math.exp(-tEnd) * np.sin(xVal) The graphs coincided. Thanks!

 

[pbuk]

Unfortunately, that doesn't change the graph.

Ah yes, the assignment to old_profile of the reference to finalProfile_numerical was not a problem because you were replacing the reference with finalProfile_numerical = np.zeros().