0
$\begingroup$

I am attempting to work through a very simple problem.

Determine the Fourier series expansion for the following heat PDE problem with ICS and BCS:

$$ u_{t} = \alpha^2u_{xx}$$

$$ u(0, t) = u(L, t) = 0$$

$$ u(x, 0) = \begin{cases} 0 & 0 \leq x \leq L/2 \\ 1 & L/2 < x \leq L\end{cases}$$

Given the form of the boundary conditions, and using separation of variables, I concluded that:

$$\kappa = \frac{n\pi}{L}$$

$$u(x, t) = \sum_{n=1}^{\infty} a_n \exp(-\alpha^2\kappa^2t) \sin(\kappa x)$$

In order to use the ICs, one uses the orthogonality of sinusoids to obtain:

$$ a_n = \frac{2}{L}\int_{0}^{L} u(x, 0) \sin(\kappa x)\;\textrm{d}x$$

In our case, with the given $u(x, 0)$, I obtained:

$$ \begin{align} a_n &= \frac{2}{L}\int_{0}^{L} u(x, 0) \sin(\kappa x)\;\textrm{d}x \\ &= \frac{2}{L}\left[\int_{0}^{L/2} 0\;\textrm{d}x + \int_{L/2}^{L} \sin(\kappa x) \;\textrm{d}x\right] \\ &= \frac{2}{L}\int_{L/2}^{L} \sin(\kappa x) \;\textrm{d}x \\ &= -\frac{2}{n\pi}\cos{n\pi} \end{align} $$

Now, I have written some very simple python code to compute the first however many terms I'd like:

import numpy as np

#=========================================

def sum_f_series(start, end_p1, t, x, L, alpha2):
    n = np.arange(start,end_p1)
    n2 = n*n

    K = (np.pi/L)

    C1 = -1*alpha2*(K**2)*t
    t_part = np.exp(C1*n2)

    C2 = K
    x_part = np.sin(C2*n)

    a_n = -1*(2*np.pi/n)*np.cos(n*np.pi)

    return np.sum(t_part*x_part*a_n)

#======================================

start = 1
end_p1 = 11 # last term + 1

a = 0
b = 1.5
L = b-a

t = 0
x = 1

alpha2 = 5

print sum_f_series(start, end_p1, t, x, L, alpha2)

The result is around $6.6$ -- a far cry from $1$, which is what the result should be (consider the initial conditions, and the fact that $1 > 0.75 = L/2$).

So, I am making an error somewhere, but I have been unable to find it. Can you help?

$\endgroup$
2
$\begingroup$

Your errors are on the lines for x_part and a_n. The argument to np.sin() should include x, and the coefficient for a_n by your calculation should be 2/(np.pi*n).

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.