PDE5

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u_t=ku_{xx}\,u(0,t) = 0\,
u(l,t) = 0\,

u(x,0) = f(x)\,


First separate variables and plug back into the original equation.

u(x,t) = X(x)T(t)\,

XT' = kX''T\,

Find the separated boundary conditions (worry about the initial condition later).

X(0) = X(l) = 0\,

Now seperate the two functions on either side of the equation. The reasoning is that if the left side is a function of t and it is equal to the right side, which is a function of x, then both sides of the equation must be equal to a constant. It's good to keep higher derivatives on top of lower ones, and keep the k with T(t). When setting the next part equal to a constant, − λ works best.

\frac{T'}{kT} = \frac{X''}{X} = - \lambda\,

This gives the two solutions to the ODEs:

T(t) = c_1 e^{-\lambda k t}\,

X(x) = c_2 \cos(\sqrt{\lambda}x) + c_3 \frac{\sin(\sqrt{\lambda}x)}{\sqrt{\lambda}}\,

In the solution of X(x) the extra square root of λ was added in the denominator so that we won't have to make a special case for \lambda=0\,, which does not give interesting solutions anyway.

Use the separated boundary conditions on the function X(x).

X(0) = c_2 = 0\,

X(l) = c_3 \frac{\sin(\sqrt{\lambda}l)}{\sqrt{\lambda}}=0\,

Nothing interesting would come from letting c_3=0\,, so \lambda_n = \frac{n^2 \pi^2}{l^2},\,\,\,n=1,2,3,...\,. n\ne 0\, because \lambda\ne 0\,. These values of lambda are the eigenvalues. The eigenfunctions are \sin(\frac{n\pi x}{l}),\,\,\,n=1,2,3,...\,.

For each value of n, a solution is then

u_n(x,t) = c_1 e^{-\lambda_n k t} c_3 \frac{\sin(\sqrt{\lambda_n} x)}{\sqrt{\lambda_n}}\,

Using the principle of superposition, the formal solution is the sum of all possible solutions:

u(x,t) = \sum_{n=1}^\infty u_n(x,t) = \sum_{n=1}^\infty A_n e^{-\lambda_n k t} \sin(\frac{n\pi x}{l}), A_n = \frac{c_1 c_3}{\frac{n\pi}{l}}\,

The initial condition gives

u(x,0) = f(x) = \sum_{n=1}^\infty A_n \sin(\frac{n\pi x}{l})\,

This series is the Fourier sine series of f(x), so the coefficient is given by

A_n = \frac{2}{l}\int_0^l f(x) \sin(\frac{n\pi x}{l}) \,dx\,

So the final answer is

u(x,t) = \sum_{n=1}^\infty A_n e^{-\left(\frac{n\pi}{l}\right)^2 k t} \sin(\frac{n\pi x}{l}), \,\,\,A_n =\frac{2}{l}\int_0^l f(x) \sin(\frac{n\pi x}{l}) \,dx\,

Or just for fun:

u(x,t) = \frac{2}{l} \sum_{n=1}^\infty \left[ \int_0^l f(\xi) \sin(\frac{n\pi \xi}{l}) \,d\xi \right] e^{-\left(\frac{n\pi}{l}\right)^2 k t} \sin(\frac{n\pi x}{l})\,


Partial Differential Equations

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