PDE6

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$u_t=ku_{xx}\,$

$u_x(0,t) = 0\,$
$u_x(1,t) = 0\,$

$u(x,0) = \phi(x)\,$

First separate variables and plug back into the original equation.

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

$XT' = X''T\,$

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

$X'(0) = X'(1) = 0\,$

When seperating variables, it's good to keep higher derivatives on top of lower ones. When setting the next part equal to a constant, $- λ$ works best.

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

This gives the two solutions to the ODEs:

$T(t) = c_1 e^{-\lambda t}\,$ and $X(x) = c_2 \cos(\sqrt{\lambda}x) + c_3 \frac{\sin(\sqrt{\lambda}x)}{\sqrt{\lambda}}\,$

In the solution of $X(x)\,$ the extra square root of $\lambda\,$ 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. In this problem the sine term will disappear, but in general the extra part in the denominator is a good idea.

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

$X'(0) = c_3 = 0\,$

$X'(1) = -c_2 \sqrt{\lambda} \sin(\sqrt{\lambda})=0\,$

Nothing interesting would come from setting $c_2=0\,$ . So,

$\sqrt{\lambda_n} = n\pi,\,\,\,n=0,1,2,3,...\,$

Then for each $n$ , a solution is

$u_n(x,t) = A_n e^{-\lambda_n t} \cos(\sqrt{\lambda_n}x)\,$

So the formal solution is the sum of all of these:

$u(x,t) = \sum_{n=0}^\infty A_n e^{n\pi t} \cos(n\pi x)\,$

Now deal with the initial condition.

$u(x,0) = \phi(x) = \sum_{n=0}^\infty A_n \cos(n\pi x)\,$

The series on the RHS is the Fourier cosine series for $φ(x)$ . Therefore, the coefficients are:

$A_n = 2 \int_0^1 \phi(x) \cos(n\pi x)\,dx,\,\,\,n=0,1,2...\,$

Partial Differential Equations

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