# 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, $-\lambda$ 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 $\phi (x)$. Therefore, the coefficients are:

$A_{n}=2\int _{0}^{1}\phi (x)\cos(n\pi x)\,dx,\,\,\,n=0,1,2...\,$