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, -\lambda works best.

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

This gives the two solutions to the ODEs:

T(t)=c_{1}e^{{-\lambda kt}}\,

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.

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\neq 0\, because \lambda \neq 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}kt}}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}kt}}\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}kt}}\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}kt}}\sin({\frac  {n\pi x}{l}})\,


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