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...\,

Partial Differential Equations

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