PDE11

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u_{{xx}}=a^{{-2}}u_{t}\,

u(0,t)=10\,
u(10,t)=30\,

u(x,0)=0\,

0<x<10,\,\,\,\,\,t>0\,



Write the solution as a sum of two functions, one independent of time.

u(x,t)=s(x)+v(x,t)\,

The new boundary conditions are:

s(0)=10,\,\,\,s(10)=30\,

v(0,t)=0,\,\,\,v(10,t)=0\,

u(x,0)=s(x)+v(x,0)=0,\,\,\,v(x,0)=-s(x)\,

Substituting the sum of functions into the DE gives a simple ODE and a PDE:

s''=0\,

s(x)=Ax+B\,

s(0)=B=10\,

s(10)=10A+10=30,\,\,\,A=2\,

  • s(x)=2x+10\,

And the PDE

v_{{xx}}=a^{{-2}}v_{t}\,

Seperate variables to get two ODEs. Set them equal to a constant that will simplify calculations later:

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

X''T=a^{{-2}}XT'\,

{X'' \over X}=a^{{-2}}{T' \over T}=-\lambda ^{2}\,

X''+\lambda ^{2}X=0\,

X(x)=c_{1}\cos(\lambda x)+c_{2}\sin(\lambda x)\,

X(0)=c_{1}=0\,

X(10)=c_{2}\sin(\lambda 10)=0\,

\lambda _{n}={n\pi  \over 10},\,\,\,n=1,2,3,...\,

And

T'+a^{2}\lambda ^{2}T=0\,

T(t)=c_{3}e^{{-a^{2}\lambda ^{2}t}}\,

  • v(x,t)=\sum _{{n=1}}^{\infty }A_{n}\sin(\lambda _{n}x)e^{{-a^{2}\lambda ^{2}t}}\,

v(x,0)=\sum _{{n=1}}^{\infty }A_{n}\sin({n\pi x \over 10})=-2x-10\,

This is a Fourier sine series for the function on the right. The coefficients are then given by:

A_{n}={-1 \over 5}\int _{0}^{{10}}(2x+10)\sin({n\pi x \over 10})\,dx,\,\,\,n=1,2,3,...\,

={20(3(-1)^{n}-1) \over n\pi }\,

The solution is

  • u(x,t)=2x+10+{20 \over \pi }\sum _{{n=1}}^{\infty }{3(-1)^{n}-1 \over n}\sin({n\pi x \over 10})e^{{-a^{2}{n^{2}\pi ^{2} \over 10^{2}}t}}\,

Partial Differential Equations

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