PDE11

From Exampleproblems

Jump to: navigation, search
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^2X=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^2T=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

Main Page

Retrieved from "http://www.exampleproblems.com/wiki/index.php/PDE11"
Argan Oil
Natural Skin Care
Organic Skin Care
visitor stats