Question

solve the equation du/dt=d^2u/dx^2 with the boundary condition u(0,t)=u(1,t)=0 and u(x,0)=3sinnπx where 0<x<1 ,t>0
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Answer 1

Answer:

du/dt = d^2u/dx^2 + d^2u/dy^2,    0 <x <a,  0<y<a, with boundary conditions

u(0,y,t)= 0

u(a,y,t)=0

du/dy(x,0,t) = 0

du/dy(x,a,t) = 0

and Initial Condition u(x,y,0) = f(x,y)

Answer 2

Answer:

Use separation of variables to transform this partial differential equation into two ordinary differential equations. Note that the boundary conditions are homogeneous, with the long-term steady state being u=0 . Assume:

u(x,t)=v(x)w(t)

Then

1w(t)w′(t)=1v(x)v′′(x)=λ

First, solve

v′′(x)=λv(x)

which has nonzero solutions when

λ=−(π2n)2

and those solutions are:

v(x)=cnsin(π2nx)

where n=1,2,3,⋯

since v(0)=v(2)=0 from the homogeneous boundary conditions.

Next, solve

w′(t)=λw(t)

with

λ=−(π2n)2

Those solutions are:

w(t)=eλt=e−(nπ2)2t

Therefore, the general solution is

u(x,t)=v(x)w(t)

u(x,t)=∑∞n=1cnsin(π2nx)e−(nπ2)2t

The coefficients can be found from the initial condition u(x,0)=x :

f(x)=x=∑∞n=1cnsin(π2nx)

These coefficients can be found by finding the corresponding odd Fourier series:

cn=4(−1)n+1πn

Therefore, the general solution is:

u(x,t)=∑∞n=14(−1)n+1πnsin(π2nx)e−(nπ2)2t

In the graph below, x is the x-axis, u is on the y-axis, with each trace showing a different time from t=0 to t=2 . By t=2 , u is almost uniformly 0 :