Heat equation - separation of variables

[Gekko]

Homework Statement

du/dt=d2u/dx2, u(0,t)=0, u(pi,t)=0

u(x,0) = sin^2(x) 0<x<pi

Find the solution

Also find the solution to the initial condition:

du/dt u(x,0) = sin^2(x) 0<x<pi

The Attempt at a Solution

From separation of variables I obtain

u(x,t) = B.e^(-L^2t).sin(Lx)

For the boundary condition u(pi,t)=0, u(x,t) = Bm . e^(-m^2t) . sin(mx)

Finally for u(x,0) = sin^2x

u(x,0) = 4(cos(m.pi)-1).e^(-m^2t).sinmx / pi(m^3-4m)
found through superposition and Fourier sum of sines

What I don't see is how to solve the initial condition because when I diff. above wrt t and set t-0 I obtain

du/dt u(x,0) = -m^2 4(cos(m.pi)-1).e^(-m^2t).sinmx / pi(m^3-4m) = sin^2x

How can this be set to sin^2x by choosing a value of m?

 

[HallsofIvy]

I'm not sure what you are talking about here. You understand that these are two different problems, don't you? You solution "for u(x,0) = sin^2x

u(x,0) = 4(cos(m.pi)-1).e^(-m^2t).sinmx / pi(m^3-4m)" is a sum isn't it?

That is
u(x, t)= \sum_{m=1}^\infty e^{-m^2t}(4(cos(m.pi)-1).e^(-m^2t).sinmx / pi(m^3-4m)).

Now, with initial condition u_t(x, 0)= sin^2(x) you must have
u_t(x, 0)= \sum_{n=1}^\infty -m^2 B_m sin(mx)= sin^2(x)

Find Fourier sine coefficients for sin^2(x) and solve for B_m.

 

[vela]

From separation of variables I obtain

u(x,t) = B.e^(-L^2t).sin(Lx)

For the boundary condition u(pi,t)=0, u(x,t) = Bm . e^(-m^2t) . sin(mx)

Just to be clear, separation of variables gives you

u(x,t) = e^{-L^2t}(A\cos Lx + B\sin Lx)

Then applying the boundary condition u(0,t)=0 gives you A=0, and applying the condition u(π,t)=0 requires L be an integer.

 

[Gekko]

Understood now. Thanks a lot for clearing that up