# Heat equation - separation of variables

1. Jul 24, 2010

### Gekko

1. The problem statement, all variables and given/known data

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

3. 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 dont 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?

2. Jul 24, 2010

### HallsofIvy

Staff Emeritus
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 $B_m$.

3. Jul 24, 2010

### vela

Staff Emeritus
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.

4. Jul 24, 2010

### Gekko

Understood now. Thanks a lot for clearing that up