PDE, heat equation with mixed boundary conditions

Hakkinen
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Homework Statement


solve the heat equation over the interval [0,1] with the following initial data and mixed boundary conditions.

Homework Equations


[itex]\partial _{t}u=2\partial _{x}^{2}u[/itex]
[itex]u(0,t)=0, \frac{\partial u}{\partial x}(1,t)=0[/itex]

with B.C

[itex]u(x,0)=f(x)[/itex]
where f is piecewise with values:
[itex]0, 0<x\leq \frac{1}{2}[/itex]
[itex]3, \frac{1}{2}<x<1[/itex]

The Attempt at a Solution

after separation of variables where [itex]u(x,t)=h(x)\phi (t)[/itex]:

[itex]h''(x)=-\frac{\lambda }{2}h(x)[/itex]
[itex]\phi'(t)=-\lambda \phi(t)[/itex]

gen. solution to h is
[itex]h(x)=a\sin \sqrt{\frac{\lambda }{2}}[/itex] the constant with the cos term is 0 from initial value

I'm stuck trying to find the eigenvalue
[itex]h'(1)=\frac{\lambda }{2}a\cos\sqrt{\frac{\lambda }{2}}=0[/itex]
[itex]\sqrt\frac{\lambda }{2}=\arccos 0[/itex]

I'm not sure what expression with n to use for arccos of 0. npi/2 won't work, or (n+1)pi/2, is this the right procedure though?

Any help is greatly appreciated!

EDIT:

I'm trying [itex]\frac{\pi }{2}+n\pi[/itex] now to solve for the eigenvalue
 
on Phys.org
So for the general solution of u I have [itex]u(x,t)=\sum_{n=1}^{\infty}A_{n}\sin [\frac{\pi}{2}(1+2n)x]\exp -2t[\frac{\pi}{2}(1+2n)]^2[/itex]

and the coefficient A_n given by

[itex]A_{n}=\frac{12}{\pi(1+2n)}\cos \pi(1+2n)[/itex]

There was another cosine term with pi/2 in the argument that was always zero for any n, but is this an acceptable way to leave the coefficient expression?
 

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