# Thread: Inhomogenous heat boundary condition problems

1. ## Inhomogenous heat boundary condition problems

I am having a lot of problems with these sorts of questions..help please!

#1
u_t = alpha^2*u_xx + sin(2x)
u(−pi, t) = u(pi, t), u_x(−pi, t) = u_x(pi, t)
u(x, 0) = cos(x)

Solve the problem by eigenfunction expansion

I don't know how to deal with the boundary conditions. I am used to seeing the BC being u(x,t)=0, u(L,t)=4 or similar but never having the BC equal to each other.

#2

I don't know how to deal with the r. Do i treat both the r and k as constants?

Help is greatly appreciated. Please be as detailed as possible, I am not very comfortable with these types of problems.

2. ## Re: Inhomogenous heat boundary condition problems

We can handle the nonhomegeneous term one of two ways.

(1) Try a substitution of the form

$\displaystyle u(x,t) = v(x,t) + f(x)$

and try and find $\displaystyle f(x)$ as to make your PDE becomes the heat equation. However, in doing so you want your BC's to remain unchanged. This tends to be difficult to do however you sometimes can get lucky.

(2) Find an eigenfunction expansion for

$\displaystyle u_t = \alpha^2 u_{xx}$

that will satisfy your BC's. Find a Fourier series for your source term (in your case it already has this form) and equate coefficients of the sine and cosine terms.

This is what I did

Spoiler:

$\displaystyle u = v + \frac{1}{4 \alpha^2} \sin 2x$

New problem

$\displaystyle v_t = \alpha^2 v_{xx}$

$\displaystyle v(-\pi,t) = v(\pi,t)$

$\displaystyle v_x(-\pi,t) = v_x(\pi,t)$

$\displaystyle v(0,t) = \cos x - \frac{1}{4 \alpha^2} \sin 2x$

Now try separation of variables.