[PDE] Transforming Nonhomogeneous BCs into Homogeneous Ones

[Arkuski]

So there's this problem in my text that's pretty challenging. I can't seem to work out the answer that is given in the back of the book, and then I found a solution manual online that contains yet another solution.

The problem is a the heat equation as follows:

PDE: u_{t} = α^2u_{xx}

BCs: u(0,t) = 1
u_x(1,t)+hu(1,t) = 1

IC: u(x,0) = \displaystyle\sin (πx)+x

Also, 0<x<1.

We assume that u(x,t) = S(x,t)+U(x,t) and that S(x,t) = A(t)[1-x]+B(t)[x]. By substituting these into the BCs, I get A(t) = 1 and B(t) = \frac{2}{1+h}. Now we have S(x,t) = 1-x+\frac{2x}{1+h} = 1+x\frac{1-h}{1+h}. With the steady state solution in place, we construct our homogenous problem as follows:

PDE: U_{t} = α^2U_{xx}

BCs: U(0,t) = 0
U_x(1,t)+hU(1,t) = 0

IC: U(x,0) = \displaystyle\sin (πx)+x\frac{2h}{1+h}-1

If I try to solve this one, it turns into an eigenvalue problem which isn't covered until the next section and the IC is a nightmare. Anyways, the book gives me the following answer:

x+e^{-(πα)^2t}\displaystyle\sin (πx)

Moreover, the solution manual I found on the top of pg 16 looks as if it's solving an entirely different problem. Any help on this problem would be greatly appreciated.

 

[LCKurtz]

It looks to me like all your work is correct as far as you went. The book's answer$$
x+e^{-(πα)^2t}\displaystyle\sin (πx)$$doesn't solve the original problem's boundary conditions, so it isn't the answer to the problem. After all, you would expect the answer to depend on $h$.