PDE with Robin boundary condition

[captaincook]

Hi guys,
I am trying to solve this equation, but I have problem with the second boundary condition. I am trying to use combination of variables but it wont work. Could you please give me some hints about how to tackle this problem, using any solving method.

(∂^2 T)/(∂x^2 )=1/∝ ∂T/∂t
IC: T(x,0) = Ti
BC1: T(x→∞, t) = Ti
BC2: -k ∂T/∂x|x=0 = h[T∞-T(0,t)]

I tried combining the variables by W= x/(4∝t)^0.5

and the equation became : (d^2 T)/(dW^2 )=-2W dT/dW

and

IC: T(W→∞) = Ti
BC1: T(W→∞) = Ti
BC2: -k dT/dW * 1/(4∝t)^0.5 |W=0 = h[T∞-T(W=0)]

As you see term "t" can not be canceled out the second boundary condition. Please give me a hint to what to do.

Thank you

 

[dsk2]

captaincook,

You are complicating the problem by introducing W. You donot need to do that. In fact you shouldn't.

You can follow the steps below:

1. Discretize the transient diffusion equation using central difference method.
2. Assuming you discretize the domain in the x-direction into 'N' points. You can treat the last point as infinity, and apply the boundary conditions.
3. Applying BC1 is straight forward since it is a constant. However when you apply BC2, try to discretize BC2 using second order backward differencing ( first order will work but the accuracy is less and probably less complicated).
4. Try and form the linear matrix [A][T]=. Try to adjust the boundary terms so that they are properly reflected in matrices A and B.
5. Use, a standard matrix inversion method to solve for T. Your matrix A might most likely be tridiagonal, in which case, you can just use a more efficient algorithm ( like Thomas tridiagonal), but dont worry about it in the beginning.

Look at some examples to see how to solve such linear systems. The tricky part is applying the boundary conditions correctly.

Also, note that you are solving for the transient solution. Hence, T is always a function of time 't'. Also, you will have the time step 'dt' when you discretize the equations. If you looking for the steady state equation, you can use a different form of equation to begin with.

Cheers,
Sai.