# Solving Ax=b with both unknowns in A and b

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Hai Nguyen on 13 Aug 2019
Commented: Torsten on 15 Aug 2019
Anyone please help me with this problem. I only know some basic when it comes to matlab and having lots of difficulties trying to code this matrix problem Ax=b that have unknown in both A and b.
The linear system obtained in the form of
where are constants, m is time step, , (initial and boundary values - W at t=0 and W at the last node and outside =0)
How to create the tridiagonal matrix A and vector b in matalb please? I tried to search similiar codes but only found the equations where A and b are known...in this case A and b have symbolic arguments in power form and they are related to the solution

Hai Nguyen on 13 Aug 2019
Mr Roberson, could you please elaborate a bit more?
Walter Roberson on 13 Aug 2019
diag(vector1, -1) + diag(vector2) + diag(vector3, 1)
builds a tridiagonal matrix.
Walter Roberson on 13 Aug 2019
All of the j=i-1 entries are constructed the same way and they all fall along the diagonal just to the left of the major diagonal

Torsten on 13 Aug 2019
Edited: Torsten on 13 Aug 2019
Now if W_{0}^{m} and W_{N+1}^{m} are 0 (I think you meant W_{0}^{m} instead of W_{N}^{m}), you have a system of N equations in W_{1}^{m},...,W_{N}^{m}.
You can use Walter's suggestion to solve for W_{1}^{m+1},...,W_{N}^{m+1}.
I don't understand why you write that there are unknowns in A and b. The W's in A and b are the W's from the last time step (m) and thus known. The unkowns are W_{1}^{m+1},...,W_{N}^{m+1}.

Torsten on 15 Aug 2019
Take (A-5) and (A-7) to calculate W_{0}^{m}.
This is the missing boundary value at x=0.
In the last program I posted, you just have to replace b(1) by this value.
Hai Nguyen on 15 Aug 2019
Thank you Sir for your kindness and patience. If I would like to reduce the time step from 1 to 0.01, which lines of the above program I need to change?
Torsten on 15 Aug 2019
From your equations, you have to change C1 and C2 accordingly, don't you ?
I wonder where the (delta_x)^2 in the denominator of C1 has gone ...