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June 12, 2017 compiled on — Monday June 12, 2017 at 01:18 AM

Solve

Initial conditions

Boundary conditions

Let

| (2) |

where is steady state solution that only needs to satisfy boundary conditions and satisfies the PDE itself but with homogenous B.C. At steady state, the PDE becomes

The solution is . Hence (2) becomes

Substituting the above in (1) gives

With boundary conditions . This is now in standard form and separation of variables can be used to solve it.

Now acts as a source term. The eigenfunctions are known to be where . Hence by eigenfunction expansion, the solution to (3) is

| (3A) |

Substituting this into (3) gives

| (4) |

Expanding using same basis (eigenfunctions) gives

Applying orthogonality

But since and the above simplifies to

But , hence

Therefore and (4) becomes

This is an ODE in whose solution is

From (3A) now becomes

| (5) |

To find , from initial conditions, at the above becomes

Hence

Therefore (5) becomes

And since then the solution is

To simulate

Here is the animation from the above

Here is the numerical solution to compare with

Here is the animation from the above

Reference: stokes second problem question and answer