如何使用MATLAB解决此PDE

Question

I have the following question on a practice exam:

I need to use MATLAB to solve it. The problem is, I have not seen a problem like this before and I'm struggling to get started.

I have my 1x1 grid, split into 10x10. I know I can calculate the whole bottom row besides the corners using 1/10 * x*2. I also know I can calculate the entire right row using (1/10)(1+t)^2. However, I cannot figure out how to get enough points to be able to fill in the values for the entire grid. I know it must have something to do with the partial derivatives given in the problem, but I'm not quite sure where they come into play (especially the u_x equation). Can someone help me get a start here?

I don't need the whole solution. Once I have enough points I can easily write a matlab program to solve the rest. Really, I think I just need the x=0 axis solved, then I just fill in the middle of the grid.

I have calculated the bottom row, minus the two corners, to be 0.001, 0.004, 0.009, 0.016, 0.025, 0.036, 0.049, 0.064, 0.081. And similarly, the entire right row is trival to calculate using the given boundry condition. I just can't piece together where to go from there.

Edit: the third boundry condition equation was mistyped. it should read:

u_x(0,t) = 1/5t, NOT u(0,t) = 1/5t

Answer 1

First realise that the equation you have to solve is the linear wave equation , and the numerical scheme you are given can be rewritten as

( u^(n+1)_m - 2u^n_m + u^(n-1)_m )/k^2 = ( u^n_(m-1) - 2u^n_m + u^n_(m+1) )/h^2

where k is the time step and h is the delta x in space.

The reformulated numerical scheme makes clear that the left- and right-hand sides are the second order centred finite difference approximations of u_tt and u_xx respectively.

To solve the problem numerically, however, you need to use the form given to you because it is the explicit update formula that you need to implement numerically: it gives you the solution at time n+1 as a function of the previous two times n and n-1 . You need to start from the initial condition and march the solution in time.

Observe that the solution is assigned on the boundaries of the domain ( x=0 and x=1 ), so the values of the discretized solution u^(n)_0 and u^(n)_10 are known for any n ( t=n*k ). At the n th time step your unknown is the vector [u^(n+1)_1, u^(n+1)_2, ..., u^(n+1)_9] .

Observe also that to use the update formula to find the solution at the n+1 step, requires the knowledge of the solution at two previous steps. So, how do you start from n=0 if you need information from two previous times? This is where the initial conditions come into play. You have the solution at n=0 ( t=0 ), but you also have u_t at t=0 . These two pieces of information combined can give you both u^0 and u^1 and get you started.

I would use the following start-up scheme:

u^0_m = u(h*m,0)  // initial condition on u
(u^2_m - u^0_m)/(2k) = u_t(h*m,0)  // initial condition on u_t

that combined with the numerical scheme used with n=1 gives you everything you need to define a linear system for both u^1_m and u^2_m for m=1,...,9 .

To summarize:

--use the start-up scheme to find solution at n=1 and n=2 simultaneously .

--from there on march in time using the numerical scheme you are given.

If you are completely lost check out things like: finite difference schemes, finite difference schemes for advection equations, finite difference schemes for hyperbolic equations, time marching.

EDITING:

For the boundary condition on u_x you typically use the ghost cell method:

• Introduce a ghost cell at m=-1 , ie a fictitious (or auxiliary) grid point that is used to deal with boundary condition, but that is not part of the solution.

• The first node m=0 is back into your unknown vector, ie you are now working with [u_0 u_1 ... u_9] .

• Use the left side boundary condition to close the system. Specifically, by writing down the centered approx of the boundary condition

  u^n_(1) - u^n_(-1) = 2*h*u_x(0,k*n)

• The above equation allows you to express the solution on the ghost node in terms on the solution on an internal, real node. Therefore you can apply the time-marching numerical scheme (the one you are given) to the m=0 node. (The numerical scheme applied to m=0 would contain contributions from the m=-1 ghost node, but now you have that expressed in terms of the m=1 node.)