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## Solution Instability for the Explicit Method

Solution Instability for the Explicit Method:As we observed in experiments using myheat.m the solution can befall unbounded unless the time steps are small.

:Text the Difference Equations in Matrix FormIf we use the boundary conditions u(0) = u(L) = 0 then the explicit method of the previous section has the form:

u

_{i,j+1}= ru_{i−1,j }+ (1 − 2r)u_{i,j}+ ru_{i+1,j}, 1 ≤ i ≤ m − 1, 0 ≤ j ≤ n − 1,where u

_{0,j}= 0 and u_{m,j}= 0. This is equal to the matrix equation:u

_{j+1}= Au_{j},where u

_{j}is the column vector (u_{1,j}, u_{2,j}, . . . , u_{m,j})′ representing the state at the jth time step and A is the matrix:Unluckily this matrix can have a property which is extremely bad in this context. Specifically it can cause exponential growth of error unless r is small. To observe how this happens suppose that Ujis the vector of correct values of u at time step tjand Ejis the error of the approximation uj, then

u

_{j}= U_{j}+ E_{j}.The estimation at the next time step will be:

u

_{j+1}= AU_{j}+ AE_{j},And if we continue for k steps,

u

_{j+k}= AkU_{j}+ AkE_{j}.The problem with this is the term AkE

_{j}. This term is precisely what we would do in the power method for finding the eigenvalue of A with the largest absolute value. If the matrix A has ew’s with total value greater than 1 then this term will grow exponentially. The largest complete value of an ew of A as a function of the parameter r for various sizes of the matrix.A. As you can observe for r >1/2 the largest absolute ew grows rapidly for any m and rapidly becomes greater than 1.

Maximum complete eigenvalue EW as a function of r for the matrix A from the explicit method for the heat equation calculated for matrices A of sizes m = 2 . . . 10. When EW >1 the method is unstable that is errors grow exponentially with each step. When by means of the explicit method r <1/2 is a safe choice.

:ConsequencesRecall that r = ck/h

^{2}. Since this should be less than 1/2, we have:k <h

^{2}/2c.The first consequence is observable k should be relatively small. The second is that h can’t be too small. Ever since h

^{2}appears in the formula making h small would force k to be extremely small!A third consequence is that we comprise a converse of this analysis. Presume r < .5. Then every eigenvalues will be less than one. Evoke that the error terms satisfy:

u

_{j+k}= AkU_{j}+ AkE_{j}.If all the eigenvalues of A are less than 1 in complete value then AkE

_{j}grows smaller and smaller as k increases. This is actually good. Rather than building up the consequence of any error diminishes as time passes! From this we get there at the following principle- If the explicit numerical solution for a parabolic equation doesn’t blow up then errors from previous steps fade away!Ultimately we note that if we have other boundary conditions then instead of equation we have:

u

_{j+1}= Au_{j}+ rb_{j}Where the first as well as last entries of bjcontain the boundary conditions and all the other entries are zero. In this case the errors perform just as before if r >1/2 then the errors grow and if r <1/2 the errors fade away.

We are able to write a function program myexppmatrix that produces the matrix A in for given inputs m and r. Without using loops we are able to use the diag command to set up the matrix

function A = myexpmatrix(m,r)% produces the matrix for the explicit method for a parabolic equation

% Inputs: m -- the size of the matrix

% r -- the main parameter, ck/h^2

% Output: A -- an m by m matrix

u = (1-2*r)*ones(m,1);

v = r*ones(m-1,1);

A = diag(u) + diag(v,1) + diag(v,-1);

Test this by means of m = 6 and r = .4.6, Check the eigenvalues as well as eigenvectors of the resulting matirices

> A = myexpmatrix

> [v e] = eig(A)

What is the ‘mode’ represented by the eigenvector with the largest absolute eigenvalue? How is that reproduce in the unstable solutions?

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