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HW #1

HW #1
For N=5, the node map looks like the figure below. The nodes start from 0 at the bottom left corner.

Node map for N=5.

Sketched solution map of analytic solution:

Solution map of U

The solution map of U has some clearly defined contour lines at theta = pi/6 and theta = pi/2, because of the cos(3*theta) term.  At theta = pi/3 and theta = 0, we have the max and min values of the solution.

 

Solution Process:

I next created arrays for x(k), y(k), r(k), and theta(k).  They are saved here:
x.dat
y.dat
r.dat
theta.dat

After verifying the analytic solution, which I did by plugging in the solution U into Laplace's Equation with cylindrical coordinates, I then took center difference equations for each term.

Verification of Analytic Solution
Center Difference and Interior Molecule
Boundary A Molecule
Boundary B and C Molecules
Boundary D and Corner CD Molecule

Next I created 2D contours for the analytic solution to compare to the initial solution. For N=5, the contour looked like the following figure.

N=25 Analytic Solution

Then, continuing on to the numerical solution, I used Matlab's left division operator to directly solve Ax = B.  The A and B matrices are shown below for N=5.

For N=5, the solution and the current vector, grad(U).

For N=10, the solution and current vector.

It's hard to see from the N=5 solutions whether or not the numerical solution agrees with the initial solution map of U.  Since the mesh with 5 nodes, including the boundaries, is so coarse, it's difficult to accurately assess the PDE across the geometry.  But as N is increased to 10, the general picture of the solution becomes clearer, and the similarity to the solution map can be seen.

 

Absolute Error:

Once I had both NxN matrices for the analytic solution and the numerical solution, it was a simple step to calculate the RMS error.  The error map is shown on the geometry for both N=5 and N=10.

They both show that the error is very close to 0 for the majority of the geometry, but as N increases, there is a corner where the error begins to look problematic.  This could be do to an issue with the molecule along that boundary.

RMS Error vs ∆r:

In my simulation, I used the single precision option in MATLAB, which still contained more than six significant figures.  I found that as N increased, the error decreased as expected, and went as O(h^2).

Another expected result was that the error continued to decrease as N increased, but hit a saturation point and started to increase thereafter.  This would be the point where the round off error begins to dominate the error calculation.

 

“Converged” (high N) :

For high N, I chose N=125 to show the converged numerical solution for the analytic solution and the current vector.  With the mesh, it becomes a little difficult to see the solution, but the colormap agrees with the initial sketch.

For "high" N, this would be at the point that where round-off starts to dominant the error calculation.  If my solution had worked correctly, I would have seen an uptick in the error, which would signal N has reached a sufficient level.