flappy_bird

I tried the following:

a = 0

b = 2

n =25

1) The 24x24 matrix that approximates the 2nd derivative is

$$U_{xx}=\begin{pmatrix} -312.5 & 156.25 & 0 & \ldots & 0 & 0 & 0\\ 156.25 & -312.5 & 156.25 & \ldots & 0 & 0 & 0\\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots\\ 0 & 0 & 0 &\ldots & 156.25 & -312.5 & 156.25\\ 0 & 0 & 0 & \ldots & 0 & 156.25 & -312.5 \end{pmatrix}$$

2) The eigenvalue of smallest magnitude and its corresponding eigenvector are:

$$\lambda = -2.4642$$

$$\vec u = \begin{pmatrix} -0.0354\\ -0.0703\\-0.1041\\-0.1362\\-0.1662\\-0.1936\\-0.2179\\-0.2388\\-0.2559\\-0.2689\\-0.2778\\-0.2822\\-0.2822\\-0.2778\\-0.2689\\-0.2559\\-0.2388\\-0.2179\\-0.1936\\-0.1662\\-0.1362\\-0.1041\\-0.0703\\-0.0354\end{pmatrix}$$

3) The eigenfunction for this problem was: $$k* sin(\sqrt{-\lambda}*x)$$.

If we drop k, we get the following plot:

But if we let $k = -0.09\pi$, we get the following:

Now, I don't remember talking about using the approximation to find $k$ in class. @mark Could you explain why this was necessary here?

My matrix was:

  [0, 1, 3, 1, 2],

  [2, 0, 1, 3, 1],

  [2, 4, 0, 1, 3],

  [2, 3, 1, 0, 1],

  [4, 2, 2, 2, 0]

My ranking was:

Team 3: rating = 0.5273129974036033
Team 5: rating = 0.5215157378625016
Team 1: rating = 0.4172591286356548
Team 4: rating = 0.3713873702087703
Team 2: rating = 0.3713873702087703

The steady-state temperature of the bar at the lower right corner is approximately $0.909$ [temperature units]

Given the boundary and initial conditions, $u(t=1s) \approx 0.92$ [temperature units] at the midpoint of the insulated boundary.

Bird face just looks ominous (。_。)