Jesus

My matrix is:

matrix = [

  [0,4,2,1,4],

  [4,0,2,3,1],

  [3,1,0,3,2],

  [1,1,1,0,4],

  [4,4,1,2,0]

]

Team 1: rating = 0.5079982820145703
Team 5: rating = 0.5032938392103756
Team 2: rating = 0.4485785148338271
Team 3: rating = 0.4101302319490532
Team 4: rating = 0.34525869395863085

That means Team 1 was the best but doesn't look like it is by too much.

For our conditions, $k=0$ and $f=0$, the steady-state of this polygon is shown above. The temperature in the lower right corner, at about $(31,1)$, is approximately $.96731.$

Given my parameters, the temperature near the midpoint of the insulated side of the triangle at $t=1s$ is about $0.52492$.

We will look at a bar of length $1$ that lies along the unit interval. Its temperature distribution is given by:

$$g(x)=3x^2 - 2x.$$

At time t=0, its left end is set to temperature 3 and its right end to -2. Here is the sketch of the temperature distribution at times $t=0, t=0.01, t=0.1$, and $t=10.$

We will look at heat flow with a constant internal heat source. This will be governed by, $$u_t=4{u_xx}+8.$$

Such that we have the initial conditions, $u(0,t)=-3$ and $u(1,t)=3.$

Given that we are looking for a steady-state distribution, we can set $u_t=0,$ such that $u(x,t)$ does not change with respect to time, $t.$

Therefore, we have the equation $$0=4{u_xx}+8.$$

Simplifying we get:

$$4{u_xx}=-8$$

$${u_xx}=-2.$$

Integrating twice we get

$${u_x}=-2x+{c_1}$$

$$u=-x^{2}+x{c_1}+{c_2}.$$

Using our initial conditions, we will solve for $c_1$ and $c_2$:

We get ${c_1}=7$ and ${c_2}=-3.$

Thus, our steady-state distribution function is:

$$u=-x^2+7x-3.$$