dualrey

My matrix was

matrix = [

   [0,   4,   4,   4,   2],

   [2,   0,   3,   4,   4],

   [3,   1,   0,   3,   3],

   [2,   1,   1,   0,   1],

   [2,   3,   2,   2,   0]

]

The eigenranking was

Team 1: rating = 0.5676428888501164
Team 2: rating = 0.5140250259852949
Team 3: rating = 0.4266803335123048
Team 5: rating = 0.4096681506358447
Team 4: rating = 0.25234048971012685

And team 1 was the best.

With zero source and the boundary conditions given in my problem, the temperature at the lower right corner of the bar is $0.91330$ temperature units.

The equation for heat flow in my problem is

$$u_t=4u_{xx}+3$$

with boundary conditions $u(0,t)=-3$ and $u(1,t)=2$.

To find the steady state solution, we set $u_t=0$. Then,

$$0=4u_{xx}+3$$

$$\Rightarrow u_{xx}=-\frac{4}{3}$$

$$\Rightarrow u_{x}=-\frac{4}{3}x+\alpha$$

$$\Rightarrow u(x)=-\frac{2}{3}x^2+\alpha x+\beta.$$

Inserting the boundary conditions, we get

$$u(0)=\beta=-3$$

and

$$u(1)=-\frac{2}{3}+\alpha-3=2$$

$$\Rightarrow \alpha=\frac{17}{3}.$$

Therefore, the steady state solution is

$$\boxed{u(x,t)=-\frac{2}{3}x^2+\frac{17}{3}x-3}$$

This is the heat flow model from the parameters given in my particular problem. At about $t=1s$, the temperature near the midpoint of the insulated edge is approximately 0.32308 heat units.

$\textbf{Theorem}$. Suppose $\omega(x,y,z)$ is a scalar field and $\vec{\varphi}(x,y,z)$ is a vector field. Then,

$$\nabla\cdot\left(\omega\vec{\varphi}\right)=\vec{\varphi}\cdot\nabla\omega+\omega\nabla\cdot\vec{\varphi}.$$

$\textit{Proof.}$

$$\nabla\cdot\left(\omega\vec{\varphi}\right)=\left\langle \frac{\partial}{\partial x},\frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right\rangle\cdot\omega\left\langle\varphi_x,\varphi_y,\varphi_z\right\rangle$$

$$=\left\langle \frac{\partial}{\partial x},\frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right\rangle\cdot\left\langle\omega\varphi_x,\omega\varphi_y,\omega\varphi_z\right\rangle$$

$$=\frac{\partial(\omega\varphi_x)}{\partial x}+\frac{\partial(\omega\varphi_y)}{\partial y}+\frac{\partial(\omega\varphi_z)}{\partial z}$$

$$=\left(\frac{\partial\omega}{\partial x}\varphi_x+\frac{\partial\varphi_x}{\partial x}\omega\right)+\left(\frac{\partial\omega}{\partial y}\varphi_y+\frac{\partial\varphi_y}{\partial y}\omega\right)+\left(\frac{\partial\omega}{\partial z}\varphi_z+\frac{\partial\varphi_z}{\partial z}\omega\right)$$

$$=\left(\varphi_x\frac{\partial\omega}{\partial x}+\varphi_y\frac{\partial\omega}{\partial y}+\varphi_z\frac{\partial\omega}{\partial z}\right)+\omega\left(\frac{\partial\varphi_x}{\partial x}+\frac{\partial\varphi_y}{\partial y}+\frac{\partial\varphi_z}{\partial z}\right)$$

$$=\vec{\varphi}\cdot\nabla\omega+\omega\nabla\cdot\vec{\varphi}. $$

$$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \square$$