Jules_Wim

Here is my Matrix:

M=[

[0,4,1,2,3],

[4,0,2,1,4],

[3,4,0,4,4],

[1,3,2,0,2],

[3,4,3,1,0]

]

And here is my team ranking:

Team 3: rating = 0.5621626778711195
Team 5: rating = 0.45596356825094614
Team 2: rating = 0.4451535544402384
Team 1: rating = 0.40012824676275205
Team 4: rating = 0.34322885535517655

The temperature at the lower right end of the bar looks to be roughly 0.9467 in the steady state.

The temperature at the midpoint of the insulated side of the triangle when we reached one second is around 0.1725.

Heat flow with a constant internal heat source is governed by u_t = 1u_{xx}+7, u(0,t)=-2, u(1,t)=4. This is how I found the steady state temperature distribution.

For the steady state temperature distribution, $u(x,t)$ does not change with respect to time so $u_t=0$, Then for my equation $$0=u_{xx}+7,$$ $$u_{xx}=-7.$$

Then I will need to integrate to find $u(x)$. We start with $$u_{xx}=-7.$$ Then after the first integral we get $$u_{x}=-7x+c_1.$$ The second integral is $$u(x)=\frac{-7}{2}x^2+c_1x+c_2.$$

With the conditions where $u(0,t)=-2$ and $u(1,t)=4$, we are able to find that $c_1=\frac{19}{2}$ and $c_2=-2$.

Therefore, the steady state temperature distribution is $$u(x)=\frac{-7}{2}x^2+\frac{19}{2}x-2.$$

The displacement from equilibrium of the midpoint of the string at time $t=1.2$ seconds into the vibration is $-0.28$.