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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…
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@mark Ah, sweet example. I now see where my "proof" went wrong. I was rather cavalier about dividing by $u^2$, as if it would never be zero, but our boundary conditions say it must be zero somewhere! Any function where my ratio doesn't blow up can't be a solution to the problem, hence the contradiction. We should be good…
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@mark I don't know what you mean. I thought I had proved that the right side $\textit{is}$ a constant, unless there is some external reason that $\nabla u\cdot\nabla u$ and $u^2$ cannot be proportional, in which case we would have a contradiction.
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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.
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@mark and @Abs That's correct. For example, if we were to integrate $x^2$ from $-1$ to $0$, $$\int_{-1}^{0}x^2dx=\left.\frac{1}{3}x^3\right|_{-1}^{0}=\frac{1}{3}(0)^3-\frac{1}{3}(-1)^3=-\left(-\frac{1}{3}\right)=\frac{1}{3}$$ or from $-2$ to $-1$:…
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I'm pretty sure a positive integrand can never give a negative integral...that would be geometrically nonsensical.
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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, $$0=4u_{xx}+3$$ $$\Rightarrow u_{xx}=-\frac{4}{3}$$ $$\Rightarrow u_{x}=-\frac{4}{3}x+\alpha$$ $$\Rightarrow u(x)=-\frac{2}{3}x^2+\alpha x+\beta.$$…
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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.
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$\textbf{Theorem}$. 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…