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The displacement of the midpoint from equilibrium at time t=1.1s is 0.010.

Heat flow with a constant internal heat source is governed by

$$u_t = 2u_{xx} + 6, u(0,t) = -1, u(1,t) = 4.$$

For a steady state temperature distribution, $u(x,t)$ does not change with respect to time, so $u_t = 0$. Hence, $2u_{xx} + 6 = 0$. Solving this equation for $u_{xx}$ yields $u_{xx} = -3$, and then integrating twice gives the solution

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

where C and D are constants. By setting $u(0,t) = -1$ and $u(1,t) = 4$, I was able to solve for C and D, and get a final solution of

$$u(x,t) = -\frac{3}{2}x^2 +\frac{13}{2}x - 1.$$

Recall our basic conservation law expressed as a PDE:

$$u_t + \varphi_x = f.$$

Here we assume the source $f = 0$ and the flux $\varphi = \frac{1}{2}u^2$. Taking the derivative of the flux with respect to $x$ by using the chain rule, we find $\varphi_x = uu_x$. Plugging these variables back into the original equation, we get

$$u_t + uu_x = 0.$$

This is an example of as advection equation of the form

$$u_t + cu_x +au = f(x,t)$$

$$u(x,0) = u_0,$$

where $c = 1$, $a = -3$, $f(x,t) = 0$, and $u_0 = x^2$. To solve this equation, we rewrite $u(x,t)$ as $U(\xi,\tau)$, where $\xi = x-ct = x-t$ and $\tau = t.$ By applying the chain rule, the PDE becomes

$$(U_{\xi}(-1) + U_{\tau}(1)) + U_{\xi}(1) = 3U.$$

SImplifying, we get

$$U_{\tau} = 3U$$

or

$$\frac{\partial U}{\partial \tau} = 3U.$$

We move all $U$ and $\tau$ terms to one side,

$$\frac{dU}{U} = 3d\tau,$$

and integrate:

$$lnU = 3\tau + \phi(\xi),$$

where $\phi$ is an unknown function of $\xi$. We then solve for U by raising $e$ to both sides:

$$U = e^{3\tau + \phi(\xi)}.$$

We simplify the right hand side using a technique familiar to students of calculus and ordinary differential equations:

$$e^{3\tau + \phi(\xi)} = e^{ \phi(\xi)}e^{3\tau} = \phi(\xi)e^{3\tau},$$

where we abuse $\phi$ to represent $e^{\phi}$ as is done with constants $C$ in calculus. it may seem odd that we can seemingly just transform an exponential function of $\xi$ into a regular one. Recall, however, that we do not yet know what the original $\phi(\xi)$ is, and it could very well be a logarithmic function, canceling out the exponential. We will get the same answer regardless when we plug in our initial conditions.

Hence our formula for U is

$$U(\xi,\tau) = \phi(\xi)e^{3\tau}.$$

Now we rewrite our equation in terms of $x$ and $t$:

$$u(x,t) = \phi(x-t)e^{3t}.$$

Finally, we plug in our initial values to solve for $\phi$:

$$u(x,0) = \phi(x) = x^2.$$

Our final solution, then, is

$$u(x,t) = (x-t)^2e^{3t}.$$

The solution to the PDE is

$$u = u_0(x - 2t),$$

so at each time interval, the graph above will shift to the right by two units. Here are sketches of the graph at each of the three times:

If we changed the PDE to $u_t + 2u_x = -u$, the solution becomes

$$u = u_0(x - 2t)e^{-t},$$

so $u$ will decay over time, approaching zero at large values of $t$.

As can be observed in the screenshot above, the temperature near the midpoint of the insulated edge of the triangle (see my randomly generated problem for details of the model) is 1.04558.

With $\kappa = 0$ and $f = 0$, the steady state temperature at the lower right corner of the bar is 0.84905.

My matrix was:

matrix = [

  [0,1,4,3,3],

  [2,0,2,4,3],

  [3,3,0,3,2],

  [2,1,1,0,2],

  [3,1,3,1,0]

]

and my ranking was:

Team 3: rating = 0.5054388063612655
Team 1: rating = 0.5016207712530641
Team 2: rating = 0.48157957363128245
Team 5: rating = 0.4108981576640922
Team 4: rating = 0.30356553355238997

Team 3 was the best.

(1) We start by separating variables, and assume $u(r,t) = R(r)T(t)$. Our PDE then becomes

$$(RT)_t = (RT)_{rr} + \frac{1}{r}(RT)_r$$ or

$$RT' = R''T + \frac{1}{r}R'T.$$

Dividing both sides by $RT$, we get

$$\frac{T'}{T} = \frac{R''}{R} +\frac{1}{r}\frac{R'}{R} = -\lambda.$$

Thus, $\frac{T'}{T} = -\lambda$, so $T' = -\lambda T.$ Hence,

$$T = e^{-\lambda t}.$$ Meanwhile,

$$\frac{R''}{R} + \frac{1}{r}\frac{R'}{R} = -\lambda,$$ so

$$R'' + \frac{1}{r}R' = -\lambda R.$$

This can be rewritten as $$-(rR'(r))' = \lambda rR(r).$$

The solutions to this radial differential equation are the Bessel functions $$R(r) = c_1J_0\left(\sqrt{\lambda}r\right) + c_2Y_0\left(\sqrt{\lambda}r\right).$$

Since $Y_0$ is unbounded at $z=0, c_2$ must equal $0,$ so $$R(r) = c_1J_0\left(\sqrt{\lambda}r\right).$$

From our boundary conditions, we know $$R(1) = c_1J_0\left(\sqrt{\lambda}\right) = 0,$$ so

$$\sqrt{\lambda} = z_n, n=1,2,3...$$

where $z_n$ are the zeros of $J_0.$ Hence, the eigenvalues are $\lambda_n = z_n^2$ and the corresponding eigenfunctions are $R_n(r) = J_0(z_nr).$ Thus, our full solution is

$$u(r,t) = \sum_{n=1}^{\infty}c_ne^{-\lambda_n^2t}J_0(z_nr).$$

(2) By our initial conditions,

$$u(r,0) = \sum_{n=1}^{\infty}c_nJ_0(z_nr) = 5r^3(1-r).$$

Since Bessel functions satisfy the orthogonality condition,

$$c_n = \frac{\int_{0}^{1}5r^3(1-r)J_0(z_nr)rdr}{\int_{0}^{1}J_0^2(z_nr)^2rdr}.$$

Plugging an initial condition $u(r,0)$ into the "Radial heat flow" ObservableHQ page gives us the normalized Bessel function $F(r)$ for each term of the solution and its constant coefficient, which we'll denote $b_n,$ where

$$u(r,0) = \sum_{n=1}^{\infty}b_nF_n(r) = \sum_{n=1}^{\infty}b_n\frac{R_n(r)}{||R_n||}.$$

Since $u(r,0) = \sum_{n=1}^{\infty}c_nR_n(r),$ it follows that

$$c_n = \frac{b_n}{||R_n||}.$$

We plug in $u(r,0) = 5r^3(1-r)$ and find that

$$u(r,t) \approx \frac{0.17333401324973097}{0.3670927157694067}e^{-z_1^2t}J_0(z_1r) + \frac{-0.18156860394077448}{0.24060355211655088}e^{-z_2^2t}J_0(z_2r) + \frac{0.0731801393963544}{0.16437392311147672}e^{-z_3^2t}J_0(z_3r)$$

$$= 0.4721804759e^{-z_1^2t}J_0(z_1r) - 0.7546380855e^{-z_2^2t}J_0(z_2r) + 0.4452052857e^{-z_3^2t}J_0(z_3r)$$

(3)

Using the same Observable page, with N=10, we graph the function at time $t = 0.02$:

(4)

Using the final tool on the Observable page with $N=10$, we find $$u(0, 0.02) = 0.086.$$