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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

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

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

$$u_t + uu_x = 0.$$

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

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

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