$\newcommand{\const}{\mathrm{const}}$ $\newcommand{\erf}{\operatorname{erf}}$

Deadline Wednesday, October 10: students of both sections can submit either in the morning class (MP202; 9:00-10:00) or in room 1008 of 215 Huron between 15:00 and 15:30.

Crucial in many problems is formula (14) rewritten as \begin{equation} u(x,t)=\int _{-\infty}^\infty G(x,y,t) g(y)\,dy. \label{eq-1} \end{equation} with \begin{equation} G(x,y,t)=\frac{1}{2\sqrt{k\pi t}}e^{-\frac{(x-y)^2}{4kt}} \label{eq-2} \end{equation} This formula solves IVP for a heat equation \begin{equation} u_t=ku_{xx} \label{eq-3} \end{equation} with the initial function $g(x)$.

In many problems below for a modified standard problem you need to derive a similar formula albeit with modified $G(x,y,t)$. Consider \begin{equation*} \erf(z)=\sqrt{\frac{2}{\pi}}\int_0^ze^{-z^2/2}\,dz \tag{Erf}\label{eq-Erf} \end{equation*} as a standard function.

Using method of continuation obtain formula similar to (\ref{eq-1})-(\ref{eq-2}) for solution of IBVP for a heat equation on ${x>0,t>0}$ with the initial function $g(x)$ and with

- Dirichlet boundary condition $u|_{x=0}=0$;
- Neumann boundary condition $u_x|_{x=0}=0$;

Consider heat equation with a convection term \begin{equation} u_t+\underbracket{v u_x}_{\text{convection term}} =ku_{xx}. \label{eq-4} \end{equation}

- Using change of variables $u(x,t)=U(x-vt,t)$ reduce it to ordinary heat equation and using (\ref{eq-1})-(\ref{eq-2}) for a latter write a formula for solution $u (x,t)$.
- Can we use the method of continuation to solve IBVP with Dirichlet or Neumann boundary condition at $x>0$ for (\ref{eq-4}) on $\{x>0,t>0\}$? Justify your answer.

Using either formula (\ref{eq-1})-(\ref{eq-2}) or its modification (if needed)

- Solve IVP for a heat equation (\ref{eq-3}) with $g(x)=e^{-\alpha |x|}$; what happens as $\alpha \to +0$?
- Solve IVP for a heat equation with convection (\ref{eq-4}) with $g(x)=e^{-\alpha |x|}$; what happens as $\alpha \to +0$?
- Solve IBVP with the Dirichlet boundary condition for a heat equation (\ref{eq-3}) with $g(x)=e^{-\alpha |x|}$; what happens as $\alpha \to +0$?
- Solve IBVP with the Neumann boundary condition for a heat equation (\ref{eq-3}) with $g(x)=e^{-\alpha |x|}$; what happens as $\alpha \to +0$?

Consider a solution of the diffusion equation $u_t=u_{xx}$ in $[0\le x \le l, 0\le t <\infty]$.

Let \begin{gather*} M(T)= \max _{[0\le x \le l, 0\le t \le T]} u(x,t),\\ m(T)= \min _{[0\le x \le l, 0\le t \le T]} u(x,t). \end{gather*}

- Does $M(T)$ increase or decrease as a function of $T$?
- Does $m(T)$ increase or decrease as a function of $T$?

The purpose of this exercise is to show that the maximum principle is not true for the equation $u_t=xu_{xx}$ which has a variable coefficient.

- Verify that $u=-2xt-x^2$ is a solution.
- Find the location of its maximum in the closed rectangle $[-2\le x\le 2, 0\le t\le 1]$.
- Where precisely does our proof of the maximum principle break down for this equation?

Consider the heat equation on $J=(-\infty,\infty)$ and prove that an

*energy*\begin{equation} E(t)=\int_J u^2 (x,t)\,dx \label{eq-5} \end{equation} does not increase; further, show that it really decreases unless $u(x,t)=\const$;Consider the heat equation on $J=(0,l)$ with the Dirichlet or Neumann boundary conditions and prove that an $E(t)$ does not increase; further, show that it really decreases unless $u(x,t)=\const$;

Consider the heat equation on $J=(0,l)$ with the Robin boundary conditions \begin{gather} u_x(0,t)-a_0u(0,t)=0,\\ u_x(l,t)+a_lu(l,t)=0. \end{gather} If $a_0>0$ and $a_l>0$, show that the endpoints contribute to the decrease of $E(t)=\int_0^l u^2 (x,t)\,dx$.

This is interpreted to mean that part of the *energy* is lost at the boundary, so we call the boundary conditions *radiating* or *dissipative*.

**Hint.** To prove decrease of $E(t)$ consider it derivative by $t$, replace $u_t$ by $ku_{xx}$ and integrate by parts.

**Remark.** In the case of heat (or diffusion) equation *energy* given by (\ref{eq-5}) is rather mathematical artefact;