Partial Differential Equations III - Epiphany Revision Class

In this revision class we will solve questions 2, 3, 7, and 8 from the Partial Differential Equations III May 2022 exam.

Question 0 (Question 2 from May 2022 Exam).

Let $v:{ℝ}^n\to ℝ$ be given. Consider the following Cauchy problem associated to the heat equation

(1)

1. Let $\lambda \in ℝ$, $\lambda \ne 0$. Show that if $u$ is a solution to (1) with initial datum $v$, then $u_{\lambda }(𝒙,t):=u(\lambda 𝒙,{\lambda }^{2}t)$ is a solution to the same PDE, with initial datum $v_{\lambda }(𝒙)=v(\lambda 𝒙).$

2. Suppose that $v$ is integrable, bounded, continuous and nonnegative. Show that (if $u$ was bounded then) ${lim}_{\lambda \to +\mathrm{\infty }}{u}_{\lambda }(𝒙,t)=0$ for all $(𝒙,t)\in {ℝ}^n\times (0,+\mathrm{\infty })$. [Hint: use the representation formula via the fundamental solution].

3. Let $v$ be twice continuously differentiable with compact support. Show that the solution $u$ to (1) is stationary (i.e. $u(𝒙,t_1)=u(𝒙,t_2)$ for any $t_1,t_2>0$ and $𝒙\in {ℝ}^n$) if and only if $v$ is harmonic.

4. Show that the situation in c can take place only if $v$ and $u$ are constant zero functions.

Solution.

1. To simplify matters we can define new variables: $\xi =\lambda x$ and $\tau ={\lambda }^{2}t$. We find that

   $$u_{\lambda }(x,t)=u(\xi ,\tau ).$$

   Using the chain rule, the fact that $𝝃$ depends only on $𝒙$ and $\tau$ only on $t$, and the fact that

   $$\frac{\partial {\xi }_j}{\partial x_i}=\lambda {\delta }_{ij}$$

   we have that

   $${\partial }_t\left(u_{\lambda }(𝒙,t)\right)={\partial }_{\tau }u(𝝃,\tau ){\partial }_t\tau ={\lambda }^2{\partial }_{\tau }u(𝝃,\tau )$$

   and

   $${\partial }_{x_i}\left(u_{\lambda }(𝒙,t)\right)=\sum _{j=1}^n{\partial }_{{\xi }_j}u(𝝃,\tau ){\partial }_{x_i}{\xi }_j=\lambda {\partial }_{{\xi }_j}u(𝝃,\tau ).$$

   Continuing on the above we find that

   $${\partial }_{x_i{x}_i}\left(u_{\lambda }(𝒙,t)\right)={\partial }_{x_i}\left(\lambda {\partial }_{{\xi }_j}u(𝝃,\tau )\right)={\lambda }^2{\partial }_{{\xi }_i{\xi }_j}u(𝝃,\tau )$$

   and as such

   $$\mathrm{\Delta }{u}_{\lambda }(𝒙,t)={\lambda }^2{\mathrm{\Delta }}_{𝝃}u(𝝃,\tau ).$$

   We conclude that

   $${\partial }_t{u}_{\lambda }(𝒙,t)-\mathrm{\Delta }{u}_{\lambda }(𝒙,t)={\lambda }^2\left({\partial }_{\tau }u(𝝃,\tau )-{\mathrm{\Delta }}_{𝝃}u(𝝃,\tau )\right)=0.$$

   Lastly, we notice that the initial condition for $u_{\lambda }$ is

   $$u_{\lambda }(𝒙,0)=u(\lambda 𝒙,0)=v\left(\lambda 𝒙\right).$$

2. As $u_{\lambda }$ solves the heat equation on ${ℝ}^n$ with a bounded and continuous initial datum, and since it is bounded due to the fact that $u$ is bounded, we know that the unique solution to the equation is given by

   $$u_{\lambda }(𝒙,t)=\left(\mathrm{\Phi }\ast u_{\lambda }(\cdot ,0)\right)(𝒙,t)=\frac{1}{{\left(4\pi t\right)}^{\frac{n}{2}}}{\int }_{{ℝ}^n}{e}^{-\frac{{\left|𝒙-𝒚\right|}^2}{4t}}v\left(\lambda 𝒚\right)𝑑𝒚.$$

   Using the change of variable $𝒛=\lambda 𝒚$ we find that $d𝒛={\lambda }^{n}d𝒚$ and

   $$u_{\lambda }(𝒙,t)=\frac{1}{{\left(4\pi t\right)}^{\frac{n}{2}}}\frac{1}{{\lambda }^n}{\int }_{{ℝ}^n}{e}^{-\frac{{\left|𝒙-\frac{𝒛}{\lambda }\right|}^2}{4t}}v\left(𝒚\right)𝑑𝒚.$$

   Since $e^{-\frac{{\left|𝒙-\frac{𝒛}{\lambda }\right|}^2}{4t}}\le 1$ for all $𝒙$ and $𝒛$ and since

   due to the non-negativity and integrability of $v$, we find that (again, using the non-negativity of $v$ and the fundamental solution)

   $$0\le u_{\lambda }(𝒙,t)\le \frac{1}{{\left(4\pi t\right)}^{\frac{n}{2}}}\frac{1}{{\lambda }^n}{\parallel v\parallel }_{L^1\left({ℝ}^n\right)}.$$

   We conclude that

   $$\underset{\lambda \to \mathrm{\infty }}{lim}{u}_{\lambda }(𝒙,t)=0.$$

3. Assuming that $u$ is stationary, we find that $u(𝒙,t)=u(𝒙,0)=v\left(𝒙\right)$. Plugging this into equation (1) using the fact that $v\in C^2$ yields

   $$0={\partial }_{t}u(𝒙,t)-\mathrm{\Delta }u(𝒙,t)=-\mathrm{\Delta }v\left(x\right),$$

   i.e. $v$ is harmonic.
   Assume now that $v$ is harmonic. From our formula for $u$ and the fact that we can interchange differentiation with integration in this case due to the compactness of the support of $v$ we find that

   $$\mathrm{\Delta }u(𝒙,t)=\frac{1}{{\left(4\pi t\right)}^{\frac{n}{2}}}{\int }_{{ℝ}^n}{\mathrm{\Delta }}_{𝒙}\mathrm{\Phi }(𝒙-𝒚,t)v\left(𝒚\right)𝑑𝒚$$

   $$=\frac{1}{{\left(4\pi t\right)}^{\frac{n}{2}}}{\int }_{{ℝ}^n}{\mathrm{\Delta }}_{𝒚}\mathrm{\Phi }(𝒙-𝒚,t)v\left(𝒚\right)𝑑𝒚$$

   $$\underset{v\in C_c\left({ℝ}^n\right)}{=}\frac{1}{{\left(4\pi t\right)}^{\frac{n}{2}}}{\int }_{{ℝ}^n}\mathrm{\Phi }(𝒙-𝒚,t)\underset{=0}{\underbrace{{\mathrm{\Delta }}_{𝒚}v\left(𝒚\right)}}𝑑𝒚=0.$$

   As $u$ satisfies the heat equation we conclude that ${\partial }_{t}u(𝒙,t)=0$ which shows that $u$ is stationary.

4. If $v$ is harmonic and compactly supported on ${ℝ}^n$ then it is bounded and according to Liouville’s theorem $v$ must be constant.

∎

Question 0 (Question 3 from May 2022 Exam).

1. State the definition of harmonic functions $v:{ℝ}^2\to ℝ$ that are twice continuously differentiable.

2. State the weak maximum principle for harmonic functions.

3. Check whether the function $v:{ℝ}^2\to ℝ$, defined as $v(x,y)=x^2-y^2$ is harmonic.

4. Let $B_1:=\{(x,y)\in {ℝ}^2:x^2+y^2\le 1\}$. Find the maximum and minimum of the function $v(x,y)=x^2-y^2$ over $B_1$.

Solution.

1. A twice differentiable function $v:{ℝ}^2\to ℝ$ is harmonic if $\mathrm{\Delta }v\left(𝒙\right)=0$.

2. The weak maximum principle: Let $\mathrm{\Omega }$ be an open and bounded subset of ${ℝ}^n$. If $v\in C^2\left(\mathrm{\Omega }\right)\cap C\left(\overline{\mathrm{\Omega }}\right)$ is a harmonic function then

   $$\underset{\overline{\mathrm{\Omega }}}{\max }u\left(𝒙\right)=\underset{\partial \mathrm{\Omega }}{\max }u\left(𝒙\right).$$

   Remark: In class we stated this with the additional condition that $\mathrm{\Omega }$ is connected. This is not needed for the weak maximum principle (as we saw with the Heat equation) but marks would not have been deducted if you’ve required this extra condition.

3. We have that

   $$v_{xx}(x,y)+v_{yy}(x,y)=2-2=0.$$

4. As $v$ is harmonic on $B_1$ and continuous on $\overline{B_1}$ we can use the weak maximum principle to conclude that

   $$\underset{\overline{B_1}}{\max }v\left(𝒙\right)=\underset{x^2+y^2=1}{\max }v\left(𝒙\right)=\underset{x^2+y^2=1}{\max }\left(x^2-y^2\right)$$

   $$=\underset{-1\le x\le 1}{\max }\left(2x^2-1\right)=1.$$

∎

Question 0 (Question 7 from May 2022 Exam).

Consider the problem

(2)

where $u_0:[0,2\pi ]\to ℂ$ is a given smooth function.

1. Using Fourier series find a candidate for the solution $u$ to (2) in terms of the Fourier coefficients of $u_0$.

2. Let $u_0(x):=v_0(x)+\frac{1}{n}\exp (inx)$ for $n\in \mathbb{N}$ and $v_0:[0,2\pi ]\to ℂ$ is another given smooth function. Find the candidate for the solution to (2) for this initial datum in terms of Fourier coefficients of $v_0$.

3. Conclude that the problem (2) is ill-posed. [Hint: think about the stability, i.e. continuous dependence of the solution on the data. Use the uniform convergence.] (ill-posedness was discussed in the Michaelmas term)

Solution.

1. Assuming that we are allowed to interchange differentiation and summation for the Fourier series associated to $u(x,t)$, given by

   $$\sum _{n\in \mathbb{Z}}{\widehat{u}}_n(t)e^{inx}$$

   with

   $${\widehat{u}}_n(t)=\frac{1}{2\pi }{\int }_0^{2\pi }u(x,t)e^{-inx}𝑑x,$$

   we find that

   $$u_t(x,t)=\sum _{n\in \mathbb{Z}}{\widehat{u}}_n^{\prime }(t)e^{inx}$$

   and

   $$u_{xx}(x,t)=\sum _{n\in \mathbb{Z}}{\widehat{u}}_n{\left(in\right)}^2{e}^{inx}=-\sum _{n\in \mathbb{Z}}{n}^2{\widehat{u}}_n(t)e^{inx}.$$

   Plugging this back into our equation we find that (again, under the assumption that we are allowed to manipulate the sums)

   $$\sum _{n\in \mathbb{Z}}\left({\widehat{u}}_n^{\prime }(t)-n^2{\widehat{u}}_n(t)\right)e^{inx}=0$$

   for $x\in (0,2\pi )$ and $t\in (0,\mathrm{\infty })$. The uniqueness of the Fourier coefficients imply that for all $n\in \mathbb{Z}$

   $${\widehat{u}}_n^{\prime }(t)=n^2{\widehat{u}}_n(t).$$

   The solution to these equations is given by

   $${\widehat{u}}_n(t)={\widehat{u}}_n(0)e^{n^t}={\widehat{u_0}}_n{e}^{n^{2}t}$$

   where we have used the fact that $u(x,0)=u_0(x)$. We conclude that a candidate for a solution to (2) is given by

   $$u(x,t)=\sum _{n\in \mathbb{Z}}{\widehat{u_0}}_n{e}^{n^{2}t}{e}^{inx}.$$

2. Since

   $${\left(\widehat{\alpha f+\beta g}\right)}_n=\alpha {\widehat{f}}_n+\beta {\widehat{g}}_n$$

   and

   $${\widehat{\left(e^{imx}\right)}}_n=\frac{1}{2\pi }{\int }_0^{2\pi }{e}^{imx}{e}^{-inx}𝑑x={\delta }_{nm}$$

   we find that

   $${\widehat{u_0}}_m={\widehat{v_0}}_m+\frac{{\delta }_{mn}}{n}$$

   and conclude that our candidate for a solution is given by

   $$u(x,t)=\sum _{m\in \mathbb{Z}}{\widehat{v_0}}_m{e}^{m^{2}t}{e}^{imx}+\frac{e^{n^{2}t}{e}^{inx}}{n}.$$

3. Let $v$ be a smooth solution to or equation with initial datum $v_0$. Let $u_n$ be a smooth solution to our equation with initial datum

   $$u_n(x,0)=v_0(x)+\frac{1}{n}{e}^{inx}.$$

   We find that, according to part (b),

   $$\left|u_n(x,t)-v(x,t)\right|=\frac{\left|e^{n^{2}t}{e}^{inx}\right|}{n}=\frac{e^{n^{2}t}}{n}\underset{n\to \mathrm{\infty }}{⟶}\mathrm{\infty }$$

   for any $t>0$. However, ${\parallel u_n(0)-v(0)\parallel }_{L^{\mathrm{\infty }}(0,2\pi )}=\frac{1}{n}\underset{n\to \mathrm{\infty }}{⟶}0$. This shows that the problem is ill-posed.

∎

Question 0 (Question 8 from May 2022 Exam).

We consider the following minimisation problem

$$\min \{{\int }_0^1{e}^t[u^{\prime }{(t)}^2+2u{(t)}^2+4u(t)]dt:\mathit{\hspace{1em}}u\in C^1([0,1]),u(0)=0\}.$$

We assume that this problem has a minimiser.

1. Explain why can we find this minimiser by taking the first variation of the functional? (We have not discussed this in class this year).

2. Take an admissible class of perturbations and derive the ODE and the boundary conditions satisfied by the minimiser.

3. Solve the problem obtained in b. Check whether this indeed satisfies the boundary condition.

4. Compute the minimal value of the original minimisation problem (it can be left in an implicit form as an integral).

Solution.

1. We haven’t discussed this in class this year. It is connected to the notion of convexity.

2. Assuming that $u$ is a minimiser an admissible functions $\phi$ must satisfy that $u_{\epsilon }=u+\epsilon \phi$ is in $C^1\left([0,1]\right)$ and $u_{\epsilon }(0)=0$. Consequently, our admissible functions must be in

   $$V=\{\phi :[0,1]\to ℝ|\phi \in C^1\left([0,1]\right),\phi (0)=0\}.$$

   Defining

   $$E[u]={\int }_0^1{e}^t[u^{\prime }{(t)}^2+2u{(t)}^2+4u(t)]𝑑t$$

   and $g(\epsilon )=E\left[u_{\epsilon }\right]$ we conclude that if $u$ is a minimum then $g^{\prime }(0)=0$, if $g$ is differentiable at $\epsilon =0$. As

   $$E\left[u_{\epsilon }\right]={\int }_0^1{e}^t[u^{\prime }{(t)}^2+2u{(t)}^2+4u(t)]𝑑t$$

   $$+\epsilon {\int }_0^1{e}^t\left[2u^{\prime }(t){\phi }^{\prime }(t)+4u(t)\phi (t)+4\phi (t)\right]𝑑t+{\epsilon }^2{\int }_0^1{e}^t\left[{\phi }^{\prime }{(t)}^2+2\phi {(t)}^2\right]𝑑t$$

   we find that

   $$0=g^{\prime }(0)={\int }_0^1{e}^t\left[2u^{\prime }(t){\phi }^{\prime }(t)+4u(t)\phi (t)+4\phi (t)\right]𝑑t=2e^1{u}^{\prime }(1)\phi (1)-2e^0{u}^{\prime }(0)\underset{=0}{\underbrace{\phi (0)}}$$

   $$+{\int }_0^1\left({\left(-2e^t{u}^{\prime }(t)\right)}^{\prime }+4e^{t}u(t)+4e^t\right)\phi (t)𝑑t$$

   For any $\phi \in V$ that satisfied $\phi (1)=0$ we find that

   $${\int }_0^1\left({\left(-e^t{u}^{\prime }(t)\right)}^{\prime }+2e^t\left(u(t)+1\right)\right)\phi (t)𝑑t=0$$

   and as $\phi$ is arbitrary

   $${\left(-e^t{u}^{\prime }(t)\right)}^{\prime }+2e^t\left(u(t)+1\right)=0$$

   or equivalently

   $$-u^{\prime \prime }(t)-u^{\prime }(t)+2u(t)+2=0.$$

   This implies, in particular, that for any $\phi \in V$

   $$0=g^{\prime }(0)=2e^1{u}^{\prime }(1)\phi (1).$$

   Since $\phi$ is arbitrary we must also have that $u^{\prime }(1)=0$.
   We conclude that the minimiser to our functional satisfies

   (3)

3. To solve (3) we start by noticing that $u(t)=-1$ is a particular solution. As such we just need to find the solution to the homogeneous equation

   $$v^{\prime \prime }(t)+v^{\prime }(t)-2v(t)=0.$$

   The characteristic equation to this second order ODE with constant coefficients is

   $$r^2+r-2=0$$

   whose roots are $r=1,-2$. Thus, the general solution to our equation is given by

   $$u(t)=A{e}^t+B{e}^{-2t}-1.$$

   As $u^{\prime }(t)=A{e}^t-2B{e}^{-2t}$ we find that since $u(0)=u^{\prime }(1)=0$

   $$A+B-1=0,$$

   $$Ae-2B{e}^{-2}=0$$

   The solution to which is $A=\frac{2}{2+e^3}$ and $B=\frac{e^3}{2+e^3}$. We conclude that the minimiser is given by

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

4. Since

   $$u^{\prime }(t)=\frac{2}{2+e^3}{e}^t-\frac{2e^3}{2+e^3}{e}^{-2t}$$

   the minimum is given by

   $${\int }_0^1{e}^t[{(\frac{2}{2+e^3}{e}^t-\frac{2e^3}{2+e^3}{e}^{-2t})}^2$$

   $$+2{(\frac{2}{2+e^3}{e}^t+\frac{e^3}{2+e^3}{e}^{-2t}-1)}^2+4(\frac{2}{2+e^3}{e}^t+\frac{e^3}{2+e^3}{e}^{-2t}-1)]dt.$$

∎