MA52300 Fall 2025

Homework Assignment 7 – Solutions

1.

Given a $C^2$ function $u$ defined on a set $U\subset {ℝ}^n\setminus \{0\}$, we define the function $𝒦[u]$ on $U^{\ast }$ by

$$𝒦[u](x)={|x|}^{2-n}u(x^{\ast }),\text{where }{x}^{\ast }=x/{|x|}^2,U^{\ast }=\{x^{\ast }:x\in U\};$$

the function $𝒦[u]$ is called the Kelvin transform of $u$. Prove that

$$\mathrm{\Delta }(𝒦[u])=𝒦[{|x|}^4\mathrm{\Delta }u]\text{in }{U}^{\ast }.$$

In particular, if $u$ is harmonic in $U$, then $𝒦[u]$ is harmonic in $U^{\ast }$.

Hint: 1) Prove that for any polynomial $p$ in ${ℝ}^n$ homogeneous of degree $m$, one has

$$\mathrm{\Delta }({|x|}^{2-n-2m}p)={|x|}^{2-n-2m}\mathrm{\Delta }p,$$

which is equivalent to $\mathrm{\Delta }(𝒦[p])=𝒦[{|x|}^4\mathrm{\Delta }p]$. Start by computing $\mathrm{\Delta }({|x|}^{k}p)$ for arbitrary $k$ and use that $xDp(x)=mp(x)$ by homogeneity. 2) Conclude that the statement of the problem holds for any polynomial $p$, not necessarily homogeneous, and then use the density of polynomials in $C^2$ norm on compact subsets of $U$.

Solution.

Let $p_m$ be a homogeneous polynomial of degree $m$. We then can compute

$$𝒦[p_m](x)={|x|}^{2-n}{p}_m(x/{|x|}^2)={|x|}^{2-n-2m}{p}_m(x).$$

Noting that ${|x|}^4\mathrm{\Delta }{p}_m$ is a homogeneous polynomial of degree $m+2$, we will also have

$$𝒦[{|x|}^4\mathrm{\Delta }{p}_m](x)={|x|}^{2-n}{|x/{|x|}^2|}^4(\mathrm{\Delta }{p}_m)(x/{|x|}^2)={|x|}^{2-n-2m}\mathrm{\Delta }{p}_m(x).$$

Next, let $k$ be a real number and note that

$$\mathrm{\Delta }({|x|}^k{p}_m(x))=\mathrm{\Delta }({|x|}^k)p_m(x)+2D({|x|}^k)D{p}_m(x)+{|x|}^k\mathrm{\Delta }{p}_m(x),$$

which is a particular case of a more general formula

$$\mathrm{\Delta }(uv)=(\mathrm{\Delta }u)v+2DuDv+u(\mathrm{\Delta }v).$$

Now, following the same computations as in the derivation of the fundamental solution, we have

$$D({|x|}^k)=k{|x|}^{k-2}x,\mathrm{\Delta }({|x|}^k)=k(k-2+n){|x|}^{k-2},x\ne 0$$

which gives

$$\begin{array}{c}\mathrm{\Delta }({|x|}^k{p}_m(x))\hfill \\ \hfill =k(k-2+n){|x|}^{k-2}{p}_m(x)+2k{|x|}^{k-2}xD{p}_m(x)+{|x|}^k\mathrm{\Delta }{p}_m(x)\\ \hfill =k(k-2+n){|x|}^{k-2}{p}_m(x)+2km{|x|}^{k-2}{p}_m(x)+{|x|}^k\mathrm{\Delta }{p}_m(x)\\ \hfill =k(k-2+n+2m){|x|}^{k-2}{p}_m(x)+{|x|}^k\mathrm{\Delta }{p}_m(x),\end{array}$$

where we have used that $xD{p}_m(x)=m{p}_m(x)$. Taking, $k=2-n-2m$, we then have

$$\mathrm{\Delta }({|x|}^{2-n-2m}{p}_m(x))={|x|}^{2-n-2m}\mathrm{\Delta }{p}_m(x),$$

which from the earlier computation is equivalent to

$$\mathrm{\Delta }(𝒦[p_m])=𝒦[{|x|}^4\mathrm{\Delta }{p}_m],x\ne 0.$$

Now, noticing that that the Kelvin transform is a linear operator and decomposing any polynomial $p$ of degree $N$ into a finite sum $p={\sum }_{m=0}^N{p}_m$, where $p_m$ is now a homogeneous polynomial of degree $m$, we will have that the statement of the problem holds for any polynomial $p$. Finally, using the density of polynomials in $C^2$ norm locally in $U$, and taking a sequence of polynomials $p^j\to u$ in $C^2(V)$ for all $V⋐U$, we will have

$$\mathrm{\Delta }(𝒦[u])=\underset{j\to \mathrm{\infty }}{lim}\mathrm{\Delta }(𝒦[p^j])=\underset{j\to \mathrm{\infty }}{lim}𝒦[{|x|}^4\mathrm{\Delta }{p}^j]=𝒦[{|x|}^4\mathrm{\Delta }u]$$

locally uniformly in $U^{\ast }$, where we have used that $𝒦[p^j]\to 𝒦[u]$ in $C^2(V^{\ast })$ for any $V⋐U$. ∎

2.

Let $\mathrm{\Omega }$ be a bounded domain with $C^1$ boundary, $g\in C(\partial \mathrm{\Omega })$ and $f\in C(\overline{\mathrm{\Omega }})$. Consider then the so-called Neumann problem

($\ast$)

where $\nu$ is the outer normal on $\partial \mathrm{\Omega }$. Show that the solution of ($\ast$ ‣ 2) in $C^2(\mathrm{\Omega })\cap C^1(\overline{\mathrm{\Omega }})$ is unique up to a constant; i.e., if $u_1$ and $u_2$ are both solutions of ($\ast$ ‣ 2), then $u_2=u_1+const$ in $\mathrm{\Omega }$.

Hint: Look at the proof of the uniqueness for the Dirichlet problem by energy methods, [E, 2.2.5a].

Solution.

The proof is almost a verbatim repetition of that of Theorem 16 in [E, 2.2.5a].

Set $w=u_2-u_1$. Then $\mathrm{\Delta }w=0$ in $\mathrm{\Omega }$ and $\partial w/\partial \nu =0$ on $\partial \mathrm{\Omega }$. Integrating by parts, we obtain

$${\int }_{\mathrm{\Omega }}{|Dw|}^2𝑑x=-{\int }_{\mathrm{\Omega }}w\mathrm{\Delta }w𝑑x+{\int }_{\partial \mathrm{\Omega }}w\frac{\partial w}{\partial \nu }𝑑S=0.$$

Thus, $Dw\equiv 0$ in $\mathrm{\Omega }$ and, since $\mathrm{\Omega }$ is connected, we deduce $w=u_2-u_1\equiv const$ in $\mathrm{\Omega }$. This completes the proof. ∎

3.

Write down an explicit formula for a solution of

where $c\in ℝ$.

[Hint: Rewrite the problem in terms of $v(x,t):=e^{ct}u(x,t)$]

Solution.

Following the hint let $v(x,t)=e^{ct}u(x,t)$. Then also

$$u(x,t)=e^{-ct}v(x,t).$$

Furthermore,

$u_t$ $=-c{e}^{-ct}v+e^{-ct}{v}_t$

$\mathrm{\Delta }u$ $=e^{-ct}\mathrm{\Delta }v$

Then the equation

$$u_t-\mathrm{\Delta }u+cu=f$$

takes the form

$$e^{-ct}(-cv+v_t-\mathrm{\Delta }v+cv)=f$$

or, equivalently,

(1) $$v_t-\mathrm{\Delta }v=e^{ct}f.$$

Besides, at the initial time we have

(2) $$v(x,0)=u(x,0)=g(x).$$

A solution to problem (1)–(2) is given then by the formula

$$v(x,t)={\int }_{{ℝ}^n}g(y)\mathrm{\Phi }(x-y,t)𝑑y+{\int }_0^t{\int }_{{ℝ}^n}{e}^{cs}f(y,s)\mathrm{\Phi }(x-y,t-s).$$

This gives

$u(x,t)$ $={\displaystyle {\int }_{{ℝ}^n}}g(y)e^{-ct}\mathrm{\Phi }(x-y,t)𝑑y+{\displaystyle {\int }_0^t}{\displaystyle {\int }_{{ℝ}^n}}f(y,s)e^{-c(t-s)}\mathrm{\Phi }(x-y,t-s)$

$={\displaystyle {\int }_{{ℝ}^n}}g(y){\mathrm{\Phi }}_c(x-y,t)𝑑y+{\displaystyle {\int }_0^t}{\displaystyle {\int }_{{ℝ}^n}}f(y,s){\mathrm{\Phi }}_c(x-y,t-s),$

where

$${\mathrm{\Phi }}_c(x,t)=e^{-ct}\mathrm{\Phi }(x,t)=\frac{1}{{(4\pi t)}^{n/2}}{e}^{-ct-\frac{{|x|}^2}{4t}}.$$

(The latter function can be actually considered as the fundamental solution of the equation $u_t-\mathrm{\Delta }u+cu=0$.) ∎