Equivalence of two harmonic problems on different domains

Question

I want to solve \begin{cases} \Delta u = 0,&\text{ in }\mathbb{R}^3\setminus B_1(0) \\ u=0,&\text{ as }\Vert x\Vert\rightarrow +\infty \\ u=1,&\text{ on }\partial B_1(0). \end{cases} I know that the solution to this problem is $$ \bar{u}(x) = \Vert x\Vert^{-1}. $$

My question is if by solving the following BVP

\begin{cases} \Delta v = 0,&\text{ in }B_R(0)\setminus B_1(0) \\ v=\bar{u}|_{\partial B_R},&\text{ on }\partial B_R(0)\\ v=1,&\text{ on } \partial B_1(0), \end{cases}

$R>1$, I get a solution $v$ which is the restriction of the solution $u$ of the original problem onto the new domain $\Omega = B_R(0)\setminus B_1(0)$. Namely, if $v = u|_{\Omega}$.

I am claiming this because : If I set $D = \big(\mathbb{R}^3\setminus B_1(0)\big)\cap \big(B_R(0)\setminus B_1(0)\big)$ I can define a third problem over $D$ which is solved by $w=u-v$ on $D$: \begin{cases} \Delta w = 0,&\text{in }D \\ w = 1-1=0,&\text{on }\partial B_1(0) \\ w = \bar{u}|_{\partial B_R(0)}-\bar{u}|_{\partial B_R(0)}=0,&\text{on }\partial B_R(0) \end{cases} so by maximum principle I can conclude that $$ 0=\min_{x\in \partial D} w \leq w(y) \leq \max_{x\in \partial D} w = 0 $$ for any $y\in D$ and hence that $w=u-v\equiv 0$.

Answer 1

Your approach to the solution of the problem by using the maximum principle for the Laplace operator is correct. However, since the maximum principle for the Laplace operator is a strong maximum principle, it can also be used to prove directly that, if $M=\max_{B_R(0)\setminus B_1(0)}w$ and $m=\min_{B_R(0)\setminus B_1(0)}w$ then $M=m=0$. Let's see how.

Maximum principle for the Laplace operator ([1], theorem 2, §2.1 p. 53). Let $$ \Delta u\ge 0\text{ in }D $$ If $u$ attains its maximum $M$ at any point of $D$, then $u\equiv M$ in $D$.
Note that

• $D$ is a connected domain (i.e. a connected open set) in $\Bbb R^n$, not necessarily bounded nor simply connect (it can have holes but every two point in it can be joined by a continuous path), with also no requirements on the boundary $\partial D$.
• as Protter and Weinberger note ([1], §2.1 p. 54), the maximum principle implies a minimum principle, just by considering $-u$ instead of $u$, i.e. let $$ \Delta u\le 0\text{ in }D $$ If $u$ attains its minimum $m$ at any point of $D$, then $u\equiv m$ in $D$.

Now, since $\Delta u=0$ implies $\Delta w\ge 0$ and $\Delta w\le 0$, if $w$ has a maximum $M$ in $B_R(0)\setminus B_1(0)$ then $w=M$ on the whole $B_R(0)\setminus B_1(0)$ and since $w=0$ on $\partial B_R(0)\cup\partial B_1(0)$ this implies $M=0$, and the same happens if we assume that $w$ has a minimum, $m=0$. Thus $w$ is constant and equal to zero on through the whole closed domain $B_R(0)\setminus B_1(0)\cup \big(\partial B_R(0)\cup\partial B_1(0)\big)\iff u=v$ on the same closed domain.

Final notes

• Saying that the solutions of a given equation satisfy a "strong maximum principle" means that if one of them reaches its maximum value at a point of the interior of its domain of definition, it is actually constant though the domain. Otherwise, when the solutions of a given equation reach their maximum value on the boundary of their domain of definition, it is said that they satisfy a weak maximum principle, since this leaves the possibility that the same maximum value could be reached at an interior point. The solution of Laplace's equation satisfy a strong maximum principle, and this the stronger statement implies $w=u-v=0$ in our case.
• The solution to your problem by using the maximum is possibly the "right" one, because it works for every connected domain $D$ and every sufficiently regular boundary value $u|_{\partial D}$. However, in this particular case, due to the spherical symmetry of the $B_R(0)\setminus B_1(0)$ domain, we can solve the boundary value problem for $w$ directly by using the expression of $\Delta w$ in spherical coordinates (written below for $n=3$ for the sake of simplicity): $$ \begin{split} \Delta w &= \frac{1}{r^2} \frac{\partial}{\partial r} \left(r^2 \frac{\partial w}{\partial r} \right) + \frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta} \left(\sin \theta \frac{\partial w}{\partial \theta} \right) + \frac{1}{r^2 \sin^2 \theta} \frac{\partial^2 w}{\partial \varphi^2} \\ &= \frac{1}{r} \frac{\partial^2}{\partial r^2} (rw) + \frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta} \left(\sin \theta \frac{\partial w}{\partial \theta} \right) + \frac{1}{r^2 \sin^2 \theta} \frac{\partial^2 w}{\partial \varphi^2} \end{split} $$ Now, since our boundary values have a spherical symmetry, we can assume that all the derivatives of $w$ respectively to the angle variables vanish, and thus Laplace's equation reduces to the following ordinary differential equation respect to the radial variable $r$ $$ \begin{split} \Delta w=\frac{1}{r} \frac{\partial^2}{\partial r^2} (rw)=0&\iff\frac{\partial^2}{\partial r^2} (rw)=0\\ &\iff \frac{\partial}{\partial r} (rw)=b\quad b=\mathrm{const.}\\ & \iff rw = a+br\!\quad a =\mathrm{const.}\\ & \iff w =\frac{a}{r} +b \end{split} $$ and the given boundary values for $w$ imply $a=b=0$ and thus $w=0$.

Reference

[1] Protter, Murray H.; Weinberger, Hans F., Maximum principles in differential equations, Corrected reprint, New York-Berlin-Heidelberg-Tokyo: Springer-Verlag, pp X+261, (1984), MR0762825, Zbl 0549.35002,