# Equivalence of two harmonic problems on different domains

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 $$w=u-v\equiv 0$$.

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,