Return to Answer

This result leads to the following uniqueness theorem which can be improved making weaker some hypotheses on the behaviour of the function on the "regular" boundary.

(It is worth stressing the the above result immediately entails the fantastic fact that harmonic functions are always $C^\infty$ and not only $C^2$, actually it is possible to prove that they are real analytic.) This result leads to the following uniqueness theorem which can be improved making weaker some hypotheses on the behaviour of the function on the "regular" boundary.

Therfore, applying the uniqueness property in $\overline{\Omega}$, we are committed to conclude that the potential $\varphi$ in the region $\Omega \cup \partial \Omega$ is the same in both cases.

Theorem. Suppose $\Omega \subset \mathbb R^n$ is non-empty open and $\overline{\Omega}$ is compact. Let $p \in \Omega$ and consider the problem: $$\Delta \varphi(x) =\rho \quad x \in \Omega \setminus \{p\}$$ with boundary conditions $$\varphi|_{\partial \Omega} = f$$ where $$\varphi \in C^2(\Omega \setminus \{p\}) \cap C^0(\partial \Omega \cup \Omega \setminus \{p\} )$$ and $f \in C^0(\partial \Omega)$ and $\rho \in C^0(\Omega \setminus \{p\})$ are assigned. If both $\varphi_1$ and $\varphi_2$ are solutions of the problem and $$ \lim_{x\to p} \varphi_1(x)- \varphi_2(x)=0\:,$$ (where the limits can diverge or not exist, if considered separately) then $$\varphi_1 = \varphi_2\:.$$

PROOF. With the given hypotheses, evidently $\phi:= \varphi_1-\varphi_2$ is continuous on $\Omega$, therefore it is a distribution for test functions, $h\in C_0^\infty (\Omega)$. If $B_\epsilon$ is a small ball around $p$ with radius $\epsilon$, using continuity of $\phi$ in particular an definign $\Omega_\epsilon := \Omega \setminus B_\epsilon$, we have $$\int_\Omega \phi \Delta h d^nx = \lim_{\epsilon \to 0^+}\int_{\Omega_\epsilon} \phi \Delta h d^nx = \lim_{\epsilon \to 0^+} \int_{\Omega_\epsilon} (\Delta \phi) h d^nx = \lim_{\epsilon \to 0^+} \int_{\Omega_\epsilon}(\rho-\rho) h d^nx$$ $$= \lim_{\epsilon \to 0^+} 0 = 0\:.$$ All that means that $\phi$ is a distribution solving $\Delta \phi =0$ in distributional sense. Therefore, in view of the mentioned elliptic regularity property, it is a smooth function up to a zero measure set. Since $\phi$ is continuous on $\Omega \setminus \{p\}$ and extends to a continuous function at $p$ (which has zero measure), $\phi = \varphi_1-\varphi_2$ is a smooth function everywhere in $\Omega$. In particular, by continuity of second derivatives the smoothly extended function $\varphi$ verifies $\Delta \varphi =0$ in the whole set $\Omega$ in the proper sense. By construction, we end up with a function $\varphi$ which is $C^\infty(\Omega) \cup C^0(\overline{\Omega})$, satisfying $\Delta \varphi =0$ in $\Omega$ and $\varphi =0$ on $\partial \Omega$. In view of the standard uniqueness result, $\phi=0$ in $\overline{\Omega}$, i.e. $\varphi_1= \varphi_2$ in $\overline{\Omega}$. QED

Therefore, applying the uniqueness property in $\overline{\Omega}$, we are committed to conclude that the potential $\varphi$ in the region $\Omega \cup \partial \Omega$ is the same in both cases.

Theorem. Suppose $\Omega \subset \mathbb R^n$ is non-empty open and $\overline{\Omega}$ is compact. Let $p \in \Omega$ and consider the problem: $$\Delta \varphi(x) =\rho \quad x \in \Omega \setminus \{p\}$$ with boundary conditions $$\varphi|_{\partial \Omega} = f$$ where $$\varphi \in C^2(\Omega \setminus \{p\}) \cap C^0(\partial \Omega \cup \Omega \setminus \{p\} )$$ and $f \in C^0(\partial \Omega)$ and $\rho \in C^0(\Omega \setminus \{p\})$ are assigned. If both $\varphi_1$ and $\varphi_2$ are solutions of the problem and $$ \lim_{x\to p} (\varphi_1(x)- \varphi_2(x))=0\:,$$ (where the limits can diverge or not exist, if considered separately in order to embody the case of a point charge at $q$) then $$\varphi_1 = \varphi_2\:.$$

PROOF. With the given hypotheses, evidently $\phi:= \varphi_1-\varphi_2$ is continuous on $\Omega$, therefore it is a distribution for test functions, $h\in C_0^\infty (\Omega)$. If $B_\epsilon$ is a small ball around $p$ with radius $\epsilon$, using continuity of $\phi$ in particular, integrating by parts and defining $\Omega_\epsilon := \Omega \setminus B_\epsilon$, we have $$\int_\Omega \phi \Delta h d^nx = \lim_{\epsilon \to 0^+}\int_{\Omega_\epsilon} \phi \Delta h d^nx = \lim_{\epsilon \to 0^+} \int_{\Omega_\epsilon} (\Delta \phi) h d^nx = \lim_{\epsilon \to 0^+} \int_{\Omega_\epsilon}(\rho-\rho) h d^nx$$ $$= \lim_{\epsilon \to 0^+} 0 = 0\:.$$ All that means that $\phi$ is a distribution solving $\Delta \phi =0$ in distributional sense. Therefore, in view of the mentioned elliptic regularity property, it is a smooth function up to a zero measure set. Since $\phi$ is continuous on $\Omega \setminus \{p\}$ and extends to a continuous function at $p$ (which has zero measure), $\phi = \varphi_1-\varphi_2$ is a smooth function everywhere in $\Omega$. In particular, by continuity of second derivatives the smoothly extended function $\phi$ verifies $\Delta \phi =0$ in the whole set $\Omega$ in the proper sense. By construction, we end up with a function $\phi$ which is $C^\infty(\Omega) \cup C^0(\overline{\Omega})$, satisfying $\Delta \phi =0$ in $\Omega$ and $\phi =0$ on $\partial \Omega$. In view of the standard uniqueness result, $\phi=0$ in $\overline{\Omega}$, i.e. $\varphi_1= \varphi_2$ in $\overline{\Omega}$. QED

Theorem. Suppose $\Omega \subset \mathbb R^n$ is non-empty open and $\overline{\Omega}$ is compact. Let $p \in \Omega$ and consider the problem: $$\Delta \varphi(x) =\rho \quad x \in \Omega \setminus \{p\}$$ with boundary conditions $$\varphi|_{\partial \Omega} = f$$ where $$\varphi \in C^2(\Omega \setminus \{p\}) \cap C^0(\partial \Omega \cup \Omega \setminus \{p\} )$$ and $f \in C^0(\partial \Omega)$ and $\rho \in C^0(\Omega \setminus \{p\})$ are assigned. If both $\varphi_1$ and $\varphi_2$ are solutions of the problem and $$ \lim_{x\to p}\varphi_1(x) = \lim_{x\to p} \varphi_2(x)\:,$$ where both limits are supposed to be finite, then $$\varphi_1 = \varphi_2\:.$$

Theorem. Suppose $\Omega \subset \mathbb R^n$ is non-empty open and $\overline{\Omega}$ is compact. Let $p \in \Omega$ and consider the problem: $$\Delta \varphi(x) =\rho \quad x \in \Omega \setminus \{p\}$$ with boundary conditions $$\varphi|_{\partial \Omega} = f$$ where $$\varphi \in C^2(\Omega \setminus \{p\}) \cap C^0(\partial \Omega \cup \Omega \setminus \{p\} )$$ and $f \in C^0(\partial \Omega)$ and $\rho \in C^0(\Omega \setminus \{p\})$ are assigned. If both $\varphi_1$ and $\varphi_2$ are solutions of the problem and $$ \lim_{x\to p} \varphi_1(x)- \varphi_2(x)=0\:,$$ (where the limits can diverge or not exist, if considered separately) then $$\varphi_1 = \varphi_2\:.$$