Exercise 1 (Maximum principle for subharmonic functions). Let \(\Omega\) be an open, bounded and connected set in \(\mathbb{R}^n\). We say that \(u\in C^2\left( \Omega\right)\cap C\left( \overline{\Omega}\right)\) is subharmonic if \[-\Delta u\leq 0,\qquad \text{in }\Omega.\]
Show that subharmonic functions satisfy the mean value formulae \[u(x) \leq \frac{1}{\left\lvert\partial B_{r}(x)\right\rvert}\int_{\partial B_{r}(x)} u(y)dS(y),\qquad u(x) \leq \frac{1}{\left\lvert B_{r}(x)\right\rvert}\int_{ B_{r}(x)} u(y)dy,\] for any \(x\in \Omega\) and \(r>0\) such that \(\overline{B_r\left( x\right)}\subset \Omega\) (recall that we denote the integral over a domain divided by the size of the domain by an integral with a line across its middle).
Show that subharmonic functions satisfy the strong maximum principle: If there exists \(x_0\in \Omega\) such that \[u\left( x_0\right)=\max_{\overline{\Omega}}u(x)\] then \(u\) is constant.
Conclude that subharmonic functions satisfy the weak maximum principle: \[\max_{\overline{\Omega}}u(x)=\max_{\partial \Omega}u(x).\]
Do subharmonic functions satisfy the minimum principle?
Remark: A function \(u\in C^2\left( \Omega\right)\cap C\left( \overline{\Omega}\right)\) is called superharmonic if \[-\Delta u\geq 0,\qquad \text{in }\Omega,\] or equivalently if \(-u\) is subharmonic. Superharmonic functions satisfy the mean value formulae \[u(x) \geq \frac{1}{\left\lvert\partial B_{r}(x)\right\rvert}\int_{\partial B_{r}(x)} u(y)dS(y),\qquad u(x) \geq \frac{1}{\left\lvert B_{r}(x)\right\rvert}\int_{ B_{r}(x)} u(y)dy,\] and the strong and weak minimum principle.
Exercise 2 (Application of the maximum principle for subharmonic functions - comparison theorem). Let \(\Omega\) be an open, bounded and connected set in \(\mathbb{R}^n\). Assume that for \(i=1,2\) we have that \(u_i\in C^2\left( \Omega\right)\cap C\left( \overline{\Omega}\right)\) satisfy \[\nonumber \left\{ \begin{array}{ll} -\Delta u_i =f_i& \text{in } \Omega,\\ u_i=g_i & \text{on }\partial \Omega, \end{array} \right.\] where \(f_i\in C\left( \Omega\right)\) and \(g_i\in C\left( \partial \Omega\right)\) for \(i=1,2\). Assume that \(f_1\leq f_2\) and \(g_1\leq g_2\) and prove that \(u_1\leq u_2\). This is known as a comparison principle.
Exercise 3 (Weak maximum principle without mean value formula). Let \(\Omega\) be an open, bounded and connected in \(\mathbb{R}^n\). Consider the equation \[-\Delta u (x) + \textbf{b}(x)\cdot \nabla u(x) = f(x),\qquad x\in \Omega\] where \(\textbf{b}=\left\lbrace b_i\right\rbrace_{i=1}^n\in C^1\left( \overline{\Omega};\mathbb{R}^n\right)\). Our goal is to show that if \(u\in C^2\left( \Omega\right)\cap C\left( \overline{\Omega}\right)\) is a solution to the equation in \(\Omega\) then if \(f\leq 0\) (subharmonic solution) \(u\) has a weak maximum principle.
Assuming \(x_0\in \Omega\) is a local maximum for \(u\), show that if \(f<0\) on \(\Omega\) then \[-\Delta u \left( x_0\right) < 0.\]
Recall that a necessary condition for a point \(x_0\) to be a local maximum for a \(C^2\) function \(\varphi:\Omega\to \mathbb{R}\) is that the Hessian matrix at \(x_0\), \(\mathrm{Hess}\;\varphi\left( x_0\right)\), is negative semi-definite, i.e. all its eigenvalues are non-positive or equivalently, for any \(\textbf{y}\in \mathbb{R}^n\) \[\textbf{y}^T \mathrm{Hess}\;\varphi\left( x_0\right)\textbf{y} \leq -\alpha\left( x_0\right)\left\lvert\textbf{y}\right\rvert^2,\] for some \(\alpha\left( x_0\right)\geq 0\). Use this to show that if \(x_0\in \Omega\) is a local maximum for \(u\) then \(\Delta u\left( x_0\right) \leq 0\).
Show that if \(f<0\) in \(\Omega\) then \(u\) can’t have a local maximum in \(\Omega\), and as such satisfies the weak maximum principle.
We want to extend the above to the case \(f\leq 0\). For any \(\lambda,\varepsilon\in \mathbb{R}\) define \[u_{\varepsilon,\lambda}(x) = u(x) + \varepsilon e^{\lambda x_1}.\]
Show that \[\nabla u_{\varepsilon,\lambda}(x) =\nabla u(x) + \varepsilon\lambda e^{\lambda x_1}\left( 1,0,0,\dots,0\right)^T,\] and \[\Delta u_{\varepsilon,\lambda}(x) = \Delta u(x) + \varepsilon\lambda ^2 e^{\lambda x_1}.\]
Show that \(u_{\varepsilon,\lambda}\) solves the equation \[-\Delta u_{\varepsilon,\lambda}(x) + \textbf{b}(x)\cdot \nabla u_{\varepsilon,\lambda}(x) = f(x) - \varepsilon\lambda \left( \lambda-b_1(x) \right)e^{\lambda x_1}.\]
Show that there exists \(\lambda_0 >0\) such that for all \(\varepsilon>0\) \[f(x) - \varepsilon\lambda_0 \left( \lambda_0-b_1(x) \right)e^{\lambda_0 x_1} <0\] on \(\Omega\).
Conclude that for any \(\varepsilon>0\) \[\max_{\overline{\Omega}}\left( u(x)+\varepsilon e^{\lambda_0 x_1}\right)=\max_{\partial\Omega}\left( u(x)+\varepsilon e^{\lambda_0 x_1}\right)\] and consequently \[u(x) \leq \max_{\partial\Omega}u(y)+\varepsilon\max_{y\in \partial \Omega}e^{\lambda_0 y_1}\] for all \(x\in \overline{\Omega}\).
Conclude the weak maximum principle for \(u\).
Remark: This could be extended to more general elliptic PDEs.