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.
Solution:
Much like in class, we will show that the second mean value theorem follows from the first. Indeed, given \(x\in \Omega\) and \(r>0\) such that \(\overline{B_r(x)}\subset \Omega\) we see that \(\overline{B_{\rho}(x)}\subset \Omega\) for all \(\rho\in (0,r]\) and \[\int_{B_r(x)}u(y)dy = \int_{0}^r \int_{\partial_{B_{\rho}(x)}}u(y)dS(y)d\rho \underset{\text{under our assumption}}{\geq }\] \[\int_{0}^{r}\left\lvert\partial B_{\rho}(x)\right\rvert u(x)d\rho = n\alpha (n) u(x) \int_{0}^r\rho^{n-1}d\rho = \left\lvert B_r(x)\right\rvert u(x),\] which implies that \[u(x) \leq \frac{1}{\left\lvert B_{r}(x)\right\rvert}\int_{ B_{r}(x)} u(y)dy.\] We turn our attention to the first mean value theorem and consider the function \[\phi(\rho) = \frac{1}{\left\lvert\partial B_{\rho}(x)\right\rvert}\int_{\partial B_\rho(x)}u(y)dS(y).\] Just like in class, since \(u\in C^2\left( \Omega\right)\) we see that for any \(x\in \Omega\), \(\phi(\rho)\) is differentiable on an open interval \(\left( 0,r\right)\) and \[\phi^\prime (\rho) = \frac{\rho}{n}\;\frac{1}{\left\lvert B_{\rho}(x)\right\rvert}\int_{B_{\rho}(x)}\Delta u(y)dy.\] As \(u\) is subharmonic we conclude that \(\phi^\prime (\rho) \geq 0\) and consequently that \(\phi\) is increasing. We conclude that for any \(r>0\) such that \(\overline{B_r(x)}\subset \Omega\) we have that \[\frac{1}{\left\lvert\partial B_{r}(x)\right\rvert}\int_{\partial B_{r}(x)} u(y)dS(y) = \phi(r) \geq \lim_{\varepsilon\to 0^+}\phi\left( \varepsilon\right) = u(x).\] Here we have used the fact that if \(f\) is continuous at \(x_0\) then \[\frac{1}{\left\lvert\partial B_{\varepsilon}(x)\right\rvert}\int_{\partial B_{\varepsilon}(x_0)}f(y)dy \underset{\varepsilon\to 0^+}{\longrightarrow}f\left( x_0\right).\]
The proof is almost identical to the strong maximum principle from class. Assume that \(x_0 \in \Omega\) such that \[u\left( x_0\right)=\max_{\overline{\Omega}}u(x)=M.\] Then, since \(\Omega\) is open, we can find \(r\left( x_0\right)>0\) such that \(\overline{B_{r\left( x_0\right)}}\left( x_0\right)\subset \Omega\). Since \(u\) is subharmonic the mean value formula tells us that \[M = u\left( x_0\right) \leq \frac{1}{\left\lvert B_{r(x_0)}(x)\right\rvert}\int_{ B_{r\left( x_0\right)}\left( x_0\right)} u(y)dy \leq \frac{1}{\left\lvert B_{r(x_0)}(x)\right\rvert}\int_{ B_{r\left( x_0\right)}\left( x_0\right)} Mdy =M.\] This implies that \(\frac{1}{\left\lvert B_{r(x_0)}(x)\right\rvert}\int_{ B_{r\left( x_0\right)}\left( x_0\right)} u(y)dy=M\) or that \[\frac{1}{\left\lvert B_{r}(x)\right\rvert}\int_{ B_{r\left( x_0\right)}\left( x_0\right)} \left( M-u(y)\right)dy = 0.\] Since \(M-u(y) \geq 0\) for all \(y\in \Omega\) and it is a continuous function on \(B_{r}\left( x_0\right)\) we must have that \(u(y)=M\) for all \(y\in B_{r\left( x_0\right)}\left( x_0\right)\). We conclude that if \(x_0\in u^{-1}\left\lbrace M\right\rbrace\cap \Omega\) then there exists \(r\left( x_0\right)>0\) such that \(B_{r\left( x_0\right)}\left( x_0\right)\subset u^{-1}\left( M\right)\cap \Omega\). i.e. the set \(u^{-1}\left\lbrace M\right\rbrace\cap \Omega\) is open in \(\Omega\). It is also closed in \(\Omega\) as the preimage of a closed set by a continuous function. Since \(\Omega\) is connected and \(u^{-1}\left\lbrace M\right\rbrace\cap \Omega\) is not empty by assumption, we conclude that it is \(\Omega\), i.e. \(u\) is constant on \(\Omega\). Due to continuity of \(u\) we conclude that it is also constant on \(\overline{\Omega}\) which shows the desired result.
This follows directly as in class - Since \(u\in C\left( \overline{\Omega}\right)\) it must have a maximum on \(\overline{\Omega}\). Denote it by \(M\). If it is attained at an internal point then \(u\) must be constant and as such \[\max_{\overline{\Omega}}u(x)=\max_{\partial \Omega}u(x).\] Otherwise, it is attained on \(\partial \Omega\) and the above still holds.
A subharmonic function does not satisfy the minimum principle necessarily. Indeed, consider the one dimensional function \(u(x)=x^2\) on \((-1,1)\). It is subharmonic but its minimum is an internal point.
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.
Solution:
We define \(u=u_1-u_2\in C^2\left(
\Omega\right)\cap C\left( \overline{\Omega}\right)\) and notice
that in \(\Omega\) \[-\Delta u = f_1-f_2 \leq 0,\] i.e. \(u\) is subharmonic. Using the weak maximum
principle we find that \[\max_{\overline{\Omega}}u = \max_{\partial
\Omega}u = \max_{\partial \Omega}\left( g_1-g_2\right) \leq 0,\]
which implies that for any \(x\in
\overline{\Omega}\) \[u_1(x)-u_2(x)=u(x)\leq 0,\] showing the
desired result.
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.
Solution:
We know that if \(x_0\in \Omega\) is a local extremum then \(\nabla u\left( x_0\right) =0\). Consequently, \[-\Delta u\left( x_0\right)=f(x_0) <0.\]
We recall that \[\Delta u (x) = \sum_{i=1}^n \partial _{x_ix_i}u(x) = \mathrm{tr}\left( \mathrm{Hess}u (x)\right).\] As the trace of a symmetric matrix is the sum of its eigenvalues, we know that if \(x_0\) is a local maximum then these eigenvalues are non-positive and consequently \[\Delta u\left( x_0\right) \leq 0.\]
This follows immediately from the last two parts. Indeed, if \(x_0\) was local maximum then \(-\Delta u(x_0) <0\) and \(\Delta u(x_0) \leq 0\) which is impossible. Consequently, \(u\) can’t attain any global maximum in an internal point and we get the weak maximum principle.
We have that \[\partial_{x_i}u_{\varepsilon,\lambda}(x) = \partial_{x_i}u(x) + \begin{cases} \varepsilon\lambda e^{\lambda x_1}, & i=1, \\ 0,& i\not=1, \end{cases}\] which shows the first statement. Similarly \[\partial_{x_i x_i}u_{\varepsilon,\lambda}(x) = \partial_{x_i x_i}u(x) + \begin{cases} \varepsilon\lambda^2 e^{\lambda x_1}, & i=1, \\ 0,& i\not=1, \end{cases}\] and as such \[\Delta u_{\varepsilon,\lambda}(x) = \sum_{i=1}^n \partial_{x_ix_i}u(x) = \Delta u(x) + \varepsilon\lambda^2 e^{\lambda x} .\]
We have that \[-\Delta u_{\varepsilon,\lambda}(x) + \textbf{b}(x)\cdot\nabla u_{\varepsilon,\lambda}(x) = -\Delta u(x) - \varepsilon\lambda^2 e^{\lambda x_1} + \textbf{b}(x)\cdot \nabla u(x)\] \[+ \varepsilon\lambda e^{\lambda x_1}\textbf{b}(x)\cdot \left( 1,0,0,\dots,0\right)^T = f(x) - \varepsilon\lambda^2 e^{\lambda x_1} + \varepsilon\lambda b_1(x)e^{\lambda x_1},\] from which we get the desired equality.
Choosing any \(\lambda > \left\lVert b_1\right\rVert_{L^\infty\left( \overline{\Omega}\right)}\) will give us the desired result. For instance we can choose \(\lambda_0= \left\lVert b_1\right\rVert_{L^\infty\left( \overline{\Omega}\right)}+1.\)
Since \[-\Delta u_{\varepsilon,\lambda_0}(x) + \textbf{b}(x)\cdot \nabla u_{\varepsilon,\lambda_0}(x) = f(x) - \varepsilon\lambda_0 \left( \lambda_0-b_1(x) \right)e^{\lambda_0 x_1}\] \[\leq 0 - \varepsilon\lambda_0 \left( \lambda_0-b_1(x) \right)e^{\lambda_0 x_1} < 0\] on \(\Omega\) we conclude from \(\eqref{item:f_less_zero}\) that \[\max_{\overline{\Omega}}\left( u(x)+\varepsilon e^{\lambda_0 x_1}\right)=\max_{\partial\Omega}\left( u(x)+\varepsilon e^{\lambda_0 x_1}\right).\] Consequently, for any \(\varepsilon>0\) and any \(x\in \overline{\Omega}\) \[u(x) \leq u(x)+\varepsilon e^{\lambda_0 x_1} \leq \max_{\overline{\Omega}}\left( u(y)+\varepsilon e^{\lambda_0 y_1}\right)\] \[=\max_{\partial\Omega}\left( u(y)+\varepsilon e^{\lambda_0 y_1}\right) \leq \max_{\partial\Omega}u(y)+\varepsilon\max_{\partial \Omega}e^{\lambda_0 y_1}.\]
As the above holds for any \(\varepsilon>0\) we can take it to zero to conclude that for any \(x\in \overline{\Omega}\) \[u(x) \leq \max_{\partial\Omega}u(y).\] Consequently \[\max_{\partial\Omega}u(x) \leq \max_{\overline{\Omega}}u(x)\leq \max_{\partial\Omega}u(x),\] which gives us the desired weak maximum principle \[\max_{\overline{\Omega}}u(x)=\max_{\partial\Omega}u(x).\]