In all our exercises in this sheet we will assume that \(\Omega\) is an open, bounded, and connected set with smooth boundary.
Exercise 1 (Lower bound for Dirichlet’s Energy). In this problem we will consider the Dirichlet Energy associated to the PDE \[\begin{equation} \label{eq:dirichlet} \left\{ \begin{array}{ll} -\Delta u =f& x\in \Omega\\ u=0 & x\in\partial \Omega \end{array} \right. \end{equation}\] \[E[u]=\frac{1}{2}\int_{\Omega}\left\lvert\nabla u(x)\right\rvert^2 dx - \int_{\Omega}f(x)u(x) dx.\]
Young’s inequality states that for any \(a,b\in\mathbb{R}\) and any Hölder conjugate numbers \(p,q\in (1,\infty)\) we have that \[\left\lvert ab\right\rvert \leq \frac{\left\lvert a\right\rvert^p}{p}+\frac{\left\lvert b\right\rvert^q}{q}.\] Consequently, for any \(\varepsilon>0\) we find that by replacing \(a\) with \(\left( p\varepsilon\right)^{\frac{1}{p}}a\) and \(b\) with \(\frac{b}{\left( p\varepsilon\right)^{\frac{1}{p}}}\) we get that \[\nonumber \left\lvert ab\right\rvert \leq \varepsilon\left\lvert a\right\rvert^p+\frac{\left\lvert b\right\rvert^q}{qp^{q-1} \varepsilon^{q-1}}\] and in particular that for any \(\varepsilon>0\) choosing \(p=q=2\) yields \[\nonumber %label{eq:specialised_young} \left\lvert ab\right\rvert \leq \varepsilon a^2 + \frac{b^2}{4\varepsilon}.\] Show that for any \(\varepsilon>0\) and any \[u\in V=\left\lbrace v\in C^1\left( \overline{\Omega}\right)\;|\; v=0\text{ on }\partial \Omega\right\rbrace\] we have that \[E[u] \geq \left( \frac{1}{2}- \varepsilon C_P\left( \Omega\right)^2\right)\left\lVert u\right\rVert_{H^1_0\left( \Omega\right)}^2 - \frac{1}{4\varepsilon}\left\lVert f\right\rVert^2_{L^2\left( \Omega\right)}\] where \(C_P\left( \Omega\right)\) is the Poincaré constant associated to the domain \(\Omega\).
Conclude that there exists a constant \(C>0\) such that \[\inf_{u\in V}E[u] \geq -C .\]
Solution.
Using the given Young inequality we find that for any \(\varepsilon>0\) \[\left\lvert \int_{\Omega}f(x)u(x) dx\right\rvert \leq \int_{\Omega}\left\lvert f(x)\right\rvert\left\lvert u(x)\right\rvert dx\] \[\leq \int_{\Omega} \left( \varepsilon\left\lvert u(x)\right\rvert^2 +\frac{1}{4\varepsilon}\left\lvert f(x)\right\rvert^2\right)dx .\] Using the Poincaré inequality (which is allowed since \(u\in V\)) we find that \[\int_{\Omega} f(x)u(x) dx \leq \left\lvert \int_{\Omega}f(x)u(x) dx\right\rvert \leq \varepsilon\left\lVert u\right\rVert_{L^2\left( \Omega\right)}^2+\frac{1}{4\varepsilon}\left\lVert f\right\rVert_{L^2\left( \Omega\right)}^2\] \[\leq \varepsilon C_P\left( \Omega\right)^2\left\lVert u\right\rVert_{H^1_0\left( \Omega\right)}^2+\frac{1}{4\varepsilon}\left\lVert f\right\rVert_{L^2\left( \Omega\right)}^2.\] Consequently \[E[u] = \frac{1}{2}\left\lVert u\right\rVert_{H^1_0\left( \Omega\right)}^2 - \int_{\Omega}f(x)u(x) dx \geq \frac{1}{2}\left\lVert u\right\rVert_{H^1_0}^2 -\left( \varepsilon C_P\left( \Omega\right)^2\left\lVert u\right\rVert_{H^1_0\left( \Omega\right)}^2+\frac{1}{4\varepsilon}\left\lVert f\right\rVert_{L^2\left( \Omega\right)}^2\right)\] \[=\left( \frac{1}{2}-\varepsilon C_P\left( \Omega\right)^2\right)\left\lVert u\right\rVert_{H^1_0\left( \Omega\right)}^2-\frac{1}{4\varepsilon}\left\lVert f\right\rVert^2_{L^2\left( \Omega\right)}.\]
For any \(\varepsilon>0\) such that \[\frac{1}{2}-\varepsilon C_P\left( \Omega\right)^2 \geq 0\] we’ll find that \[E[u] \geq -\frac{1}{4\varepsilon}\left\lVert f\right\rVert^2_{L^2\left( \Omega\right)}.\] In particular, choosing \(\varepsilon= \frac{1}{2C_P\left( \Omega\right)^2}\) gives us that \[E[u] \geq -\frac{C_P\left( \Omega\right)^2}{2}\left\lVert f\right\rVert^2_{L^2\left( \Omega\right)}.\]
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Exercise 2 (Uniqueness of weak solutions). Show that if \(u\) and \(v\) are weak solutions, in the sense defined in class, for \[\nonumber \left\{ \begin{array}{ll} -\Delta u =f& x\in \Omega\\ u=0 & x\in\partial \Omega \end{array} \right.\] then \(u=v\).
Solution. We say that \(u\) is a weak solution to the Dirichlet problem above if \(u\in V=\left\lbrace v\in C^1\left( \overline{\Omega}\right)\;|\; v=0\text{ on }\partial \Omega\right\rbrace\) and for any \(\varphi\in V\) we have that \[\int_{\Omega}\nabla u(x) \cdot \nabla \varphi(x) dx = \int_{\Omega}f(x) \varphi(x) dx.\] If \(u\) and \(v\) are both weak solutions to the equation than for any \(\varphi\in V\) \[\int_{\Omega}\nabla u(x) \cdot \nabla \varphi(x) dx = \int_{\Omega}f(x) \varphi(x) dx\] and \[\int_{\Omega}\nabla v(x) \cdot \nabla \varphi(x) dx = \int_{\Omega}f(x) \varphi(x) dx.\] Consequently, \[\int_{\Omega}\nabla \left( u(x)-v(x)\right) \cdot \nabla \varphi(x) dx = 0.\] Since this holds for any \(\varphi\in V\) we must have that \(u-v=0\). Indeed, choose \(\varphi=u-v\) (or \(\varphi= \overline{u-v}\) if the functions are complex valued) we find that \[\int_{\Omega}\left\lvert\nabla \left( u(x)-v(x)\right)\right\rvert^2 dx = 0,\] which implies that \(\nabla\left( u-v\right)=0\). As the domain is connected (which implies that \(u-v\) is a constant) and the function \(u-v\) is zero on \(\partial \Omega\) we conclude that \(u=v\). ◻
Exercise 3 (Uniqueness for a more general elliptic problem). Consider the linear, second-order, elliptic PDE \[\label{Q9} \begin{align} - \mathrm{div} (A \, \nabla u) + \textbf{b} \cdot \nabla u + c u = f \quad & \textrm{in } \Omega, \\ u = g \quad & \textrm{on } \partial \Omega, \end{align}\] where \(\Omega \subset \mathbb{R}^n\) is open and bounded with smooth boundary, \(A \in C^1(\overline{\Omega};\mathbb{R}^{n \times n})\), \(\textbf{b} \in C^1(\overline{\Omega};\mathbb{R}^n)\), and \(c,f,g \in C(\overline{\Omega})\). Assume that \(c\) is non-negatives, \(\mathrm{div}\textbf{b}=0\), and \(A\) is uniformly positive definite, i.e., there exists a constant \(\alpha>0\) such that \(y^T A(x) y \ge \alpha |y|^2\) for all \(y \in \mathbb{R}^n\), \(x \in \Omega\). Prove that \(\eqref{Q9}\) has at most one solution \(u \in C^2(\overline{\Omega})\).
Solution. Defining \(u=u_1-u_2\) we see that \[-\mathrm{div} (A \nabla u) + \textbf{b} \cdot \nabla u + c u = -\mathrm{div} (A \nabla u_1) + \textbf{b} \cdot \nabla u_1 + c u_1\] \[- \left( -\mathrm{div} (A \nabla u_2) + \textbf{b} \cdot \nabla u_2 + c u_2 \right)= f -f=0 \quad \text{in }\Omega\] and \[u=g-g=0,\qquad \text{on }\partial \Omega.\] Thus, in order to show uniqueness it is enough for us to show that if \(u\) solves our equation with \(f=g=0\) then \(u\) must be the zero function. Motivated by the energy method we multiply the our equation with \(u\) (or \(\overline{u}\) if the function has complex values) and integrating over \(\Omega\). We find that \[\nonumber -\int_{\Omega}u(x)\mathrm{div} \left( A(x) \nabla u(x)\right) dx + \int_{\Omega}u(x) \textbf{b}(x)\cdot \nabla u(x) dx + \int_{\Omega}c(x) u(x)^2dx = 0.\] We notice that \[-u(x)\mathrm{div} \left( A(x) \nabla u(x)\right) = -\mathrm{div}\left( u(x) A(x) \nabla u(x)\right) + \nabla u(x) \cdot A(x) \nabla u(x).\] Similarly we have that \[u(x) \textbf{b}(x)\cdot \nabla u(x) = \mathrm{div}\left( \left( u(x)\textbf{b}(x)\right) u(x)\right) - u(x)\mathrm{div}\left( u(x)\textbf{b}(x)\right)\] \[=\mathrm{div}\left( \textbf{b}(x) u(x)^2\right) - u(x)\left( u(x)\mathrm{div}\textbf{b}(x) + \textbf{b}(x)\cdot \nabla u(x)\right)\] \[= \mathrm{div}\left( \textbf{b}(x) u(x)^2\right) - u(x)\textbf{b}(x)\cdot \nabla u(x),\] as \(\mathrm{div}\textbf{b}=0\). Consequently \[u(x) \textbf{b}(x)\cdot \nabla u(x) = \frac{1}{2}\mathrm{div}\left( \textbf{b}(x) u(x)^2\right).\] Using the fact that \(u\vert_{\partial \Omega}=0\) we conclude that \[-\int_{\Omega}u(x)\mathrm{div} \left( A (x) \nabla u(x)\right) dx=-\int_{\partial \Omega}u(y) A(y) \nabla u(y)\cdot \hat{\textbf{n}}(y)dS(y)\] \[+ \int_{\Omega}\nabla u(x) \cdot A(x) \nabla u(x)dx=\int_{\Omega}\nabla u(x) \cdot A(x) \nabla u(x)dx\] and \[\int_{\Omega}u(x) \textbf{b}(x)\cdot \nabla u(x) dx =\frac{1}{2}\int_{\partial\Omega}u(y)^2 \textbf{b}(y)\cdot \hat{\textbf{n}}(y)dS(y)=0.\] Plugging these identities into our original integral yields \[\begin{equation} \label{eq:middle_eq} \int_{\Omega}\nabla u(x) \cdot A(x) \nabla u(x)dx + \int_{\Omega} c(x)u(x)^2dx =0. \end{equation}\] Using the fact that \(c\geq 0\) and the uniform positive definiteness of \(A\) we see that \[0 \leq \alpha \int_{\Omega}\left\lvert\nabla u(x)\right\rvert^2dx \leq \int_{\Omega}\nabla u(x) \cdot A(x) \nabla u(x)dx + \int_{\Omega} c(x)u(x)^2dx =0,\] which, as each term is non-negative, implies that that \[\int_{\Omega}\left\lvert\nabla u(x)\right\rvert^2 udx=0.\] As this implies that \(\nabla u =0\) and since \(u\vert_{\partial \Omega}=0\) we conclude that \(u=0\). ◻
Exercise 4. Consider the space \[V = \{ \varphi \in C^1(\overline{\Omega}) \, : \, \varphi=0 \textrm{ on } \partial \Omega, \; \varphi \ne 0\}\] and the functional \(E:V \to \mathbb{R}\) defined by \[E[v] = \frac{\displaystyle \int_\Omega |\nabla v(x)|^2 \, d x}{\displaystyle \int_\Omega |v(x)|^2 \, d x}.\] Suppose that \(u \in C^2(\overline{\Omega}) \cap V\) minimises \(E\) and show that \[\begin{aligned} - \Delta u & = \lambda u \quad \textrm{in } \Omega, \\ u & = 0 \quad \; \, \; \; \, \textrm{on } \partial \Omega, \end{aligned}\] where \(\lambda = E[u]\).
Solution. For any \(\varphi\in
V\) and any \(\varepsilon>0\)
we find that \(u_\varepsilon=u+\varepsilon\varphi \in C^1\left(
\overline{\Omega}\right)\) and \(u_\varepsilon\vert_{\partial \Omega}=0\).
We claim that \(u_\varepsilon\ne 0\)
for \(\varepsilon\) small enough.
Indeed we have that \[\left\lvert
u(x)+\varepsilon\varphi(x)\right\rvert \geq \left\lvert u(x)\right\rvert
- \varepsilon\left\lvert\varphi(x)\right\rvert \geq \left\lvert
u(x)\right\rvert -
\varepsilon\left\lVert\varphi\right\rVert_{L^\infty\left(
\Omega\right)}.\] Since \(u\ne=0\) we can find \(x_0\) such that \(\left\lvert u(x_0)\right\rvert>0\).
Consequently, for \(\varepsilon<
\frac{\left\lvert
u(x_0)\right\rvert}{\left\lVert\varphi\right\rVert_{L^\infty\left(
\Omega\right)}}\) we have that \(\left\lvert
u_\varepsilon(x_0)\right\rvert>0\), showing that \(u_\varepsilon\ne0\).
Next we consider the function \[g(\varepsilon) = E\left[
u_\varepsilon\right]\] We know that \[g(0) = E[u_0] = E[u] \leq E\left[
u_\varepsilon\right]=g(\varepsilon)\] for any \(\varepsilon>0\) small enough.
Consequently, if \(g\) is
differentiable at \(\varepsilon=0\) we
must have that \(g'(0)=0\). By the
definition of \(E\) we see that \[g(\varepsilon) =
\frac{\int_{\Omega}\left\lvert\nabla u(x) + \varepsilon\nabla
\varphi(x)\right\rvert^2 dx}{\int_{\Omega}\left\lvert
u(x)+\varepsilon\varphi(x)\right\rvert^2 dx}\] \[=\frac{\int_{\Omega}\left( \left\lvert\nabla
u(x)\right\rvert^2 + 2\varepsilon\nabla u(x)\cdot \nabla \varphi(x) +
\varepsilon^2 \left\lvert\nabla
\varphi(x)\right\rvert^2\right)dx}{\int_{\Omega}\left( u(x)^2 +
2\varepsilon u(x)\varphi(x) + \varepsilon^2
\varphi(x)^2\right)dx},\] which shows the required
differentiability. We see that \[g'(\varepsilon) =\frac{1}{\left(
\int_{\Omega}\left\lvert u(x)+\varepsilon\varphi(x)\right\rvert^2
dx\right)^2}\Big(\left( \int_{\Omega}\left( 2 \nabla u(x)\cdot \nabla
\varphi(x) +2\varepsilon\left\lvert\nabla
\varphi(x)\right\rvert^2\right)dx\right)\int_{\Omega}\left\lvert
u(x)+\varepsilon\varphi(x)\right\rvert^2 dx\] \[-\left( \int_{\Omega}\left( 2u(x)\varphi(x) +
2\varepsilon\varphi(x)^2\right)dx\right)\int_{\Omega}\left\lvert\nabla
u(x) + \varepsilon\nabla \varphi(x)\right\rvert^2 dx\Big)\] and
as such \[g'(0) = \frac{2\left(
\int_{\Omega}\nabla u(x)\cdot \nabla
\varphi(x)\right)\int_{\Omega}u(x)^2 dx-2\left(
\int_{\Omega}u(x)\varphi(x)dx\right)\int_{\Omega}\left\lvert\nabla
u(x)\right\rvert^2dx}{\left( \int_{\Omega}u(x)^2dx\right)^2}\]
\[=\frac{2}{\int_{\Omega}u(x)^2dx}\left(
\int_{\Omega}\left( \nabla u(x)\cdot \nabla \varphi(x) - E[u]
u(x)\varphi(x)\right)dx\right).\] Since \(g'(0)=0\) we conclude that for any
\(\varphi\in V\) \[\int_{\Omega}\left( \nabla u(x)\cdot \nabla
\varphi(x) - E[u] u(x)\varphi(x)\right)dx=0.\] Since \(u\in C^2\left( \overline{\Omega}\right)\)
we have that \[\nabla u(x)\cdot \nabla
\varphi(x) = \mathrm{div}\left( \varphi(x) \nabla u(x)\right) -
\varphi(x) \Delta u(x).\] Using the divergence theorem and the
fact that \(\varphi\vert_{\partial
\Omega}=0\) we find that \[0=\int_{\Omega}\left( \mathrm{div}\left(
\varphi(x) \nabla u(x)\right) - \varphi(x) \Delta u(x) - E[u]
u(x)\varphi(x)\right)dx\] \[=\int_{\partial \Omega}\varphi(y)\nabla u(y)\cdot
\hat{n}(y)dS(y) - \int_{\Omega}\left( \Delta u(x)+E[u]
u(x)\right)\varphi(x)dx\] \[= -\int_{\Omega}\left( \Delta u(x)+E[u]
u(x)\right)\varphi(x)dx.\] This implies that \[\int_{\Omega}\left( \Delta u(x)+E[u]
u(x)\right)\varphi(x)dx=0\] for any \(\varphi\in V\). Consequently, (the
fundamental theorem of the calculus of variation) we must have that
\[-\Delta u(x)=E[u]u(x),\qquad x\in
\Omega.\] As \(u\in V\) we have
that \(u\vert_{\Omega}=0\) and we
conclude the desired result. ◻