Green’s functions. By the Fundamental Theorem of Calculus, integrating \(u''(y)=f(y)\) over \([0,z]\), for any \(z \in [0,1]\), gives \[\int_0^z u''(y) \, dy = - \int_0^z f(y) \, dy \quad \Longleftrightarrow %\quad %u'(z) - u'(a) = - \int_a^z f(y) \, dy %\quad %\Longleftrightarrow \quad u'(z) = u'(0) - \int_0^z f(y) \, dy =- \int_0^z f(y) \, dy,\] where we have used the boundary condition \(u'(0)=0\). Integrating again, this time over \([0,x]\), gives \[\begin{aligned} \int_0^x u'(z) \, dz = - \int_0^x \int_0^z f(y) \, dy \, dz \quad \Longleftrightarrow \quad u(x) = u(0) - \int_0^x \int_0^z f(y) \, dy dz. \end{aligned}\] Taking \(x=1\) and using the boundary condition \(u(1)=0\) yields \[0 = u(0) - \int_0^1 \int_0^z f(y) \, dy dz \quad \Longleftrightarrow \quad u(0) = \int_0^1 \int_0^z f(y) \, dy dz.\] Therefore \[u(x) = \int_0^1 \int_0^z f(y) \, dydz - \int_0^x \int_0^z f(y) \, dy dz.\] By interchanging the order of integration we can write this as \[\begin{aligned} u(x) & = \int_0^1 \int_y^1 f(y) \, dz dy - \int_0^x \int_y^x f(y) \, dz dy \\ & = \int_0^1 (1-y) f(y) \, dy - \int_0^x (x-y) f(y) \, dy \\ & = \int_0^x (1-y) f(y) \, dy + \int_x^1 (1-y) f(y) \, dy - \int_0^x (x-y) f(y) \, dy \\ & = \int_0^x (1-x) f(y) \, dy + \int_x^1 (1-y) f(y) \, dy. \end{aligned}\] Therefore \[u(x) = \int_0^1 G(x,y) f(y) \, dy\] with \[\boxed{ G(x,y) = \left\{ \begin{array}{cl} 1-x & \textrm{if } y \le x, \medskip \\ 1-y & \textrm{if } y \ge x. \end{array} \right. }\]
Homogenization.
Integrate \((a_\varepsilon(y) u_\varepsilon(y))'=-f(y)\) over \(y \in [0,z]\): \[\begin{aligned} \int_0^z (a_{\varepsilon}(y) u'_{\varepsilon}(y))' \, dy = - \int_0^z f(y) \, dy \quad & \Longleftrightarrow \quad a_{\varepsilon}(z)u'_\varepsilon(z) - a_{\varepsilon}(0)u_\varepsilon'(0) = - \int_0^z f(y) \, dy \\ & \Longleftrightarrow \quad u'_\varepsilon(z) = \frac{a_{\varepsilon}(0)u'_\varepsilon(0)}{a_{\varepsilon}(z)} - \frac{1}{{a_{\varepsilon}(z)}}\int_0^z f(y) \, dy. \end{aligned}\] Now integrate over \(z \in [0,x]\): \[\begin{equation} \begin{gathered} \nonumber \int_0^x u'_\varepsilon(z) \, dz = \int_0^x \left[ \frac{a_{\varepsilon}(0)u'_\varepsilon(0)}{a_{\varepsilon}(z)} - \frac{1}{{a_{\varepsilon}(z)}} \int_0^z f(y) \, dy \right] \, d z \quad \Longleftrightarrow \\ \label{Q2_1} u_\varepsilon(x) = \underbrace{u_\varepsilon(0)}_{=0} + a_\varepsilon(0)u'_\varepsilon(0) \int_0^x \frac{1}{a_{\varepsilon}(z)} \, dz - \int_0^x \frac{1}{a_{\varepsilon}(z)} \int_0^z f(y) \, dy \, dz. \end{gathered} \end{equation}\] We determine \(u'_\varepsilon(0)\) by evaluating this expression at \(x=1\): \[\begin{gathered} \underbrace{u_\varepsilon(1)}_{=0} = a_\varepsilon(0)u'_\varepsilon(0) \int_0^1 \frac{1}{a_{\varepsilon}(z)} \, dz - \int_0^1 \frac{1}{a_{\varepsilon}(z)} \int_0^z f(y) \, dy \, dz \quad \Longleftrightarrow \\ a_\varepsilon(0) u'_\varepsilon(0) = \left( \int_0^1 \frac{1}{a_{\varepsilon}(z)} \, dz \right)^{-1} \int_0^1 \frac{1}{a_{\varepsilon}(z)} \int_0^z f(y) \, dy \, dz. \end{gathered}\] Substituting this into \(\eqref{Q2_1}\) gives \[\boxed{ u_\varepsilon(x) = \left( \int_0^1 \frac{1}{a_{\varepsilon}(z)} \, dz \right)^{-1} \int_0^1 \frac{1}{a_{\varepsilon}(z)} \int_0^z f(y) \, dy \, dz \int_0^x \frac{1}{a_{\varepsilon}(z)} \, dz - \int_0^x \frac{1}{a_{\varepsilon}(z)} \int_0^z f(y) \, dy \, dz }\] as required.
Taking \(\varepsilon=\varepsilon_n=\frac 1n\) gives \[u_{\varepsilon_n}(x) = \left( \int_0^1 \frac{1}{a(nz)} \, dz \right)^{-1} \int_0^1 \frac{1}{a(nz)} \int_0^z f(y) \, dy \, dz \int_0^x \frac{1}{a(nz)} \, dz - \int_0^x \frac{1}{a(nz)} \int_0^z f(y) \, dy \, dz.\] We are told in the hint to use the Riemann-Lebesgue Lemma, which states that if \(g \in L^\infty(\mathbb{R})\) is \(1\)–periodic, then for any interval \([c,d] \subseteq \mathbb{R}\), \[\begin{equation} \label{ws} \lim_{n \to \infty} \int_c^d g(nz) h(z) \, dz = \int_c^d \overline{g} \, h(z) \, dz \qquad \forall \; h \in L^1(\mathbb{R}) \cap C^1(\mathbb{R}) \end{equation}\] Applying \(\eqref{ws}\) with \(c=0\), \(d=1\), \(g(z)=1/a(z)\), and \(h(z)=1\) on \([c,d]\) gives \[\lim_{n \to \infty} \int_0^1 \frac{1}{a(nz)} \, dz = \int_0^1 \overline{\left( \frac 1a \right)} \, dz = \overline{\left( \frac 1a \right)}.\] (Technical remark: We cannot take \(h(z)=1\) for all \(z \in \mathbb{R}\), else \(h \notin L^1(\mathbb{R})\). But we can take \(h\) to be any function in \(L^1(\mathbb{R}) \cap C^1(\mathbb{R})\) such that \(h=1\) on \([c,d]\). The choice of \(h\) outside \([c,d]\) does not matter since it does not affect the integrals in \(\eqref{ws}\).)
Applying \(\eqref{ws}\) with \(c=0\), \(d=x\), \(g(z)=1/a(z)\) (since \(a\) is periodic and bounded below by a positive constant, \(g\) is periodic and bounded), and \(h(z)=1\) on \([c,d]\) gives \[\lim_{n \to \infty} \int_0^x \frac{1}{a(nz)} \, dz = \int_0^x \overline{\left( \frac 1a \right)} \, dz = x \, \overline{\left( \frac 1a \right)}.\] Applying \(\eqref{ws}\) with \(c=0\), \(d=1\), \(g(z)=1/a(z)\), and \(h(z)= \int_0^z f(y) \, dy\) on \([c,d]\) gives \[\lim_{n \to \infty} \int_0^1 \frac{1}{a(nz)} \int_0^z f(y) \, dy \, dz = \overline{\left( \frac 1a \right)} \int_0^1 \int_0^z f(y) \, dy \, dz.\] Finally, applying \(\eqref{ws}\) with \(c=0\), \(d=x\), \(g(z)=1/a(z)\), and \(h(z)= \int_0^z f(y) \, dy\) on \([c,d]\) gives \[\lim_{n \to \infty} \int_0^x \frac{1}{a(nz)} \int_0^z f(y) \, dy \, dz = \overline{\left( \frac 1a \right)} \int_0^x \int_0^z f(y) \, dy \, dz.\] \[\begin{equation} \label{Q2_2} \boxed{ \lim_{n \to \infty} u_{\varepsilon_n}(x) = u_0(x) := x \overline{\left( \frac 1a \right)} \int_0^1 \int_0^z f(y) \, dy \, dz - \overline{\left( \frac 1a \right)} \int_0^x \int_0^z f(y) \, dy \, dz. } \end{equation}\]
This is simply a matter of interchanging the order of integration: \[\begin{aligned} u_0(x) & = x \overline{\left( \frac 1a \right)} \int_0^1 \int_0^z f(y) \, dy \, dz - \overline{\left( \frac 1a \right)} \int_0^x \int_0^z f(y) \, dy \, dz \\ & = x \overline{\left( \frac 1a \right)} \int_0^1 \int_y^1 f(y) \, dz \, dy - \overline{\left( \frac 1a \right)} \int_0^x \int_y^x f(y) \, dz \, dy \\ & = x \overline{\left( \frac 1a \right)} \int_0^1 (1-y) f(y) dy - \overline{\left( \frac 1a \right)} \int_0^x (x-y) f(y) \, dy \\ & = \overline{\left( \frac 1a \right)} \left\{ \int_0^x [x(1-y)-(x-y)]f(y) \, dy + \int_x^1 x(1-y)f(y) \, dy \right\} \\ & = \overline{\left( \frac 1a \right)} \left\{ \int_0^x y(1-x)f(y) \, dy + \int_x^1 x(1-y)f(y) \, dy \right\} \\ & = \int_0^1 G(x,y) f(y) \, dy \end{aligned}\] with \[\boxed{ G(x,y) = \left\{ \begin{array}{cl} \displaystyle \overline{\left( \frac 1a \right)} y(1-x) & \textrm{if } y \le x, \phantom{\Bigg|} \\ \displaystyle \overline{\left( \frac 1a \right)} x(1-y) & \textrm{if } y \ge x. \phantom{\Bigg|} \end{array} \right. }\]
Clearly \(u_0\) satisfies the boundary conditions. By the Fundamental Theorem of Calculus, differentiating equation \(\eqref{Q2_2}\) gives \[u'_0(x) = \overline{\left( \frac 1a \right)} \int_0^1 \int_0^z f(y) \, dy \, dz - \overline{\left( \frac 1a \right)} \int_0^x f(y) \, dy.\] Differentiating again gives \[u''_0(x) = - \overline{\left( \frac 1a \right)}f(x).\] Therefore \[- a_0 u''_0(x) = - \frac{1}{\, \overline{\left( \frac{1}{a} \right)} \,} \left[ - \overline{\left( \frac 1a \right)}f(x)\right] = f(x)\] as required.
By definition, \[\overline{a} = \int_0^1 a(x) \, dx = \int_0^{\frac 12} \frac12 \, dx + \int_{\frac 12}^1 1 \, dx = \frac 12 \cdot \frac 12 + \frac 12 \cdot 1 = \boxed{ \frac 34}\] On the other hand, \[a_0 = \left( \int_0^1 \frac{1}{a(x)} \, dx \right)^{-1} = \left( \int_0^{\frac 12} 2 \, dx + \int_{\frac 12}^1 1 \, dx \right)^{-1} = \left( \frac 12 \cdot 2 + \frac 12 \cdot 1 \right)^{-1} = \left( \frac 32 \right)^{-1} = \boxed{ \frac 23 }\] Therefore \(a_0 \ne \overline{a}\). In fact the Cauchy-Schwarz inequality can be used to show that \[a_0 \le \overline{a}\] for any choice of \(a\).
Without loss of generality we can assume that \(c>0\). Using the hint and integration by parts gives \[\begin{align} \nonumber \int_c^d g(nz) h(z) \, dz & = \int_c^d \left( \frac{1}{n} \int_0^{nz} g(y) \, dy \right)' h(z) \, dz \\ \label{Q2_3} & = \left. \frac{1}{n} \int_0^{nz} g(y) \, dy \, h(z) \right|_c^d - \int_c^d \frac{1}{n} \int_0^{nz} g(y) \, dy \, h'(z) \, dz. \end{align}\] Let \(z \in [c,d], n \in \mathbb{N}\) and let \(\lfloor nz \rfloor \in (nz-1,nz]\) denote floor\((nz)\), which is the largest integer less than or equal to \(nz\). Since \(a\) is 1–periodic, \[\begin{equation} \label{Q2_4} \int_0^{nz} g(y) \, dy = \int_0^{\lfloor nz \rfloor} g(y) \, dy + \int_{\lfloor nz \rfloor}^{nz} g(y) \, dy = \lfloor nz \rfloor \int_0^1 g(y) \, dy + \int_{\lfloor nz \rfloor}^{nz} g(y) \, dy. \end{equation}\] Observe that \[z - \frac 1n = \frac{nz-1}{n} < \frac{\lfloor nz \rfloor}{n} \le \frac{nz}{n} = z.\] Therefore by the Pinching Lemma (Squeezing Lemma) \[\begin{equation} \label{Q2_5} \lim_{n \to \infty} \frac{\lfloor nz \rfloor}{n} = z. \end{equation}\] Also \[\left| \frac1n \int_{\lfloor nz \rfloor}^{nz} g(y) \, dy \right| \le \frac 1n (nz - \lfloor nz \rfloor) \| g \|_{L^\infty(\mathbb{R})} \le \frac{1}{n} \, \| g \|_{L^\infty(\mathbb{R})}.\] Therefore \[\begin{equation} \label{Q2_6} \lim_{n \to \infty} \, \frac1n \int_{\lfloor nz \rfloor}^{nz} g(y) \, dy = 0. \end{equation}\] By combining equations \(\eqref{Q2_4}\), \(\eqref{Q2_5}\), \(\eqref{Q2_6}\) we find that \[\begin{equation} \label{Q2_7} \lim_{n \to \infty} \frac{1}{n} \int_0^{nz} g(y) \, dy = z \int_0^1 g(y) \, dy. \end{equation}\] Therefore the limit of the first term on the right-hand side of equation \(\eqref{Q2_3}\) is \[\begin{equation} \label{Q2_8} \lim_{n \to \infty} \left. \frac{1}{n} \int_0^{nz} g(y) \, dy \, h(z) \right|_c^d = \left. z \int_0^1 g(y) \, dy \, h(z) \, \right|_c^d. \end{equation}\] Now we find the limit of the second term on the right-hand side of \(\eqref{Q2_3}\). By the computations above \[\begin{aligned} & \left | \int_c^d \frac{1}{n} \int_0^{nz} g(y) \, dy \, h'(z) \, dz - \int_c^d z \int_0^1 g(y) \, dy \, h'(z) \, dz \right| \\ & \le \int_c^d \left| \frac{1}{n} \int_0^{nz} g(y) \, dy - z \int_0^1 g(y) \, dy \right| |h'(z)| \, dz \\ & \le \int_c^d \left| \frac{\lfloor nz \rfloor}{n} \int_0^1 g(y) \, dy + \frac{1}{n} \int_{\lfloor nz \rfloor}^{nz} g(y) \, dy - z \int_0^1 g(y) \, dy \right| |h'(z)| \, dz \\ & \le \int_c^d \left( \left| \frac{\lfloor nz \rfloor}{n} \int_0^1 g(y) \, dy - z \int_0^1 g(y) \, dy \right| + \frac{1}{n} \int_{\lfloor nz \rfloor}^{nz} |g(y)| \, dy \right) |h'(z)| \, dz \\ & \le \int_c^d \left| \frac{\lfloor nz \rfloor - nz}{n} \int_0^1 g(y) \, dy \right| |h'(z)| \, dz + \frac{1}{n} \, \| g \|_{L^\infty(\mathbb{R})} \| h' \|_{L^1([c,d])} \\ & \le \int_c^d \frac{1}{n} \int_0^1 |g(y)| \, dy \, |h'(z)| \, dz + \frac{1}{n} \, \| g \|_{L^\infty(\mathbb{R})} \| h' \|_{L^1([c,d])} \\ & \le \frac{2}{n} \, \| g \|_{L^\infty(\mathbb{R})} \| h' \|_{L^1([c,d])} \to 0 \textrm{ as } n \to \infty. \end{aligned}\] Therefore \[\begin{equation} \label{Q2_9} \lim_{n \to \infty} \int_c^d \frac{1}{n} \int_0^{nz} g(y) \, dy \, h'(z) \, dz = \int_c^d z \int_0^1 g(y) \, dy \, h'(z) \, dz. \end{equation}\] Combining \(\eqref{Q2_3}\), \(\eqref{Q2_8}\), \(\eqref{Q2_9}\) and then integrating by parts yields \[\begin{aligned} \lim_{n \to \infty} \int_c^d g(nz) h(z) \, dz & = \left. z \int_0^1 g(y) \, dy \, h(z) \, \right|_c^d - \int_c^d z \int_0^1 g(y) \, dy \, h'(z) \, dz \\ & = \int_c^d \int_0^1 g(y) \, dy \, h(z) \, dz \\ & = \int_c^d \overline{g} \, h(z) \, dz \end{aligned}\] as required.
Radial symmetry of Laplace’s equation on \(\mathbb{R}^n\). Let \(v:\mathbb{R}^n \to \mathbb{R}\) be a harmonic function. Let \(R \in O(n,\mathbb{R})\) and define \(w:\mathbb{R}^n \to \mathbb{R}\) by \(w(\boldsymbol{x}):=v(R \boldsymbol{x})\). Then \[\begin{aligned} w_{x_i} & = \sum_{j=1}^n \frac{\partial v}{\partial x_j} \frac{\partial (R \boldsymbol{x})_j}{\partial x_i} = \sum_{j=1}^n \frac{\partial v}{\partial x_j} \frac{\partial}{\partial x_i} \sum_{k=1}^n R_{jk} x_k \\ & = \sum_{j=1}^n \frac{\partial v}{\partial x_j} \sum_{k=1}^n R_{jk} \frac{\partial x_k}{\partial x_i} = \sum_{j=1}^n \frac{\partial v}{\partial x_j} \sum_{k=1}^n R_{jk} \delta_{ki} \\ & = \sum_{j=1}^n \frac{\partial v}{\partial x_j} R_{ji}. \end{aligned}\] To be precise \[w_{x_i}(\boldsymbol{x}) = \sum_{j=1}^n v_{x_j}(R \boldsymbol{x}) R_{ji}.\] (This can also be written as \(\nabla w (\boldsymbol{x}) = R^T \nabla v(R \boldsymbol{x})\).) Now we compute the second partial derivatives: \[\begin{aligned} w_{x_i x_i} & = \frac{\partial}{\partial x_i} \sum_{j=1}^n v_{x_j}(R \boldsymbol{x}) R_{ji} \\ & = \sum_{j=1}^n \sum_{k=1}^n \frac{\partial v_{x_j}}{\partial x_k} \frac{\partial (R \boldsymbol{x})_k}{\partial x_i} R_{ji} \\ & = \sum_{j=1}^n \sum_{k=1}^n v_{x_j x_k} R_{ji} \frac{\partial}{\partial x_i} \sum_{l=1}^n R_{kl} x_l \\ & = \sum_{j=1}^n \sum_{k=1}^n v_{x_j x_k} R_{ji} \sum_{l=1}^n R_{kl} \frac{\partial x_l}{\partial x_i} \\ & = \sum_{j=1}^n \sum_{k=1}^n v_{x_j x_k} R_{ji} \sum_{l=1}^n R_{kl} \delta_{il} \\ & = \sum_{j=1}^n \sum_{k=1}^n v_{x_j x_k} R_{ji} R_{ki}. \end{aligned}\] Therefore \[\begin{align} \nonumber \Delta w & = \sum_{i=1}^n w_{x_i x_i} \\ \nonumber & = \sum_{i=1}^n \sum_{j=1}^n \sum_{k=1}^n v_{x_j x_k} R_{ji} R_{ki} \\ \nonumber & = \sum_{j=1}^n \sum_{k=1}^n v_{x_j x_k} \sum_{i=1}^n R_{ji} R_{ki} \\ \nonumber & = \sum_{j=1}^n \sum_{k=1}^n v_{x_j x_k} \sum_{i=1}^n R_{ji} (R^T)_{ik} \\ \nonumber & = \sum_{j=1}^n \sum_{k=1}^n v_{x_j x_k} (R R^T)_{jk} \\ \label{Q2:1} & = \sum_{j=1}^n \sum_{k=1}^n v_{x_j x_k} I_{jk} \end{align}\] since \(R\) is an orthogonal matrix. There are two ways to conclude from here: If are are familiar with the matrix inner product, then \(\eqref{Q2:1}\) gives \[\Delta w = D^2v : I = \mathrm{trace} (D^2 v) = \Delta v = 0\] since \(v\) is harmonc. Otherwise we can continue from \(\eqref{Q2:1}\) using indices: \[\Delta w = \sum_{j=1}^n \sum_{k=1}^n v_{x_j x_k} I_{jk} = \sum_{j=1}^n \sum_{k=1}^n v_{x_j x_k} \delta_{jk} = \sum_{j=1}^n v_{x_j x_j} = \Delta v = 0,\] as required.
Fundamental solution of Poisson’s equation in 3D.
One way of computing \(\| \Phi \|_{L^1(B_R(\mathbf{0}))}\) is using spherical polar coordinates: \[\begin{aligned} \| \Phi \|_{L^1(B_R(\mathbf{0}))} & = \int_{B_R(\mathbf{0})} |\Phi(\boldsymbol{x})| \, d \boldsymbol{x} \\ & = \frac{1}{4 \pi} \int_{B_R(\mathbf{0})} \frac{1}{|\boldsymbol{x}|} \, d \boldsymbol{x} \\ & = \frac{1}{4 \pi} \int_{\phi = 0}^{2 \pi} \int_{\theta = 0}^\pi \int_{r=0}^R \frac{1}{r} \, r^2 \sin \theta \, dr d \theta d \phi \\ & = \frac{1}{4 \pi} \int_{0}^{2 \pi} 1 \, d \phi \int_{0}^\pi \sin \theta \, d \theta \int_{0}^R r \, dr \\ & = \boxed{ \frac{R^2}{2} } \end{aligned}\] Another way of computing \(\| \Phi \|_{L^1(B_R(\mathbf{0}))}\) is as follows: \[\begin{aligned} \| \Phi \|_{L^1(B_R(\mathbf{0}))} & = \int_{B_R(\mathbf{0})} |\Phi(\boldsymbol{x})| \, d \boldsymbol{x} \\ & = \int_{0}^R \left( \int_{\partial B_r(\mathbf{0})} |\Phi(\boldsymbol{y})| \, dS(\boldsymbol{y}) \right) dr \\ & = \int_0^R \left( \int_{\partial B_r(\mathbf{0})} \frac{1}{4 \pi} \frac{1}{|\boldsymbol{y}|} \, dS(\boldsymbol{y}) \right) dr \\ & = \frac{1}{4 \pi} \int_0^R \left( \int_{\partial B_r(\mathbf{0})} \frac{1}{r} \, dS(\boldsymbol{y}) \right) dr \\ & = \frac{1}{4 \pi} \int_0^R \left( \textrm{area}(\partial B_r(\mathbf{0})) \, \frac{1}{r} \right) dr \\ & = \frac{1}{4 \pi} \int_0^R \left( 4 \pi r^2 \, \frac{1}{r} \right) dr \\ & = \int_0^R r \, dr \\ & = \boxed{ \frac{R^2}{2} } \end{aligned}\]
Let \(K \subset \mathbb{R}^3\) be compact. Since \(K\) is bounded, there exists \(R>0\) such that \(K \subset B_R(\mathbf{0})\). Therefore \[\int_K |\Phi(\boldsymbol{x})| \, d \boldsymbol{x}\le \int_{B_R(\mathbf{0})} |\Phi(\boldsymbol{x})| \, d \boldsymbol{x}= \frac{R^2}{2} < \infty.\] Therefore \(\Phi \in L^1_{\mathrm{loc}}(\mathbb{R}^3)\).
By part (i), \[\lim_{R \to \infty} \| \Phi \|_{L^1(B_R(\mathbf{0}))} = \lim_{R \to \infty} \frac{R^2}{2} = + \infty.\] Therefore \(\Phi \notin L^1(\mathbb{R}^3)\).
By the Chain Rule \[\nabla \Phi(\boldsymbol{x}) = \frac{1}{4 \pi} \left( - \frac{1}{|\boldsymbol{x}|^2} \right) \nabla |\boldsymbol{x}| = \frac{1}{4 \pi} \left( - \frac{1}{|\boldsymbol{x}|^2} \right) \frac{\boldsymbol{x}}{|\boldsymbol{x}|} = - \frac{1}{4 \pi} \frac{\boldsymbol{x}}{|\boldsymbol{x}|^3}.\] Let \(K \subset \mathbb{R}^3\) be compact. Since \(K\) is bounded, there exists \(R>0\) such that \(K \subset B_R(\mathbf{0})\). Therefore \[\begin{aligned} \int_K |\nabla \Phi(\boldsymbol{x})| \, d \boldsymbol{x}& \le \int_{B_R(\mathbf{0})} |\nabla \Phi(\boldsymbol{x})| \, d \boldsymbol{x} \\ & = \int_{B_R(\mathbf{0})} \frac{1}{4 \pi} \frac{1}{|\boldsymbol{x}|^2} \, d \boldsymbol{x} \\ & = \frac{1}{4 \pi} \int_{\phi = 0}^{2 \pi} \int_{\theta = 0}^\pi \int_{r=0}^R \frac{1}{r^2} \, r^2 \sin \theta \, dr d \theta d \phi \\ & = \frac{1}{4 \pi} \int_{0}^{2 \pi} 1 \, d \phi \int_{0}^\pi \sin \theta \, d \theta \int_{0}^R 1 \, dr \\ & = R \\ & < \infty. \end{aligned}\] Therefore \(\nabla \Phi \in L^1_{\mathrm{loc}}(\mathbb{R}^3)\).
Fundamental solution of Poisson’s equation in 1D. We compute \[\begin{aligned} u''(x) & = (\Phi*f)''(x) \\ & = (f*\Phi)''(x) \quad & \textrm{(symmetry of convolution)} \\ & = \frac{d^2}{dx^2} \int_{- \infty}^\infty \Phi(y) f(x-y) \, dy \\ & = \int_{-\infty}^\infty \Phi(y) \, \frac{d^2}{dx^2} f(x-y) \, dy \\ & = \int_{-\infty}^0 y \, \frac{d^2}{dx^2} f(x-y) \, dy \\ & = \int_{-\infty}^0 y \, \frac{d^2}{dy^2} f(x-y) \, dy \\ & = y \, \frac{d}{dy} f(x-y) \Big|_{-\infty}^{0} - \int_{-\infty}^0 \frac{d}{dy} f(x-y) \, dy \quad & \textrm{(integration by parts)} \\ & = - f(x) \quad & \textrm{(Fundamental Theorem of Calculus)} \end{aligned}\] as required.
The function spaces \(L^1\) and \(L^1_{\mathrm{loc}}\). Let \(f:\mathbb{R} \to \mathbb{R}\), \(f(x)=|x|^k\), \(k \in \mathbb{R}\). By integrating we see that
\(f \in L^1((-R,R))\) for \(k>-1\),
\(f \in L^1((R,\infty))\) for \(k<-1\),
\(f \in L^1_{\mathrm{loc}}(\mathbb{R})\) for \(k>-1\),
\(f \notin L^1(\mathbb{R})\) for any \(k\) (by parts (i),(ii)).
Properties of the convolution.
Let \(\varphi \in L^1_{\mathrm{loc}}(\mathbb{R})\), \(f \in C_c(\mathbb{R})\) and let \(K=\mathrm{supp}(f)\). Choose \(R>0\) such that \(K \subset [-R,R]\). In particular, \(f=0\) outside the interval \([-R,R]\). Therefore \[\begin{aligned} |(\varphi * f)(x)| & = \left | \int_{-\infty}^\infty \varphi(x-y)f(y) \, dy \right| \\ & = \left| \int_{-R}^R \varphi(x-y)f(y) \, dy \right| \\ & \le \int_{-R}^R |\varphi(x-y)| |f(y)| \, dy \\ & \le \max_{[-R,R]} |f| \int_{-R}^R |\varphi(x-y)| \, dy \\ & = \max_{[-R,R]} |f| \int_{-R-x}^{R-x} |\varphi(z)| \, dz \\ & < \infty \end{aligned}\] since \(\varphi \in L^1_{\mathrm{loc}}(\mathbb{R})\) and \([-R-x,R-x]\) is compact.
Now assume that \(\varphi \in L^1(\mathbb{R})\). By Lemma 4.12, \(f \in L^\infty(\mathbb{R})\). Therefore \[\begin{aligned} |(\varphi * f)(x)| & \le \int_{-\infty}^\infty |\varphi(x-y)| |f(y)| \, dy \\ & \le \sup_{y \in \mathbb{R}} |f(y)| \int_{-\infty}^\infty |\varphi(x-y)| \, dy \\ & = \sup_{y \in \mathbb{R}} |f(y)| \int_{-\infty}^\infty |\varphi(z)| \, dz \\ & = \| f \|_{L^\infty(\mathbb{R})} \| \varphi \|_{L^1(\mathbb{R})}. \end{aligned}\] Therefore \[\| \varphi * f \|_{L^\infty(\mathbb{R})} = \sup_{x \in \mathbb{R}} |(\varphi * f)(x)| \le \| f \|_{L^\infty(\mathbb{R})} \| \varphi \|_{L^1(\mathbb{R})} < \infty\] and so \(\varphi * f \in L^\infty(\mathbb{R})\), as required.
The convolution is commutative since \[\begin{aligned} (\varphi * f)(x) & = \int_{-\infty}^\infty \varphi(x-y)f(y) \, dy \\ & = \int_{\infty}^{-\infty} \varphi(z) f(x-z) (-1) dz & (z=x-y) \\ & = \int_{-\infty}^{\infty} \varphi(z) f(x-z) dz \\ & = (f * \varphi)(x) \end{aligned}\] as required.
The Poincaré inequality for functions that vanish on the boundary. Let \(f \in C^1([a,b])\) satisfy \(f(a)=f(b)=0\). Then \[f(x) = f(a) + \int_a^x f'(y) \, d y = \int_a^x f'(y) \, dy\] since \(f(a)=0\). Therefore \[\begin{aligned} | f(x) | & = \left| \int_a^x f'(y) \, dy \right| \\ & = \left| \int_a^x 1 \cdot f'(y) \, dy \right| \\ & \le \left| \int_a^x 1^2 \, d y \right|^{1/2} \left| \int_a^x |f'(y)|^2 \, dy \right|^{1/2} \; \; & \textrm{(Cauchy-Schwarz)} \\ & \le (x-a)^{1/2} \left( \int_a^b |f'(y)|^2 \, dy \right)^{1/2}. \end{aligned}\] Squaring and integrating gives \[\begin{aligned} \int_a^b | f(x) |^2 \, dx & \le \int_a^b (x-a) \int_a^b |f'(y)|^2 \, dy \, dx \\ & = \int_a^b (x-a) \, dx \, \int_a^b |f'(y)|^2 \, dy \\ & = \frac12 (x-a)^2 \Big|_a^b \int_a^b |f'(y)|^2 \, dy \\ & = \frac12 (b-a)^2 \int_a^b |f'(y)|^2 \, dy. \end{aligned}\] This is the Poincaré inequality with \(C = \frac12 (b-a)^2\).
The Poincaré inequality on unbounded domains.
For \(n \in \mathbb{N}\), define \(f_n:\mathbb{R} \to \mathbb{R}\) by \[f_n(x) = \left\{ \begin{array}{cl} 0 & \textrm{if } x \in (-\infty,-n-1], \\ (x-(-n-1))^2 (x-(-n+1))^2 & \textrm{if } x \in [-n-1,-n], \\ 1 & \textrm{if } x \in [-n,n], \\ (x-(n+1))^2 (x-(n-1))^2 & \textrm{if } x \in [n,n+1], \\ 0 & \textrm{if } x \in [n+1,\infty). \end{array} \right.\] (Exercise: Sketch \(f_n\) to get a better understanding of the example.) Observe that \[\begin{gathered} f_n(-n-1) = f_n(n+1) = 0, \\f_n(-n) = f_n(n) = 1, \\ f'_n(-n-1) = f_n'(-n) = f_n'(n) = f_n'(n+1) = 0. \end{gathered}\] Therefore \(f_n \in C^1(\mathbb{R})\). We also have \(f_n \in L^2(\mathbb{R})\) since \[\begin{aligned} \| f_n \|^2_{L^2(\mathbb{R})} = \int_{-\infty}^\infty |f_n(x)|^2 \, dx < \int_{-n-1}^{n+1} 1 \, dx = 2(n+1). \end{aligned}\] We compute \[\begin{aligned} \| f_n' \|^2_{L^2(\mathbb{R})} & = \int_{-\infty}^\infty |f'_n(x)|^2 \, dx \\ & = 2 \int_{n}^{n+1} \left[ \frac{d}{dx} (x-(n+1))^2 (x-(n-1))^2 \right]^2 \, dx \\ & = 2 \int_n^{n+1} \left[ 2(x-(n+1)) (x-(n-1))^2 + 2 (x-(n+1))^2 (x-(n-1)) \right]^2 \, dx \\ & = 2 \int_0^1 \left[ 2(y-1) (y+1)^2 + 2 (y-1)^2 (y+1) \right]^2 \, dy & (y=x-n) \end{aligned}\] which is independent of \(n\). But \[\begin{aligned} \| f_n \|^2_{L^2(\mathbb{R})} = \int_{-\infty}^\infty |f_n(x)|^2 \, dx > \int_{-n}^n |f_n(x)|^2 \, dx = 2n. % \stackrel{n \to \infty}{\longrightarrow} \infty. \end{aligned}\] Therefore \[\| f_n' \|_{L^2(\mathbb{R})} = \textrm{constant}, \qquad \| f_n \|_{L^2(\mathbb{R})} \stackrel{n \to \infty}{\longrightarrow} \infty\] as required. This means that, given any \(C>0\), we can choose \(N\) large enough so that \[\int_{-\infty}^\infty |f_N(x)|^2 \, dx > C \int_{-\infty}^\infty |f_N'(x)|^2 \, dx,\] which means that the Poincaré inequality on \(\mathbb{R}\) does not hold. We constructed this counter example using spreading; the support of \(f_n\) spreads as \(n \to \infty\) without changing the \(L^2\)–norm of \(f_n'\).
Let \(\Omega = (a,b) \times (-\infty,\infty)\). Let \(f \in C^1(\overline{\Omega}) \cap L^2(\Omega)\) with \(\nabla f \in L^2(\Omega)\) and with \(f(a,y)=f(b,y)=0\) for all \(y \in \mathbb{R}\). Then \[\begin{aligned} \int_\Omega |f(\boldsymbol{x})|^2 \, d \boldsymbol{x}& = \int_{-\infty}^\infty \left( \int_a^b |f(x,y)|^2 \, dx \right) dy \\ & \le \int_{-\infty}^\infty \left( C \int_a^b |f_x(x,y)|^2 \, dx \right) dy \quad & \textrm{(Poincar\'e inequality in } x) \\ & \le C \int_{-\infty}^\infty \int_a^b (|f_x(x,y)|^2 + |f_y(x,y)|^2)\, dx dy \\ & = C \int_\Omega |\nabla f(\boldsymbol{x})|^2 \, d \boldsymbol{x} \end{aligned}\] as required.
The Poincaré constant depends on the domain. There exits \(C_1 > 0\) such that \[\begin{equation} \label{eq:Q9} \int_0^1 |f(x)|^2 \, d x \le C_1 \int_0^1 |f'(x)|^2 \, d x \end{equation}\] for all \(f \in C^1([0,1])\) with \(f(0)=f(1)=0\). Let \(g \in C^1([0,L])\) with \(g(0)=g(L)=0\). Then \[\begin{align} %\label{eq:Q92} \int_0^L |g(x)|^2 \, d x & = \int_0^1 |g(Ly)|^2 L \, dy & (y=x/L) \\ & = L \int_0^1 |f(y)|^2 \, dy & (f(y):=g(Ly)) \\ & \le L C_1 \int_0^1 |f'(y)|^2 \, d y & (\textrm{equation } \eqref{eq:Q9}) \\ & = L C_1 \int_0^1 |L g'(Ly)|^2 \, d y & (f'(y)=L g'(Ly)) \\ & = L^3 C_1 \int_0^1 |g'(Ly)|^2 \, d y \\ & = L^2 C_1 \int_0^L |g'(x)|^2 \, dx & (y=x/L) \\ & = C_L \int_0^L |g'(x)|^2 \, d x \end{align}\] with \(C_L = L^2 C_1\), as desired.
Eigenvalues of \(-\Delta\): Can you hear the shape of a drum? Multiply the PDE \(-\Delta u = \lambda u\) by \(\overline{u}\) (the complex conjugate of \(u\)) and integrate over \(\Omega\): \[\begin{aligned} - \int_{\Omega} \overline{u} \, \Delta u \, d \boldsymbol{x}= \lambda \int_{\Omega} \overline{u} u \, d \boldsymbol{x} \quad \Longleftrightarrow \quad & -\int_{\partial \Omega} \overline{u} \, \nabla u \cdot \boldsymbol{n}\, dL + \int_\Omega \nabla \overline{u} \cdot \nabla u \, d \boldsymbol{x} = \lambda \int_{\Omega} |u|^2 \, d \boldsymbol{x}. \end{aligned}\] The boundary condition \(u=0\) on \(\partial \Omega\) implies that \(\overline{u}=0\) on \(\partial \Omega\) and so \[\begin{aligned} \int_\Omega \overline{\nabla u} \cdot \nabla u \, d \boldsymbol{x} = \lambda \int_{\Omega} |u|^2 \, d \boldsymbol{x} \quad & \Longleftrightarrow \quad \int_\Omega |\nabla u|^2 \, d \boldsymbol{x} = \lambda \int_{\Omega} |u|^2 \, d \boldsymbol{x} \\ & \Longleftrightarrow \quad \lambda = \frac{\displaystyle \int_\Omega |\nabla u|^2 \, d \boldsymbol{x}}{\displaystyle \int_{\Omega} |u|^2 \, d \boldsymbol{x}} \; > \; 0 \end{aligned}\] as required.
The optimal Poincaré constant and eigenvalues of \(-\Delta\).
Multiply the PDE \(-\Delta u = \lambda u\) by \(u\) and integrate over \(\Omega\): \[- \int_\Omega u \Delta u \, d \boldsymbol{x}= \lambda \int_\Omega u^2 \, d \boldsymbol{x} \quad \Longleftrightarrow \quad \int_\Omega |\nabla u|^2 \, d \boldsymbol{x}= \lambda \int_\Omega u^2 \, d \boldsymbol{x}\] since \(u=0\) on \(\partial \Omega\). Rearranging gives \[\lambda = \frac{\displaystyle \int_\Omega |\nabla u|^2 \, d \boldsymbol{x}}{\displaystyle \int_\Omega |u|^2 \, d \boldsymbol{x}}.\]
Let \(u \in C^2(\overline{\Omega}) \cap V\) minimise \(E\). Let \(\varphi \in V\). Define \(u_\varepsilon = u + \varepsilon \varphi \in V\) and define \(g(\varepsilon)=E[u_\varepsilon]\), \(\varepsilon \in \mathbb{R}\). Since \(E\) is minimised by \(u\), then \(g\) is minimised by \(0\). It follows that \[\begin{aligned} 0 & = g'(0) \\ & = \frac{d}{d \varepsilon} \Big|_{\varepsilon=0} E[u_\varepsilon] \\ & = \frac{d}{d \varepsilon} \Big|_{\varepsilon=0} \frac{\displaystyle \int_\Omega |\nabla u_\varepsilon|^2 \, d \boldsymbol{x}}{\displaystyle \int_\Omega |u_\varepsilon|^2 \, d \boldsymbol{x}} \\ & = \frac{\displaystyle 2 \int_\Omega \nabla u \cdot \nabla \varphi \, d \boldsymbol{x}\int_\Omega |u|^2 \, d \boldsymbol{x} - 2 \int_\Omega |\nabla u|^2 \, d \boldsymbol{x}\int_{\Omega} u \varphi \, d \boldsymbol{x}}{\displaystyle \left( \int_\Omega |u|^2 \, d \boldsymbol{x}\right)^2}. \end{aligned}\] The numerator must be zero. Rearranging gives \[%2 \int_\Omega \nabla u \cdot \nabla \varphi \, d \bx \int_\Omega |u|^2 \, d \bx %- 2 \int_\Omega |\nabla u|^2 \, d \bx \int_{\Omega} u \varphi \, d \bx = 0 %\quad \Longleftrightarrow %\quad \int_\Omega \nabla u \cdot \nabla \varphi \, d \boldsymbol{x} = \left( \frac{\displaystyle \int_\Omega |\nabla u|^2 \, d \boldsymbol{x}}{\displaystyle \int_\Omega |u|^2 \, d \boldsymbol{x}} \right) \int_{\Omega} u \varphi \, d \boldsymbol{x}.\] Integrating by parts gives \[- \int_\Omega \Delta u \, \varphi \, d \boldsymbol{x}= \left( \frac{\displaystyle \int_\Omega |\nabla u|^2 \, d \boldsymbol{x}}{\displaystyle \int_\Omega |u|^2 \, d \boldsymbol{x}} \right) \int_{\Omega} u \varphi \, d \boldsymbol{x}.\] Since this holds for all \(\varphi \in V\), the Fundamental Lemma of the Calculus of Variations implies that \[- \Delta u = \left( \frac{\displaystyle \int_\Omega |\nabla u|^2 \, d \boldsymbol{x}}{\displaystyle \int_\Omega |u|^2 \, d \boldsymbol{x}} \right) u \quad \textrm{in } \Omega.\] If we define \[\lambda = \frac{\displaystyle \int_\Omega |\nabla u|^2 \, d \boldsymbol{x}}{\displaystyle \int_\Omega |u|^2 \, d \boldsymbol{x}},\] then \[- \Delta u = \lambda u \quad \textrm{in } \Omega.\] In other words, \(u\) is an eigenfunction of \(-\Delta\). By definition \[E[u] = \frac{\displaystyle \int_\Omega |\nabla u|^2 \, d \boldsymbol{x}}{\displaystyle \int_\Omega |u|^2 \, d \boldsymbol{x}} = \lambda.\] Since \(u\) minimises \(E\), then \(\lambda\) must be the smallest eigenvalue of \(-\Delta\) on \(V\), i.e., \(\lambda = \lambda_1\), otherwise we obtain a contradiction. Therefore \(E[u]=\lambda_1\), as required.
Let \(C>0\) satisfy \[\| f \|_{L^2(\Omega)} \le C \| \nabla f \|_{L^2(\Omega)}\] for all \(f \in C^1(\overline{\Omega})\) with \(f=0\) on \(\partial \Omega\). Then \[\frac{1}{C} \le \frac{\| \nabla f \|_{L^2(\Omega)}}{\| f \|_{L^2(\Omega)}}\] for all \(f \in V\) and so \[\frac{1}{C} \le \inf_{f \in V} \frac{\| \nabla f \|_{L^2(\Omega)}}{\| f \|_{L^2(\Omega)}}.\] The smallest value of \(C\) satisfying this inequality is \(C=C_{\textrm{P}}\) where \[\frac{1}{C_{\textrm{P}}} = \inf_{f \in V} \frac{\| \nabla f \|_{L^2(\Omega)}}{\| f \|_{L^2(\Omega)}}.\]
Combining parts (ii) and (iii) gives \[\frac{1}{C_{\textrm{P}}} = \inf_{f \in V} \frac{\| \nabla f \|_{L^2(\Omega)}}{\| f \|_{L^2(\Omega)}} = \inf_{f \in V} E[v]^{1/2} = \left( \inf_{f \in V} E[v] \right)^{1/2} = \sqrt{\lambda_1}.\] Therefore \[C_{\textrm{P}} = \frac{1}{\sqrt{\lambda_1}}\] as desired.
If \(\Omega = (0,2 \pi)\), then the corresponding eigenvalue problem is \[- u'' = \lambda u \quad \textrm{in } (0,2 \pi), \qquad u(0) = u(2 \pi) = 0.\] The eigenfunctions are \(u_n(x) = \sin\left(\tfrac{nx}{2}\right)\) (see Exercise Sheet 5, Q16) and the corresponding eigenvalues are \(\lambda_n = n^2/4\), \(n \in \mathbb{N}\). Therefore \(\lambda_1 = 1/4\) and \(C_{\textrm{P}} =1/\sqrt{1/4} = 2\). In Q8 we obtained the Poincaré constant \((b-a)/\sqrt{2} = \sqrt{2} \pi\), which is obviously much bigger than the optimal constant \(C_{\textrm{P}} =2\).
Uniqueness for Poisson’s equation with Robin boundary conditions. Let \(u_1\) and \(u_2\) be solutions of \[\begin{aligned} - \Delta u = f \quad & \textrm{in } \Omega, \\ \nabla u \cdot \boldsymbol{n}+ \alpha u = g \quad & \textrm{on } \partial \Omega. \end{aligned}\] Let \(w=u_1 - u_2\). Since the PDE is linear, subtracting the equations satisfied by \(u_1\) and \(u_2\) gives \[\begin{aligned} - \Delta w = 0 \quad & \textrm{in } \Omega, \\ \nabla w \cdot \boldsymbol{n}+ \alpha w = 0 \quad & \textrm{on } \partial \Omega. \end{aligned}\] Multiply \(-\Delta w = 0\) by \(w\) and integrate by parts over \(\Omega\): \[\begin{aligned} -\int_\Omega w \Delta w \, d \boldsymbol{x}= 0 \quad & \Longleftrightarrow \quad - \int_{\partial \Omega} w \, \nabla w \cdot \boldsymbol{n}\, d S + \int_\Omega | \nabla w |^2 \, d \boldsymbol{x}= 0 \\ & \Longleftrightarrow \quad \alpha \int_{\partial \Omega} w^2 \, dS + \int_\Omega | \nabla w |^2 \, d \boldsymbol{x}= 0 \end{aligned}\] since \(\nabla w \cdot \boldsymbol{n}= - \alpha w\) on \(\partial \Omega\). But \(\alpha > 0\). Therefore \[\int_{\partial \Omega} w^2 \, dS = 0, \qquad \int_\Omega | \nabla w |^2 \, d \boldsymbol{x}= 0.\] The second equation implies that \(\nabla w = \textbf{0}\) and hence \(w=\) constant (or at least constant on each connected component of \(\Omega\)). The first equation implies that this constant must be zero. Therefore \(w=0\) and \(u_1 = u_2\), as required.
Uniqueness for a more general elliptic problem. Consider the linear, second-order, elliptic PDE \[\label{Q9} \begin{align} - \mathrm{div} (A \, \nabla u) + \boldsymbol{b}\cdot \nabla u + c u = f \quad & \textrm{in } \Omega, \\ u = g \quad & \textrm{on } \partial \Omega. \end{align}\] (i) Suppose that \(u_1,u_2 \in C^2(\overline{\Omega})\) satisfy \(\eqref{Q9}\). Let \(w=u_1 - u_2\). Since the PDE is linear, subtracting the equations satisfied by \(u_1\) and \(u_2\) gives \[\label{Q9:2} \begin{align} - \mathrm{div} (A \, \nabla w) + \boldsymbol{b}\cdot \nabla w + c w = 0 \quad & \textrm{in } \Omega, \\ w = 0 \quad & \textrm{on } \partial \Omega. \end{align}\] Clearly \(w=0\) satisfies \(\eqref{Q9:2}\). We want to show that it is the only solution. Multiply the PDE for \(w\) by \(w\) and integrate over \(\Omega\): \[\begin{align} \nonumber 0 & = \int_{\Omega} w (- \mathrm{div} (A \, \nabla w) + \boldsymbol{b}\cdot \nabla w + c w) \, d \boldsymbol{x} \\ \nonumber & = - \int_{\Omega} w \, \mathrm{div} (A \, \nabla w) \, d \boldsymbol{x}+ \int_{\Omega} w \, \boldsymbol{b}\cdot \nabla w \, d \boldsymbol{x}+ \int_{\Omega} c w^2 \, d \boldsymbol{x} \\ \nonumber & = - \int_{\partial \Omega} w (A \, \nabla w) \cdot \boldsymbol{n}\, dS + \int_{\Omega} \nabla w \cdot (A \, \nabla w) \, d \boldsymbol{x} + \int_{\Omega} w \, \boldsymbol{b}\cdot \nabla w \, d \boldsymbol{x}+ \int_{\Omega} c w^2 \, d \boldsymbol{x} \\ \label{Q9:3} & = \int_{\Omega} \nabla w \cdot (A \, \nabla w) \, d \boldsymbol{x} + \int_{\Omega} w \, \boldsymbol{b}\cdot \nabla w \, d \boldsymbol{x}+ \int_{\Omega} c w^2 \, d \boldsymbol{x} \end{align}\] since \(w=0\) on \(\partial \Omega\). Observe that \[\begin{equation} \label{Q9:4} \int_{\Omega} \nabla w \cdot (A \, \nabla w) \, d \boldsymbol{x}= \int_{\Omega} (\nabla w)^{\mathrm{T}} A \, \nabla w \, d \boldsymbol{x}\ge \alpha \int_{\Omega} | \nabla w |^2 \, d \boldsymbol{x} \end{equation}\] by the assumption that \(A\) is uniformly positive definite (take \(\boldsymbol{y}= \nabla w\) in \(\boldsymbol{y}^{\mathrm{T}} A(\boldsymbol{x}) \boldsymbol{y}\ge \alpha |\boldsymbol{y}|^2\)). Integrating by parts gives \[\begin{aligned} \int_{\Omega} w \, \boldsymbol{b}\cdot \nabla w \, d \boldsymbol{x}& = \int_{\partial \Omega} w^2 \, \boldsymbol{b}\cdot \boldsymbol{n}\, dS - \int_{\Omega} w \, \mathrm{div} (w \boldsymbol{b}) \, d \boldsymbol{x} \\ & = - \int_{\Omega} w \, \mathrm{div} (w \boldsymbol{b}) \, d \boldsymbol{x} & (w=0 \textrm{ on } \partial \Omega) \\ & = - \int_{\Omega} w \, (\nabla w \cdot \boldsymbol{b}+ w \, \mathrm{div} \boldsymbol{b}) \, d \boldsymbol{x} & \textrm{(product rule)} \\ & = - \int_{\Omega} w \, \boldsymbol{b}\cdot \nabla w \, d \boldsymbol{x} \end{aligned}\] by the assumption that \(\mathrm{div} \, \boldsymbol{b}= 0\). Therefore \[\begin{equation} \label{Q9:5} \int_{\Omega} w \, \boldsymbol{b}\cdot \nabla w \, d \boldsymbol{x}= - \int_{\Omega} w \, \boldsymbol{b}\cdot \nabla w \, d \boldsymbol{x} \quad \Longleftrightarrow \quad \int_{\Omega} w \, \boldsymbol{b}\cdot \nabla w \, d \boldsymbol{x}= 0. \end{equation}\] Combining \(\eqref{Q9:3}\), \(\eqref{Q9:4}\), \(\eqref{Q9:5}\) yields \[\alpha \int_{\Omega} | \nabla w |^2 \, d \boldsymbol{x}+ \int_{\Omega} c w^2 \, d \boldsymbol{x} \le 0.\] But \(c \ge 0\) by assumption. Therefore \[\alpha \int_{\Omega} | \nabla w |^2 \, d \boldsymbol{x}= 0\] and so \(\nabla w = \mathbf{0}\) in \(\Omega\). Hence \(w\) is constant (or at least constant on each connected component of \(\Omega\)). But \(w=0\) on \(\partial \Omega\). Therefore \(w=0\), as required.
(ii) The idea is the same as for (i). Let \(u_n\) be the unique solution to the PDE with \(A_n\) and let \(u\) be the unique solution to the PDE with the matrix \(A\). Define \(w_n:=u_n-u\). We need to show that \(\nabla w_n\to 0\) in \(L^2(\Omega)\) as \(n\to+\infty\). Taking the two PDEs, subtracting them and multiplying the resulting PDE by \(w_n\), we obtain \[\begin{align} \label{eq:stability} 0 & = \int_{\Omega} w_n (- \mathrm{div} (A_n \, \nabla u_n) + \mathrm{div} (A \, \nabla u) + \boldsymbol{b}\cdot \nabla w_n + c w_n) \, d \boldsymbol{x}. \end{align}\] Proceeding exactly as in (i), we find \[\int_{\Omega} w_n \boldsymbol{b}\cdot \nabla w_n\, d \boldsymbol{x}=0.\] Moreover, we compute (using integration by parts, since \(w_n=0\) on \(\partial\Omega\)) \[\begin{aligned} &\int_{\Omega} w_n [- \mathrm{div} (A_n \, \nabla u_n) + \mathrm{div} (A \, \nabla u)] \, d \boldsymbol{x}\\ &= \int_{\Omega} w_n [- \mathrm{div} (A_n \, \nabla u_n) + \mathrm{div} (A_n \, \nabla u) - \mathrm{div} (A_n \, \nabla u) + \mathrm{div} (A \, \nabla u)] \, d \boldsymbol{x}\\ &= \int_{\Omega} w_n [- \mathrm{div} (A_n \, (\nabla u_n-\nabla u)) - \mathrm{div} ((A_n-A) \, \nabla u) ] \, d \boldsymbol{x}\\ &=\int_{\Omega} [ \nabla w_n \cdot (A_n \, \nabla w_n) + \nabla w_n\cdot((A_n-A) \, \nabla u) ] \, d \boldsymbol{x}\\ &\ge \int_{\Omega}[ \alpha |\nabla w_n|^2 + \nabla w_n\cdot((A_n-A) \, \nabla u) ] \, d \boldsymbol{x} \end{aligned}\] So, all these arguments yield \[\int_{\Omega}[ \alpha |\nabla w_n|^2 + \nabla w_n\cdot((A_n-A) \, \nabla u) + c w_n^2 ] \, d \boldsymbol{x}\le 0.\] Now, for any \(\varepsilon>0\), Young’s inequality yields \[\int_{\Omega}\nabla w_n\cdot((A_n-A) \, \nabla u)\, d \boldsymbol{x}= -\frac{\varepsilon}{2}\int_\Omega|\nabla w_n|^2 \, d \boldsymbol{x}-\int_\Omega\frac{1}{2\varepsilon}|A_n-A|^2|\nabla u|^2\, d \boldsymbol{x}.\] By setting \(\varepsilon:=\alpha\), the previous two identities imply \[\int_{\Omega}\left[ \frac{\alpha}{2} |\nabla w_n|^2 + c w_n^2 \right] \, d \boldsymbol{x}\le \frac{1}{2\alpha}\|A_n-A\|^2_{L^\infty}\int_{\Omega}|\nabla u|^2\,d\boldsymbol{x}.\] And by the non-negative property of \(c\), one has \[\int_{\Omega}\frac{\alpha}{2} |\nabla w_n|^2 \, d \boldsymbol{x}\le \frac{1}{2\alpha}\|A_n-A\|^2_{L^\infty}\int_{\Omega}|\nabla u|^2\,d\boldsymbol{x}.\] We conclude by the facts that \(\int_{\Omega}|\nabla u|^2\,d\boldsymbol{x}\) is bounded and \(\|A_n-A\|_{L^\infty}\to 0\), as \(n\to+\infty\).
Uniqueness for a degenerate diffusion equation. Clearly \(u = \pi\) satisfies \[\begin{aligned} \Delta u^m = 0 \quad & \textrm{in } \Omega, \\ u = \pi \quad & \textrm{on } \partial \Omega. \end{aligned}\] We use the energy method to show that it is the only positive solution. Let \(v\) be any positive solution. Subtracting the PDE for \(u\) from the PDE for \(v\) and multiplying by \((v-u)\) gives \[0 = (v-u)(\Delta v^m - \Delta u^m) = (v - \pi)(\Delta v^m - \Delta \pi^m) = (v- \pi) \Delta v^m.\] Now integrate over \(\Omega\): \[\begin{aligned} 0 & = \int_\Omega (v- \pi) \Delta v^m \, d \boldsymbol{x} \\ & = \int_{\Omega} (v - \pi) \, \mathrm{div} \nabla (v^m) \, d \boldsymbol{x} & (\Delta = \mathrm{div} \nabla) \\ & = \int_{\Omega} (v - \pi) \, \mathrm{div} (m v^{m-1} \nabla v) \, d \boldsymbol{x} & \textrm{(Chain Rule)} \\ & = \int_{\partial \Omega} \underbrace{(v - \pi)}_{=0} \, m \, v^{m-1} \nabla v \cdot \boldsymbol{n}\, dS - \int_{\Omega} \underbrace{\nabla(v - \pi)}_{=\nabla v} \cdot \, m v^{m-1} \nabla v \, d \boldsymbol{x}& \textrm{(Integration by parts)} \\ & = - \int_{\Omega} m v^{m-1} |\nabla v|^2 \, d \boldsymbol{x}. \end{aligned}\] Therefore \[\int_{\Omega} m v^{m-1} |\nabla v|^2 \, d \boldsymbol{x}= 0.\] But \(v > 0\), by assumption. Hence \(\nabla v= \mathbf{0}\) in \(\Omega\) and so \(v\) is constant in \(\Omega\). Since \(v=\pi\) on \(\partial \Omega\), we conclude that \(v=\pi\) everywhere, as required.
The \(H_0^1\) and \(H^1\) norms.
We need to check that \(\| \cdot \|_{L^2([a,b])}\) satisfies the three properties of a norm: positivity, \(1\)–homogeneity, and the triangle inequality. First we prove positivity. Let \(f \in C([a,b])\). Clearly \(\| f \|_{L^2([a,b])} \ge 0\). Suppose that \(\| f \|_{L^2([a,b])} = 0\) and assume for contradiction that \(f \ne 0\). Since \(f\) is continuous, then there exists \(x_0 \in (a,b)\), \(h>0\) and \(\varepsilon > 0\) such that \(|f(x)|>\varepsilon\) for all \(x \in (x_0-h,x_0+h)\). Therefore \[\| f \|_{L^2([a,b])}^2 \ge \int_{x_0-h}^{x_0+h} |f(x)|^2 \, dx \ge \int_{x_0-h}^{x_0+h} \varepsilon^2 \, dx = 2 h \varepsilon^2 > 0,\] which is a contradiction. Second we check that \(\| \cdot \|_{L^2([a,b])}\) is \(1\)–homogeneous. Let \(\lambda \in \mathbb{R}\). Then \[\| \lambda f \|_{L^2([a,b])} = \left( \int_a^b |\lambda f(x)|^2 \right)^{1/2} = |\lambda| \left( \int_a^b |f(x)|^2 \right)^{1/2} = |\lambda| \, \| f \|_{L^2([a,b])}\] as required. Finally, we prove the triangle inequality. Let \(f,g \in C([a,b])\). Then \[\begin{aligned} \| f + g \|^2_{L^2([a,b])} & = \int_a^b (f(x)+g(x))^2 \, dx \\ & = \int_a^b f(x)^2 \, dx + 2 \int_a^b f(x) g(x) \, dx + \int_a^b g(x)^2 \, dx \\ & \le \int_a^b f(x)^2 \, dx + 2 \left( \int_a^b f(x)^2 \, dx \right)^{1/2} \left( \int_a^b g(x)^2 \, dx \right)^{1/2} + \int_a^b g(x)^2 \, dx \quad \\ & \qquad \textrm{(by the Cauchy-Schwarz inequality)} \\ & = \| f \|^2_{L^2([a,b])} + 2 \| f \|_{L^2([a,b])} \, \| g \|_{L^2([a,b])} + \| g \|^2_{L^2([a,b])} \\ & = \left( \| f \|_{L^2([a,b])} + \| g \|_{L^2([a,b])} \right)^2. \end{aligned}\] Taking the square root gives the triangle inequality.
Remark: An alternative proof is to prove that the function \((\cdot, \cdot):C([a,b]) \times C([a,b]) \to \mathbb{R}\), \[(f,g) = \int_a^b f(x) g(x) \, dx,\] is an inner product on \(C([a,b])\). It then follows that \(\| f \|:= \sqrt{(f,f)}\) is a norm on \(C([a,b])\) (the norm induced by the inner product; see Definition A.16 in the lecture notes). But this is just the \(L^2\)–norm \(\| \cdot \|_{L^2([a,b])}\).
Remark: The Cauchy-Schwarz inequality can be proved by considering the quadratic polynomial \[t \mapsto p(t) := \| f + tg \|^2_{L^2([a,b])}.\] Since \(p\) is non-negative, then it must have non-positive discriminant, i.e., if \(p(t)=\alpha t^2 + \beta t + \gamma\), then \(\beta^2 - 4 \alpha \gamma \le 0\). It is easy to check that this condition is exactly the Cauchy-Schwarz inequality.
We will prove that the function \((\cdot,\cdot)_{H^1}:C^1([a,b]) \times C^1([a,b]) \to \mathbb{R}\) defined by \[(f,g)_{H^1} := \int_a^b f(x) g(x) \, dx + \int_a^b f'(x) g'(x) \, dx\] is an inner product on \(C^1([a,b])\). It then follows that \[\| f \|_{H^1([a,b])} = \sqrt{(f,f)_{H^1}}\] is a norm on \(C^1([a,b])\) (see Definition A.16 in the lecture notes). It is clear that \((\cdot,\cdot)_{H^1}\) is symmetric and bilinear and that \((f,f)_{H^1} \ge 0\) for all \(f \in C^1([a,b])\). Suppose that \((f,f)_{H^1} = 0\). Then \(\| f \|_{H^1([a,b])}=0\) and in particular \(\| f \|_{L^2([a,b])}=0\). Therefore \(f=0\) by part (i).
This is similar to part (ii). We will prove that the function \((\cdot,\cdot)_{H_0^1}:V \times V \to \mathbb{R}\) defined by \[(f,g)_{H_0^1} := \int_a^b f'(x) g'(x) \, dx\] is an inner product on \(V\). It is clear that \((\cdot,\cdot)_{H_0^1}\) is symmetric and bilinear and that \((f,f)_{H_0^1} \ge 0\) for all \(f \in V\). Suppose that \((f,f)_{H_0^1} = 0\). Then \(\| f \|_{H_0^1([a,b])}=0\) and in particular \(\| f' \|_{L^2([a,b])}=0\). Therefore \(f'=0\) by part (i) and so \(f\) is a constant function. But \(f(a)=f(b)=0\) and hence \(f=0\), as required.
We need to find constants \(c,C>0\) such that \[c \| f \|_{H_0^1([a,b])} \le \| f \|_{H^1([a,b])} \le C \| f \|_{H_0^1([a,b])} \quad \forall \, f \in V.\] Let \(f \in V\). We have \[\| f \|_{H_0^1([a,b])} = \| f' \|_{L^2([a,b])} \le \left( \| f \|^2_{L^2([a,b])} + \| f' \|^2_{L^2([a,b])} \right)^{1/2} = \| f \|_{H^1([a,b])}.\] Therefore \(c=1\). On the other hand, \[\| f \|_{H^1([a,b])}^2 = \| f \|_{L^2([a,b])}^2 + \| f' \|_{L^2([a,b])}^2 \le C_{\mathrm{P}}^2 \| f' \|_{L^2([a,b])}^2 + \| f' \|_{L^2([a,b])}^2\] where \(C_{\mathrm{P}}\) is the Poincaré constant. Therefore we can take \(C=(C_{\mathrm{P}}^2 +1)^{1/2}\).
Continuous dependence. Let \(u \in C^2(\overline{\Omega})\) satisfy \[\begin{aligned} - \mathrm{div} (A \, \nabla u) + c u = f \quad & \textrm{in } \Omega, \\ u = 0 \quad & \textrm{on } \partial \Omega. \end{aligned}\] Multiplying the PDE by \(u\) and integrating over \(\Omega\) gives \[\begin{aligned} \int_{\Omega} fu \, d \boldsymbol{x}& = \int_{\Omega} u \, ( - \mathrm{div} (A \, \nabla u) +c u ) \, d \boldsymbol{x} \\ & = - \int_{\Omega} u \, \mathrm{div} (A \, \nabla u) \, d \boldsymbol{x}+c \int_{\Omega} u^2 \, d \boldsymbol{x} \\ & = - \int_{\partial \Omega} u \, (A \, \nabla u) \cdot \boldsymbol{n}\, d S + \int_{\Omega} \nabla u \cdot (A \, \nabla u) \, d \boldsymbol{x} +c \int_{\Omega} u^2 \, d \boldsymbol{x} \\ & = \int_{\Omega} (\nabla u)^\mathrm{T} A \, \nabla u \, d \boldsymbol{x} +c \int_{\Omega} u^2 \, d \boldsymbol{x}& (u=0 \textrm{ on } \partial \Omega) \\ & \ge \alpha \int_{\Omega} |\nabla u|^2 \, d \boldsymbol{x}+c \int_{\Omega} u^2 \, d \boldsymbol{x} & (A \textrm{ is uniformly positive definite)} \\ & \ge \min \{ \alpha , c \} \left( \int_{\Omega} |\nabla u|^2 \, d \boldsymbol{x}+ \int_{\Omega} u^2 \, d \boldsymbol{x}\right) \\ & = \min \{ \alpha , c \} \, \| u \|^2_{H^1(\Omega)} \end{aligned}\] by definition of the \(H^1\)–norm. Therefore \[\min \{ \alpha , c \} \, \| u \|^2_{H^1(\Omega)} \le \int_{\Omega} fu \, d \boldsymbol{x} \le \| f \|_{L^2(\Omega)} \| u \|_{L^2(\Omega)} \le \| f \|_{L^2(\Omega)} \| u \|_{H^1(\Omega)}\] where we have used the Cauchy-Schwarz inequality and the fact that \(\| v \|_{L^2(\Omega)} \le \| v \|_{H^1(\Omega)}\) for all \(v \in C^1(\overline{\Omega})\). Cancelling one power of \(\| u \|_{H^1(\Omega)}\) from both sides gives the desired result: \[\| u \|_{H^1(\Omega)} \le C \| f \|_{L^2(\Omega)}\] with \(C = 1/ \min \{ \alpha , c \}\).
Remark: Note that this estimate degenerates as \(c\) tends to \(0\) (\(C \to +\infty\) as \(c \to 0\)). If \(c=0\) or \(c\) is small then a better estimate can be obtained using the Poincaré inequality: As above \[\int_{\Omega} fu \, d \boldsymbol{x}\ge \alpha \int_{\Omega} |\nabla u|^2 \, d \boldsymbol{x}+c \int_{\Omega} u^2 \, d \boldsymbol{x}\ge \alpha \int_{\Omega} |\nabla u|^2 \, d \boldsymbol{x} = \alpha \| u \|^2_{H_0^1(\Omega)} .\] Therefore \[\alpha \| u \|^2_{H_0^1(\Omega)} \le \int_{\Omega} fu \, d \boldsymbol{x} \le \| f \|_{L^2(\Omega)} \| u \|_{L^2(\Omega)} \le C_{\mathrm{P}} \| f \|_{L^2(\Omega)} \| \nabla u \|_{L^2(\Omega)} = C_{\mathrm{P}} \| f \|_{L^2(\Omega)} \| u \|_{H_0^1(\Omega)}\] where \(C_{\mathrm{P}}(\Omega)\) is the Poincaré constant. Cancelling one power of \(\| u \|_{H_0^1(\Omega)}\) from both sides gives \[\| u \|_{H_0^1(\Omega)} \le C \| f \|_{L^2(\Omega)}\] with \(C = C_{\mathrm{P}}/\alpha\).
Continuous dependence with a first-order term.
Let \(u \in C^2(\overline{\Omega})\) satisfy \[\label{Q17a} \begin{align} - k \Delta u + \boldsymbol{b}\cdot \nabla u + c u = f \quad & \textrm{in } \Omega, \\ u = 0 \quad & \textrm{on } \partial \Omega. \end{align}\] Multiply the PDE by \(u\) and integrate over \(\Omega\): \[\begin{aligned} & - k \int_{\Omega} u \, \Delta u \, d \boldsymbol{x}+ \int_{\Omega} u \, \boldsymbol{b}\cdot \nabla u \, d \boldsymbol{x}+ \int_{\Omega} c u^2 \, d \boldsymbol{x}= \int_{\Omega} fu \, d \boldsymbol{x} \\ & \Longleftrightarrow \quad - k \left[ \int_{\partial \Omega} u \, \nabla u \cdot \boldsymbol{n}\, dS - \int_{\Omega} |\nabla u|^2 \, d \boldsymbol{x}\right] + \int_{\Omega} u \, \boldsymbol{b}\cdot \nabla u \, d \boldsymbol{x}+ \int_{\Omega} c u^2 \, d \boldsymbol{x}= \int_{\Omega} fu \, d \boldsymbol{x} \\ & \Longleftrightarrow \quad k \| \nabla u \|^2_{L^2(\Omega)} + \int_{\Omega} (\boldsymbol{b}\cdot \nabla u) \, u \, d \boldsymbol{x}+ c \| u \|^2_{L^2(\Omega)} = \int_{\Omega} fu \, d \boldsymbol{x} \le \| f \|_{L^2(\Omega)} \| u \|_{L^2(\Omega)} \end{aligned}\] by the Cauchy-Schwarz inequality.
Let \(\varepsilon > 0\). Then \[\begin{aligned} \left| \int_{\Omega} (\boldsymbol{b}\cdot \nabla u) \, u \, d \boldsymbol{x}\right| & \le \| \boldsymbol{b}\|_{L^\infty(\Omega)} \int_{\Omega} |\nabla u| \, |u| \, d \boldsymbol{x} \\ & \le \| \boldsymbol{b}\|_{L^\infty(\Omega)} \| \nabla u \|_{L^2(\Omega)} \| u \|_{L^2(\Omega)} & \quad \textrm{(Cauchy-Schwarz)} \\ & = \| \boldsymbol{b}\|_{L^\infty(\Omega)} \left( \sqrt{2 \varepsilon} \, \| \nabla u \|_{L^2(\Omega)} \right) \left( \frac{1}{\sqrt{2 \varepsilon}} \, \| u \|_{L^2(\Omega)} \right) \\ & \le \| \boldsymbol{b}\|_{L^\infty(\Omega)} \left( \varepsilon \| \nabla u \|^2_{L^2(\Omega)} + \frac{1}{4 \varepsilon} \| u \|^2_{L^2(\Omega)} \right) \end{aligned}\] by the Young inequality.
Combining parts (a) and (b) gives \[\begin{aligned} \| f \|_{L^2(\Omega)} \| u \|_{L^2(\Omega)} & \ge k \| \nabla u \|^2_{L^2(\Omega)} + \int_{\Omega} (\boldsymbol{b}\cdot \nabla u) \, u \, d \boldsymbol{x}+ c \| u \|^2_{L^2(\Omega)} \\ & \ge k \| \nabla u \|^2_{L^2(\Omega)} - \left| \int_{\Omega} (\boldsymbol{b}\cdot \nabla u) \, u \, d \boldsymbol{x}\right| + c \| u \|^2_{L^2(\Omega)} \\ & \ge k \| \nabla u \|^2_{L^2(\Omega)} - \| \boldsymbol{b}\|_{L^\infty(\Omega)} \left( \varepsilon \| \nabla u \|^2_{L^2(\Omega)} + \frac{1}{4 \varepsilon} \| u \|^2_{L^2(\Omega)} \right) + c \| u \|^2_{L^2(\Omega)} \\ & = \left( k - \varepsilon \| \boldsymbol{b}\|_{L^\infty(\Omega)} \right) \| \nabla u \|^2_{L^2(\Omega)} + \left( c - \frac{\| \boldsymbol{b}\|_{L^\infty(\Omega)}}{4 \varepsilon} \right) \| u \|^2_{L^2(\Omega)}. \end{aligned}\]
Let \(\varepsilon > 0\) satisfy \(k - \varepsilon \| \boldsymbol{b}\|_{L^\infty(\Omega)} > 0\), i.e., let \[\begin{equation} \label{epeq} 0 < \varepsilon < \frac{k}{\| \boldsymbol{b}\|_{L^\infty(\Omega)}}. \end{equation}\] Let \[c_0 = \frac{\| \boldsymbol{b}\|_{L^\infty(\Omega)}}{4 \varepsilon}.\] If \(c > c_0\), then \[c - \frac{\| \boldsymbol{b}\|_{L^\infty(\Omega)}}{4 \varepsilon} > 0.\] Therefore if \(c > c_0\) and \(\varepsilon\) satisfies \(\eqref{epeq}\), then \[k - \varepsilon \| \boldsymbol{b}\|_{L^\infty(\Omega)} >0, \qquad c - \frac{\| \boldsymbol{b}\|_{L^\infty(\Omega)}}{4 \varepsilon} > 0\] and so by part (c) \[\begin{aligned} \| f \|_{L^2(\Omega)} \| u \|_{H^1(\Omega)} & \ge \| f \|_{L^2(\Omega)} \| u \|_{L^2(\Omega)} \\ & \ge \min \left\{ k - \varepsilon \| \boldsymbol{b}\|_{L^\infty(\Omega)} , c - \frac{\| \boldsymbol{b}\|_{L^\infty(\Omega)}}{4 \varepsilon} \right\} \left( \| \nabla u \|^2_{L^2(\Omega)} + \| u \|^2_{L^2(\Omega)}\right) \\ & = \min \left\{ k - \varepsilon \| \boldsymbol{b}\|_{L^\infty(\Omega)} , c - \frac{\| \boldsymbol{b}\|_{L^\infty(\Omega)}}{4 \varepsilon} \right\} \| u \|^2_{H^1(\Omega)}. \end{aligned}\] Therefore if \(c>c_0\) and \(\varepsilon\) satisfies \(\eqref{epeq}\), then \[\| u \|_{H^1(\Omega)} \le M \| f \|_{L^2(\Omega)}\] with \[M = \frac{1}{\min \left\{ k - \varepsilon \| \boldsymbol{b}\|_{L^\infty(\Omega)} , c - \frac{\| \boldsymbol{b}\|_{L^\infty(\Omega)}}{4 \varepsilon} \right\}}.\] For example, if we choose \[\varepsilon = \frac 12 \frac{k}{\| \boldsymbol{b}\|_{L^\infty(\Omega)}},\] then \[c_0 = \frac{\| \boldsymbol{b}\|_{L^\infty(\Omega)}}{2k}, \qquad M = \frac{1}{\min \left\{ k/2 , c - c_0 \right\}} = \max \left\{ \frac{2}{k}, \frac{1}{c-c_0} \right\} .\]
Let \(v \in C^2(\overline{\Omega})\) satisfy \(\eqref{Q17a}\). Then \(w=u-v\) satisfies \(\eqref{Q17a}\) with \(f=0\). Therefore by part (d) \[\| w \|_{H^1(\Omega)} \le 0\] and so \(w=0\) and \(u=v\), as required.
Neumann boundary conditions for variational problems.
Let \(u \in C^1(\overline{\Omega})\) be a minimiser of \(E\). For any \(\varphi \in V\), \(\varepsilon \in \mathbb{R}\), define \(u_\varepsilon = u + \varepsilon \varphi.\) Then \(u_\varepsilon \in C^1(\overline{\Omega})\) since the sum of \(C^1\) functions is \(C^1\). Let \(g(\varepsilon) = E[u_\varepsilon]\). Note that \(u_\varepsilon = u\) when \(\varepsilon = 0\). Therefore \(g\) is minimised by \(\varepsilon=0\) since \(E\) is minimised by \(u\). Hence \[\begin{aligned} 0 & = g'(0) = \left. \frac{d}{d \varepsilon} \right|_{\varepsilon=0} E[u_\varepsilon] \phantom{\Bigg|} \\ & = \left. \frac{d}{d \varepsilon} \right|_{\varepsilon=0} \left[ \frac12 \int_\Omega |\nabla u_\varepsilon|^2 \, d \boldsymbol{x}- \int_{\Omega} f u_\varepsilon \, d \boldsymbol{x}\right] \phantom{\Bigg|} \\ & = \left. \frac{d}{d \varepsilon} \right|_{\varepsilon=0} \left[ \frac12 \int_\Omega (\nabla u + \varepsilon \nabla \varphi) \cdot (\nabla u + \varepsilon \nabla \varphi) \, d \boldsymbol{x}- \int_{\Omega} f (u + \varepsilon \varphi) \, d \boldsymbol{x}\right] \phantom{\Bigg|} \\ & = \frac12 \int_\Omega \left. \frac{d}{d \varepsilon} \right|_{\varepsilon=0}[(\nabla u + \varepsilon \nabla \varphi) \cdot (\nabla u + \varepsilon \nabla \varphi)] \, d \boldsymbol{x}- \int_{\Omega} f \left. \frac{d}{d \varepsilon} \right|_{\varepsilon=0}(u + \varepsilon \varphi) \, d \boldsymbol{x}\phantom{\Bigg|} \\ & = \frac12 \int_\Omega [\nabla \varphi \cdot (\nabla u + \varepsilon \nabla \varphi) + (\nabla u + \varepsilon \nabla \varphi)\cdot \nabla \varphi] \bigg|_{\varepsilon=0} \, d \boldsymbol{x}- \int_{\Omega} f \varphi \, d \boldsymbol{x}\phantom{\Bigg|} \\ & = \int_{\Omega} \nabla u \cdot \nabla \varphi \, d \boldsymbol{x}- \int_{\Omega} f \varphi \, d \boldsymbol{x}. \end{aligned}\] Therefore \[\begin{equation} \label{eq:sol1} \int_\Omega \nabla u \cdot \nabla \varphi \, d \boldsymbol{x}= \int_\Omega f \varphi \, d \boldsymbol{x}\quad \textrm{for all } \varphi \in C^1(\overline{\Omega}) \end{equation}\] as required.
First choose a test function \(\varphi \in C^1(\overline{\Omega})\) such that \(\varphi = 0\) on \(\partial \Omega\). Since \(u \in C^2(\overline{\Omega})\), we can integrate by parts in \(\eqref{eq:sol1}\) to obtain \[\int_{\partial \Omega}\nabla u \, \varphi \cdot \boldsymbol{n}\, dS - \int_\Omega\underbrace{\mathrm{div} \nabla u}_{=\Delta u} \, \varphi \, d \boldsymbol{x}= \int_\Omega f \varphi \, d \boldsymbol{x} \quad \Longleftrightarrow \quad \int_\Omega (-\Delta u - f) \varphi \, d \boldsymbol{x}= 0\] because \(\varphi = 0\) on \(\partial \Omega\). Since this holds for all test functions \(\varphi \in C^1(\overline{\Omega})\) such that \(\varphi = 0\) on \(\partial \Omega\), the Fundamental Lemma of the Calculus of Variations implies that \[\begin{equation} \label{eq:sol2} -\Delta u - f = 0 \quad \textrm{in } \Omega \end{equation}\] as required. We still need to show that \(u\) satisfies the Neumann boundary condition. Now take any test function \(\varphi \in C^1(\overline{\Omega})\) in \(\eqref{eq:sol1}\) and integrate by parts as before to obtain \[\begin{aligned} \int_{\partial \Omega}\nabla u \, \varphi \cdot \boldsymbol{n}\, dS - \int_\Omega \Delta u \, \varphi \, d \boldsymbol{x}= \int_\Omega f \varphi \, d \boldsymbol{x} \quad & \Longleftrightarrow \quad \int_{\partial \Omega}\nabla u \, \varphi \cdot \boldsymbol{n}\, dS + \int_\Omega \underbrace{(-\Delta u - f)}_{=0 \textrm{ by } \eqref{eq:sol2}} \varphi \, d \boldsymbol{x}= 0 \\ & \Longleftrightarrow \quad \int_{\partial \Omega}\nabla u \cdot \boldsymbol{n}\, \varphi \, dS = 0. \end{aligned}\] Since this holds for all \(\varphi \in C^1(\overline{\Omega})\), then \(\nabla u \cdot \boldsymbol{n}= 0\) on \(\partial \Omega\), as required.
The \(p\)–Laplacian operator.
Let \(u \in C^2(\overline{\Omega}) \cap V\) minimise \(E_p\). For any \(\varphi \in V\), \(\varepsilon \in \mathbb{R}\), define \(u_\varepsilon = u + \varepsilon \varphi.\) Observe that \(u_\varepsilon\) vanishes on the boundary of \(\Omega\) since both \(u\) and \(\varphi\) vanish there. Also \(u_\varepsilon \in C^1(\overline{\Omega})\) since the sum of \(C^1\) functions is \(C^1\). Hence \(u_\varepsilon \in V\). Define \(g(\varepsilon) = E_p[u_\varepsilon]\). Now \(u_\varepsilon = u\) when \(\varepsilon = 0\). Therefore \(g\) is minimised by \(\varepsilon=0\) since \(E_p\) is minimised by \(u\). We have reduced the problem of minimising the functional \(E_p\) to minimising the function of one variable \(g\). Since \(g\) is minimised at \(\varepsilon=0\), \[\begin{align} \nonumber 0 & = g'(0) \\ \nonumber & = \left. \frac{d}{d \varepsilon} \right|_{\varepsilon=0} E_p[u_\varepsilon] \phantom{\Bigg|} \\ \nonumber & = \left. \frac{d}{d \varepsilon} \right|_{\varepsilon=0} \left[ \frac1p \int_\Omega |\nabla u_\varepsilon|^p \, d \boldsymbol{x}- \int_{\Omega} f u_\varepsilon \, d \boldsymbol{x}\right] \phantom{\Bigg|} \\ \nonumber & = \frac1p \int_\Omega \left. \frac{d}{d \varepsilon} \right|_{\varepsilon=0} |\nabla u + \varepsilon \nabla \varphi|^p \, d \boldsymbol{x}- \int_{\Omega} f \left. \frac{d}{d \varepsilon} \right|_{\varepsilon=0}(u + \varepsilon \varphi) \, d \boldsymbol{x}\phantom{\Bigg|} \\ \nonumber & = \frac1p \int_\Omega p|\nabla u + \varepsilon \nabla \varphi|^{p-1} \frac{\nabla u + \varepsilon \nabla \varphi}{|\nabla u + \varepsilon \nabla \varphi|} \cdot \nabla \varphi \bigg|_{\varepsilon=0} \, d \boldsymbol{x}- \int_{\Omega} f \varphi \, d \boldsymbol{x}\phantom{\Bigg|} \\ \label{Q12_1} & = \int_\Omega |\nabla u|^{p-2} \nabla u \cdot \nabla \varphi \, d \boldsymbol{x}- \int_{\Omega} f \varphi \, d \boldsymbol{x} \end{align}\] where the differentiation was performed using the Chain Rule and the fact that \[\frac{d}{dx} \, x^p = p x^{p-1}, \quad \qquad \nabla_{\boldsymbol{y}} \, |\boldsymbol{y}| = \frac{\boldsymbol{y}}{|\boldsymbol{y}|}, \qquad \quad \frac{d}{d \varepsilon} (\nabla u + \varepsilon \nabla \varphi) = \nabla \varphi.\] Recall the integration by parts formula \[\int_\Omega \boldsymbol{g}\cdot \nabla h \, d \boldsymbol{x}= \int_{\partial \Omega} \boldsymbol{g}h \cdot \boldsymbol{n}\, dS - \int_\Omega h \, \mathrm{div} \, \boldsymbol{g}\, d \boldsymbol{x}.\] By applying this with \(h = \varphi\), \(\boldsymbol{g}= |\nabla u|^{p-2} \nabla u\), we can rewrite equation \(\eqref{Q12_1}\) as \[0 = \int_{\partial \Omega} |\nabla u|^{p-2} \nabla u \, \varphi \cdot \boldsymbol{n}\, dS - \int_\Omega \varphi \, \mathrm{div} (|\nabla u|^{p-2} \nabla u) \, d \boldsymbol{x} - \int_{\Omega} f \varphi \, d \boldsymbol{x}.\] But \(\varphi = 0\) on \(\partial \Omega\) since \(\varphi \in V\). Therefore \[0 = \int_\Omega [\mathrm{div} (|\nabla u|^{p-2} \nabla u) +f] \, \varphi \, d \boldsymbol{x}\quad \textrm{for all } \varphi \in V.\] Since \(\varphi\) is arbitrary, the Fundamental Lemma of the Calculus of Variations gives \[\mathrm{div} (|\nabla u|^{p-2} \nabla u) +f = 0 \quad \textrm{in } \Omega.\] Therefore \[- \underbrace{\mathrm{div} (|\nabla u|^{p-2} \nabla u)}_{=\Delta_p u} = f \quad \textrm{in } \Omega\] as required. Note that \(u = 0\) on \(\partial \Omega\) by definition of \(V\).
Multiply the PDE \(- \mathrm{div} (|\nabla u|^{p-2} \nabla u) = f\) by \(u\) and integrate by parts over \(\Omega\) to obtain \[\begin{align} \nonumber & - \int_{\Omega} u \, \mathrm{div} (|\nabla u|^{p-2} \nabla u) \, d \boldsymbol{x} = \int_{\Omega} fu \, d \boldsymbol{x} \\ \nonumber & \Longleftrightarrow \quad - \int_{\partial \Omega} u (|\nabla u|^{p-2} \nabla u) \cdot \boldsymbol{n}\, dS + \int_{\Omega} |\nabla u|^{p-2} \nabla u \cdot \nabla u \, d \boldsymbol{x}= \int_{\Omega} fu \, d \boldsymbol{x} \\ \label{Q12_2} & \Longleftrightarrow \quad \int_{\Omega} |\nabla u|^{p}\, d \boldsymbol{x}= \int_{\Omega} fu \, d \boldsymbol{x} \end{align}\] since \(u=0\) on \(\partial \Omega\). Therefore \[\begin{aligned} E_p[u] & = \frac{1}{p}\int_\Omega |\nabla u |^p \, d \boldsymbol{x}- \int_\Omega f u \, d \boldsymbol{x} \\ & = \frac{1}{p}\int_\Omega |\nabla u |^p \, d \boldsymbol{x}-\int_{\Omega} |\nabla u|^{p}\, d \boldsymbol{x} \quad & \textrm{(by equation \eqref{Q12_2})} \\ & = \frac{1-p}{p} \int_\Omega |\nabla u |^p \, d \boldsymbol{x} \\ & = \frac{1-p}{p} \int_\Omega f u \, d \boldsymbol{x} \quad & \textrm{(by equation \eqref{Q12_2})} \end{aligned}\] as required.
The minimal surface equation: PDEs and soap films. Let \(u \in C^2(\overline{\Omega}) \cap V\) be a minimiser of \(A\). Let \(\varepsilon \in \mathbb{R}\) and \(\varphi \in C^1(\overline{\Omega})\) with \(\varphi=0\) on \(\partial \Omega\). Define \(u_{\varepsilon} = u + \varepsilon \varphi\). Then \(u_\varepsilon \in V\) since the sum of continuously differential functions is continuously differentiable and, if \(\boldsymbol{x}\in \partial \Omega\), then \[u_{\varepsilon}(\boldsymbol{x}) = u(\boldsymbol{x}) + \varepsilon \varphi(\boldsymbol{x}) = g(\boldsymbol{x}) + \varepsilon \cdot 0 = g(\boldsymbol{x})\] as required. Define \(h:\mathbb{R} \to \mathbb{R}\) by \(h(\varepsilon)=A[u_{\varepsilon}]\). Then \(h(0)=A[u]\) and so \(0\) is a minimum point of \(h\) since \(u\) is a minimum point of \(A\). Therefore \[\begin{aligned} 0 & = h'(0) \\ & = \left. \frac{d}{d \varepsilon} \right|_{\varepsilon=0} A[u_\varepsilon] \phantom{\Bigg|} \\ & = \left. \frac{d}{d \varepsilon} \right|_{\varepsilon=0} \int_\Omega \sqrt{1+|\nabla u_{\varepsilon}|^2} \, d \boldsymbol{x}\phantom{\Bigg|} \\ & = \int_\Omega \left. \frac{d}{d \varepsilon} \right|_{\varepsilon=0} \sqrt{1+|\nabla u + \varepsilon \nabla \varphi|^2} \, d \boldsymbol{x}\phantom{\Bigg|} \\ & = \int_\Omega \left. \tfrac{1}{2} (1+|\nabla u + \varepsilon \nabla \varphi|^2)^{-1/2} \, 2 |\nabla u + \varepsilon \nabla \varphi| \, \frac{\nabla u + \varepsilon \nabla \varphi}{|\nabla u + \varepsilon \nabla \varphi|} \cdot \nabla \varphi \right|_{\varepsilon=0} \, d \boldsymbol{x}\phantom{\Bigg|} \\ & = \int_\Omega \frac{\nabla u}{\sqrt{1+|\nabla u|^2}} \cdot \nabla \varphi \, d \boldsymbol{x}.% \phantom{\Bigg|} \end{aligned}\] This means that \(u\) is a weak solution of the minimal surface equation. Since \(u \in C^2(\overline{\Omega})\), then we can integrate by parts to obtain \[\begin{aligned} 0 & = \int_\Omega \frac{\nabla u}{\sqrt{1+|\nabla u|^2}} \cdot \nabla \varphi \, d \boldsymbol{x} \\ & = \int_{\partial \Omega} \varphi \, \frac{\nabla u}{\sqrt{1+|\nabla u|^2}} \cdot \boldsymbol{n}\, dS - \int_\Omega \mathrm{div} \left( \frac{\nabla u}{\sqrt{1+|\nabla u|^2}} \right) \varphi \, d \boldsymbol{x} \\ & = - \int_\Omega \mathrm{div} \left( \frac{\nabla u}{\sqrt{1+|\nabla u|^2}} \right) \varphi \, d \boldsymbol{x} \end{aligned}\] since \(\varphi = 0\) on \(\partial \Omega\). This holds for all \(\varphi \in C^1(\overline{\Omega})\) with \(\varphi=0\). Therefore \(u\) satisfies the minimal surface equation \[\mathrm{div} \left( \frac{\nabla u}{\sqrt{1+|\nabla u |^2}} \right) = 0 \quad \textrm{in } \Omega.\] by the Fundamental Lemma of the Calculus of Variations (Lemma 3.20).
Homogenization and the calculus of variations.
Let \(u \in C^2([0,1]) \cap V\) minimise \(E\). For any \(\varepsilon \in \mathbb{R}\) and any \(\varphi \in C^1([0,1])\) such that \(\varphi(0)=\varphi(1)=0\), define \(u_\varepsilon = u + \varepsilon \varphi.\) Then \[u_\varepsilon(0) = u(0) + \varepsilon \varphi(0) = l + \varepsilon \cdot 0 = l\] and similarly \(u_\varepsilon(1)=r\). Therefore \(u_\varepsilon \in V\). Define \(F(\varepsilon) = E[u_\varepsilon]\). Now \(u_\varepsilon = u\) when \(\varepsilon = 0\). Therefore the minimum of \(F\) is attained at \(0\) since the minimum of \(E\) is attained at \(u\). We have reduced the problem of minimising the functional \(E\) to minimising the function of one variable \(F\). Since \(F\) is minimised at \(0\), \[\begin{align} \nonumber 0 & = F'(0) \\ \nonumber & = \left. \frac{d}{d \varepsilon} \right|_{\varepsilon=0} E[u_\varepsilon] \phantom{\Bigg|} \\ \nonumber & = \left. \frac{d}{d \varepsilon} \right|_{\varepsilon=0} \left[ \frac 12 \int_0^1 a(x)|u_\varepsilon'(x)|^2 \, dx - \int_0^1 f(x)u_\varepsilon(x) \, dx\right] \phantom{\Bigg|} \\ \nonumber & = \left. \frac{d}{d \varepsilon} \right|_{\varepsilon=0} \left[ \frac 12 \int_0^1 a(x)(u'(x)+\varepsilon \varphi'(x))^2 \, dx - \int_0^1 f(x)(u(x)+\varepsilon \varphi(x)) \, dx\right] \phantom{\Bigg|} \\ \label{Q15:1} & = \int_0^1 a(x)u'(x)\varphi'(x) \, dx - \int_0^1 f(x) \varphi(x) \, dx. \end{align}\] Since \(u \in C^2([0,1])\), we can use integration by parts to rewrite equation \(\eqref{Q15:1}\) as \[0 = a(x)u'(x)\varphi(x)\Big|_0^1 - \int_0^1 (a(x)u'(x))'\varphi(x) \, dx - \int_0^1 f(x) \varphi(x) \, dx = - \int_0^1 [(a(x)u'(x))'+f(x)] \varphi(x) \, dx.\] But this holds for all \(\varphi \in C^1([0,1])\) such that \(\varphi(0)=\varphi(1)=0\). Therefore by the Fundamental Lemma of the Calculus of Variations \[(a(x)u'(x))'+f(x) = 0, \quad x \in (0,1),\] as required. Note that \(u\) satisfies the Dirichlet boundary conditions by definition of \(V\).
Recall from Q2(ii) that if \(g \in L^\infty(\mathbb{R})\) is \(1\)–periodic, then for any interval \([c,d] \subseteq \mathbb{R}\), \[\begin{equation} \label{Q15:2} \lim_{n \to \infty} \int_c^d g(nx) h(x) \, dx = \int_c^d \overline{g} \, h(x) \, dx \qquad \forall \; h \in L^1(\mathbb{R}). \end{equation}\] Applying \(\eqref{Q15:2}\) with \(c=0\), \(d=1\), \(g(x)=a(x)\), \(h(x)=\frac12 |v'(x)|^2\) on \([0,1]\), gives the desired result: \[\lim_{n \to \infty} E_n[v] = \frac 12 \int_0^1 \overline{a} \, |v'(x)|^2 \, dx - \int_0^1 f(x)v(x) \, dx =: E_\infty[v].\]
Observe that \(E_\infty\) is just the one-dimensional Dirichlet energy with an additional constant \(\overline{a}\) in the first term. It follows from Dirichlet’s Principle (see the lecture notes) that \(u_\infty\) satisfies the Poisson equation \[\begin{equation} \begin{gathered} %\label{Q15:3} - \, \overline{a} \, u_\infty''(x) = f(x), \quad x \in (0,1), \\ u_\infty(0)= u_\infty(1)=0. \end{gathered} \end{equation}\] In Q2 we showed that \(\lim_{n \to \infty} u_n(x) = u_0(x)\), where \(u_0\) satisfies \[\begin{equation} \begin{gathered} %\label{Q15:4} -\, a_0 u_0''(x) = f(x), \quad x \in (0,1), \\ u_0(0)=u_0(1)=0, \end{gathered} \end{equation}\] where \[a_0 = \frac{1}{\, \overline{\left( \frac{1}{a} \right)} \,}.\] Since \(a_0 \ne \overline{a}\) in general, it follows that \(u_0 \ne u_\infty\) and hence \[\lim_{n \to \infty} u_n(x) = u_0(x) \ne u_\infty(x),\] as required.
In fact it can be shown that \(a_0 \le \overline{a}\) as follows: \[1 = \left[ \int_0^1 \sqrt{a(x)} \frac{1}{\sqrt{a(x)}} \, dx \right]^2 \le \left[ \left( \int_0^1 a(x) \, dx \right)^{1/2} \left( \int_0^1 \frac{1}{a(x)} \, dx \right)^{1/2} \right]^2 = \overline{a} \, \overline{\left( \frac{1}{a} \right)} = \overline{a} \, a_0^{-1}\] where we have used the Cauchy-Schwarz inequality. It follows that the \(\Gamma\)–limit \(E_0\) is less than or equal to the pointwise limit \(E_\infty\).