Partial Differential Equations III/IV

[6pt] Exercise Sheet 4: Solutions

1. Green’s functions. By the Fundamental Theorem of Calculus, integrating $u^{\prime \prime }(y)=f(y)$ over $[0,z]$, for any $z\in [0,1]$, gives

   $${\int }_0^z{u}^{\prime \prime }(y)𝑑y=-{\int }_0^{z}f(y)𝑑y\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}{u}^{\prime }(z)=u^{\prime }(0)-{\int }_0^{z}f(y)𝑑y=-{\int }_0^{z}f(y)𝑑y,$$

   where we have used the boundary condition $u^{\prime }(0)=0$. Integrating again, this time over $[0,x]$, gives

   ${\displaystyle {\int }_0^x}{u}^{\prime }(z)𝑑z=-{\displaystyle {\int }_0^x}{\displaystyle {\int }_0^z}f(y)𝑑y𝑑z\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}u(x)=u(0)-{\displaystyle {\int }_0^x}{\displaystyle {\int }_0^z}f(y)𝑑y𝑑z.$

   Taking $x=1$ and using the boundary condition $u(1)=0$ yields

   $$0=u(0)-{\int }_0^1{\int }_0^{z}f(y)𝑑y𝑑z\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}u(0)={\int }_0^1{\int }_0^{z}f(y)𝑑y𝑑z.$$

   Therefore

   $$u(x)={\int }_0^1{\int }_0^{z}f(y)𝑑y𝑑z-{\int }_0^x{\int }_0^{z}f(y)𝑑y𝑑z.$$

   By interchanging the order of integration we can write this as

   $u(x)$ $={\displaystyle {\int }_0^1}{\displaystyle {\int }_y^1}f(y)𝑑z𝑑y-{\displaystyle {\int }_0^x}{\displaystyle {\int }_y^x}f(y)𝑑z𝑑y$

   $={\displaystyle {\int }_0^1}(1-y)f(y)𝑑y-{\displaystyle {\int }_0^x}(x-y)f(y)𝑑y$

   $={\displaystyle {\int }_0^x}(1-y)f(y)𝑑y+{\displaystyle {\int }_x^1}(1-y)f(y)𝑑y-{\displaystyle {\int }_0^x}(x-y)f(y)𝑑y$

   $={\displaystyle {\int }_0^x}(1-x)f(y)𝑑y+{\displaystyle {\int }_x^1}(1-y)f(y)𝑑y.$

   Therefore

   $$u(x)={\int }_0^{1}G(x,y)f(y)𝑑y$$

   with

2. Homogenization.

   • Integrate ${(a_{\epsilon }(y)u_{\epsilon }(y))}^{\prime }=-f(y)$ over $y\in [0,z]$:

     ${\displaystyle {\int }_0^z}{(a_{\epsilon }(y)u_{\epsilon }^{\prime }(y))}^{\prime }𝑑y=-{\displaystyle {\int }_0^z}f(y)𝑑y$ $⟺\mathit{\hspace{1em}}{a}_{\epsilon }(z)u_{\epsilon }^{\prime }(z)-a_{\epsilon }(0)u_{\epsilon }^{\prime }(0)=-{\displaystyle {\int }_0^z}f(y)𝑑y$

     $⟺\mathit{\hspace{1em}}{u}_{\epsilon }^{\prime }(z)={\displaystyle \frac{a_{\epsilon }(0)u_{\epsilon }^{\prime }(0)}{a_{\epsilon }(z)}}-{\displaystyle \frac{1}{a_{\epsilon }(z)}}{\displaystyle {\int }_0^z}f(y)𝑑y.$

     Now integrate over $z\in [0,x]$:

     $${\int }_0^x{u}_{\epsilon }^{\prime }(z)𝑑z={\int }_0^x\left[\frac{a_{\epsilon }(0)u_{\epsilon }^{\prime }(0)}{a_{\epsilon }(z)}-\frac{1}{a_{\epsilon }(z)}{\int }_0^{z}f(y)𝑑y\right]𝑑z\mathit{\hspace{1em}}⟺$$

     $$u_{\epsilon }(x)=\underset{=0}{\underbrace{u_{\epsilon }(0)}}+a_{\epsilon }(0)u_{\epsilon }^{\prime }(0){\int }_0^x\frac{1}{a_{\epsilon }(z)}𝑑z-{\int }_0^x\frac{1}{a_{\epsilon }(z)}{\int }_0^{z}f(y)𝑑y𝑑z.$$ (1)

     We determine $u_{\epsilon }^{\prime }(0)$ by evaluating this expression at $x=1$:

     $$\underset{=0}{\underbrace{u_{\epsilon }(1)}}=a_{\epsilon }(0)u_{\epsilon }^{\prime }(0){\int }_0^1\frac{1}{a_{\epsilon }(z)}𝑑z-{\int }_0^1\frac{1}{a_{\epsilon }(z)}{\int }_0^{z}f(y)𝑑y𝑑z\mathit{\hspace{1em}}⟺$$

     $$a_{\epsilon }(0)u_{\epsilon }^{\prime }(0)={\left({\int }_0^1\frac{1}{a_{\epsilon }(z)}𝑑z\right)}^{-1}{\int }_0^1\frac{1}{a_{\epsilon }(z)}{\int }_0^{z}f(y)𝑑y𝑑z.$$

     Substituting this into (1) gives

     $$\boxed{u_{\epsilon }(x)={\left({\int }_0^1\frac{1}{a_{\epsilon }(z)}𝑑z\right)}^{-1}{\int }_0^1\frac{1}{a_{\epsilon }(z)}{\int }_0^{z}f(y)𝑑y𝑑z{\int }_0^x\frac{1}{a_{\epsilon }(z)}𝑑z-{\int }_0^x\frac{1}{a_{\epsilon }(z)}{\int }_0^{z}f(y)𝑑y𝑑z}$$

     as required.

   • Taking $\epsilon ={\epsilon }_n=\frac{1}{n}$ gives

     $$u_{{\epsilon }_n}(x)={\left({\int }_0^1\frac{1}{a(nz)}𝑑z\right)}^{-1}{\int }_0^1\frac{1}{a(nz)}{\int }_0^{z}f(y)𝑑y𝑑z{\int }_0^x\frac{1}{a(nz)}𝑑z-{\int }_0^x\frac{1}{a(nz)}{\int }_0^{z}f(y)𝑑y𝑑z.$$

     We are told in the hint to use the Riemann-Lebesgue Lemma, which states that if $g\in L^{\mathrm{\infty }}(ℝ)$ is $1$–periodic, then for any interval $[c,d]\subseteq ℝ$,

     $$\underset{n\to \mathrm{\infty }}{lim}{\int }_c^{d}g(nz)h(z)𝑑z={\int }_c^d\overline{g}h(z)𝑑z\mathit{\hspace{1em}\hspace{1em}}\forall h\in L^1(ℝ)\cap C^1(ℝ)$$ (2)

     Applying (2) with $c=0$, $d=1$, $g(z)=1/a(z)$, and $h(z)=1$ on $[c,d]$ gives

     $$\underset{n\to \mathrm{\infty }}{lim}{\int }_0^1\frac{1}{a(nz)}𝑑z={\int }_0^1\overline{\left(\frac{1}{a}\right)}𝑑z=\overline{\left(\frac{1}{a}\right)}.$$

     (Technical remark: We cannot take $h(z)=1$ for all $z\in ℝ$, else $h\notin L^1(ℝ)$. But we can take $h$ to be any function in $L^1(ℝ)\cap C^1(ℝ)$ 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 (2).)

     Applying (2) 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

     $$\underset{n\to \mathrm{\infty }}{lim}{\int }_0^x\frac{1}{a(nz)}𝑑z={\int }_0^x\overline{\left(\frac{1}{a}\right)}𝑑z=x\overline{\left(\frac{1}{a}\right)}.$$

     Applying (2) with $c=0$, $d=1$, $g(z)=1/a(z)$, and $h(z)={\int }_0^{z}f(y)𝑑y$ on $[c,d]$ gives

     $$\underset{n\to \mathrm{\infty }}{lim}{\int }_0^1\frac{1}{a(nz)}{\int }_0^{z}f(y)𝑑y𝑑z=\overline{\left(\frac{1}{a}\right)}{\int }_0^1{\int }_0^{z}f(y)𝑑y𝑑z.$$

     Finally, applying (2) with $c=0$, $d=x$, $g(z)=1/a(z)$, and $h(z)={\int }_0^{z}f(y)𝑑y$ on $[c,d]$ gives

     $$\underset{n\to \mathrm{\infty }}{lim}{\int }_0^x\frac{1}{a(nz)}{\int }_0^{z}f(y)𝑑y𝑑z=\overline{\left(\frac{1}{a}\right)}{\int }_0^x{\int }_0^{z}f(y)𝑑y𝑑z.$$

     $$\boxed{\underset{n\to \mathrm{\infty }}{lim}{u}_{{\epsilon }_n}(x)=u_0(x):=x\overline{\left(\frac{1}{a}\right)}{\int }_0^1{\int }_0^{z}f(y)dydz-\overline{\left(\frac{1}{a}\right)}{\int }_0^x{\int }_0^{z}f(y)dydz.}$$ (3)

   • This is simply a matter of interchanging the order of integration:

     $u_0(x)$ $=x\overline{\left({\displaystyle \frac{1}{a}}\right)}{\displaystyle {\int }_0^1}{\displaystyle {\int }_0^z}f(y)𝑑y𝑑z-\overline{\left({\displaystyle \frac{1}{a}}\right)}{\displaystyle {\int }_0^x}{\displaystyle {\int }_0^z}f(y)𝑑y𝑑z$

     $=x\overline{\left({\displaystyle \frac{1}{a}}\right)}{\displaystyle {\int }_0^1}{\displaystyle {\int }_y^1}f(y)𝑑z𝑑y-\overline{\left({\displaystyle \frac{1}{a}}\right)}{\displaystyle {\int }_0^x}{\displaystyle {\int }_y^x}f(y)𝑑z𝑑y$

     $=x\overline{\left({\displaystyle \frac{1}{a}}\right)}{\displaystyle {\int }_0^1}(1-y)f(y)𝑑y-\overline{\left({\displaystyle \frac{1}{a}}\right)}{\displaystyle {\int }_0^x}(x-y)f(y)𝑑y$

     $=\overline{\left({\displaystyle \frac{1}{a}}\right)}\left\{{\displaystyle {\int }_0^x}[x(1-y)-(x-y)]f(y)𝑑y+{\displaystyle {\int }_x^1}x(1-y)f(y)𝑑y\right\}$

     $=\overline{\left({\displaystyle \frac{1}{a}}\right)}\left\{{\displaystyle {\int }_0^x}y(1-x)f(y)𝑑y+{\displaystyle {\int }_x^1}x(1-y)f(y)𝑑y\right\}$

     $={\displaystyle {\int }_0^1}G(x,y)f(y)𝑑y$

     with

   • Clearly $u_0$ satisfies the boundary conditions. By the Fundamental Theorem of Calculus, differentiating equation (3) gives

     $$u_0^{\prime }(x)=\overline{\left(\frac{1}{a}\right)}{\int }_0^1{\int }_0^{z}f(y)𝑑y𝑑z-\overline{\left(\frac{1}{a}\right)}{\int }_0^{x}f(y)𝑑y.$$

     Differentiating again gives

     $$u_0^{\prime \prime }(x)=-\overline{\left(\frac{1}{a}\right)}f(x).$$

     Therefore

     $$-a_0{u}_0^{\prime \prime }(x)=-\frac{1}{\overline{\left(\frac{1}{a}\right)}}\left[-\overline{\left(\frac{1}{a}\right)}f(x)\right]=f(x)$$

     as required.

   • By definition,

     $$\overline{a}={\int }_0^{1}a(x)𝑑x={\int }_0^{\frac{1}{2}}\frac{1}{2}𝑑x+{\int }_{\frac{1}{2}}^{1}1𝑑x=\frac{1}{2}\cdot \frac{1}{2}+\frac{1}{2}\cdot 1=\boxed{\frac{3}{4}}$$

     On the other hand,

     $$a_0={\left({\int }_0^1\frac{1}{a(x)}𝑑x\right)}^{-1}={\left({\int }_0^{\frac{1}{2}}2𝑑x+{\int }_{\frac{1}{2}}^{1}1𝑑x\right)}^{-1}={\left(\frac{1}{2}\cdot 2+\frac{1}{2}\cdot 1\right)}^{-1}={\left(\frac{3}{2}\right)}^{-1}=\boxed{\frac{2}{3}}$$

     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

     ${\displaystyle {\int }_c^d}g(nz)h(z)𝑑z$ $={\displaystyle {\int }_c^d}{\left({\displaystyle \frac{1}{n}}{\displaystyle {\int }_0^{nz}}g(y)𝑑y\right)}^{\prime }h(z)𝑑z$

     $={{\displaystyle \frac{1}{n}}{\displaystyle {\int }_0^{nz}}g(y)𝑑yh(z)|}_c^d-{\displaystyle {\int }_c^d}{\displaystyle \frac{1}{n}}{\displaystyle {\int }_0^{nz}}g(y)𝑑y{h}^{\prime }(z)𝑑z.$ (4)

     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,

     $${\int }_0^{nz}g(y)𝑑y={\int }_0^{\lfloor nz\rfloor }g(y)𝑑y+{\int }_{\lfloor nz\rfloor }^{nz}g(y)𝑑y=\lfloor nz\rfloor {\int }_0^{1}g(y)𝑑y+{\int }_{\lfloor nz\rfloor }^{nz}g(y)𝑑y.$$ (5)

     Observe that

     Therefore by the Pinching Lemma (Squeezing Lemma)

     $$\underset{n\to \mathrm{\infty }}{lim}\frac{\lfloor nz\rfloor }{n}=z.$$ (6)

     Also

     $$\left|\frac{1}{n}{\int }_{\lfloor nz\rfloor }^{nz}g(y)𝑑y\right|\le \frac{1}{n}(nz-\lfloor nz\rfloor ){\parallel g\parallel }_{L^{\mathrm{\infty }}(ℝ)}\le \frac{1}{n}{\parallel g\parallel }_{L^{\mathrm{\infty }}(ℝ)}.$$

     Therefore

     $$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}{\int }_{\lfloor nz\rfloor }^{nz}g(y)𝑑y=0.$$ (7)

     By combining equations (5), (6), (7) we find that

     $$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}{\int }_0^{nz}g(y)𝑑y=z{\int }_0^{1}g(y)𝑑y.$$ (8)

     Therefore the limit of the first term on the right-hand side of equation (4) is

     $${\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}{\int }_0^{nz}g(y)𝑑yh(z)|}_c^d={z{\int }_0^{1}g(y)𝑑yh(z)|}_c^d.$$ (9)

     Now we find the limit of the second term on the right-hand side of (4). By the computations above

     $\left|{\displaystyle {\int }_c^d}{\displaystyle \frac{1}{n}}{\displaystyle {\int }_0^{nz}}g(y)𝑑y{h}^{\prime }(z)𝑑z-{\displaystyle {\int }_c^d}z{\displaystyle {\int }_0^1}g(y)𝑑y{h}^{\prime }(z)𝑑z\right|$

     $\le {\displaystyle {\int }_c^d}\left|{\displaystyle \frac{1}{n}}{\displaystyle {\int }_0^{nz}}g(y)𝑑y-z{\displaystyle {\int }_0^1}g(y)𝑑y\right||h^{\prime }(z)|𝑑z$

     $\le {\displaystyle {\int }_c^d}\left|{\displaystyle \frac{\lfloor nz\rfloor }{n}}{\displaystyle {\int }_0^1}g(y)𝑑y+{\displaystyle \frac{1}{n}}{\displaystyle {\int }_{\lfloor nz\rfloor }^{nz}}g(y)𝑑y-z{\displaystyle {\int }_0^1}g(y)𝑑y\right||h^{\prime }(z)|𝑑z$

     $\le {\displaystyle {\int }_c^d}\left(\left|{\displaystyle \frac{\lfloor nz\rfloor }{n}}{\displaystyle {\int }_0^1}g(y)𝑑y-z{\displaystyle {\int }_0^1}g(y)𝑑y\right|+{\displaystyle \frac{1}{n}}{\displaystyle {\int }_{\lfloor nz\rfloor }^{nz}}|g(y)|𝑑y\right)|h^{\prime }(z)|𝑑z$

     $\le {\displaystyle {\int }_c^d}\left|{\displaystyle \frac{\lfloor nz\rfloor -nz}{n}}{\displaystyle {\int }_0^1}g(y)𝑑y\right||h^{\prime }(z)|𝑑z+{\displaystyle \frac{1}{n}}{\parallel g\parallel }_{L^{\mathrm{\infty }}(ℝ)}{\parallel h^{\prime }\parallel }_{L^1([c,d])}$

     $\le {\displaystyle {\int }_c^d}{\displaystyle \frac{1}{n}}{\displaystyle {\int }_0^1}|g(y)|𝑑y|h^{\prime }(z)|𝑑z+{\displaystyle \frac{1}{n}}{\parallel g\parallel }_{L^{\mathrm{\infty }}(ℝ)}{\parallel h^{\prime }\parallel }_{L^1([c,d])}$

     $\le {\displaystyle \frac{2}{n}}{\parallel g\parallel }_{L^{\mathrm{\infty }}(ℝ)}{\parallel h^{\prime }\parallel }_{L^1([c,d])}\to 0\text{ as }n\to \mathrm{\infty }.$

     Therefore

     $$\underset{n\to \mathrm{\infty }}{lim}{\int }_c^d\frac{1}{n}{\int }_0^{nz}g(y)𝑑y{h}^{\prime }(z)𝑑z={\int }_c^{d}z{\int }_0^{1}g(y)𝑑y{h}^{\prime }(z)𝑑z.$$ (10)

     Combining (4), (9), (10) and then integrating by parts yields

     $\underset{n\to \mathrm{\infty }}{lim}{\displaystyle {\int }_c^d}g(nz)h(z)𝑑z$ $={z{\displaystyle {\int }_0^1}g(y)𝑑yh(z)|}_c^d-{\displaystyle {\int }_c^d}z{\displaystyle {\int }_0^1}g(y)𝑑y{h}^{\prime }(z)𝑑z$

     $={\displaystyle {\int }_c^d}{\displaystyle {\int }_0^1}g(y)𝑑yh(z)𝑑z$

     $={\displaystyle {\int }_c^d}\overline{g}h(z)𝑑z$

     as required.

3. Radial symmetry of Laplace’s equation on ${ℝ}^n$. Let $v:{ℝ}^n\to ℝ$ be a harmonic function. Let $R\in O(n,ℝ)$ and define $w:{ℝ}^n\to ℝ$ by $w(𝒙):=v(R𝒙)$. Then

   $w_{x_i}$ $={\displaystyle \sum _{j=1}^n}{\displaystyle \frac{\partial v}{\partial x_j}}{\displaystyle \frac{\partial {(R𝒙)}_j}{\partial x_i}}={\displaystyle \sum _{j=1}^n}{\displaystyle \frac{\partial v}{\partial x_j}}{\displaystyle \frac{\partial }{\partial x_i}}{\displaystyle \sum _{k=1}^n}{R}_{jk}{x}_k$

   $={\displaystyle \sum _{j=1}^n}{\displaystyle \frac{\partial v}{\partial x_j}}{\displaystyle \sum _{k=1}^n}{R}_{jk}{\displaystyle \frac{\partial x_k}{\partial x_i}}={\displaystyle \sum _{j=1}^n}{\displaystyle \frac{\partial v}{\partial x_j}}{\displaystyle \sum _{k=1}^n}{R}_{jk}{\delta }_{ki}$

   $={\displaystyle \sum _{j=1}^n}{\displaystyle \frac{\partial v}{\partial x_j}}{R}_{ji}.$

   To be precise

   $$w_{x_i}(𝒙)=\sum _{j=1}^n{v}_{x_j}(R𝒙)R_{ji}.$$

   (This can also be written as $\nabla w(𝒙)=R^T\nabla v(R𝒙)$.)

   Now we compute the second partial derivatives:

   $w_{x_i{x}_i}$ $={\displaystyle \frac{\partial }{\partial x_i}}{\displaystyle \sum _{j=1}^n}{v}_{x_j}(R𝒙)R_{ji}$

   $={\displaystyle \sum _{j=1}^n}{\displaystyle \sum _{k=1}^n}{\displaystyle \frac{\partial v_{x_j}}{\partial x_k}}{\displaystyle \frac{\partial {(R𝒙)}_k}{\partial x_i}}{R}_{ji}$

   $={\displaystyle \sum _{j=1}^n}{\displaystyle \sum _{k=1}^n}{v}_{x_j{x}_k}{R}_{ji}{\displaystyle \frac{\partial }{\partial x_i}}{\displaystyle \sum _{l=1}^n}{R}_{kl}{x}_l$

   $={\displaystyle \sum _{j=1}^n}{\displaystyle \sum _{k=1}^n}{v}_{x_j{x}_k}{R}_{ji}{\displaystyle \sum _{l=1}^n}{R}_{kl}{\displaystyle \frac{\partial x_l}{\partial x_i}}$

   $={\displaystyle \sum _{j=1}^n}{\displaystyle \sum _{k=1}^n}{v}_{x_j{x}_k}{R}_{ji}{\displaystyle \sum _{l=1}^n}{R}_{kl}{\delta }_{il}$

   $={\displaystyle \sum _{j=1}^n}{\displaystyle \sum _{k=1}^n}{v}_{x_j{x}_k}{R}_{ji}{R}_{ki}.$

   Therefore

   $\mathrm{\Delta }w$ $={\displaystyle \sum _{i=1}^n}{w}_{x_i{x}_i}$

   $={\displaystyle \sum _{i=1}^n}{\displaystyle \sum _{j=1}^n}{\displaystyle \sum _{k=1}^n}{v}_{x_j{x}_k}{R}_{ji}{R}_{ki}$

   $={\displaystyle \sum _{j=1}^n}{\displaystyle \sum _{k=1}^n}{v}_{x_j{x}_k}{\displaystyle \sum _{i=1}^n}{R}_{ji}{R}_{ki}$

   $={\displaystyle \sum _{j=1}^n}{\displaystyle \sum _{k=1}^n}{v}_{x_j{x}_k}{\displaystyle \sum _{i=1}^n}{R}_{ji}{(R^T)}_{ik}$

   $={\displaystyle \sum _{j=1}^n}{\displaystyle \sum _{k=1}^n}{v}_{x_j{x}_k}{(R{R}^T)}_{jk}$

   $={\displaystyle \sum _{j=1}^n}{\displaystyle \sum _{k=1}^n}{v}_{x_j{x}_k}{I}_{jk}$ (11)

   since $R$ is an orthogonal matrix. There are two ways to conclude from here: If are are familiar with the matrix inner product, then (11) gives

   $$\mathrm{\Delta }w=D^{2}v:I=\mathrm{trace}(D^{2}v)=\mathrm{\Delta }v=0$$

   since $v$ is harmonc. Otherwise we can continue from (11) using indices:

   $$\mathrm{\Delta }w=\sum _{j=1}^n\sum _{k=1}^n{v}_{x_j{x}_k}{I}_{jk}=\sum _{j=1}^n\sum _{k=1}^n{v}_{x_j{x}_k}{\delta }_{jk}=\sum _{j=1}^n{v}_{x_j{x}_j}=\mathrm{\Delta }v=0,$$

   as required.

4. Fundamental solution of Poisson’s equation in 3D.

   • One way of computing ${\parallel \mathrm{\Phi }\parallel }_{L^1(B_R(\mathrm{𝟎}))}$ is using spherical polar coordinates:

     ${\parallel \mathrm{\Phi }\parallel }_{L^1(B_R(\mathrm{𝟎}))}$ $={\displaystyle {\int }_{B_R(\mathrm{𝟎})}}|\mathrm{\Phi }(𝒙)|𝑑𝒙$

     $={\displaystyle \frac{1}{4\pi }}{\displaystyle {\int }_{B_R(\mathrm{𝟎})}}{\displaystyle \frac{1}{|𝒙|}}𝑑𝒙$

     $={\displaystyle \frac{1}{4\pi }}{\displaystyle {\int }_{\varphi =0}^{2\pi }}{\displaystyle {\int }_{\theta =0}^{\pi }}{\displaystyle {\int }_{r=0}^R}{\displaystyle \frac{1}{r}}{r}^2\sin \theta drd\theta d\varphi$

     $={\displaystyle \frac{1}{4\pi }}{\displaystyle {\int }_0^{2\pi }}1𝑑\varphi {\displaystyle {\int }_0^{\pi }}\sin \theta d\theta {\displaystyle {\int }_0^R}r𝑑r$

     $=\boxed{{\displaystyle \frac{R^2}{2}}}$

     Another way of computing ${\parallel \mathrm{\Phi }\parallel }_{L^1(B_R(\mathrm{𝟎}))}$ is as follows:

     ${\parallel \mathrm{\Phi }\parallel }_{L^1(B_R(\mathrm{𝟎}))}$ $={\displaystyle {\int }_{B_R(\mathrm{𝟎})}}|\mathrm{\Phi }(𝒙)|𝑑𝒙$

     $={\displaystyle {\int }_0^R}\left({\displaystyle {\int }_{\partial B_r(\mathrm{𝟎})}}|\mathrm{\Phi }(𝒚)|𝑑S(𝒚)\right)𝑑r$

     $={\displaystyle {\int }_0^R}\left({\displaystyle {\int }_{\partial B_r(\mathrm{𝟎})}}{\displaystyle \frac{1}{4\pi }}{\displaystyle \frac{1}{|𝒚|}}𝑑S(𝒚)\right)𝑑r$

     $={\displaystyle \frac{1}{4\pi }}{\displaystyle {\int }_0^R}\left({\displaystyle {\int }_{\partial B_r(\mathrm{𝟎})}}{\displaystyle \frac{1}{r}}𝑑S(𝒚)\right)𝑑r$

     $={\displaystyle \frac{1}{4\pi }}{\displaystyle {\int }_0^R}\left(\text{area}(\partial B_r(\mathrm{𝟎})){\displaystyle \frac{1}{r}}\right)𝑑r$

     $={\displaystyle \frac{1}{4\pi }}{\displaystyle {\int }_0^R}\left(4\pi r^2{\displaystyle \frac{1}{r}}\right)𝑑r$

     $={\displaystyle {\int }_0^R}r𝑑r$

     $=\boxed{{\displaystyle \frac{R^2}{2}}}$

   • Let $K\subset {ℝ}^3$ be compact. Since $K$ is bounded, there exists $R>0$ such that $K\subset B_R(\mathrm{𝟎})$. Therefore

     Therefore $\mathrm{\Phi }\in L_{\mathrm{loc}}^1({ℝ}^3)$.

   • By part (i),

     $$\underset{R\to \mathrm{\infty }}{lim}{\parallel \mathrm{\Phi }\parallel }_{L^1(B_R(\mathrm{𝟎}))}=\underset{R\to \mathrm{\infty }}{lim}\frac{R^2}{2}=+\mathrm{\infty }.$$

     Therefore $\mathrm{\Phi }\notin L^1({ℝ}^3)$.

   • By the Chain Rule

     $$\nabla \mathrm{\Phi }(𝒙)=\frac{1}{4\pi }\left(-\frac{1}{{|𝒙|}^2}\right)\nabla |𝒙|=\frac{1}{4\pi }\left(-\frac{1}{{|𝒙|}^2}\right)\frac{𝒙}{|𝒙|}=-\frac{1}{4\pi }\frac{𝒙}{{|𝒙|}^3}.$$

     Let $K\subset {ℝ}^3$ be compact. Since $K$ is bounded, there exists $R>0$ such that $K\subset B_R(\mathrm{𝟎})$. Therefore

     ${\displaystyle {\int }_K}|\nabla \mathrm{\Phi }(𝒙)|𝑑𝒙$ $\le {\displaystyle {\int }_{B_R(\mathrm{𝟎})}}|\nabla \mathrm{\Phi }(𝒙)|𝑑𝒙$

     $={\displaystyle {\int }_{B_R(\mathrm{𝟎})}}{\displaystyle \frac{1}{4\pi }}{\displaystyle \frac{1}{{|𝒙|}^2}}𝑑𝒙$

     $={\displaystyle \frac{1}{4\pi }}{\displaystyle {\int }_{\varphi =0}^{2\pi }}{\displaystyle {\int }_{\theta =0}^{\pi }}{\displaystyle {\int }_{r=0}^R}{\displaystyle \frac{1}{r^2}}{r}^2\sin \theta drd\theta d\varphi$

     $={\displaystyle \frac{1}{4\pi }}{\displaystyle {\int }_0^{2\pi }}1𝑑\varphi {\displaystyle {\int }_0^{\pi }}\sin \theta d\theta {\displaystyle {\int }_0^R}1𝑑r$

     $=R$

     Therefore $\nabla \mathrm{\Phi }\in L_{\mathrm{loc}}^1({ℝ}^3)$.

5. Fundamental solution of Poisson’s equation in 1D. We compute

   $u^{\prime \prime }(x)$ $={(\mathrm{\Phi }*f)}^{\prime \prime }(x)$

   $={(f*\mathrm{\Phi })}^{\prime \prime }(x)$ (symmetry of convolution)

   $={\displaystyle \frac{d^2}{d{x}^2}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}\mathrm{\Phi }(y)f(x-y)𝑑y$

   $={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}\mathrm{\Phi }(y){\displaystyle \frac{d^2}{d{x}^2}}f(x-y)𝑑y$

   $={\displaystyle {\int }_{-\mathrm{\infty }}^0}y{\displaystyle \frac{d^2}{d{x}^2}}f(x-y)𝑑y$

   $={\displaystyle {\int }_{-\mathrm{\infty }}^0}y{\displaystyle \frac{d^2}{d{y}^2}}f(x-y)𝑑y$

   $={y{\displaystyle \frac{d}{dy}}f(x-y)|}_{-\mathrm{\infty }}^0-{\displaystyle {\int }_{-\mathrm{\infty }}^0}{\displaystyle \frac{d}{dy}}f(x-y)𝑑y$ (integration by parts)

   $=-f(x)$ (Fundamental Theorem of Calculus)

   as required.

6. The function spaces $L^{\mathrm{1}}$ and $L_{\mathrm{loc}}^{\mathrm{1}}$. Let $f:ℝ\to ℝ$, $f(x)={|x|}^k$, $k\in ℝ$. By integrating we see that

   • $f\in L^1((-R,R))$ for $k>-1$,

   • $f\in L^1((R,\mathrm{\infty }))$ for ,

   • $f\in L_{\mathrm{loc}}^1(ℝ)$ for $k>-1$,

   • $f\notin L^1(ℝ)$ for any $k$ (by parts (i),(ii)).

7. Properties of the convolution.

   • Let $\phi \in L_{\mathrm{loc}}^1(ℝ)$, $f\in C_c(ℝ)$ 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

     $|(\phi *f)(x)|$ $=\left|{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}\phi (x-y)f(y)𝑑y\right|$

     $=\left|{\displaystyle {\int }_{-R}^R}\phi (x-y)f(y)𝑑y\right|$

     $\le {\displaystyle {\int }_{-R}^R}|\phi (x-y)||f(y)|𝑑y$

     $\le \underset{[-R,R]}{\max }|f|{\displaystyle {\int }_{-R}^R}|\phi (x-y)|𝑑y$

     $=\underset{[-R,R]}{\max }|f|{\displaystyle {\int }_{-R-x}^{R-x}}|\phi (z)|𝑑z$

     since $\phi \in L_{\mathrm{loc}}^1(ℝ)$ and $[-R-x,R-x]$ is compact.

   • Now assume that $\phi \in L^1(ℝ)$. By Lemma 4.12, $f\in L^{\mathrm{\infty }}(ℝ)$. Therefore

     $|(\phi *f)(x)|$ $\le {\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}|\phi (x-y)||f(y)|𝑑y$

     $\le \underset{y\in ℝ}{sup}|f(y)|{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}|\phi (x-y)|𝑑y$

     $=\underset{y\in ℝ}{sup}|f(y)|{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}|\phi (z)|𝑑z$

     $={\parallel f\parallel }_{L^{\mathrm{\infty }}(ℝ)}{\parallel \phi \parallel }_{L^1(ℝ)}.$

     Therefore

     and so $\phi *f\in L^{\mathrm{\infty }}(ℝ)$, as required.

   • The convolution is commutative since

     $(\phi *f)(x)$ $={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}\phi (x-y)f(y)𝑑y$

     $={\displaystyle {\int }_{\mathrm{\infty }}^{-\mathrm{\infty }}}\phi (z)f(x-z)(-1)𝑑z$ $(z=x-y)$

     $={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}\phi (z)f(x-z)𝑑z$

     $=(f*\phi )(x)$

     as required.

8. The Poincaré inequality for functions that vanish on the boundary. Let $f\in C^1([a,b])$ satisfy $f(a)=f(b)=0$. Then

   $$f(x)=f(a)+{\int }_a^x{f}^{\prime }(y)𝑑y={\int }_a^x{f}^{\prime }(y)𝑑y$$

   since $f(a)=0$. Therefore

   $|f(x)|$ $=\left|{\displaystyle {\int }_a^x}{f}^{\prime }(y)𝑑y\right|$

   $=\left|{\displaystyle {\int }_a^x}1\cdot f^{\prime }(y)𝑑y\right|$

   $\le {\left|{\displaystyle {\int }_a^x}{1}^2𝑑y\right|}^{1/2}{\left|{\displaystyle {\int }_a^x}{|f^{\prime }(y)|}^2𝑑y\right|}^{1/2}$ (Cauchy-Schwarz)

   $\le {(x-a)}^{1/2}{\left({\displaystyle {\int }_a^b}{|f^{\prime }(y)|}^2𝑑y\right)}^{1/2}.$

   Squaring and integrating gives

   ${\displaystyle {\int }_a^b}{|f(x)|}^2𝑑x$ $\le {\displaystyle {\int }_a^b}(x-a){\displaystyle {\int }_a^b}{|f^{\prime }(y)|}^2𝑑y𝑑x$

   $={\displaystyle {\int }_a^b}(x-a)𝑑x{\displaystyle {\int }_a^b}{|f^{\prime }(y)|}^2𝑑y$

   $={{\displaystyle \frac{1}{2}}{(x-a)}^2|}_a^b{\displaystyle {\int }_a^b}{|f^{\prime }(y)|}^2𝑑y$

   $={\displaystyle \frac{1}{2}}{(b-a)}^2{\displaystyle {\int }_a^b}{|f^{\prime }(y)|}^2𝑑y.$

   This is the Poincaré inequality with $C=\frac{1}{2}{(b-a)}^2$.

9. The Poincaré inequality on unbounded domains.

   • For $n\in \mathbb{N}$, define $f_n:ℝ\to ℝ$ by

     (Exercise: Sketch $f_n$ to get a better understanding of the example.) Observe that

     $$f_n(-n-1)=f_n(n+1)=0,$$

     $$f_n(-n)=f_n(n)=1,$$

     $$f_n^{\prime }(-n-1)=f_n^{\prime }(-n)=f_n^{\prime }(n)=f_n^{\prime }(n+1)=0.$$

     Therefore $f_n\in C^1(ℝ)$. We also have $f_n\in L^2(ℝ)$ since

     We compute

     ${\parallel f_n^{\prime }\parallel }_{L^2(ℝ)}^2$ $={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{|f_n^{\prime }(x)|}^2𝑑x$

     $=2{\displaystyle {\int }_n^{n+1}}{\left[{\displaystyle \frac{d}{dx}}{(x-(n+1))}^2{(x-(n-1))}^2\right]}^2𝑑x$

     $=2{\displaystyle {\int }_n^{n+1}}{\left[2(x-(n+1)){(x-(n-1))}^2+2{(x-(n+1))}^2(x-(n-1))\right]}^2𝑑x$

     $=2{\displaystyle {\int }_0^1}{\left[2(y-1){(y+1)}^2+2{(y-1)}^2(y+1)\right]}^2𝑑y$ $(y=x-n)$

     which is independent of $n$. But

     ${\parallel f_n\parallel }_{L^2(ℝ)}^2={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{|f_n(x)|}^2𝑑x>{\displaystyle {\int }_{-n}^n}{|f_n(x)|}^2𝑑x=2n.$

     Therefore

     $${\parallel f_n^{\prime }\parallel }_{L^2(ℝ)}=\text{constant},{\parallel f_n\parallel }_{L^2(ℝ)}\stackrel{n\to \mathrm{\infty }}{⟶}\mathrm{\infty }$$

     as required. This means that, given any $C>0$, we can choose $N$ large enough so that

     $${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|f_N(x)|}^2𝑑x>C{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|f_N^{\prime }(x)|}^2𝑑x,$$

     which means that the Poincaré inequality on $ℝ$ does not hold. We constructed this counter example using spreading; the support of $f_n$ spreads as $n\to \mathrm{\infty }$ without changing the $L^2$–norm of $f_n^{\prime }$.

   • Let $\mathrm{\Omega }=(a,b)\times (-\mathrm{\infty },\mathrm{\infty })$. Let $f\in C^1(\overline{\mathrm{\Omega }})\cap L^2(\mathrm{\Omega })$ with $\nabla f\in L^2(\mathrm{\Omega })$ and with $f(a,y)=f(b,y)=0$ for all $y\in ℝ$. Then

     ${\displaystyle {\int }_{\mathrm{\Omega }}}{|f(𝒙)|}^2𝑑𝒙$ $={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}\left({\displaystyle {\int }_a^b}{|f(x,y)|}^2𝑑x\right)𝑑y$

     $\le {\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}\left(C{\displaystyle {\int }_a^b}{|f_x(x,y)|}^2𝑑x\right)𝑑y$ $\text{(Poincaré inequality in }x)$

     $\le C{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{\displaystyle {\int }_a^b}({|f_x(x,y)|}^2+{|f_y(x,y)|}^2)𝑑x𝑑y$

     $=C{\displaystyle {\int }_{\mathrm{\Omega }}}{|\nabla f(𝒙)|}^2𝑑𝒙$

     as required.

10. The Poincaré constant depends on the domain. There exits $C_1>0$ such that

    $${\int }_0^1{|f(x)|}^2𝑑x\le C_1{\int }_0^1{|f^{\prime }(x)|}^2𝑑x$$ (12)

    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

    ${\displaystyle {\int }_0^L}{|g(x)|}^2𝑑x$ $={\displaystyle {\int }_0^1}{|g(Ly)|}^{2}L𝑑y$ $(y=x/L)$

    $=L{\displaystyle {\int }_0^1}{|f(y)|}^2𝑑y$ $(f(y):=g(Ly))$

    $\le L{C}_1{\displaystyle {\int }_0^1}{|f^{\prime }(y)|}^2𝑑y$ $(\text{equation }(\mathit{\text{<a href="\#S0.E12" title="(12) ‣ item 10" class="ltx\_ref ltx\_font\_italic"><span class="ltx\_text ltx\_ref\_tag">12</span></a>}}))$

    $=L{C}_1{\displaystyle {\int }_0^1}{|L{g}^{\prime }(Ly)|}^2𝑑y$ $(f^{\prime }(y)=L{g}^{\prime }(Ly))$

    $=L^3{C}_1{\displaystyle {\int }_0^1}{|g^{\prime }(Ly)|}^2𝑑y$

    $=L^2{C}_1{\displaystyle {\int }_0^L}{|g^{\prime }(x)|}^2𝑑x$ $(y=x/L)$

    $=C_L{\displaystyle {\int }_0^L}{|g^{\prime }(x)|}^2𝑑x$

    with $C_L=L^2{C}_1$, as desired.

11. Eigenvalues of $\mathrm{-}\mathrm{\Delta }$: Can you hear the shape of a drum? Multiply the PDE $-\mathrm{\Delta }u=\lambda u$ by $\overline{u}$ (the complex conjugate of $u$) and integrate over $\mathrm{\Omega }$:

    $-{\displaystyle {\int }_{\mathrm{\Omega }}}\overline{u}\mathrm{\Delta }u𝑑𝒙=\lambda {\displaystyle {\int }_{\mathrm{\Omega }}}\overline{u}u𝑑𝒙\mathit{\hspace{1em}}⟺$ $-{\displaystyle {\int }_{\partial \mathrm{\Omega }}}\overline{u}\nabla u\cdot 𝒏dL+{\displaystyle {\int }_{\mathrm{\Omega }}}\nabla \overline{u}\cdot \nabla ud𝒙=\lambda {\displaystyle {\int }_{\mathrm{\Omega }}}{|u|}^2𝑑𝒙.$

    The boundary condition $u=0$ on $\partial \mathrm{\Omega }$ implies that $\overline{u}=0$ on $\partial \mathrm{\Omega }$ and so

    ${\displaystyle {\int }_{\mathrm{\Omega }}}\overline{\nabla u}\cdot \nabla ud𝒙=\lambda {\displaystyle {\int }_{\mathrm{\Omega }}}{|u|}^2𝑑𝒙$ $⟺\mathit{\hspace{1em}}{\displaystyle {\int }_{\mathrm{\Omega }}}{|\nabla u|}^2𝑑𝒙=\lambda {\displaystyle {\int }_{\mathrm{\Omega }}}{|u|}^2𝑑𝒙$

    $⟺\mathit{\hspace{1em}}\lambda ={\displaystyle \frac{{\displaystyle {\int }_{\mathrm{\Omega }}}{|\nabla u|}^2𝑑𝒙}{{\displaystyle {\int }_{\mathrm{\Omega }}}{|u|}^2𝑑𝒙}}>\mathrm{\hspace{0.33em}0}$

    as required.

12. The optimal Poincaré constant and eigenvalues of $\mathrm{-}\mathrm{\Delta }$.

    • Multiply the PDE $-\mathrm{\Delta }u=\lambda u$ by $u$ and integrate over $\mathrm{\Omega }$:

      $$-{\int }_{\mathrm{\Omega }}u\mathrm{\Delta }u𝑑𝒙=\lambda {\int }_{\mathrm{\Omega }}{u}^2𝑑𝒙\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}{\int }_{\mathrm{\Omega }}{|\nabla u|}^2𝑑𝒙=\lambda {\int }_{\mathrm{\Omega }}{u}^2𝑑𝒙$$

      since $u=0$ on $\partial \mathrm{\Omega }$. Rearranging gives

      $$\lambda =\frac{{\displaystyle {\int }_{\mathrm{\Omega }}}{|\nabla u|}^2𝑑𝒙}{{\displaystyle {\int }_{\mathrm{\Omega }}}{|u|}^2𝑑𝒙}.$$

    • Let $u\in C^2(\overline{\mathrm{\Omega }})\cap V$ minimise $E$. Let $\phi \in V$. Define $u_{\epsilon }=u+\epsilon \phi \in V$ and define $g(\epsilon )=E[u_{\epsilon }]$, $\epsilon \in ℝ$. Since $E$ is minimised by $u$, then $g$ is minimised by $0$. It follows that

      $0$ $=g^{\prime }(0)$

      $={{\displaystyle \frac{d}{d\epsilon }}|}_{\epsilon =0}E[u_{\epsilon }]$

      $={{\displaystyle \frac{d}{d\epsilon }}|}_{\epsilon =0}{\displaystyle \frac{{\displaystyle {\int }_{\mathrm{\Omega }}}{|\nabla u_{\epsilon }|}^2𝑑𝒙}{{\displaystyle {\int }_{\mathrm{\Omega }}}{|u_{\epsilon }|}^2𝑑𝒙}}$

      $={\displaystyle \frac{2{\displaystyle {\int }_{\mathrm{\Omega }}}\nabla u\cdot \nabla \phi d𝒙{\displaystyle {\int }_{\mathrm{\Omega }}}{|u|}^2𝑑𝒙-2{\displaystyle {\int }_{\mathrm{\Omega }}}{|\nabla u|}^2𝑑𝒙{\displaystyle {\int }_{\mathrm{\Omega }}}u\phi 𝑑𝒙}{{\left({\displaystyle {\int }_{\mathrm{\Omega }}}{|u|}^2𝑑𝒙\right)}^2}}.$

      The numerator must be zero. Rearranging gives

      $${\int }_{\mathrm{\Omega }}\nabla u\cdot \nabla \phi d𝒙=\left(\frac{{\displaystyle {\int }_{\mathrm{\Omega }}}{|\nabla u|}^2𝑑𝒙}{{\displaystyle {\int }_{\mathrm{\Omega }}}{|u|}^2𝑑𝒙}\right){\int }_{\mathrm{\Omega }}u\phi 𝑑𝒙.$$

      Integrating by parts gives

      $$-{\int }_{\mathrm{\Omega }}\mathrm{\Delta }u\phi 𝑑𝒙=\left(\frac{{\displaystyle {\int }_{\mathrm{\Omega }}}{|\nabla u|}^2𝑑𝒙}{{\displaystyle {\int }_{\mathrm{\Omega }}}{|u|}^2𝑑𝒙}\right){\int }_{\mathrm{\Omega }}u\phi 𝑑𝒙.$$

      Since this holds for all $\phi \in V$, the Fundamental Lemma of the Calculus of Variations implies that

      $$-\mathrm{\Delta }u=\left(\frac{{\displaystyle {\int }_{\mathrm{\Omega }}}{|\nabla u|}^2𝑑𝒙}{{\displaystyle {\int }_{\mathrm{\Omega }}}{|u|}^2𝑑𝒙}\right)u\mathit{\hspace{1em}}\text{in }\mathrm{\Omega }.$$

      If we define

      $$\lambda =\frac{{\displaystyle {\int }_{\mathrm{\Omega }}}{|\nabla u|}^2𝑑𝒙}{{\displaystyle {\int }_{\mathrm{\Omega }}}{|u|}^2𝑑𝒙},$$

      then

      $$-\mathrm{\Delta }u=\lambda u\mathit{\hspace{1em}}\text{in }\mathrm{\Omega }.$$

      In other words, $u$ is an eigenfunction of $-\mathrm{\Delta }$. By definition

      $$E[u]=\frac{{\displaystyle {\int }_{\mathrm{\Omega }}}{|\nabla u|}^2𝑑𝒙}{{\displaystyle {\int }_{\mathrm{\Omega }}}{|u|}^2𝑑𝒙}=\lambda .$$

      Since $u$ minimises $E$, then $\lambda$ must be the smallest eigenvalue of $-\mathrm{\Delta }$ on $V$, i.e., $\lambda ={\lambda }_1$, otherwise we obtain a contradiction. Therefore $E[u]={\lambda }_1$, as required.

    • Let $C>0$ satisfy

      $${\parallel f\parallel }_{L^2(\mathrm{\Omega })}\le C{\parallel \nabla f\parallel }_{L^2(\mathrm{\Omega })}$$

      for all $f\in C^1(\overline{\mathrm{\Omega }})$ with $f=0$ on $\partial \mathrm{\Omega }$. Then

      $$\frac{1}{C}\le \frac{{\parallel \nabla f\parallel }_{L^2(\mathrm{\Omega })}}{{\parallel f\parallel }_{L^2(\mathrm{\Omega })}}$$

      for all $f\in V$ and so

      $$\frac{1}{C}\le \underset{f\in V}{inf}\frac{{\parallel \nabla f\parallel }_{L^2(\mathrm{\Omega })}}{{\parallel f\parallel }_{L^2(\mathrm{\Omega })}}.$$

      The smallest value of $C$ satisfying this inequality is $C=C_{\text{P}}$ where

      $$\frac{1}{C_{\text{P}}}=\underset{f\in V}{inf}\frac{{\parallel \nabla f\parallel }_{L^2(\mathrm{\Omega })}}{{\parallel f\parallel }_{L^2(\mathrm{\Omega })}}.$$

    • Combining parts (ii) and (iii) gives

      $$\frac{1}{C_{\text{P}}}=\underset{f\in V}{inf}\frac{{\parallel \nabla f\parallel }_{L^2(\mathrm{\Omega })}}{{\parallel f\parallel }_{L^2(\mathrm{\Omega })}}=\underset{f\in V}{inf}E{[v]}^{1/2}={\left(\underset{f\in V}{inf}E[v]\right)}^{1/2}=\sqrt{{\lambda }_1}.$$

      Therefore

      $$C_{\text{P}}=\frac{1}{\sqrt{{\lambda }_1}}$$

      as desired.

    • If $\mathrm{\Omega }=(0,2\pi )$, then the corresponding eigenvalue problem is

      $$-u^{\prime \prime }=\lambda u\mathit{\hspace{1em}}\text{in }(0,2\pi ),u(0)=u(2\pi )=0.$$

      The eigenfunctions are $u_n(x)=\sin \left(\frac{nx}{2}\right)$ (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 $C_{\text{P}}=1/\sqrt{1/4}=2$. In Q8 we obtained the Poincaré constant $(b-a)/\sqrt{2}=\sqrt{2}\pi$, which is obviously much bigger than the optimal constant $C_{\text{P}}=2$.

13. Uniqueness for Poisson’s equation with Robin boundary conditions. Let $u_1$ and $u_2$ be solutions of

    $-\mathrm{\Delta }u=f$ $\text{in }\mathrm{\Omega },$

    $\nabla u\cdot 𝒏+\alpha u=g$ $\text{on }\partial \mathrm{\Omega }.$

    Let $w=u_1-u_2$. Since the PDE is linear, subtracting the equations satisfied by $u_1$ and $u_2$ gives

    $-\mathrm{\Delta }w=0$ $\text{in }\mathrm{\Omega },$

    $\nabla w\cdot 𝒏+\alpha w=0$ $\text{on }\partial \mathrm{\Omega }.$

    Multiply $-\mathrm{\Delta }w=0$ by $w$ and integrate by parts over $\mathrm{\Omega }$:

    $-{\displaystyle {\int }_{\mathrm{\Omega }}}w\mathrm{\Delta }w𝑑𝒙=0$ $⟺\mathit{\hspace{1em}}-{\displaystyle {\int }_{\partial \mathrm{\Omega }}}w\nabla w\cdot 𝒏dS+{\displaystyle {\int }_{\mathrm{\Omega }}}{|\nabla w|}^2𝑑𝒙=0$

    $⟺\mathit{\hspace{1em}}\alpha {\displaystyle {\int }_{\partial \mathrm{\Omega }}}{w}^2𝑑S+{\displaystyle {\int }_{\mathrm{\Omega }}}{|\nabla w|}^2𝑑𝒙=0$

    since $\nabla w\cdot 𝒏=-\alpha w$ on $\partial \mathrm{\Omega }$. But $\alpha >0$. Therefore

    $${\int }_{\partial \mathrm{\Omega }}{w}^2𝑑S=0,{\int }_{\mathrm{\Omega }}{|\nabla w|}^2𝑑𝒙=0.$$

    The second equation implies that $\nabla w=\text{𝟎}$ and hence $w=$ constant (or at least constant on each connected component of $\mathrm{\Omega }$). The first equation implies that this constant must be zero. Therefore $w=0$ and $u_1=u_2$, as required.

14. Uniqueness for a more general elliptic problem. Consider the linear, second-order, elliptic PDE

    $-\mathrm{div}(A\nabla u)+𝒃\cdot \nabla u+cu=f$ $\text{in }\mathrm{\Omega },$ (13)

    $u=g$ $\text{on }\partial \mathrm{\Omega }.$

    (i) Suppose that $u_1,u_2\in C^2(\overline{\mathrm{\Omega }})$ satisfy (13). Let $w=u_1-u_2$. Since the PDE is linear, subtracting the equations satisfied by $u_1$ and $u_2$ gives

    $-\mathrm{div}(A\nabla w)+𝒃\cdot \nabla w+cw=0$ $\text{in }\mathrm{\Omega },$ (14)

    $w=0$ $\text{on }\partial \mathrm{\Omega }.$

    Clearly $w=0$ satisfies (14). We want to show that it is the only solution. Multiply the PDE for $w$ by $w$ and integrate over $\mathrm{\Omega }$:

    $0$ $={\displaystyle {\int }_{\mathrm{\Omega }}}w(-\mathrm{div}(A\nabla w)+𝒃\cdot \nabla w+cw)𝑑𝒙$

    $=-{\displaystyle {\int }_{\mathrm{\Omega }}}w\mathrm{div}(A\nabla w)𝑑𝒙+{\displaystyle {\int }_{\mathrm{\Omega }}}w𝒃\cdot \nabla wd𝒙+{\displaystyle {\int }_{\mathrm{\Omega }}}c{w}^2𝑑𝒙$

    $=-{\displaystyle {\int }_{\partial \mathrm{\Omega }}}w(A\nabla w)\cdot 𝒏𝑑S+{\displaystyle {\int }_{\mathrm{\Omega }}}\nabla w\cdot (A\nabla w)𝑑𝒙+{\displaystyle {\int }_{\mathrm{\Omega }}}w𝒃\cdot \nabla wd𝒙+{\displaystyle {\int }_{\mathrm{\Omega }}}c{w}^2𝑑𝒙$

    $={\displaystyle {\int }_{\mathrm{\Omega }}}\nabla w\cdot (A\nabla w)𝑑𝒙+{\displaystyle {\int }_{\mathrm{\Omega }}}w𝒃\cdot \nabla wd𝒙+{\displaystyle {\int }_{\mathrm{\Omega }}}c{w}^2𝑑𝒙$ (15)

    since $w=0$ on $\partial \mathrm{\Omega }$. Observe that

    $${\int }_{\mathrm{\Omega }}\nabla w\cdot (A\nabla w)𝑑𝒙={\int }_{\mathrm{\Omega }}{(\nabla w)}^{\mathrm{T}}A\nabla wd𝒙\ge \alpha {\int }_{\mathrm{\Omega }}{|\nabla w|}^2𝑑𝒙$$ (16)

    by the assumption that $A$ is uniformly positive definite (take $𝒚=\nabla w$ in ${𝒚}^{\mathrm{T}}A(𝒙)𝒚\ge \alpha {|𝒚|}^2$). Integrating by parts gives

    ${\displaystyle {\int }_{\mathrm{\Omega }}}w𝒃\cdot \nabla wd𝒙$ $={\displaystyle {\int }_{\partial \mathrm{\Omega }}}{w}^2𝒃\cdot 𝒏𝑑S-{\displaystyle {\int }_{\mathrm{\Omega }}}w\mathrm{div}(w𝒃)𝑑𝒙$

    $=-{\displaystyle {\int }_{\mathrm{\Omega }}}w\mathrm{div}(w𝒃)𝑑𝒙$ $(w=0\text{ on }\partial \mathrm{\Omega })$

    $=-{\displaystyle {\int }_{\mathrm{\Omega }}}w(\nabla w\cdot 𝒃+w\mathrm{div}𝒃)𝑑𝒙$ (product rule)

    $=-{\displaystyle {\int }_{\mathrm{\Omega }}}w𝒃\cdot \nabla wd𝒙$

    by the assumption that $\mathrm{div}𝒃=0$. Therefore

    $${\int }_{\mathrm{\Omega }}w𝒃\cdot \nabla wd𝒙=-{\int }_{\mathrm{\Omega }}w𝒃\cdot \nabla wd𝒙\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}{\int }_{\mathrm{\Omega }}w𝒃\cdot \nabla wd𝒙=0.$$ (17)

    Combining (15), (16), (17) yields

    $$\alpha {\int }_{\mathrm{\Omega }}{|\nabla w|}^2𝑑𝒙+{\int }_{\mathrm{\Omega }}c{w}^2𝑑𝒙\le 0.$$

    But $c\ge 0$ by assumption. Therefore

    $$\alpha {\int }_{\mathrm{\Omega }}{|\nabla w|}^2𝑑𝒙=0$$

    and so $\nabla w=\mathrm{𝟎}$ in $\mathrm{\Omega }$. Hence $w$ is constant (or at least constant on each connected component of $\mathrm{\Omega }$). But $w=0$ on $\partial \mathrm{\Omega }$. 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 $L^2(\mathrm{\Omega })$ as $n\to +\mathrm{\infty }$. Taking the two PDEs, subtracting them and multiplying the resulting PDE by $w_n$, we obtain

    $0$ $={\displaystyle {\int }_{\mathrm{\Omega }}}{w}_n(-\mathrm{div}(A_n\nabla u_n)+\mathrm{div}(A\nabla u)+𝒃\cdot \nabla w_n+c{w}_n)𝑑𝒙.$ (18)

    Proceeding exactly as in (i), we find

    $${\int }_{\mathrm{\Omega }}{w}_n𝒃\cdot \nabla w_{n}d𝒙=0.$$

    Moreover, we compute (using integration by parts, since $w_n=0$ on $\partial \mathrm{\Omega }$)

    ${\displaystyle {\int }_{\mathrm{\Omega }}}{w}_n[-\mathrm{div}(A_n\nabla u_n)+\mathrm{div}(A\nabla u)]𝑑𝒙$

    $={\displaystyle {\int }_{\mathrm{\Omega }}}{w}_n[-\mathrm{div}(A_n\nabla u_n)+\mathrm{div}(A_n\nabla u)-\mathrm{div}(A_n\nabla u)+\mathrm{div}(A\nabla u)]𝑑𝒙$

    $={\displaystyle {\int }_{\mathrm{\Omega }}}{w}_n[-\mathrm{div}(A_n(\nabla u_n-\nabla u))-\mathrm{div}((A_n-A)\nabla u)]𝑑𝒙$

    $={\displaystyle {\int }_{\mathrm{\Omega }}}[\nabla w_n\cdot (A_n\nabla w_n)+\nabla w_n\cdot ((A_n-A)\nabla u)]𝑑𝒙$

    $\ge {\displaystyle {\int }_{\mathrm{\Omega }}}[\alpha {|\nabla w_n|}^2+\nabla w_n\cdot ((A_n-A)\nabla u)]𝑑𝒙$

    So, all these arguments yield

    $${\int }_{\mathrm{\Omega }}[\alpha {|\nabla w_n|}^2+\nabla w_n\cdot ((A_n-A)\nabla u)+c{w}_n^2]𝑑𝒙\le 0.$$

    Now, for any $\epsilon >0$, Young’s inequality yields

    $${\int }_{\mathrm{\Omega }}\nabla w_n\cdot ((A_n-A)\nabla u)𝑑𝒙=-\frac{\epsilon }{2}{\int }_{\mathrm{\Omega }}{|\nabla w_n|}^2𝑑𝒙-{\int }_{\mathrm{\Omega }}\frac{1}{2\epsilon }{|A_n-A|}^2{|\nabla u|}^2𝑑𝒙.$$

    By setting $\epsilon :=\alpha$, the previous two identities imply

    $${\int }_{\mathrm{\Omega }}\left[\frac{\alpha }{2}{|\nabla w_n|}^2+c{w}_n^2\right]𝑑𝒙\le \frac{1}{2\alpha }{\parallel A_n-A\parallel }_{L^{\mathrm{\infty }}}^2{\int }_{\mathrm{\Omega }}{|\nabla u|}^2𝑑𝒙.$$

    And by the non-negative property of $c$, one has

    $${\int }_{\mathrm{\Omega }}\frac{\alpha }{2}{|\nabla w_n|}^2𝑑𝒙\le \frac{1}{2\alpha }{\parallel A_n-A\parallel }_{L^{\mathrm{\infty }}}^2{\int }_{\mathrm{\Omega }}{|\nabla u|}^2𝑑𝒙.$$

    We conclude by the facts that ${\int }_{\mathrm{\Omega }}{|\nabla u|}^2𝑑𝒙$ is bounded and ${\parallel A_n-A\parallel }_{L^{\mathrm{\infty }}}\to 0$, as $n\to +\mathrm{\infty }$.

15. Uniqueness for a degenerate diffusion equation. Clearly $u=\pi$ satisfies

    $\mathrm{\Delta }{u}^m=0$ $\text{in }\mathrm{\Omega },$

    $u=\pi$ $\text{on }\partial \mathrm{\Omega }.$

    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

    $$0=(v-u)(\mathrm{\Delta }{v}^m-\mathrm{\Delta }{u}^m)=(v-\pi )(\mathrm{\Delta }{v}^m-\mathrm{\Delta }{\pi }^m)=(v-\pi )\mathrm{\Delta }{v}^m.$$

    Now integrate over $\mathrm{\Omega }$:

    $0$ $={\displaystyle {\int }_{\mathrm{\Omega }}}(v-\pi )\mathrm{\Delta }{v}^m𝑑𝒙$

    $={\displaystyle {\int }_{\mathrm{\Omega }}}(v-\pi )\mathrm{div}\nabla (v^m)d𝒙$ $(\mathrm{\Delta }=\mathrm{div}\nabla )$

    $={\displaystyle {\int }_{\mathrm{\Omega }}}(v-\pi )\mathrm{div}(m{v}^{m-1}\nabla v)𝑑𝒙$ (Chain Rule)

    $={\displaystyle {\int }_{\partial \mathrm{\Omega }}}\underset{=0}{\underbrace{(v-\pi )}}m{v}^{m-1}\nabla v\cdot 𝒏dS-{\displaystyle {\int }_{\mathrm{\Omega }}}\underset{=\nabla v}{\underbrace{\nabla (v-\pi )}}\cdot m{v}^{m-1}\nabla vd𝒙$ (Integration by parts)

    $=-{\displaystyle {\int }_{\mathrm{\Omega }}}m{v}^{m-1}{|\nabla v|}^2𝑑𝒙.$

    Therefore

    $${\int }_{\mathrm{\Omega }}m{v}^{m-1}{|\nabla v|}^2𝑑𝒙=0.$$

    But $v>0$, by assumption. Hence $\nabla v=\mathrm{𝟎}$ in $\mathrm{\Omega }$ and so $v$ is constant in $\mathrm{\Omega }$. Since $v=\pi$ on $\partial \mathrm{\Omega }$, we conclude that $v=\pi$ everywhere, as required.

16. The $H_{\mathrm{0}}^{\mathrm{1}}$ and $H^{\mathrm{1}}$ norms.

    • We need to check that $\parallel \cdot {\parallel }_{L^2([a,b])}$ 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 ${\parallel f\parallel }_{L^2([a,b])}\ge 0$. Suppose that ${\parallel f\parallel }_{L^2([a,b])}=0$ and assume for contradiction that $f\ne 0$. Since $f$ is continuous, then there exists $x_0\in (a,b)$, $h>0$ and $\epsilon >0$ such that $|f(x)|>\epsilon$ for all $x\in (x_0-h,x_0+h)$. Therefore

      $${\parallel f\parallel }_{L^2([a,b])}^2\ge {\int }_{x_0-h}^{x_0+h}{|f(x)|}^2𝑑x\ge {\int }_{x_0-h}^{x_0+h}{\epsilon }^2𝑑x=2h{\epsilon }^2>0,$$

      which is a contradiction. Second we check that $\parallel \cdot {\parallel }_{L^2([a,b])}$ is $1$–homogeneous. Let $\lambda \in ℝ$. Then

      $${\parallel \lambda f\parallel }_{L^2([a,b])}={\left({\int }_a^b{|\lambda f(x)|}^2\right)}^{1/2}=|\lambda |{\left({\int }_a^b{|f(x)|}^2\right)}^{1/2}=|\lambda |{\parallel f\parallel }_{L^2([a,b])}$$

      as required. Finally, we prove the triangle inequality. Let $f,g\in C([a,b])$. Then

      ${\parallel f+g\parallel }_{L^2([a,b])}^2$ $={\displaystyle {\int }_a^b}{(f(x)+g(x))}^2𝑑x$

      $={\displaystyle {\int }_a^b}f{(x)}^2𝑑x+2{\displaystyle {\int }_a^b}f(x)g(x)𝑑x+{\displaystyle {\int }_a^b}g{(x)}^2𝑑x$

      $\le {\displaystyle {\int }_a^b}f{(x)}^2𝑑x+2{\left({\displaystyle {\int }_a^b}f{(x)}^2𝑑x\right)}^{1/2}{\left({\displaystyle {\int }_a^b}g{(x)}^2𝑑x\right)}^{1/2}+{\displaystyle {\int }_a^b}g{(x)}^2𝑑x$

      $\mathrm{\hspace{1em}\hspace{1em}}\text{(by the Cauchy-Schwarz inequality)}$

      $={\parallel f\parallel }_{L^2([a,b])}^2+2{\parallel f\parallel }_{L^2([a,b])}{\parallel g\parallel }_{L^2([a,b])}+{\parallel g\parallel }_{L^2([a,b])}^2$

      $={\left({\parallel f\parallel }_{L^2([a,b])}+{\parallel g\parallel }_{L^2([a,b])}\right)}^2.$

      Taking the square root gives the triangle inequality.

      Remark: An alternative proof is to prove that the function $(\cdot ,\cdot ):C([a,b])\times C([a,b])\to ℝ$,

      $$(f,g)={\int }_a^{b}f(x)g(x)𝑑x,$$

      is an inner product on $C([a,b])$. It then follows that $\parallel f\parallel :=\sqrt{(f,f)}$ 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 $\parallel \cdot {\parallel }_{L^2([a,b])}$.

      Remark: The Cauchy-Schwarz inequality can be proved by considering the quadratic polynomial

      $$t\mapsto p(t):={\parallel f+tg\parallel }_{L^2([a,b])}^2.$$

      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 ${(\cdot ,\cdot )}_{H^1}:C^1([a,b])\times C^1([a,b])\to ℝ$ defined by

      $${(f,g)}_{H^1}:={\int }_a^{b}f(x)g(x)𝑑x+{\int }_a^b{f}^{\prime }(x)g^{\prime }(x)𝑑x$$

      is an inner product on $C^1([a,b])$. It then follows that

      $${\parallel f\parallel }_{H^1([a,b])}=\sqrt{{(f,f)}_{H^1}}$$

      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 ${\parallel f\parallel }_{H^1([a,b])}=0$ and in particular ${\parallel f\parallel }_{L^2([a,b])}=0$. Therefore $f=0$ by part (i).

    • This is similar to part (ii). We will prove that the function ${(\cdot ,\cdot )}_{H_0^1}:V\times V\to ℝ$ defined by

      $${(f,g)}_{H_0^1}:={\int }_a^b{f}^{\prime }(x)g^{\prime }(x)𝑑x$$

      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 ${\parallel f\parallel }_{H_0^1([a,b])}=0$ and in particular ${\parallel f^{\prime }\parallel }_{L^2([a,b])}=0$. Therefore $f^{\prime }=0$ 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

      $$c{\parallel f\parallel }_{H_0^1([a,b])}\le {\parallel f\parallel }_{H^1([a,b])}\le C{\parallel f\parallel }_{H_0^1([a,b])}\mathit{\hspace{1em}}\forall f\in V.$$

      Let $f\in V$. We have

      $${\parallel f\parallel }_{H_0^1([a,b])}={\parallel f^{\prime }\parallel }_{L^2([a,b])}\le {\left({\parallel f\parallel }_{L^2([a,b])}^2+{\parallel f^{\prime }\parallel }_{L^2([a,b])}^2\right)}^{1/2}={\parallel f\parallel }_{H^1([a,b])}.$$

      Therefore $c=1$. On the other hand,

      $${\parallel f\parallel }_{H^1([a,b])}^2={\parallel f\parallel }_{L^2([a,b])}^2+{\parallel f^{\prime }\parallel }_{L^2([a,b])}^2\le C_{\mathrm{P}}^2{\parallel f^{\prime }\parallel }_{L^2([a,b])}^2+{\parallel f^{\prime }\parallel }_{L^2([a,b])}^2$$

      where $C_{\mathrm{P}}$ is the Poincaré constant. Therefore we can take $C={(C_{\mathrm{P}}^2+1)}^{1/2}$.

17. Continuous dependence. Let $u\in C^2(\overline{\mathrm{\Omega }})$ satisfy

    $-\mathrm{div}(A\nabla u)+cu=f$ $\text{in }\mathrm{\Omega },$

    $u=0$ $\text{on }\partial \mathrm{\Omega }.$

    Multiplying the PDE by $u$ and integrating over $\mathrm{\Omega }$ gives

    ${\displaystyle {\int }_{\mathrm{\Omega }}}fu𝑑𝒙$ $={\displaystyle {\int }_{\mathrm{\Omega }}}u(-\mathrm{div}(A\nabla u)+cu)𝑑𝒙$

    $=-{\displaystyle {\int }_{\mathrm{\Omega }}}u\mathrm{div}(A\nabla u)𝑑𝒙+c{\displaystyle {\int }_{\mathrm{\Omega }}}{u}^2𝑑𝒙$

    $=-{\displaystyle {\int }_{\partial \mathrm{\Omega }}}u(A\nabla u)\cdot 𝒏𝑑S+{\displaystyle {\int }_{\mathrm{\Omega }}}\nabla u\cdot (A\nabla u)𝑑𝒙+c{\displaystyle {\int }_{\mathrm{\Omega }}}{u}^2𝑑𝒙$

    $={\displaystyle {\int }_{\mathrm{\Omega }}}{(\nabla u)}^{\mathrm{T}}A\nabla ud𝒙+c{\displaystyle {\int }_{\mathrm{\Omega }}}{u}^2𝑑𝒙$ $(u=0\text{ on }\partial \mathrm{\Omega })$

    $\ge \alpha {\displaystyle {\int }_{\mathrm{\Omega }}}{|\nabla u|}^2𝑑𝒙+c{\displaystyle {\int }_{\mathrm{\Omega }}}{u}^2𝑑𝒙$ $(A\text{ is uniformly positive definite)}$

    $\ge \min \{\alpha ,c\}\left({\displaystyle {\int }_{\mathrm{\Omega }}}{|\nabla u|}^2𝑑𝒙+{\displaystyle {\int }_{\mathrm{\Omega }}}{u}^2𝑑𝒙\right)$

    $=\min \{\alpha ,c\}{\parallel u\parallel }_{H^1(\mathrm{\Omega })}^2$

    by definition of the $H^1$–norm. Therefore

    $$\min \{\alpha ,c\}{\parallel u\parallel }_{H^1(\mathrm{\Omega })}^2\le {\int }_{\mathrm{\Omega }}fu𝑑𝒙\le {\parallel f\parallel }_{L^2(\mathrm{\Omega })}{\parallel u\parallel }_{L^2(\mathrm{\Omega })}\le {\parallel f\parallel }_{L^2(\mathrm{\Omega })}{\parallel u\parallel }_{H^1(\mathrm{\Omega })}$$

    where we have used the Cauchy-Schwarz inequality and the fact that ${\parallel v\parallel }_{L^2(\mathrm{\Omega })}\le {\parallel v\parallel }_{H^1(\mathrm{\Omega })}$ for all $v\in C^1(\overline{\mathrm{\Omega }})$. Cancelling one power of ${\parallel u\parallel }_{H^1(\mathrm{\Omega })}$ from both sides gives the desired result:

    $${\parallel u\parallel }_{H^1(\mathrm{\Omega })}\le C{\parallel f\parallel }_{L^2(\mathrm{\Omega })}$$

    with $C=1/\min \{\alpha ,c\}$.

    Remark: Note that this estimate degenerates as $c$ tends to $0$ ($C\to +\mathrm{\infty }$ as $c\to 0$). If $c=0$ or $c$ is small then a better estimate can be obtained using the Poincaré inequality: As above

    $${\int }_{\mathrm{\Omega }}fu𝑑𝒙\ge \alpha {\int }_{\mathrm{\Omega }}{|\nabla u|}^2𝑑𝒙+c{\int }_{\mathrm{\Omega }}{u}^2𝑑𝒙\ge \alpha {\int }_{\mathrm{\Omega }}{|\nabla u|}^2𝑑𝒙=\alpha {\parallel u\parallel }_{H_0^1(\mathrm{\Omega })}^2.$$

    Therefore

    $$\alpha {\parallel u\parallel }_{H_0^1(\mathrm{\Omega })}^2\le {\int }_{\mathrm{\Omega }}fu𝑑𝒙\le {\parallel f\parallel }_{L^2(\mathrm{\Omega })}{\parallel u\parallel }_{L^2(\mathrm{\Omega })}\le C_{\mathrm{P}}{\parallel f\parallel }_{L^2(\mathrm{\Omega })}{\parallel \nabla u\parallel }_{L^2(\mathrm{\Omega })}=C_{\mathrm{P}}{\parallel f\parallel }_{L^2(\mathrm{\Omega })}{\parallel u\parallel }_{H_0^1(\mathrm{\Omega })}$$

    where $C_{\mathrm{P}}(\mathrm{\Omega })$ is the Poincaré constant. Cancelling one power of ${\parallel u\parallel }_{H_0^1(\mathrm{\Omega })}$ from both sides gives

    $${\parallel u\parallel }_{H_0^1(\mathrm{\Omega })}\le C{\parallel f\parallel }_{L^2(\mathrm{\Omega })}$$

    with $C=C_{\mathrm{P}}/\alpha$.

18. Continuous dependence with a first-order term.

    • Let $u\in C^2(\overline{\mathrm{\Omega }})$ satisfy

      $-k\mathrm{\Delta }u+𝒃\cdot \nabla u+cu=f$ $\text{in }\mathrm{\Omega },$ (19)

      $u=0$ $\text{on }\partial \mathrm{\Omega }.$

      Multiply the PDE by $u$ and integrate over $\mathrm{\Omega }$:

      $-k{\displaystyle {\int }_{\mathrm{\Omega }}}u\mathrm{\Delta }u𝑑𝒙+{\displaystyle {\int }_{\mathrm{\Omega }}}u𝒃\cdot \nabla ud𝒙+{\displaystyle {\int }_{\mathrm{\Omega }}}c{u}^2𝑑𝒙={\displaystyle {\int }_{\mathrm{\Omega }}}fu𝑑𝒙$

      $⟺\mathit{\hspace{1em}}-k\left[{\displaystyle {\int }_{\partial \mathrm{\Omega }}}u\nabla u\cdot 𝒏dS-{\displaystyle {\int }_{\mathrm{\Omega }}}{|\nabla u|}^2𝑑𝒙\right]+{\displaystyle {\int }_{\mathrm{\Omega }}}u𝒃\cdot \nabla ud𝒙+{\displaystyle {\int }_{\mathrm{\Omega }}}c{u}^2𝑑𝒙={\displaystyle {\int }_{\mathrm{\Omega }}}fu𝑑𝒙$

      $⟺\mathit{\hspace{1em}}k{\parallel \nabla u\parallel }_{L^2(\mathrm{\Omega })}^2+{\displaystyle {\int }_{\mathrm{\Omega }}}(𝒃\cdot \nabla u)u𝑑𝒙+c{\parallel u\parallel }_{L^2(\mathrm{\Omega })}^2={\displaystyle {\int }_{\mathrm{\Omega }}}fu𝑑𝒙\le {\parallel f\parallel }_{L^2(\mathrm{\Omega })}{\parallel u\parallel }_{L^2(\mathrm{\Omega })}$

      by the Cauchy-Schwarz inequality.

    • Let $\epsilon >0$. Then

      $\left|{\displaystyle {\int }_{\mathrm{\Omega }}}(𝒃\cdot \nabla u)u𝑑𝒙\right|$ $\le {\parallel 𝒃\parallel }_{L^{\mathrm{\infty }}(\mathrm{\Omega })}{\displaystyle {\int }_{\mathrm{\Omega }}}|\nabla u||u|𝑑𝒙$

      $\le {\parallel 𝒃\parallel }_{L^{\mathrm{\infty }}(\mathrm{\Omega })}{\parallel \nabla u\parallel }_{L^2(\mathrm{\Omega })}{\parallel u\parallel }_{L^2(\mathrm{\Omega })}$ $\mathrm{\hspace{1em}}\text{(Cauchy-Schwarz)}$

      $={\parallel 𝒃\parallel }_{L^{\mathrm{\infty }}(\mathrm{\Omega })}\left(\sqrt{2\epsilon }{\parallel \nabla u\parallel }_{L^2(\mathrm{\Omega })}\right)\left({\displaystyle \frac{1}{\sqrt{2\epsilon }}}{\parallel u\parallel }_{L^2(\mathrm{\Omega })}\right)$

      $\le {\parallel 𝒃\parallel }_{L^{\mathrm{\infty }}(\mathrm{\Omega })}\left(\epsilon {\parallel \nabla u\parallel }_{L^2(\mathrm{\Omega })}^2+{\displaystyle \frac{1}{4\epsilon }}{\parallel u\parallel }_{L^2(\mathrm{\Omega })}^2\right)$

      by the Young inequality.

    • Combining parts (a) and (b) gives

      ${\parallel f\parallel }_{L^2(\mathrm{\Omega })}{\parallel u\parallel }_{L^2(\mathrm{\Omega })}$ $\ge k{\parallel \nabla u\parallel }_{L^2(\mathrm{\Omega })}^2+{\displaystyle {\int }_{\mathrm{\Omega }}}(𝒃\cdot \nabla u)u𝑑𝒙+c{\parallel u\parallel }_{L^2(\mathrm{\Omega })}^2$

      $\ge k{\parallel \nabla u\parallel }_{L^2(\mathrm{\Omega })}^2-\left|{\displaystyle {\int }_{\mathrm{\Omega }}}(𝒃\cdot \nabla u)u𝑑𝒙\right|+c{\parallel u\parallel }_{L^2(\mathrm{\Omega })}^2$

      $\ge k{\parallel \nabla u\parallel }_{L^2(\mathrm{\Omega })}^2-{\parallel 𝒃\parallel }_{L^{\mathrm{\infty }}(\mathrm{\Omega })}\left(\epsilon {\parallel \nabla u\parallel }_{L^2(\mathrm{\Omega })}^2+{\displaystyle \frac{1}{4\epsilon }}{\parallel u\parallel }_{L^2(\mathrm{\Omega })}^2\right)+c{\parallel u\parallel }_{L^2(\mathrm{\Omega })}^2$

      $=\left(k-\epsilon {\parallel 𝒃\parallel }_{L^{\mathrm{\infty }}(\mathrm{\Omega })}\right){\parallel \nabla u\parallel }_{L^2(\mathrm{\Omega })}^2+\left(c-{\displaystyle \frac{{\parallel 𝒃\parallel }_{L^{\mathrm{\infty }}(\mathrm{\Omega })}}{4\epsilon }}\right){\parallel u\parallel }_{L^2(\mathrm{\Omega })}^2.$

    • Let $\epsilon >0$ satisfy $k-\epsilon {\parallel 𝒃\parallel }_{L^{\mathrm{\infty }}(\mathrm{\Omega })}>0$, i.e., let

      (20)

      Let

      $$c_0=\frac{{\parallel 𝒃\parallel }_{L^{\mathrm{\infty }}(\mathrm{\Omega })}}{4\epsilon }.$$

      If $c>c_0$, then

      $$c-\frac{{\parallel 𝒃\parallel }_{L^{\mathrm{\infty }}(\mathrm{\Omega })}}{4\epsilon }>0.$$

      Therefore if $c>c_0$ and $\epsilon$ satisfies (20), then

      $$k-\epsilon {\parallel 𝒃\parallel }_{L^{\mathrm{\infty }}(\mathrm{\Omega })}>0,c-\frac{{\parallel 𝒃\parallel }_{L^{\mathrm{\infty }}(\mathrm{\Omega })}}{4\epsilon }>0$$

      and so by part (c)

      ${\parallel f\parallel }_{L^2(\mathrm{\Omega })}{\parallel u\parallel }_{H^1(\mathrm{\Omega })}$ $\ge {\parallel f\parallel }_{L^2(\mathrm{\Omega })}{\parallel u\parallel }_{L^2(\mathrm{\Omega })}$

      $\ge \min \{k-\epsilon {\parallel 𝒃\parallel }_{L^{\mathrm{\infty }}(\mathrm{\Omega })},c-{\displaystyle \frac{{\parallel 𝒃\parallel }_{L^{\mathrm{\infty }}(\mathrm{\Omega })}}{4\epsilon }}\}\left({\parallel \nabla u\parallel }_{L^2(\mathrm{\Omega })}^2+{\parallel u\parallel }_{L^2(\mathrm{\Omega })}^2\right)$

      $=\min \{k-\epsilon {\parallel 𝒃\parallel }_{L^{\mathrm{\infty }}(\mathrm{\Omega })},c-{\displaystyle \frac{{\parallel 𝒃\parallel }_{L^{\mathrm{\infty }}(\mathrm{\Omega })}}{4\epsilon }}\}{\parallel u\parallel }_{H^1(\mathrm{\Omega })}^2.$

      Therefore if $c>c_0$ and $\epsilon$ satisfies (20), then

      $${\parallel u\parallel }_{H^1(\mathrm{\Omega })}\le M{\parallel f\parallel }_{L^2(\mathrm{\Omega })}$$

      with

      $$M=\frac{1}{\min \{k-\epsilon {\parallel 𝒃\parallel }_{L^{\mathrm{\infty }}(\mathrm{\Omega })},c-\frac{{\parallel 𝒃\parallel }_{L^{\mathrm{\infty }}(\mathrm{\Omega })}}{4\epsilon }\}}.$$

      For example, if we choose

      $$\epsilon =\frac{1}{2}\frac{k}{{\parallel 𝒃\parallel }_{L^{\mathrm{\infty }}(\mathrm{\Omega })}},$$

      then

      $$c_0=\frac{{\parallel 𝒃\parallel }_{L^{\mathrm{\infty }}(\mathrm{\Omega })}}{2k},M=\frac{1}{\min \{k/2,c-c_0\}}=\max \{\frac{2}{k},\frac{1}{c-c_0}\}.$$

    • Let $v\in C^2(\overline{\mathrm{\Omega }})$ satisfy (19). Then $w=u-v$ satisfies (19) with $f=0$. Therefore by part (d)

      $${\parallel w\parallel }_{H^1(\mathrm{\Omega })}\le 0$$

      and so $w=0$ and $u=v$, as required.

19. Neumann boundary conditions for variational problems.

    • Let $u\in C^1(\overline{\mathrm{\Omega }})$ be a minimiser of $E$. For any $\phi \in V$, $\epsilon \in ℝ$, define $u_{\epsilon }=u+\epsilon \phi .$ Then $u_{\epsilon }\in C^1(\overline{\mathrm{\Omega }})$ since the sum of $C^1$ functions is $C^1$. Let $g(\epsilon )=E[u_{\epsilon }]$. Note that $u_{\epsilon }=u$ when $\epsilon =0$. Therefore $g$ is minimised by $\epsilon =0$ since $E$ is minimised by $u$. Hence

      $0$ $=g^{\prime }(0)={{\displaystyle \frac{d}{d\epsilon }}|}_{\epsilon =0}E[u_{\epsilon }]$

      $={{\displaystyle \frac{d}{d\epsilon }}|}_{\epsilon =0}\left[{\displaystyle \frac{1}{2}}{\displaystyle {\int }_{\mathrm{\Omega }}}{|\nabla u_{\epsilon }|}^2𝑑𝒙-{\displaystyle {\int }_{\mathrm{\Omega }}}f{u}_{\epsilon }𝑑𝒙\right]$

      $={{\displaystyle \frac{d}{d\epsilon }}|}_{\epsilon =0}\left[{\displaystyle \frac{1}{2}}{\displaystyle {\int }_{\mathrm{\Omega }}}(\nabla u+\epsilon \nabla \phi )\cdot (\nabla u+\epsilon \nabla \phi )𝑑𝒙-{\displaystyle {\int }_{\mathrm{\Omega }}}f(u+\epsilon \phi )𝑑𝒙\right]$

      $={{\displaystyle \frac{1}{2}}{\displaystyle {\int }_{\mathrm{\Omega }}}{\displaystyle \frac{d}{d\epsilon }}|}_{\epsilon =0}[(\nabla u+\epsilon \nabla \phi )\cdot (\nabla u+\epsilon \nabla \phi )]d𝒙-{{\displaystyle {\int }_{\mathrm{\Omega }}}f{\displaystyle \frac{d}{d\epsilon }}|}_{\epsilon =0}(u+\epsilon \phi )d𝒙$

      $={{\displaystyle \frac{1}{2}}{\displaystyle {\int }_{\mathrm{\Omega }}}[\nabla \phi \cdot (\nabla u+\epsilon \nabla \phi )+(\nabla u+\epsilon \nabla \phi )\cdot \nabla \phi ]|}_{\epsilon =0}d𝒙-{\displaystyle {\int }_{\mathrm{\Omega }}}f\phi 𝑑𝒙$

      $={\displaystyle {\int }_{\mathrm{\Omega }}}\nabla u\cdot \nabla \phi d𝒙-{\displaystyle {\int }_{\mathrm{\Omega }}}f\phi 𝑑𝒙.$

      Therefore

      $${\int }_{\mathrm{\Omega }}\nabla u\cdot \nabla \phi d𝒙={\int }_{\mathrm{\Omega }}f\phi 𝑑𝒙\mathit{\hspace{1em}}\text{for all }\phi \in C^1(\overline{\mathrm{\Omega }})$$ (21)

      as required.

    • First choose a test function $\phi \in C^1(\overline{\mathrm{\Omega }})$ such that $\phi =0$ on $\partial \mathrm{\Omega }$. Since $u\in C^2(\overline{\mathrm{\Omega }})$, we can integrate by parts in (21) to obtain

      $${\int }_{\partial \mathrm{\Omega }}\nabla u\phi \cdot 𝒏dS-{\int }_{\mathrm{\Omega }}\underset{=\mathrm{\Delta }u}{\underbrace{\mathrm{div}\nabla u}}\phi 𝑑𝒙={\int }_{\mathrm{\Omega }}f\phi 𝑑𝒙\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}{\int }_{\mathrm{\Omega }}(-\mathrm{\Delta }u-f)\phi 𝑑𝒙=0$$

      because $\phi =0$ on $\partial \mathrm{\Omega }$. Since this holds for all test functions $\phi \in C^1(\overline{\mathrm{\Omega }})$ such that $\phi =0$ on $\partial \mathrm{\Omega }$, the Fundamental Lemma of the Calculus of Variations implies that

      $$-\mathrm{\Delta }u-f=0\mathit{\hspace{1em}}\text{in }\mathrm{\Omega }$$ (22)

      as required. We still need to show that $u$ satisfies the Neumann boundary condition. Now take any test function $\phi \in C^1(\overline{\mathrm{\Omega }})$ in (21) and integrate by parts as before to obtain

      ${\displaystyle {\int }_{\partial \mathrm{\Omega }}}\nabla u\phi \cdot 𝒏dS-{\displaystyle {\int }_{\mathrm{\Omega }}}\mathrm{\Delta }u\phi 𝑑𝒙={\displaystyle {\int }_{\mathrm{\Omega }}}f\phi 𝑑𝒙$ $⟺\mathit{\hspace{1em}}{\displaystyle {\int }_{\partial \mathrm{\Omega }}}\nabla u\phi \cdot 𝒏dS+{\displaystyle {\int }_{\mathrm{\Omega }}}\underset{=0\text{ by }(\mathit{\text{<a href="\#S0.E22" title="(22) ‣ item (ii) ‣ item 19" class="ltx\_ref ltx\_font\_italic" style="font-size:70\%;"><span class="ltx\_text ltx\_ref\_tag">22</span></a>}})}{\underbrace{(-\mathrm{\Delta }u-f)}}\phi 𝑑𝒙=0$

      $⟺\mathit{\hspace{1em}}{\displaystyle {\int }_{\partial \mathrm{\Omega }}}\nabla u\cdot 𝒏\phi dS=0.$

      Since this holds for all $\phi \in C^1(\overline{\mathrm{\Omega }})$, then $\nabla u\cdot 𝒏=0$ on $\partial \mathrm{\Omega }$, as required.

20. The $p$–Laplacian operator.

    • Let $u\in C^2(\overline{\mathrm{\Omega }})\cap V$ minimise $E_p$. For any $\phi \in V$, $\epsilon \in ℝ$, define $u_{\epsilon }=u+\epsilon \phi .$ Observe that $u_{\epsilon }$ vanishes on the boundary of $\mathrm{\Omega }$ since both $u$ and $\phi$ vanish there. Also $u_{\epsilon }\in C^1(\overline{\mathrm{\Omega }})$ since the sum of $C^1$ functions is $C^1$. Hence $u_{\epsilon }\in V$. Define $g(\epsilon )=E_p[u_{\epsilon }]$. Now $u_{\epsilon }=u$ when $\epsilon =0$. Therefore $g$ is minimised by $\epsilon =0$ 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 $\epsilon =0$,

      $0$ $=g^{\prime }(0)$

      $={{\displaystyle \frac{d}{d\epsilon }}|}_{\epsilon =0}{E}_p[u_{\epsilon }]$

      $={{\displaystyle \frac{d}{d\epsilon }}|}_{\epsilon =0}\left[{\displaystyle \frac{1}{p}}{\displaystyle {\int }_{\mathrm{\Omega }}}{|\nabla u_{\epsilon }|}^p𝑑𝒙-{\displaystyle {\int }_{\mathrm{\Omega }}}f{u}_{\epsilon }𝑑𝒙\right]$

      $={{\displaystyle \frac{1}{p}}{\displaystyle {\int }_{\mathrm{\Omega }}}{\displaystyle \frac{d}{d\epsilon }}|}_{\epsilon =0}{|\nabla u+\epsilon \nabla \phi |}^{p}d𝒙-{{\displaystyle {\int }_{\mathrm{\Omega }}}f{\displaystyle \frac{d}{d\epsilon }}|}_{\epsilon =0}(u+\epsilon \phi )d𝒙$

      $={{\displaystyle \frac{1}{p}}{\displaystyle {\int }_{\mathrm{\Omega }}}p{|\nabla u+\epsilon \nabla \phi |}^{p-1}{\displaystyle \frac{\nabla u+\epsilon \nabla \phi }{|\nabla u+\epsilon \nabla \phi |}}\cdot \nabla \phi |}_{\epsilon =0}d𝒙-{\displaystyle {\int }_{\mathrm{\Omega }}}f\phi 𝑑𝒙$

      $={\displaystyle {\int }_{\mathrm{\Omega }}}{|\nabla u|}^{p-2}\nabla u\cdot \nabla \phi d𝒙-{\displaystyle {\int }_{\mathrm{\Omega }}}f\phi 𝑑𝒙$ (23)

      where the differentiation was performed using the Chain Rule and the fact that

      $$\frac{d}{dx}{x}^p=p{x}^{p-1},{\nabla }_{𝒚}|𝒚|=\frac{𝒚}{|𝒚|},\frac{d}{d\epsilon }(\nabla u+\epsilon \nabla \phi )=\nabla \phi .$$

      Recall the integration by parts formula

      $${\int }_{\mathrm{\Omega }}𝒈\cdot \nabla hd𝒙={\int }_{\partial \mathrm{\Omega }}𝒈h\cdot 𝒏𝑑S-{\int }_{\mathrm{\Omega }}h\mathrm{div}𝒈𝑑𝒙.$$

      By applying this with $h=\phi$, $𝒈={|\nabla u|}^{p-2}\nabla u$, we can rewrite equation (23) as

      $$0={\int }_{\partial \mathrm{\Omega }}{|\nabla u|}^{p-2}\nabla u\phi \cdot 𝒏dS-{\int }_{\mathrm{\Omega }}\phi \mathrm{div}({|\nabla u|}^{p-2}\nabla u)𝑑𝒙-{\int }_{\mathrm{\Omega }}f\phi 𝑑𝒙.$$

      But $\phi =0$ on $\partial \mathrm{\Omega }$ since $\phi \in V$. Therefore

      $$0={\int }_{\mathrm{\Omega }}[\mathrm{div}({|\nabla u|}^{p-2}\nabla u)+f]\phi 𝑑𝒙\mathit{\hspace{1em}}\text{for all }\phi \in V.$$

      Since $\phi$ is arbitrary, the Fundamental Lemma of the Calculus of Variations gives

      $$\mathrm{div}({|\nabla u|}^{p-2}\nabla u)+f=0\mathit{\hspace{1em}}\text{in }\mathrm{\Omega }.$$

      Therefore

      $$-\underset{={\mathrm{\Delta }}_{p}u}{\underbrace{\mathrm{div}({|\nabla u|}^{p-2}\nabla u)}}=f\mathit{\hspace{1em}}\text{in }\mathrm{\Omega }$$

      as required. Note that $u=0$ on $\partial \mathrm{\Omega }$ by definition of $V$.

    • Multiply the PDE $-\mathrm{div}({|\nabla u|}^{p-2}\nabla u)=f$ by $u$ and integrate by parts over $\mathrm{\Omega }$ to obtain

      $-{\displaystyle {\int }_{\mathrm{\Omega }}}u\mathrm{div}({|\nabla u|}^{p-2}\nabla u)𝑑𝒙={\displaystyle {\int }_{\mathrm{\Omega }}}fu𝑑𝒙$

      $⟺\mathit{\hspace{1em}}-{\displaystyle {\int }_{\partial \mathrm{\Omega }}}u({|\nabla u|}^{p-2}\nabla u)\cdot 𝒏𝑑S+{\displaystyle {\int }_{\mathrm{\Omega }}}{|\nabla u|}^{p-2}\nabla u\cdot \nabla ud𝒙={\displaystyle {\int }_{\mathrm{\Omega }}}fu𝑑𝒙$

      $⟺\mathit{\hspace{1em}}{\displaystyle {\int }_{\mathrm{\Omega }}}{|\nabla u|}^p𝑑𝒙={\displaystyle {\int }_{\mathrm{\Omega }}}fu𝑑𝒙$ (24)

      since $u=0$ on $\partial \mathrm{\Omega }$. Therefore

      $E_p[u]$ $={\displaystyle \frac{1}{p}}{\displaystyle {\int }_{\mathrm{\Omega }}}{|\nabla u|}^p𝑑𝒙-{\displaystyle {\int }_{\mathrm{\Omega }}}fu𝑑𝒙$

      $={\displaystyle \frac{1}{p}}{\displaystyle {\int }_{\mathrm{\Omega }}}{|\nabla u|}^p𝑑𝒙-{\displaystyle {\int }_{\mathrm{\Omega }}}{|\nabla u|}^p𝑑𝒙$ (by equation (24))

      $={\displaystyle \frac{1-p}{p}}{\displaystyle {\int }_{\mathrm{\Omega }}}{|\nabla u|}^p𝑑𝒙$

      $={\displaystyle \frac{1-p}{p}}{\displaystyle {\int }_{\mathrm{\Omega }}}fu𝑑𝒙$ (by equation (24))

      as required.

21. The minimal surface equation: PDEs and soap films. Let $u\in C^2(\overline{\mathrm{\Omega }})\cap V$ be a minimiser of $A$. Let $\epsilon \in ℝ$ and $\phi \in C^1(\overline{\mathrm{\Omega }})$ with $\phi =0$ on $\partial \mathrm{\Omega }$. Define $u_{\epsilon }=u+\epsilon \phi$. Then $u_{\epsilon }\in V$ since the sum of continuously differential functions is continuously differentiable and, if $𝒙\in \partial \mathrm{\Omega }$, then

    $$u_{\epsilon }(𝒙)=u(𝒙)+\epsilon \phi (𝒙)=g(𝒙)+\epsilon \cdot 0=g(𝒙)$$

    as required. Define $h:ℝ\to ℝ$ by $h(\epsilon )=A[u_{\epsilon }]$. Then $h(0)=A[u]$ and so $0$ is a minimum point of $h$ since $u$ is a minimum point of $A$. Therefore

    $0$ $=h^{\prime }(0)$

    $={{\displaystyle \frac{d}{d\epsilon }}|}_{\epsilon =0}A[u_{\epsilon }]$

    $={{\displaystyle \frac{d}{d\epsilon }}|}_{\epsilon =0}{\displaystyle {\int }_{\mathrm{\Omega }}}\sqrt{1+{|\nabla u_{\epsilon }|}^2}𝑑𝒙$

    $={{\displaystyle {\int }_{\mathrm{\Omega }}}{\displaystyle \frac{d}{d\epsilon }}|}_{\epsilon =0}\sqrt{1+{|\nabla u+\epsilon \nabla \phi |}^2}d𝒙$

    $={{\displaystyle {\int }_{\mathrm{\Omega }}}\frac{1}{2}{(1+{|\nabla u+\epsilon \nabla \phi |}^2)}^{-1/2}\mathrm{\hspace{0.17em}2}|\nabla u+\epsilon \nabla \phi |{\displaystyle \frac{\nabla u+\epsilon \nabla \phi }{|\nabla u+\epsilon \nabla \phi |}}\cdot \nabla \phi |}_{\epsilon =0}d𝒙$

    $={\displaystyle {\int }_{\mathrm{\Omega }}}{\displaystyle \frac{\nabla u}{\sqrt{1+{|\nabla u|}^2}}}\cdot \nabla \phi d𝒙.$

    This means that $u$ is a weak solution of the minimal surface equation. Since $u\in C^2(\overline{\mathrm{\Omega }})$, then we can integrate by parts to obtain

    $0$ $={\displaystyle {\int }_{\mathrm{\Omega }}}{\displaystyle \frac{\nabla u}{\sqrt{1+{|\nabla u|}^2}}}\cdot \nabla \phi d𝒙$

    $={\displaystyle {\int }_{\partial \mathrm{\Omega }}}\phi {\displaystyle \frac{\nabla u}{\sqrt{1+{|\nabla u|}^2}}}\cdot 𝒏𝑑S-{\displaystyle {\int }_{\mathrm{\Omega }}}\mathrm{div}\left({\displaystyle \frac{\nabla u}{\sqrt{1+{|\nabla u|}^2}}}\right)\phi 𝑑𝒙$

    $=-{\displaystyle {\int }_{\mathrm{\Omega }}}\mathrm{div}\left({\displaystyle \frac{\nabla u}{\sqrt{1+{|\nabla u|}^2}}}\right)\phi 𝑑𝒙$

    since $\phi =0$ on $\partial \mathrm{\Omega }$. This holds for all $\phi \in C^1(\overline{\mathrm{\Omega }})$ with $\phi =0$. Therefore $u$ satisfies the minimal surface equation

    $$\mathrm{div}\left(\frac{\nabla u}{\sqrt{1+{|\nabla u|}^2}}\right)=0\mathit{\hspace{1em}}\text{in }\mathrm{\Omega }.$$

    by the Fundamental Lemma of the Calculus of Variations (Lemma 3.20).

22. Homogenization and the calculus of variations.

    • Let $u\in C^2([0,1])\cap V$ minimise $E$. For any $\epsilon \in ℝ$ and any $\phi \in C^1([0,1])$ such that $\phi (0)=\phi (1)=0$, define $u_{\epsilon }=u+\epsilon \phi .$ Then

      $$u_{\epsilon }(0)=u(0)+\epsilon \phi (0)=l+\epsilon \cdot 0=l$$

      and similarly $u_{\epsilon }(1)=r$. Therefore $u_{\epsilon }\in V$. Define $F(\epsilon )=E[u_{\epsilon }]$. Now $u_{\epsilon }=u$ when $\epsilon =0$. 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 $0$,

      $0$ $=F^{\prime }(0)$

      $={{\displaystyle \frac{d}{d\epsilon }}|}_{\epsilon =0}E[u_{\epsilon }]$

      $={{\displaystyle \frac{d}{d\epsilon }}|}_{\epsilon =0}\left[{\displaystyle \frac{1}{2}}{\displaystyle {\int }_0^1}a(x){|u_{\epsilon }^{\prime }(x)|}^2𝑑x-{\displaystyle {\int }_0^1}f(x)u_{\epsilon }(x)𝑑x\right]$

      $={{\displaystyle \frac{d}{d\epsilon }}|}_{\epsilon =0}\left[{\displaystyle \frac{1}{2}}{\displaystyle {\int }_0^1}a(x){(u^{\prime }(x)+\epsilon {\phi }^{\prime }(x))}^2𝑑x-{\displaystyle {\int }_0^1}f(x)(u(x)+\epsilon \phi (x))𝑑x\right]$

      $={\displaystyle {\int }_0^1}a(x)u^{\prime }(x){\phi }^{\prime }(x)𝑑x-{\displaystyle {\int }_0^1}f(x)\phi (x)𝑑x.$ (25)

      Since $u\in C^2([0,1])$, we can use integration by parts to rewrite equation (25) as

      $$0={a(x)u^{\prime }(x)\phi (x)|}_0^1-{\int }_0^1{(a(x)u^{\prime }(x))}^{\prime }\phi (x)𝑑x-{\int }_0^{1}f(x)\phi (x)𝑑x=-{\int }_0^1[{(a(x)u^{\prime }(x))}^{\prime }+f(x)]\phi (x)𝑑x.$$

      But this holds for all $\phi \in C^1([0,1])$ such that $\phi (0)=\phi (1)=0$. Therefore by the Fundamental Lemma of the Calculus of Variations

      $${(a(x)u^{\prime }(x))}^{\prime }+f(x)=0,x\in (0,1),$$

      as required. Note that $u$ satisfies the Dirichlet boundary conditions by definition of $V$.

    • Recall from Q2(ii) that if $g\in L^{\mathrm{\infty }}(ℝ)$ is $1$–periodic, then for any interval $[c,d]\subseteq ℝ$,

      $$\underset{n\to \mathrm{\infty }}{lim}{\int }_c^{d}g(nx)h(x)𝑑x={\int }_c^d\overline{g}h(x)𝑑x\mathit{\hspace{1em}\hspace{1em}}\forall h\in L^1(ℝ).$$ (26)

      Applying (26) with $c=0$, $d=1$, $g(x)=a(x)$, $h(x)=\frac{1}{2}{|v^{\prime }(x)|}^2$ on $[0,1]$, gives the desired result:

      $$\underset{n\to \mathrm{\infty }}{lim}{E}_n[v]=\frac{1}{2}{\int }_0^1\overline{a}|v^{\prime }(x){|}^{2}dx-{\int }_0^{1}f(x)v(x)dx=:E_{\mathrm{\infty }}[v].$$

    • Observe that $E_{\mathrm{\infty }}$ 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 $u_{\mathrm{\infty }}$ satisfies the Poisson equation

      $$-\overline{a}{u}_{\mathrm{\infty }}^{\prime \prime }(x)=f(x),x\in (0,1),$$

      $$u_{\mathrm{\infty }}(0)=u_{\mathrm{\infty }}(1)=0.$$

      In Q2 we showed that ${lim}_{n\to \mathrm{\infty }}{u}_n(x)=u_0(x)$, where $u_0$ satisfies

      $$-a_0{u}_0^{\prime \prime }(x)=f(x),x\in (0,1),$$

      $$u_0(0)=u_0(1)=0,$$

      where

      $$a_0=\frac{1}{\overline{\left(\frac{1}{a}\right)}}.$$

      Since $a_0\ne \overline{a}$ in general, it follows that $u_0\ne u_{\mathrm{\infty }}$ and hence

      $$\underset{n\to \mathrm{\infty }}{lim}{u}_n(x)=u_0(x)\ne u_{\mathrm{\infty }}(x),$$

      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 $\mathrm{\Gamma }$–limit $E_0$ is less than or equal to the pointwise limit $E_{\mathrm{\infty }}$.