# 偏微分方程习题选解: Evans’ PDE Chapter 5 Sobolev Spaces

Abstract. 这是一个习题答案个人总结, 习题来自 Evans 的偏微分方程教材第二版第五章. 仅供参考, 也欢迎提出问题和建议. This is a personal collection of solutions to Evan's PDE book (2ed). Any comment is welcome to improve. 题号 No. [3, 4, 8, 11, 14, 15, 17, 18]

## Problems and solutions

3. Denote by $$U$$ the open square $$\{x\in \mathbb{R}^2: |x_1|<1,\ |x_2|<1\}$$. Define$\begin{equation*} u(x)=\begin{cases} 1-x_1 \quad x_1>0,\ |x_2|< x_1 \\ 1+x_1 \quad x_1 < 0,\ |x_2| < -x_1 \\ 1-x_2 \quad x_2>0,\ |x_1|< x_2 \\ 1+x_2 \quad x_2 < 0,\ |x_1| < -x_2 . \end{cases}\end{equation*}$For which $$1\leq p\leq \infty$$ does $$u$$ belong to $$W^{1,p}(U)$$?

Proof. Let $$U_i,i=1,2,3,4$$ be the four subdomains. By a direct computation, for any $$\phi\in C_c^\infty(U)$$,$\int_U u \frac{\partial \phi}{\partial x_1} =\sum_{i=1}^{4} \int_{U_i} u \frac{\partial \phi}{\partial x_1} =\sum_{i=1}^{4} \left( -\int_{U_i} \frac{\partial u}{\partial x_1} \phi dx + \int_{\partial U_i} u\phi \nu_1 ds \right)$Since $$u$$ is continuous over $$U$$ and $$\phi=0$$ on $$\partial U$$, we have$\sum_{i=1}^{4}\int_{\partial U_i} u\phi \nu_1 ds =0.$Hence, from the definition of weak derivative, piecewise derivative, $$\frac{\partial u}{\partial x_1}=-1,1,0,0$$ respectively, on the subdomains, is exactly its weak derivative.
Since $$u,\nabla u\in L^{\infty}$$, we know $$u\in W^{1,p}$$ for all $$1\leq p\leq \infty$$. $$\heartsuit$$

4. Assume $$n=1$$ and $$u\in W^{1,p}(0,1)$$ for some $$1\leq p<\infty$$.

1. Show that $$u$$ is equal a.e. to an absolutely continuous function and $$u’$$ (which exists a.e.) belongs to $$L^p(0,1)$$. (Note by the fundamental theorem of Lebesgue integral calculus, we only need to show there exists a Lebesgue integral function $$v(t)$$ s.t. $$u(x)=u(a+\int_{a}^x v(t)dt)$$.)
2. Prove that if $$0<p<\infty$$, then$|u(x)-u(y)|\leq |x-y|^{1-1/p}\left( \int_0^1 |u’|^p dt\right)^{1/p}$for a.e. $$x,y \in [0,1]$$.

Proof. (Sec 5.3.2, Thm 3) Let $$\phi_j\in C^\infty([0,1])$$ s.t. $$\phi_j\to u$$ in $$W^{1,p}(0,1)$$, we have$\phi_j\to u \quad \text{a.e. }x\in (0,1). \tag{4.1}$Actually, we can set $$\phi_j$$ to be the mollification of $$u$$ and apply Appendix C.4 to get (4.1).Hence, for a.e. $$x,y\in (0,1)$$$u(y)-u(x) = \lim_{j} [\phi_j(y)-\phi_j(x)] =\lim_{j} \int_x^y \phi_j’ = \int_x^y u’ dt$and Holder inequality yields$|u(y)-u(x)|=\left| \int_x^y u’ dt\right| \leq \left( \int_x^y 1^q dt \right)^{1/q} \left( \int_x^y |u’|^p dt \right)^{1/p},$where $$\frac{1}{q}=1-\frac{1}{p}$$. $$\heartsuit$$

8. Let $$U$$ be bounded, with a $$C^1$$ boundary. Show that a “typical” function $$u\in L^p(U)(1\leq p<\infty)$$ does not have a trace on $$\partial U$$. More precisely, prove there does not exist a bounded linear operator$T:L^p(U)\to L^p(U)$such that $$Tu = u_{|\partial U}$$ whenever $$u\in C(\bar{U})\cap L^p(U)$$.

Proof. If such $$T$$ exists, for $$p=2$$ example, we have bound$\|Tu\|_{L^2(\partial U)}\leq C \|u\|_{L^2(U)} \quad \forall u\in C(\bar{U}).$Construct a sequence of $$u_n\in C(\bar{U})$$ s.t. $\|u_n\|_{L^2(U)} \to 0, \quad \text{but}\quad \|u_{|\partial U}\|_{L^2(\partial U)}\to \text{const.}>0.$Let $$u_n=\frac{1}{1+n \text{dist}(x,\partial U)}$$. $$\heartsuit$$

11. Suppose $$U$$ is connected and $$u\in W^{1,p}(U)$$ satisfies$Du=0 \quad \text{a.e. in } U.$Prove $$u$$ is constant a.e. in $$U$$.

Proof. Let $$U_\epsilon=\{x\in U:\ \mathrm{dist}(x,\partial \Omega)>\epsilon\}, \epsilon>0$$, by [Sec 5.3.1, Theorem 1, (1)], there holds $D u^{\epsilon} = (D u) ^{\epsilon} \quad \text{in }U_{\epsilon}$Since $$Du=0$$ a.e., we obtain by the assumption$D u^{\epsilon} = 0 \quad \text{in }U_{\epsilon}$that is $$u^\epsilon$$ is constant in $$U_\epsilon$$. Then, use $$u^\epsilon \to u$$ a.e. as $$\epsilon\to 0$$. $$\heartsuit$$

14. Verify that if $$n>1$$, the bounded function $$u=\log \log(1+\frac{1}{|x|})$$ belongs to $$W^{1,n}(U)$$, $$U=B^{0}(0,1)$$.

Proof. Clearly, for any fixed integer $$n$$ there holds $$\lim_{t \rightarrow \infty}\frac{\log(t)^{n}}{t}=0$$. Hence $\frac{[\log\log(1+t)]^{n}}{t}=\frac{[\log\log(1+t)]^{n}}{[\log(1+t)]^{n}} \frac{[\log(1+t)]^{n}}{t} \rightarrow 0, \text{ as } t \rightarrow 0.$We claim $$u\in L^{n}(U)$$. In fact,\begin{align*} \|u\|_{L^{n}(U)}^{n} &=\int_{U}[\log\log(1+\frac{1}{|x|})]^{n}dx \\ &=\int_{0}^{1}[\log\log(1+\frac{1}{r})]^{n}n \alpha(n)r^{n-1}dr \\ &=n \alpha(n)\int_{1}^{\infty}\frac{[\log\log(1+t)]^{n}}{r}\frac{1}{r^{n}}dt <\infty.\end{align*}We know from fundamental calculus $\frac{\partial u}{\partial x_{i}}=\frac{1}{\log(1+1/|x|)}\frac{1}{1+1/|x|}\frac{-x_{i}}{|x|^{3}},$and\begin{align*}\left\|\frac{\partial u}{\partial x_{i}}\right\|_{L^{n}(U)}^{n} & \leq n \alpha(n)\int_{0}^{1} \frac{1}{\log(1+1/r)}\frac{1}{1+1/r} \frac{1}{r^{2}} r^{n-1} dr\\& = n \alpha(n)\int_{1}^{\infty} \frac{1}{\log(1+t)}\frac{t}{1+t} \frac{1}{t^{n}} dt <\infty.\\\end{align*}Hence $$u\in W^{1,n}(U)$$. $$\heartsuit$$

15. Fix $$\alpha>0$$ and let $$U=B_1^0(U)$$. Show that there exists a constant $$C>0$$, depending only on $$n$$ and $$\alpha$$, such that $\int_{U}u^2 dx \leq C \int_{U}|Du|^2 dx,$provided$|\{x\in U\ |\ u(x)=0\}|\geq \alpha , \quad u\in H^1(U).$

Proof. Let $$(u)_{U}$$ be the average of $$u$$ over $$U$$. By using Poincare inequality [Sec 5.8, Thm 1], we have\begin{align*} \|u\|_{L^{2}(U)}\leq \|u-(u)_{U}\|_{L^{2}(U)}+\|(u)_{U}\|_{L^{2}(U)} \leq C(n,U) \|Du\|_{L^{2}(U)}+\|(u)_{U}\|_{L^{2}(U)}. \end{align*}Since\begin{align*} \|(u)_{U}\|_{L^{2}(U)} &\leq \frac{1}{|U|^{1/2}}\int_{U\setminus \{ u=0 \}}|u| \\ &\leq \frac{1}{|U|^{1/2}} \|1\|_{L^{2}(U\setminus \{ u=0 \})}\|u\|_{L^{2}(U\setminus \{ u=0 \})} \\ &\leq \frac{(|U|-\alpha)^{1/2}}{|U|^{1/2}}\|u\|_{L^{2}(U)},\end{align*}substituting into the above inequality, it holds true$\begin{equation*} \left( 1- \frac{(|U|-\alpha)^{1/2}}{|U|^{1/2}}\right)\|u\|_{L^{2}(U)} \leq C(n,U) \|Du\|_{L^{2}(U)}.\end{equation*}$$$\heartsuit$$

17. Let $$U \in \mathbb{R}^n$$ be a bounded domain, $$1\leq p\leq \infty$$ and $$f\in C^1(\mathbb{R})$$ with $$f’\in L^{\infty}(\mathbb{R}^n)$$. Prove that $$v=f\circ u\in W^{1,p}(U)$$ whenever $$u\in W^{1,p}(U)$$, and that $v_{x_i}=f'(u)u_{x_i},\quad 1\leq i\leq n.$

Proof. Take smooth sequence $$u_m\in C^{\infty}(U)$$ such that $$u_m\to u$$ in $$W^{1,p}(U)$$ as $$m\to \infty$$, for example we can take $$u_m$$ to be the mollification of $$u$$, and hence we know $\tag{17.1} u_m \to u,\ D_{x_i}u_{m}\to D_{x_i}u \quad \text{a.e. on } U.$To show $$f(u)\in W^{1,p}(U)$$, we need to show $$f(u)\in L^p(U)$$ and its weak derivatives are in $$L^p(U)$$. Let $$x_0\in U$$ be a Lebesgue point of $$u$$. Because $$f\in C^1(\mathbb{R})$$ and $$f’\in L^{\infty}(\mathbb{R})$$, we have for a.e. $$x\in U$$$f(u(x)) = f(u(x_0)) + \int_{0}^{u(x)} f'(t) dt \leq |f(u(x_0))| + \|f’\|_{L^\infty(U)} |u(x)|$Because $$|f(u(x_0))|$$ and $$\|f’\|_{L^\infty(U)}$$ are constant, we deduce $$f(u)\in L^p(U)$$ by the boundedness of $$U$$ and $$u\in L^p(U)$$.
Secondly, since $$f'(u)\in L^\infty(U)$$ and $$D_{x_i}u\in L^p(U)$$, we have $$f'(u)D_{x_i}u\in L^p(U)$$. For any $$\phi\in C^\infty_0(U)$$, there holds \begin{align*} \int_{U}f(u)\phi’ &= \int_{U}f(u_m)D_{x_i}\phi + \int_{U}[f(u_m)-f(u)]D_{x_i}\phi \\ &= – \int_{U}[f'(u_m) D_{x_i}u_m]\phi + \int_{U}[f(u_m)-f(u)]D_{x_i}\phi \\ &\to – \int_{U}[f'(u) D_{x_i}u]\phi, \end{align*}where we have used (17.1) and dominated convergence and the following limit as $$m$$ tends to infinity$\int_{U}[f(u_m)-f(u)]D_{x_i}\phi \leq \|f(u_m)-f(u)\|_{L^p(U)} \|D_{x_i}\phi\|_{L^q(U)} \leq C \|f’\|_{L^\infty(\mathbb{R})} \|u_m-u\|_{L^p(U)} \to 0$$$\heartsuit$$

18. Assume $$1\leq p\leq \infty$$ and $$U$$ a bounded domain.
1. Prove that if $$u\in W^{1,p}(U)$$, then $$|u|\in W^{1,p}(U)$$.
2. Prove $$u\in W^{1,p}(U)$$ implies $$u^+=\max \left\{ u,0 \right\}$$ and $$u^-=-\min \left\{ u,0 \right\}$$ in $$W^{1,p}(U)$$ with $Du^{\pm}=\begin{cases} \pm Du & \text{a.e. on } \left\{ \pm u>0 \right\},\\ \pm 0 & \text{a.e. on } \left\{ \pm u\leq 0 \right\}. \end{cases}$(Hint). $$u^+=\lim_{\epsilon \to 0} F_\epsilon(u)$$ for $F_\epsilon(z)=\begin{cases} (z^2+\epsilon^2)^{1/2}-\epsilon & \text{if } z\geq 0,\\ 0 & \text{if } z< 0. \end{cases}$
3. Prove that if $$u\in W^{1,p}(U)$$, then$Du=0\quad \text{a.e. on } \left\{ u=0 \right\}.$

Proof. First, prove 2). It is easy to show for any fixed $$\epsilon>0$$, $$F_\epsilon \in C^1(\mathbb{R})$$ and $$F_\epsilon’ \in L^{\infty}(\mathbb{R})$$. In fact,$F_\epsilon'(z) = \frac{z}{\sqrt{z^2+\epsilon^2}}, z> 0;\ \text{ and } F_\epsilon'(z) = 0,\ z\leq0.$By using the definition of weak derivative and problem 2, it holds for $$\phi\in [C^\infty_0(U)]^d$$$\int_{U}F_\epsilon(u) D\phi = -\int_{U}[F_\epsilon(u)’Du] \phi = -\int_{\{u\geq 0\}}Du \frac{u}{\sqrt{u^2+\epsilon^2}} \phi$Since $$|Du \frac{u}{\sqrt{u^2+\epsilon^2}}|\leq |Du|$$ and $$|Du|$$ is integrable, the dominated convergence theorem applies to the right-hand side of the latter identity. Similarly, since $$F_\epsilon(u)\to u^+$$ a.e. as $$\epsilon\to 0$$ and $$|F_\epsilon(u)|\leq |u|$$, the dominated convergence theorem also applies the the left-hand side. By taking $$\epsilon\to 0$$ we thus have$\int_{U}u^+ D\phi = – \int_{\{u> 0\}}Du \phi,$which says $$Du^+=\begin{cases} Du, \quad & \text{on }\{u> 0\} \\ 0,& \text{on }\{u\leq 0\}\end{cases}$$. The case of $$u^-$$ is similar.
1) Since $$|u|=u^+ + u^-$$ and from b) we know $$u^{\pm}\in W^{1,p}(U)$$, thus, also $$|u|\in W^{1,p}(U)$$.
3) We already know $$Du^{\pm}=0$$ a.e. on $$\{\pm u\leq 0\}$$, hence $$Du = Du^+ + Du^-=0$$ a.e. on $$\{u=0\}=\{u\leq 0\}\cap\{u\geq 0\}$$. $$\heartsuit$$

## References

 Evans, L. C. (2010). Partial differential equations. In Graduate studies in mathematics (2nd ed.). American Mathematical Society.

0 0 投票数 0 评论