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I am attempting Exercise 5 in Chapter 9 of PDE Evans, 2nd edition:

  1. Consider the nonlinear boundary-value problem $$\begin{cases}-\Delta u + b(Du)=f & \text{in }U \\ \qquad \qquad \quad \, \, \, u=0 & \text{on }\partial U. \end{cases}$$ Use Banach's Fixed Point Theorem to show there exists a unique weak solution $u \in H^2(U) \cap H_0^1(U)$ provided $b : \mathbb{R}^n \to \mathbb{R}$ is Lipschitz continuous, with $\text{Lip}(b)$ small enough.

Here are my thoughts so far on this problem:

I am tempted to follow Theorem 2 on page 536 in the textbook, which says (and Theorem 2 in the textbook refers to $(2)$ on page 535 which I wrote down as well):

THEOREM 2 (Existence). There exists a unique weak solution of the initial-boundary value problem for the reaction-diffusion system $$\begin{cases}\mathbf{u}_t-\Delta\mathbf{u}=\mathbf{f}(\mathbf{u}) &\text{in }U_T \\ \qquad \quad \mathbf{u}=\mathbf{0} & \text{on }\partial U \times [0,T] \\\qquad \quad \mathbf{u}=\mathbf{g} & \text{on }\partial U \times \{t=0\}.\end{cases}\tag{2}$$ Here $\mathbf{u}=(u^1,\ldots,u^m)$, $\mathbf{g}=(g^1,\ldots,g^m)$, and as usual $U_T = U\times (0,T]$, where $U \subset \mathbb{R}^n$ is open and bounded, with smooth boundary. The time $T > 0$ is fixed. We assume that the initial function $\mathbf{g}$ belongs to $H_0^1(U;\mathbb{R}^m)$.

It seems that Exercise 5 calls for the reader to construct a proof similar to that of Theorem 2. (Please comment if you need me to reproduce the latter proof here.) Anyway, the proof of Theorem 2 applies Banach's Fixed Point Theorem, just like what Exercise 5 asks to.

Somewhere in the beginning of the proof of Theorem 2, one sentence reads "Given a function $\mathbf{u} \in X$, set $\mathbf{h}(t):=\mathbf{f}(\mathbf{u}(t))$ ($0 \le t \le T)$." For Exercise 5, however, $f$ does not depend on $u$, but $b(Du)$ does.

Cookie
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1 Answers1

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I think you have the right notion, and in Exercise 5, things are 'nicer' for us in that there is no time variable to deal with.

In the following, I am assuming that the given $f \in L^2(U)$.

  1. Given a function $u \in H^1_0(U)$, set $g := f - b(Du)$, and consider the linear PDE

\begin{align} -\Delta w &= g \quad \text{in } U\\ w &= 0 \quad \text{on } \partial U. \end{align}

  1. By appealing to existence and regularity results (see Chapter 6 in Evans), we know that there exists a unique weak solution $w \in H^1_0(U)\cap H^2(U)$ to the above linear PDE.

  2. Define an operator $A: H^1_0(U) \to H^1_0(U)$ by \begin{equation} A[u] = w \end{equation} Let $u,\tilde{u} \in H^1_0(U)$, with $A[u] = w, A[\tilde{u}] = \tilde{w}$. To show: $A$ is a strict contraction. (This is a good place to try yourself first without reading further). Then

\begin{align} \left\|A[u] - A[\tilde{u}]\right\|_{H^1_0(U)} &= \|w-\tilde{w}\|_{H^1_0(U)}\\ &\leq C\|D(w - \tilde{w})\|_{L^2(U)}\quad(*).\end{align} In the above, $(*)$ follows by a Sobolev estimate. Now \begin{align} \|D(w - \tilde{w})\|_{L^2(U)}^2 &= \int_{U} D(w - \tilde{w})\cdot D(w - \tilde{w}) dx\\ &= -\int_{U}[\Delta(w - \tilde{w})] (w - \tilde{w}) dx\\ &= \int_{U} (b(D\tilde{u}) - b(Du))(w - \tilde{w}) dx\\ &\leq \|b(D\tilde{u}) - b(Du)\|_{L^2(U)}\|w - \tilde{w}\|_{L^2(U)}\quad (**)\\ &\leq C\|b(D\tilde{u}) - b(Du)\|_{L^2(U)}\|D(w - \tilde{w})\|_{L^2(U)} \quad (***) \end{align} In $(**)$, I have used the fact that $b$ is Lipschitz so $b(Du), b(D\tilde{u}) \in L^2(U)$. In $(***)$, I have used a Sobolev estimate. Our conclusion, having reached $(***)$, is that \begin{equation} \|D(w - \tilde{w})\|_{L^2(U)} \leq C\|b(D\tilde{u}) - b(Du)\|_{L^2(U)}. \end{equation} Hence returning to $(*)$, we see that \begin{align} \left\|A[u] - A[\tilde{u}]\right\|_{H^1_0(U)} &\leq C\|b(D\tilde{u}) - b(Du)\|_{L^2(U)}\\ &\leq C\text{Lip}(b)\|D\tilde{u} - Du\|_{L^2(U)}\\ &\leq C\text{Lip}(b)\|u - \tilde{u}\|_{H^1_0(U)}. \end{align} Provided $\text{Lip}(b)$ is small enough, $A$ is a strict contraction (here the constant $C$ is from the Sobolev estimates, so only depends on things like the dimension $n$, $U$, etc.).

Thus by the Banach fixed point theorem, we deduce the existence of a unique fixed point $u_* \in H^1_0(U)$, such that $A[u_*] = u_*$. The regularity theory for the PDE in 1. gives $u_* \in H^2(U)\cap H^1_0(U)$.

Lost in a Maze
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  • Finally, why does regularity theory assert $u_* \in H^2(U) \cap H_0^1(U)$? I'm asking because §6.3 of Evans asserts only $u \in H^2(U)$. – Cookie May 31 '15 at 00:06
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    For your second question, we have $u_* \in H^1_0(U)$ from the Banach fixed point theorem, because our operator $A$ is a mapping from $H^1_0(U)$ to $H^1_0(U)$. – Lost in a Maze May 31 '15 at 00:30
  • Ohh, I see. So $u_* \in H_0^1(U)$ from Banach's theorem and $u_* \in H^2(U)$ from regularity theory, and that's why $u_* \in H_0^1(U) \cap H^2(U)$, as required in the hypothesis in the exercise. – Cookie May 31 '15 at 00:36
  • How do we justify $|D(w - \tilde{w})|{L^2(U)} \leq C|b(D\tilde{u}) - b(Du)|{L^2(U)}$? I wanted to say that Poincaré's inequality was used, but that's not the case here. – Cookie May 31 '15 at 05:06
  • You are right, I used (a form of) Poincare's inequality, namely, that if a function $u \in W^{1,p}{0}(U)$ for some bounded open subset $U$, then $|u|{L^p(U)} \leq C|D u|_{L^p(U)}$. This estimate was applied to $w - \tilde{w}$ and $H^1_0$ in $()$ to arrive at $(*)$. – Lost in a Maze May 31 '15 at 17:00
  • Yes, I understood how you got $(***)$, using Poincaré's inequality (or at least some form of it), but my question was referring to three lines below that. I still want to understand why $$|D(w - \tilde{w})|{L^2(U)} \leq C|b(D\tilde{u}) - b(Du)|{L^2(U)}.$$ All I know so far is that $$\Delta(w-\tilde{w})=b(D\tilde{u})-b(Du).$$ – Cookie May 31 '15 at 18:15
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    That inequality follows through the series of estimates ending in $(*)$ - if we let $x = |D(w - \tilde{w})|{L^2(U)}$ and $y = |b(Du) - b(D\tilde{u})|{L^2(U)}$, then $(*)$ says $x^2 \leq C xy$, so $x \leq Cy$. (Did I misunderstand your question again?) – Lost in a Maze May 31 '15 at 21:20
  • This explained it perfectly, actually. You were only dividing both sides by $|D(w-\tilde{w})|_{L^2(U)}$, and I was overlooking this the whole time. – Cookie Jun 01 '15 at 03:05
  • I'm sorry, but I have one last question: When you used $$|D\tilde{u} - Du|{L^2(U)}\le C|u - \tilde{u}|{H^1_0(U)},$$ how was this justified? Once again, I keep thinking Poincaré's inequality, but then we went from $L^2(U)$ to $H_0^1(U)$. The relationship between these two spaces has always baffled me: The book says on page 299 that $H_0^1(U) \subset L^2(U)$; does this play a role in going from $L^2(U)$ to $H_0^1(U)$ for the norms? I have seen this scenario multiple times in the textbook as well. – Cookie Jun 01 '15 at 04:05
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    Don't forget that the $H^1_0$ norm for a function $u$ is defined by $|u|{H^1_0(U)}^2 = |u|{L^2(U)}^2 + |Du|^2_{L^2(U)}$. So we immediately see that $|Du|{L^2(U)} \leq |u|{H^1_0(U)}$ ($|Du|_{L^2(U)}$ is "only half of the $H^1_0$ norm"). – Lost in a Maze Jun 01 '15 at 16:38