Let $D$ be an open bounded subset in $\mathbb{R}^{n}$, with sufficiently smooth boundary. Prove that there is a weak solution in $W^{1,2}_0$$(D)$ to following equation $$\Delta u+\cos u=0.$$

Help me some hints to start.

Thanks in advanced.

  • $\begingroup$ Have you tried a fixed point method? Or freezing coefficients? $\endgroup$
    – J.R.
    Jan 23 '14 at 10:31

Existence of a weak solution. An alternative approach using a fixed point method:

Let $v\in L^2(D)$ and $\varphi(v)\in W^{1,2}_0(D)$ be a weak solution of $$ \Delta u+\cos v=0. $$ Such a weak solution exists as this means for $\varphi(v)$ that $$ \int_D \nabla\varphi(v)\cdot\nabla w\,dx= \int_{D}w\cos v\,dx \quad\text{for all $w\in W^{1,2}_0(D)$}. $$ And since $\ell(w)=\int_{D}w\cos v\,dx$ is a bounded linear functional on $W^{1,2}_0(D)$, then there exists a $\varphi(v)\in W^{1,2}_0(D)$, such that $$ \ell(w)=\langle w,\varphi(v)\rangle_{W^{1,2}_0(\Omega)}=\int_\Omega \nabla\varphi(v)\cdot\nabla w\,dx. $$ Note that $W^{1,2}_0(D)$ is a Hilbert space with inner product $\langle w,w'\rangle_{W^{1,2}_0(D)}=\int_D \nabla w\cdot\nabla w'\,dx$.

Also, for every $v\in L^2(D)$: $$ \|\varphi(v)\|_{W^{1,2}_0(D)}=\|\ell\|\le \left(\int_D \cos^2 v\,dx|\right)^{1/2} \le |D|^{1/2}=:M. $$ Hence the nonlinear functional $\varphi$ maps $L^2(D)$ into $$ B=\{w\in {W^{1,2}_0(D)}: \|w\|_{{W^{1,2}_0(D)}}\le M\}. $$ Now $B\subset \{u\in L^2(D) : \|u\|_{L^2}\le N\}$, for some $N>0$, due to Poincaré inequality. In particular, $\varphi$ maps $B$ into $B$, and $\varphi[B]\subset B$. But $B$ is compact subset of $L^2(D)$, due to Rellich compactness theorem. Hence, Schauder fixed point theorem guarantees a fixed point $u$ for $\varphi$. Clearly $u\in L^2(D)$, but $u=\varphi(u)\in B\subset W^{1,2}_0(D)$.

The weak solution is also a classical solution. We have obtained a function $u\in W_0^{1,2}(D)$ satisfying $$\Delta u=-\cos u,$$ is the sense of distributions. But if $u\in W_0^{1,2}(D)$, then $\cos u\in W^{1,2}(D)$ and hence $\Delta u\in W^{1,2}(D)$, which implies that $u\in W^{3,2}(D)$. This in turn implies that $\cos u\in W^{3,2}(D)$ and recursively we obtain that $$u\in W^{k,2}(D), \quad \text{for all $k\in\mathbb N$},$$ and thus $u$ is a classical solution.


Another way to solve this problem is the following: first, let's prove a more general theorem. Assume that $f:\mathbb{R}\to\mathbb{R}$ is a bounded continuous function. Consider the problem $$-\Delta u=f(u),\ u\in H_0^1(D)\tag{1}$$

Let $F(x)=\int_0^x f(s)ds$ and $I:H_0^1(D)\to\mathbb{R}$ the energy functional associated with $(1)$, i.e. $$I(u)=\int_D |\nabla u|^2-\int_D F(u)$$

Note that $$|F(x)|\le \|f\|_\infty |x|,\ \forall x,\tag{w}$$

therefore $$-F(u)\ge -\|f\|_\infty |u\tag{2}|.$$

We conclude from $(2)$ and the continuous embedding $H_0^1(D)\hookrightarrow L^1(D)$ that $$I(u)\ge \|u\|_{1,2}^2-c\|u\|_{1,2},\tag{3}$$

where $c$ is a positive constant, hence, from $(3)$ we conclude that $I$ is coercive (if $\|u_n\|_{1,2}\to \infty$ then $I(u_n)\to\infty$). On the other hand, assume that $u_n\to u$ weakly in $H_0^1(D)$. We can assume without loss of generality that $u_n\to u$ strogly in $L^2(D)$, $u_n\to u$ a.e. in $D$ and $|u_n|\le g\in L^2$. We combine these facts with (w) and Lebesgue theorem to conclude that $$\int_D F(u_n)\to \int_D F(u) \tag{4}$$

As $\|\cdot \|_{1,2}$ is weakly sequentially lower semicontinuous (w.s.l.s.c.), i.e. if $u_n\to u$ weakly in $H_0^1(D)$ then, $\|u\|_{1,2}\le\liminf\|u_n\|_{1,2}$, we conclude from $(4)$ that $I$ is (w.s.l.s.c.).

By combining coerciviness and (w.s.l.s.c.), we conclude that $I$ has a critcal point (weak solution for $(1)$) in $H_0^1(D)$, i.e. there $u\in H_0^1(D)$ such that $$\langle I'(u),v\rangle =\int_D \nabla u\cdot \nabla v-\int_D f(u)v=0\ \forall \ v\in H_0^1(D)$$

Now, just take $f(x)=\cos{x}$.

  • $\begingroup$ +1 vote Thank you. Where can I find this theorem? $\endgroup$
    – Misa
    Jan 23 '14 at 13:12
  • $\begingroup$ Sorry, but I don't know a specific place to find it, however, if you have some question, please, feel free to ask. $\endgroup$
    – Tomás
    Jan 23 '14 at 13:18
  • 1
    $\begingroup$ @chuyenvien94 This is Example 3, Chapter 8.1 in the book "Partial differential equations" by L. Evans (p. 435 in my edition). $\endgroup$
    – J.R.
    Jan 25 '14 at 15:46
  • $\begingroup$ Thanks for your comment @TooOldForMath. It is not the same thing, but it still a useful reference. $\endgroup$
    – Tomás
    Jan 25 '14 at 15:58
  • $\begingroup$ Thank you very much @TooOldForMath $\endgroup$
    – Misa
    Jan 26 '14 at 0:50

Fixed point idea for the operator $Tv=\Delta^{-1}\cos(v)$.

So let $v\in H^1_0(D)=W^{1,2}_0(D)$, define $\ell(w)=\int_D w\cos(v)\,dx$ and $a(u,w)=\int_D \nabla u \cdot \nabla w\, dx$; solve the variational problem $a(u,w)=\ell(w), \forall w\in H^1_0$ and call the weak solution $u=:Tv \in H^1_0$; we have reached the fixed point if $u=v$.

The functional $\ell$ is uniformly (for all $v$) bounded in all kinds of useful norms. Now try to prove that a fixed point exists.

  • $\begingroup$ What is $\triangle^{-1}\cos(v)$? $\endgroup$
    – Misa
    Jan 23 '14 at 12:08
  • $\begingroup$ Just a shorthand notation for "solution $u\in H^1_0$ of the equation $\triangle u = \cos(v)$" $\endgroup$
    – uvs
    Jan 23 '14 at 12:17
  • $\begingroup$ $△u=cos(v)$ or $△u=-cos(v)$? $\endgroup$
    – Misa
    Jan 23 '14 at 12:18
  • $\begingroup$ Sure, with the minus sign $\endgroup$
    – uvs
    Jan 23 '14 at 12:48
  • $\begingroup$ And define $\ell(w)=-\int_D w\cos(v)dx$ right? $\endgroup$
    – Misa
    Jan 23 '14 at 13:00

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