# The equation $\,\,\Delta u+\cos u=0\,\,$ possesses a weak solution in $\,W^{1,2}_0(D)$

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.$$Help me some hints to start. Thanks in advanced. • Have you tried a fixed point method? Or freezing coefficients? – J.R. Jan 23 '14 at 10:31 ## 3 Answers Existence of a weak solution. An alternative approach using a fixed point method: Let v\in L^2(D) and \varphi(v)\in W^{1,2}_0(D) be a weak solution of$$ \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$f(x)=\cos{x}$. • +1 vote Thank you. Where can I find this theorem? – Misa Jan 23 '14 at 13:12 • Sorry, but I don't know a specific place to find it, however, if you have some question, please, feel free to ask. Jan 23 '14 at 13:18 • @chuyenvien94 This is Example 3, Chapter 8.1 in the book "Partial differential equations" by L. Evans (p. 435 in my edition). – J.R. Jan 25 '14 at 15:46 • Thanks for your comment @TooOldForMath. It is not the same thing, but it still a useful reference. Jan 25 '14 at 15:58 • Thank you very much @TooOldForMath – 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. • What is$\triangle^{-1}\cos(v)$? – Misa Jan 23 '14 at 12:08 • Just a shorthand notation for "solution$u\in H^1_0$of the equation$\triangle u = \cos(v)$" – uvs Jan 23 '14 at 12:17 •$△u=cos(v)$or$△u=-cos(v)$? – Misa Jan 23 '14 at 12:18 • Sure, with the minus sign – uvs Jan 23 '14 at 12:48 • And define$\ell(w)=-\int_D w\cos(v)dx\$ right?
– Misa
Jan 23 '14 at 13:00