Existence of solution of a Non-linear PDE via Fixed point theorem - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T18:01:55Z http://mathoverflow.net/feeds/question/106931 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106931/existence-of-solution-of-a-non-linear-pde-via-fixed-point-theorem Existence of solution of a Non-linear PDE via Fixed point theorem unknown (google) 2012-09-11T17:15:44Z 2012-09-11T20:20:58Z <p>Hi all<br> I've the following non-linear PDE</p> <p>$-\Delta Y + Y^3 =U$ on $\Omega \subset R^n$, open, bounded, Lipschitz boundary domain</p> <p>$Y=0 ,$ on $\partial\Omega$ <br></p> <p><strong>1</strong>.Let $Y\in H_0^1$ and as $H_0^1 \hookrightarrow \hookrightarrow L^5$ can we define a compact operator $T:L^5 \times [0,1] \rightarrow L^5$<br>and use the <strong>Leray-Schauder Fixed Point Theorem</strong> to prove the existence of a solution of above PDE for a general $U\in L^2$?</p> <p><strong>2</strong>.Or if not then how can we apply <strong>Leray-Schauder Fixed Point Theorem</strong> to proove existence of a solution in $Y\in H_0^1$? </p> http://mathoverflow.net/questions/106931/existence-of-solution-of-a-non-linear-pde-via-fixed-point-theorem/106950#106950 Answer by Denis Serre for Existence of solution of a Non-linear PDE via Fixed point theorem Denis Serre 2012-09-11T20:01:12Z 2012-09-11T20:01:12Z <p>The Leray-Schauder is a fixed point theorem in the spirit of Brouwer's. When it gives an existence result, it says nothing about uniqueness. Your problem is much better than that, because it does have a unique solution in $H^1_0(\Omega)\cap L^4(\Omega)$, whenever $U$ belongs to the dual space $X=H^{-1}(\Omega)+L^{4/3}(\Omega)$. The reason is that $Y$ is a critical point of the functional $$E[Y]=\int_\Omega(\frac12|\nabla Y|^2+\frac14Y^4-UY)dx.$$ This function turns out to be continuous and coercive over $X$, and strictly convex. Therefore the standard arguments of the so-called <em>direct method of calculus of variations</em> yields existence and uniqueness of a critical point, which is the point $Y$ at which $E$ achieves its minimum.</p>