# How do I apply Brouwer fixed-point theorem in this claim?

1. Let $$\zeta:\mathbb{R}\to [0,+\infty)$$ be a continuous non-negative function such that $$\zeta(0)=0$$ and $$\tau\mapsto \zeta(\tau)\tau$$ is a non-decreasing differentiable function whose derivative is bounded on every compact subset of $$\mathbb{R}$$.

2. Let $$\{\phi_{k}, \lambda_{k}\}_{k \in \Bbb N}$$ be the the Dirichlet eigenpairs of the n Laplace operator on an open bounded set $$\Omega\subset \Bbb R^N$$, i.e., $$\phi_k\in H_0^1(\Omega)$$ and $$-\Delta \phi_k= \lambda_k\phi_k$$. Recall $$\{\phi_k\}_{k}$$ is an orthonormal basis in $$L^2(\Omega)$$.

3. Question: Let $$\mathcal{V}_{k}= \operatorname{span}\{\phi_1,\dotsc, \phi_{k}\}$$. According to page 5 Eq (3.3) of Starovoitov - Boundary value problem for a global in time parabolic equation, the Brouwer fixed-point theorem implies the existence of a vector $$v_k\in \mathcal{V}_k$$ such that $$$$\label{Star-3.3} \int_\Omega \nabla v_{k}\cdot \nabla \phi dx + \int_\Omega\zeta(v_k)v_k\phi dx=\int_\Omega f\phi dx\quad\text{for all}\quad\phi\in \mathcal{V}_{k}.$$$$

How can one justify this claim?

My Taught and ideas

In fact, that $$\phi_k\in L^\infty (\Omega)$$ is the only important property needed from $$\phi_k$$. So that by assumption the function $$\zeta(v_k)v_k$$ is bounded. Since we are in finite-dimensional space and $$\int_\Omega \nabla \phi_{i}\cdot \nabla \phi_j dx=\lambda_i\delta_{ij}$$, the above equation reduces into finding $$v_k=(v_{k,1}\phi_1+ \dotsb+v_{k,k}\phi_k)$$ satisfying $$$$\label{Star-3.v} \sum_{i=1}^k\lambda_iv_{k,i}\phi_i + \zeta(v_k)v_k = f_k\quad\text{in} \quad \mathcal{V}_{k},$$$$ where $$f_k=(f_{k,1}\phi_1+ \dotsb+f_{k,k}\phi_k)$$ is the projection of $$f$$ on $$\mathcal V_k$$. Note that by abuse of notation we again write $$\zeta(v_k)v_k$$ to denote its own projection on $$\mathcal{V}_k$$.

Recall Brouwer fixed-point theorem: Every continuous function from a closed ball of a Euclidean space into itself has a fixed point.

• Is $\varphi=\phi$ and does $\phi(v_k)$ mean the composition then? And what is the product $v_k\cdot b_k$ of a function with a vector? Feb 7, 2021 at 17:27
• @MartinVäth $v_k\cdot b_k= \sum v_{k,i}\lambda_i$ is just the standard scalar product. Feb 7, 2021 at 22:00
• @MartinVäth I have changed the notations.. sorry for the confusion Feb 7, 2021 at 22:34
• How can you reduce to the last equation? Shouldn't you have the projection of $\zeta(v_k)v_k$ on ${\mathcal V}_k$ in it? Feb 8, 2021 at 4:07
• @PietroMajer You are right. it is just an abuse of notation. the whole equation is projected in $\mathcal V_k$ Feb 8, 2021 at 8:57

What is needed is an a-priori $$L_\infty$$ bound for the solution $$v_k$$. If you know such an a-priori bound, you can modify $$\zeta$$ outside of this bound, and you can assume without of generality that $$\zeta(u)=0$$ for large $$|u|$$. (More precisely, you need the same a-priori bounded for the modified equation, that is, you have to know that any solution of the modified equation is also a solution of the original equation.)

Then the finite-dimensional equation is of the form $$Av+F(v)=0$$ where $$A$$ is linear and positive definite, and $$F$$ is continuous and bounded. In particular, $$A^{-1}$$ exists, and the equation thus is equivalent to $$v=-A^{-1}F(v)\text.$$ The range of the map $$G=-A^{-1}F$$ is contained in some ball. In particular, $$G$$ maps this ball into itself, and so Brouwer's fixed point theorem implies that $$G$$ has a fixed point which thus is a solution of the finite-dimensional equation.

• The assumption says the derivative of $\zeta(t)t$ is locally bounded and not the function itself. But yes this assumption implies the local boundedness $\zeta(t)t$. Feb 10, 2021 at 22:25
• I know, but I guess that some assumption was forgotten, because it was used in the question e.g. that $\zeta(v_k)v_k$ is bounded. Moreover, I am quite sure that the assertion is not provable from the finite-dimensional reduction if practically nothing more than the continuity of $u\mapsto z(u)u$ is assumed - local boundedness of the derivative is practically an empty hypothesis concerning Brouwer. It is sufficient that $\frac{\lVert F(u)\rVert}{\lVert u\rVert}\to0$ as $\rVert u\rVert\to\infty$, though. This is a bit weaker than global boundedness. Feb 10, 2021 at 22:38
• How to you get that the range of G is in a ball? I tough we have to find $G$ and $R>0$ such that $\|v\|\leq R\implies \|G(v)\|\leq R$? This would solve the problem. Feb 10, 2021 at 22:40
• If $u\mapsto\zeta(u)u$ is (globally) bounded, then $F$ (and thus $G$) is globally bounded, that is, there is some $R$ such that $\lVert G(v)\rVert\le R$ for every $v$. As mentioned in the previous comment, instead of the boundedness, sublinear growth near $\infty$ is suffiicient. Feb 10, 2021 at 22:42
• Note that $\phi_k's\in L^\infty$ and thus $v_k\in L^\infty$ because $v_k\in \mathcal V_k$ that is why we have that $\zeta(v_k)v_k$ is bounded by assumption. Feb 10, 2021 at 22:44

Only now I realize the condition that $$\zeta$$ is nonnegative. (Was it really there in the first formulation of the question?)

With this condition, it is possible to get the required a-priori bound required for my other reply by a simple sign argument:

Choose the test function $$\varphi=v_k$$ in the equation.

Then the first summmand in that equation is bounded from below by $$c\lVert v_k\rVert_{L_2}^2$$ where $$c>0$$ is a constant coming from Poincaré's inequality, the second summand is nonnegative, and the absolute value of the last summand is bounded from above by $$\lVert v_k\rVert_{L_2}$$ by Cauchy-Schwarz. Hence, the equation cannot hold if $$\lVert v_k\rVert_{L_2}\ge R$$, where $$R>0$$ is independent of the particular form of $$\zeta$$.

Hence, you can replace $$\zeta$$ by $$\widetilde\zeta(v)=\lambda(\lVert v\rVert_{L_2})\zeta(v)$$ where $$\lambda\colon[0,\infty)\to[0,1]$$ is continuous with $$\lambda|_{[0,R]}=1$$ and $$\lambda_{[R+1,\infty)}=0$$, and for both equations the solutions have $$L_2$$-norm at most $$R$$, where the equations coincide. In other words: The original equation with $$\zeta$$ and the modified equation with $$\widetilde\zeta$$ have the same solutions.

For the modified equation $$v\mapsto\widetilde\zeta(v)v$$ is globally bounded, and the argument from the other comment applies.

First using Brouwer

Let $$w\in \mathcal{V}_k$$, necessarily $$\zeta(w)$$ is a bounded function since $$\phi_k$$'s are also bounded. The Lax-Milgram lemma implies there is a unique function $$\widehat{w}\in \mathcal{V}_k$$ such that \begin{align} \int_\Omega\nabla\widehat{w}\nabla\phi +\zeta(w)\widehat{w}\phi -f\psi dx=0\quad\text{for all}\quad\phi\in \mathcal{V}_{k}. \end{align}

The Poincar'{e}--Friedrichs inequality yields $$$$\int_\Omega|\nabla\widehat{w}|^2dx + \int_\Omega \zeta(w )\widehat{w}^2\, d x\leq \|f\|_{L^{2}(\Omega)}\,\|\widehat{w}\|_{L^{2}(\Omega)}\leq C\|f\|_{L^{2}(\Omega)}\, \|\widehat{w}\|_{H_0^1(\Omega )}$$$$

Thus, letting $$R=C\,\|f\|_{L^{2}(\Omega)}$$, since $$\varphi\geq0$$ we obtain the following estimates \begin{align}\label{eq:boundedmapT} &\|\widehat{w}\|_{H_0^1(\Omega )}\leq R \quad\text{ and }\quad \int_\Omega \zeta(w )\widehat{w}^2\,dx\leq R^2. \end{align} We let $$\mathcal{B}_R=\big\{ w\in \mathcal{V}_k: \|w\|_{H_0^1(\Omega)} \leq R\big\}$$, be the closed ball in $$\mathcal{V}_k$$ of radius $$R$$ centered at the origin. Clearly, the mapping $$T:\mathcal{V}_k\to \mathcal{B}_R$$ with $$Tw=\widehat{w}$$ is well defined.

It remains to prove that $$T$$ is a continuous mapping. Indeed, let $$\{w_n\}$$ be a sequence in $$\mathcal{V}_k$$ with $$w_n= \lambda_{1,n}\phi_1+\cdots+ \lambda_{k,n}\phi_k$$ converging in $$\mathcal{V}_k$$ to a function $$w= \lambda_1\phi_1+\cdots+ \lambda_k\phi_k$$; i.e., $$\lambda_{\ell,n}\xrightarrow{n\to\infty } \lambda_\ell$$, $$\ell=1,2,\cdots,k$$. By continuity we have $$\varphi(w_n)\xrightarrow{n\to \infty}\varphi(w)$$ almost everywhere. In addition, the convergence in $$L^2(\Omega)$$ also holds, since the continuity gives $$\sup_{n\geq 0} \|\varphi(w_n) \|_{L^\infty(\Omega)} <\infty$$ because $$\sup_{n\geq 0}\|w_n\|_{L^\infty(\Omega) } <\infty$$.

On the other side, in virtue of the first estimate above the sequence $$\{Tw_n\}$$ is bounded in finite dimensional space $$\mathcal{V}_k$$ and thus converges $$\mathcal{V}_k$$ up to a subsequence to some $$w_*\in \mathcal{V}_k$$. Altogether, it follows that, for all $$\phi\in \mathcal{V}_k\subset L^\infty(\Omega)$$ % \begin{align*} (f,\phi)= \lim_{n\to \infty} \int_\Omega\nabla\widehat{w}_n\nabla\phi +\zeta(w_n)\widehat{w}_n\phi dx= \int_\Omega\nabla\widehat{w}\nabla\phi +\zeta(w)w_* \phi. \end{align*}

The uniqueness of $$\widehat{w}$$ entails $$w_*=\widehat{w}=Tw$$ and hence the whole sequence $$\{Tw_n\}$$ converges in $$Tw$$ in $$\mathcal V_k$$, which gives the continuity of $$T$$.

Therefore, by the Brouwer fixed-point theorem, $$T$$ has a fixed point $$v_k\in \mathcal V_k$$, i.e., $$v_k=Tv_k$$ which clearly satisfies the announced relation.

An alternative.

The given problem is equivalent to the minimization problem \begin{align} \mathcal{J}(v_k)= \min_{v\in \mathcal{V}_k} \mathcal{J}(v)\quad \text{with}\quad \mathcal{J}(v):= \frac12 \int_\Omega|\nabla v|^2 dx + \int_\Omega G(v)d x + \int_\Omega fvd x \end{align} and we define the function $$G(v)= \int_0^v\zeta(\tau)\tau d \tau$$. Note that $$G$$ is non-negative since $$\zeta(\tau)\geq 0$$ and that $$\mathcal J$$ is continuous on $$\mathcal V_k$$. Using the Poincaré-Friedrichs inequality we find that $$\mathcal{J}(v)\to \infty$$, as $$\|v\|_{L^2(\Omega)} \to \infty$$ and $$v\in \mathcal{V}_k$$. Which implies the existence of a minimizer $$v_k\in \mathcal{V}_k$$ of \mathcal{J}, since we are in finite dimension space.

• Why is $G$ non-negative? $v$ can be negative. Feb 14, 2021 at 13:17
• for the negative $v \leq 0$ you have to reverse the integral \int_v^0 you will get $G\geq0$ Feb 14, 2021 at 18:13
• I haven't calculated carefully, but I think that if you have $G(v)=G(|v|)$ the the derivative for e.g. strictly negative $v$ has the wrong sign. My reason to assume this without calculation is that in case $\zeta(0)\ne0$ the derivative at $0$ does not exist while I think that this should be the case for the "correct" definition of $G$. Feb 15, 2021 at 20:26
• If $v\geq 0$ then $G(v)\geq 0$ this is clear. If $v\leq 0$ then $\zeta(t)t\leq 0$ for all $v\leq t\leq 0$ which implies $\int_v^0 \zeta(t)td t\leq 0$ that is $G(v)=\int_0^v \zeta(t)tdt\geq 0$ Feb 15, 2021 at 22:59
• Oh, stupid me! Thank you! I was already so confused that the positive definitness of the derivative (which I had used in my proof) and the boundedness of the functional from below (which you had used) seemed unrelated to me at a first glance... Of course, the former must imply the latter (and actual convexity of the functional). Feb 16, 2021 at 18:55