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 \begin{equation}\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}. \end{equation}

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 \begin{equation}\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}, \end{equation} 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.

  • 1
    $\begingroup$ 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? $\endgroup$ Commented Feb 7, 2021 at 17:27
  • $\begingroup$ @MartinVäth $v_k\cdot b_k= \sum v_{k,i}\lambda_i$ is just the standard scalar product. $\endgroup$
    – Guy Fsone
    Commented Feb 7, 2021 at 22:00
  • $\begingroup$ @MartinVäth I have changed the notations.. sorry for the confusion $\endgroup$
    – Guy Fsone
    Commented Feb 7, 2021 at 22:34
  • $\begingroup$ 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? $\endgroup$ Commented Feb 8, 2021 at 4:07
  • $\begingroup$ @PietroMajer You are right. it is just an abuse of notation. the whole equation is projected in $\mathcal V_k$ $\endgroup$
    – Guy Fsone
    Commented Feb 8, 2021 at 8:57

3 Answers 3


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.

  • $\begingroup$ 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 $\zeta(t)t$. $\endgroup$
    – Guy Fsone
    Commented Feb 10, 2021 at 22:25
  • $\begingroup$ 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. $\endgroup$ Commented Feb 10, 2021 at 22:38
  • $\begingroup$ 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. $\endgroup$
    – Guy Fsone
    Commented Feb 10, 2021 at 22:40
  • $\begingroup$ 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. $\endgroup$ Commented Feb 10, 2021 at 22:42
  • $\begingroup$ 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. $\endgroup$
    – Guy Fsone
    Commented 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 \begin{equation} \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 )} \end{equation}

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.

  • $\begingroup$ Why is $G$ non-negative? $v$ can be negative. $\endgroup$ Commented Feb 14, 2021 at 13:17
  • $\begingroup$ for the negative $v \leq 0$ you have to reverse the integral \int_v^0 you will get $G\geq0$ $\endgroup$
    – Guy Fsone
    Commented Feb 14, 2021 at 18:13
  • $\begingroup$ 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 $G$. $\endgroup$ Commented Feb 15, 2021 at 20:26
  • $\begingroup$ 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 $G(v)=\int_0^v \zeta(t)tdt\geq 0$ $\endgroup$
    – Guy Fsone
    Commented Feb 15, 2021 at 22:59
  • $\begingroup$ 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). $\endgroup$ Commented Feb 16, 2021 at 18:55

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