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 $\mathbb{R}$.

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 $L^2(\Omega)$.

**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.

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