A nonlinear Poisson equation#

Authors: Anders Logg and Hans Petter Langtangen

We shall now address how to solve non-linear PDEs. We will see that non-linear problems introduce some subtle differences on how we define the variational form.

The PDE problem#

As a model for the solution of non-linear PDEs, we take the following non-linear Poisson equation

(28)#\begin{align} - \nabla \cdot (q(u) \nabla u)&=f && \text{in } \Omega,\\ u&=u_D && \text{on } \partial \Omega, \end{align}

and the coefficients $$q(u)$$ makes the problem non-linear (unless q(u) is constant in $$u$$).

Variational formulation#

As usual, we multiply the PDE by a test function $$v\in \hat{V}$$, integrate over the domain, and integrate second-order derivatives by parts. The boundary integrals arising from integration by parts vanishes wherever we employ Dirichlet conditions. The resulting variational formulation of our model problem becomes:

Find $$u\in V$$ such that

(29)#\begin{align} F(u; v)&=0 && \forall v \in \hat{V}, \end{align}

where

(30)#\begin{align} F(u; v)&=\int_{\Omega}(q(u)\nabla u \cdot \nabla v - fv)\mathrm{d}x, \end{align}

and

(31)#\begin{align} V&=\left\{v\in H^1(\Omega)\vert v=u_D \text{on } \partial \Omega \right\}\\ \hat{V}&=\left\{v\in H^1(\Omega)\vert v=0 \text{on } \partial \Omega \right\} \end{align}

The discrete problem arises as usual by restricting $$V$$ and $$\hat{V}$$ to a pair of discrete spaces. The discrete non-linear problem can therefore be written as:

Find $$u_h \in V_h$$ such that

(32)#\begin{align} F(u_h, v) &=0 \quad \forall v \in \hat{V}_h, \end{align}

with $$u_h=\sum_{j=1}^N U_j\phi_j$$. Since $$F$$ is non-linear in $$u$$, the variational statement gives rise to a system of non-linear algebraic equation in the unknowns $$U_1,\dots,U_N$$.