$ \newcommand{\Ua}{\mathrm{a}} \newcommand{\Ub}{\mathrm{b}} \newcommand{\Uc}{\mathrm{c}} \newcommand{\Ud}{\mathrm{d}} \newcommand{\Ue}{\mathrm{e}} \newcommand{\Uf}{\mathrm{f}} \newcommand{\Ug}{\mathrm{g}} \newcommand{\Uh}{\mathrm{h}} \newcommand{\Ui}{\mathrm{i}} \newcommand{\Uj}{\mathrm{j}} \newcommand{\Uk}{\mathrm{k}} \newcommand{\Ul}{\mathrm{l}} \newcommand{\Um}{\mathrm{m}} \newcommand{\Un}{\mathrm{n}} \newcommand{\Uo}{\mathrm{o}} \newcommand{\Up}{\mathrm{p}} \newcommand{\Uq}{\mathrm{q}} \newcommand{\Ur}{\mathrm{r}} \newcommand{\Us}{\mathrm{s}} \newcommand{\Ut}{\mathrm{t}} \newcommand{\Uu}{\mathrm{u}} \newcommand{\Uv}{\mathrm{v}} \newcommand{\Uw}{\mathrm{w}} \newcommand{\Ux}{\mathrm{x}} \newcommand{\Uy}{\mathrm{y}} \newcommand{\Uz}{\mathrm{z}} \newcommand{\UA}{\mathrm{A}} \newcommand{\UB}{\mathrm{B}} \newcommand{\UC}{\mathrm{C}} \newcommand{\UD}{\mathrm{D}} \newcommand{\UE}{\mathrm{E}} \newcommand{\UF}{\mathrm{F}} \newcommand{\UG}{\mathrm{G}} \newcommand{\UH}{\mathrm{H}} 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\newcommand{\JN}{\mathit{N}} \newcommand{\JO}{\mathit{O}} \newcommand{\JP}{\mathit{P}} \newcommand{\JQ}{\mathit{Q}} \newcommand{\JR}{\mathit{R}} \newcommand{\JS}{\mathit{S}} \newcommand{\JT}{\mathit{T}} \newcommand{\JU}{\mathit{U}} \newcommand{\JV}{\mathit{V}} \newcommand{\JW}{\mathit{W}} \newcommand{\JX}{\mathit{X}} \newcommand{\JY}{\mathit{Y}} \newcommand{\JZ}{\mathit{Z}} % \newcommand{\Jzero }{\mathit{0}} \newcommand{\Jone }{\mathit{1}} \newcommand{\Jtwo }{\mathit{2}} \newcommand{\Jthree}{\mathit{3}} \newcommand{\Jfour }{\mathit{4}} \newcommand{\Jfive }{\mathit{5}} \newcommand{\Jsix }{\mathit{6}} \newcommand{\Jseven}{\mathit{7}} \newcommand{\Jeight}{\mathit{8}} \newcommand{\Jnine }{\mathit{9}} % \newcommand{\BA}{\boldsymbol{A}} \newcommand{\BB}{\boldsymbol{B}} \newcommand{\BC}{\boldsymbol{C}} \newcommand{\BD}{\boldsymbol{D}} \newcommand{\BE}{\boldsymbol{E}} \newcommand{\BF}{\boldsymbol{F}} \newcommand{\BG}{\boldsymbol{G}} \newcommand{\BH}{\boldsymbol{H}} 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% $

This post builds on the formulations I showed in my previous posts by introducing their nonlinear versions.

In a typical nonlinear problem, the variational setting leads to the weak formulation

Find $u\in V$ such that
\[\begin{equation} F(u,v) = 0 \label{eq:femnonlinear2} \end{equation}\]
for all $v\in V$ where the semilinear form $F$ is nonlinear in terms of $u$ and linear in terms of $v$.

We linearize $F$:

\[\begin{equation} \Lin [F(u,v)]_{u=\bar{u}} = F(\bar{u}, v) + \varn{F(u,v)}{u}{\Var u}\evat_{u=\bar{u}} \label{eq:femnonlinear4} \end{equation}\]

Equating \eqref{eq:femnonlinear4} to zero yields a linear system in terms of $\Var u$

\[\begin{equation} \boxed{ a(\Var u, v) = b(v) } \label{eq:femnonlinear6} \end{equation}\]

where

\[\begin{equation} \boxed{ \begin{aligned} a(\Var u, v) &= \varn{F(u,v)}{u}{\Var u}\evat_{u=\bar{u}} \\ b(v) &= -F(\bar{u}, v) . \end{aligned} } \label{eq:femnonlinear5} \end{equation}\]

We can compute the components of the matrices and vectors according to \eqref{eq:femnonlinear6}

\[\begin{equation} \boxed{ \begin{alignedat}{3} \Aelid{I\!J}{} &= a(N^J,N^I) &&= \varn{F(u,N^I)}{u}{N^J}\evat_{u=\bar{u}} \\ b^{I} &= b(N^I) &&= -F(\bar{u}, N^I). \end{alignedat} } \label{eq:femnonlinear9} \end{equation}\]

Then the update vector $\Var \Bu = [\Var u^1, \Var u^2, \dots, \Var u^\nnode]\tra$ is obtained by solving

\[\begin{equation} \BA \Var \Bu = \Bb \end{equation}\]

Letting $\Var u$ be the difference between consequent iterates, we obtain the update equation as

\[\begin{equation} \boxed{ \Bu \leftarrow \bar{\Bu} + \Var\Bu } \end{equation}\]

Example: Nonlinear Poisson’s Equation

Consider the following nonlinear Poisson’s equation

\[\begin{equation} \begin{alignedat}{4} - \nabla \dtp (g(u)\nabla u) &= f \quad && \text{in} \quad && \Omega \\ u &= 0 \quad && \text{on} \quad && \del\Omega \end{alignedat} \label{eq:femnonlinear8} \end{equation}\]

The weak formulation reads

Find $u\in V$ such that
\[\begin{equation} - \int_\Omega \nabla \dtp (g(u)\nabla u) v \dv= \int_\Omega f v \dv \end{equation}\]
for all $v\in V$ where $V=H^1_0(\Omega)$.

Applying integration by parts and divergence theorem on the left-hand side

\[\begin{equation} \begin{aligned} \int_\Omega \nabla \dtp (g(u)\nabla u) v \dv &= \int_\Omega \nabla \dtp (g(u)\nabla (u) v) \dv - \int_\Omega g(u)\nabla u\dtp\nabla v \dv \\ &= \underbrace{\int_{\del\Omega} g(u) v (\nabla u\dtp\Bn) \da}_{v = 0 \text{ on } \del\Omega} - \int_\Omega g(u)\nabla u\dtp\nabla v \dv \\ \end{aligned} \end{equation}\]

Thus we have the semilinear form

\[\begin{equation} F(u,v) = \int_{\Omega} g(u) \nabla u \dtp \nabla v \dv - \int_{\Omega} f \, v \dv = 0 \end{equation}\]

The linearized version of this problem is then with \eqref{eq:femnonlinear5}

\[\begin{equation} \begin{aligned} a(\Var u,v) &= \int_{\Omega} \rbr{\deriv{g}{u}\evat_{\bar{u}} \Var u\, \nabla \bar{u} + g(\bar{u})\nabla(\Var u)} \dtp \nabla v \dv \\ b(v) &= \int_{\Omega} [f \, v - g(\bar{u}) \nabla \bar{u} \dtp \nabla v] \dv\\ \end{aligned} \end{equation}\]

and the matrix and vector components are with \eqref{eq:femnonlinear9}

\[\begin{equation} \begin{aligned} \Aelid{I\!J}{} &= \int_{\Omega} \rbr{\deriv{g}{u}\evat_{\bar{u}} N^J \, \nabla \bar{u} + g(\bar{u})\BB^J} \dtp \BB^I \dv \\ b^{I} &= \int_{\Omega} [f \, N^I - g(\bar{u}) \nabla \bar{u} \dtp \BB^I ] \dv\\ \end{aligned} \end{equation}\]

where the previous solution and its gradient are computed as

\[\begin{equation} \bar{u} = \suml{I=1}{\nnode} \bar{u}^I N^I \eqand \nabla \bar{u} = \suml{I=1}{\nnode} \bar{u}^I \BB^I . \end{equation}\]

Nonlinear Time-Dependent Problems

In the case of a nonlinear time-dependent problem, we have the following weak form:

Find $u \in V$ such that
\[\begin{equation} m(\dot{u}, v; t) + F(u,v; t) = 0 \label{eq:nonlineartimedependentweak1} \end{equation}\]
for all $v \in V$ and $t \in [0,\infty)$ where $F$ is a semilinear form.

Discretization yields the following nonlinear system of equations

\[\begin{equation} \BM(t)\Bu + \Bf(u; t) = \Bzero \end{equation}\]

where

\[\begin{equation} \begin{aligned} M^{I\!J}(t) &= m(N^J, N^I; t) \\ f^{I}(u;t) &= F(u, N^I; t). \end{aligned} \end{equation}\]

Explicit Euler Scheme

We discretize in time with the finite difference $\dot{u} \approx [u_{n+1}-u_n]/{\Delta t}$ and linearity allows us to write

\begin{equation} \boxed{ m(\dot{u}, v; t) \approx \frac{1}{\Delta t} [m(u_{n+1}, v; t_{n+1}) - m(u_n, v; t_n)] } \label{eq:discretetimedependent1} \end{equation}

We discretize the variational forms in time according to \eqref{eq:discretetimedependent1}, and evaluate the remaining terms at $t_n$:

\[\begin{equation} \frac{1}{\Delta t} [m(u_{n+1},v;t_{n+1}) - m(u_{n},v;t_{n})] + F(u_n, v; t_n) = 0 \end{equation}\]

The corresponding system of equations is

\[\begin{equation} \frac{1}{\Delta t} [\BM_{n+1}\Bu_{n+1} - \BM_n\Bu_n] + \Bf_n = \Bzero \end{equation}\]

where $\Bf_n = \Bf(u_n, t_n)$. This yields the following update equation

\[\begin{equation} \boxed{ \Bu_{n+1} = \BM_{n+1}\inv [\BM_n\Bu_n - \Delta t \Bf_n] } \end{equation}\]

For a time-independent $m$, this becomes

\[\begin{equation} \Bu_{n+1} = \Bu_n - \Delta t \BM\inv\Bf_n \end{equation}\]

Implicit Euler Scheme

For the implicit scheme, we evaluate the remaining terms at $t_{n+1}$ and let the result be equal to

\[\begin{equation} G(u_{n+1}, v) := \frac{1}{\Delta t} [m(u_{n+1},v;t_{n+1}) - m(u_{n},v;t_{n})] + F(u_{n+1}, v; t_{n+1}) = 0 \end{equation}\]

We will hereon replace $u_{n+1}$ with $u$ for brevity. The update of this nonlinear system requires the linearization of $G(u, v)$:

\[\begin{equation} \Lin[G(u,v)]_{u=\bar{u}} = G(\bar{u}, v) + \varn{G}{u}{\Var u}\evat_{u=\bar{u}} = 0 \end{equation}\]

We thus have the following linear setting for the Newton update $\Var u$:

\[\begin{equation} a(\Var u, v) = b(v) \end{equation}\]

where

\[\begin{equation} \begin{aligned} a(\Var u, v) &:= \varn{G}{u}{\Var u} \evat_{u=\bar{u}} = \frac{1}{\Delta t} m(\Var u, v; t_{n+1}) + \varn{F(u, v; t_{n+1})}{u}{\Var u} \evat_{u=\bar{u}} \\ b(v) &:= -G(\bar{u}, v) = - F(\bar{u}, v; t_{n+1}) -\frac{1}{\Delta t} [m(\bar{u},v;t_{n+1}) - m(u_{n},v;t_{n})] \end{aligned} \end{equation}\]

Discretization yields

\[\begin{equation} \rbr{\frac{1}{\Delta t} \BM_{n+1} + \tilde{\BA}}\Var \Bu = \Bb \end{equation}\]

where

\[\begin{equation} \tilde{A}^{I\!J} := \varn{F(u, N^I;t_{n+1})}{u}{N^J} \evat_{u=\bar{u}} \eqand b^I := b(N^I) \end{equation}\]

The Newton update is rendered

\[\begin{equation} \boxed{ \Bu \leftarrow \bar{\Bu} + \Var\Bu \eqwith \Var \Bu = [\frac{1}{\Delta t} \BM_{n+1} + \tilde{\BA}]\inv\Bb } \end{equation}\]

which is repeated until the solution for the next timestep $\Bu$ converges to a satisfactory value.

Nonlinear Coupled Problems

For a nonlinear coupled problem, the weak formulation is as follows

Find $u\in V_1$, $y\in V_2$ such that
\[\begin{equation} \begin{aligned} F(u, y, v) &= 0 \\ G(u, y, w) &= 0 \\ \end{aligned} \label{eq:nonlinearcoupled1} \end{equation}\]
for all $v\in V_1$, $w \in V_2$ where $F(\cdot,\cdot, \cdot)$, $G(\cdot, \cdot, \cdot)$ are nonlinear in terms of $u$ and $y$ and linear in terms of $v$ and $w$.

We linearize the semilinear forms about the nonlinear terms:

\[\begin{equation} \begin{alignedat}{4} \Lin[F(u, y, v)]_{\bar{u},\bar{y}} &= F(\bar{u},\bar{y},v) &&+ \varn{F(u, y, v)}{u}{\Var u} \evat_{\bar{u},\bar{y}} &&+ \varn{F(u, y, v)}{y}{\Var y} \evat_{\bar{u},\bar{y}} \\ \Lin[G(u, y, w)]_{\bar{u},\bar{y}} &= G(\bar{u},\bar{y},w) &&+ \varn{G(u, y, w)}{u}{\Var u} \evat_{\bar{u},\bar{y}} &&+ \varn{G(u, y, w)}{y}{\Var y} \evat_{\bar{u},\bar{y}} \end{alignedat} \label{eq:nonlinearcoupled2} \end{equation}\]

where the evaluations take place at $u=\bar{u}$ and $y=\bar{y}$.

Equating the linearized residuals to zero, we obtain a linear system of the form

\[\begin{equation} \begin{alignedat}{3} a(u, v) &+ b(y, v) &&= c(v) \\ d(u, w) &+ e(y, w) &&= f(w) \\ \end{alignedat} \label{eq:coupledweakform1} \end{equation}\]

with the bilinear forms $a$, $b$, $d$, $e$ and the linear forms $c$, $f$ which are defined as

\[\begin{equation} \begin{gathered} \begin{alignedat}{4} a(\Var u, v) &:= \varn{F(u, y, v)}{u}{\Var u} \evat_{\bar{u},\bar{y}} \quad & b(\Var y, v) &:= \varn{F(u, y, v)}{y}{\Var y} \evat_{\bar{u},\bar{y}} \\ d(\Var u, w) &:= \varn{G(u, y, w)}{u}{\Var u} \evat_{\bar{u},\bar{y}} \quad & e(\Var y, w) &:= \varn{G(u, y, w)}{y}{\Var y} \evat_{\bar{u},\bar{y}} \end{alignedat} \\ \text{and} \\ \begin{aligned} c(v) &:= -F(\bar{u},\bar{y},v) \\ f(w) &:= -G(\bar{u}, \bar{y}, w) \end{aligned} \end{gathered} \end{equation}\]

Discretizing as done in the previous section, we obtain the following linear system of equations

\[\begin{equation} \begin{bmatrix} \BA & \BB \\ \BD & \BE \end{bmatrix} \begin{bmatrix} \Var \Bu \\ \Var \By \end{bmatrix} = \begin{bmatrix} \Bc \\ \Bf \end{bmatrix} \end{equation}\]

whose solution yields the update values $\Var \Bu$ and $\Var \By$. Thus the Newton update equations are

\[\begin{equation} \begin{alignedat}{3} \Bu &\leftarrow \bar{\Bu} &&+ \Var\Bu \\ \By &\leftarrow \bar{\By} &&+ \Var\By . \end{alignedat} \end{equation}\]

Example: Cahn-Hilliard Equation

The Cahn-Hilliard equation describes the process of phase separation, by which the two components of a binary fluid spontaneously separate and form domains pure in each component. The problem is nonlinear, coupled and time-dependent. The IBVP reads

\[\begin{equation} \begin{alignedat}{4} \partd{c}{t} &= \nabla\dtp(\BM\nabla \mu) \qquad&& \text{in} \qquad&& \Omega\times I \\ \nabla c\dtp\Bn &= 0 && \text{on} && \del\Omega\times I\\ \nabla \mu\dtp\Bn &= 0 && \text{on} && \del\Omega\times I\\ c &= c_0 && \text{in} && \Omega, t = 0 \\ \mu &= 0 && \text{in} && \Omega, t = 0 \\ \end{alignedat} \label{eq:cahnhilliard1} \end{equation}\]

where

\[\begin{equation} \mu = \deriv{f}{c} - \nabla\dtp(\BLambda\nabla c) \label{eq:cahnhilliard2} \end{equation}\]

and $t\in I = [0,\infty)$. Here,

$c$ is the scalar variable for concentration,
$\mu$ is the scalar variable for the chemical potential,
$f: c \mapsto f(c)$ is the function representing chemical free energy,
$\BM$ is a second-order tensor describing the mobility of the chemical,
$\BLambda$ is a second-order tensor describing both the interface thickness and direction of phase transition.

The fourth-order PDE governing the problem can be formulated as a coupled system of two second-order PDEs with the variables $c$ and $\mu$, as demonstrated in \eqref{eq:cahnhilliard1} and \eqref{eq:cahnhilliard2}.

The weak formulation then reads

Find $c \in V_1$, $\mu\in V_2$ such that
\[\begin{equation} \begin{aligned} \int_\Omega \partd{c}{t} v \dx - \int_\Omega \nabla\dtp(\BM\nabla \mu) v \dx &=0 \\ \int_\Omega \sbr{\mu - \deriv{f}{c}} w \dx + \int_\Omega \nabla\dtp(\BLambda\nabla c) w\dx &= 0 \end{aligned} \end{equation}\]
for all $v \in V_1$, $w \in V_2$ and $t \in I$.

We discretize in time implicitly with $\del c/\del t \approx (c_{n+1}-c_n)/\Var t$. We also denote the values for the next timestep $c_{n+1}$ and $\mu_{n+1}$ as $c$ and $\mu$ for brevity. Using integration-by-parts, the divergence theorem, and the given boundary conditions, we arrive at the following nonlinear forms

\[\begin{equation} \begin{alignedat}{3} F(c,\mu,v) &= \int_\Omega \frac{1}{\Var t} (c-c_n) v \dx + \int_\Omega (\BM\nabla \mu)\dtp \nabla v \dx &&= 0 \\ G(c,\mu,w) &= \int_\Omega \sbr{\mu - \deriv{f}{c}} w \dx - \int_\Omega (\BLambda\nabla c)\dtp \nabla w\dx &&= 0 \end{alignedat} \end{equation}\]

which is a nonlinear coupled system of the form \eqref{eq:nonlinearcoupled1}.

We linearize the forms according to \eqref{eq:nonlinearcoupled2} and obtain the following variations

\[\begin{align*} \varn{F}{c}{\Var c} &= \int_\Omega \frac{1}{\Var t} \Var c\, v \dx \\ \varn{F}{\mu}{\Var \mu} &= \int_\Omega (\BM\nabla (\Var\mu))\dtp \nabla v \dx \\ \varn{G}{c}{\Var c} &= - \int_\Omega \dderiv{f}{c}\Var c \, w \dx - \int_\Omega (\BLambda\nabla (\Var c))\dtp \nabla w\dx \\ \varn{G}{\mu}{\Var \mu} &= \int_\Omega \Var\mu \, w \dx \end{align*}\]

We substitute basis functions and obtain our system matrix and vectors

\[\begin{align*} P^{I\!J} &= \int_\Omega \frac{1}{\Var t} N^JN^I \dx \\ Q^{IL} &= \int_\Omega (\BM\BB^L)\dtp\BB^I \dx \\ r^{I} &= \int_\Omega \frac{1}{\Var t}(\bar{c}-c_n)N^I \dx + \int_\Omega (\BM\nabla\bar{\mu})\dtp\BB^I\dx \\ S^{K\!J} &= - \int_\Omega \dderiv{f}{c}\evat_{c=\bar{c}} N^J N^K \dx - \int_\Omega (\BLambda \BB^J)\dtp \BB^K\dx \\ T^{K\!L} &= \int_\Omega N^L N^K \dx \\ u^{K} &= \int_\Omega \sbr{\bar{\mu} - \deriv{f}{c}\evat_{c=\bar{c}}} N^K \dx - \int_\Omega (\BLambda\nabla \bar{c})\dtp \BB^K\dx \end{align*}\]

which constitute the system

\[\begin{equation} \begin{bmatrix} \BP & \BQ \\ \BS & \BT \end{bmatrix} \begin{bmatrix} \Var \Bc \\ \Var \Bmu \end{bmatrix} = \begin{bmatrix} \Br \\ \Bu \end{bmatrix} \end{equation}\]

Solution yields the update values $\Var \Bc$ and $\Var \Bmu$. The Newton update equations are then

\[\begin{equation} \begin{alignedat}{3} \Bc &\leftarrow \bar{\Bc} &&+ \Var\Bc \\ \Bmu &\leftarrow \bar{\Bmu} &&+ \Var\Bmu . \end{alignedat} \end{equation}\]

The system is solved for $c_{n+1}$ and $\mu_{n+1}$ at each $t=t_n$ to obtain the evolutions of the concentration and chemical potential.