6. Nonlinear problems
Consider a nonlinear (partial) differential equation \(P[u]=0\), e.g. the Burgers equation (Johannes Martinus Burgers 1895–1981)
(6.1)\[
\begin{equation}
\frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x} = 0 \ , \ 0\leq x < 2\pi
\end{equation}
\]
with periodic boundary conditions on a domain \(L=2\pi\). Due to the periodicity, a Fourier Galerkin scheme is an obvious choice for the spatial discretisation. The fundamental wave number \(\alpha=2\pi/L=1\); the trial functions are thus \(\phi_k=\Phi_k = e^{\mathrm{i} kx}\) and the test functions \(\phi_l=\Phi_l = e^{\mathrm{i} lx}\). The approximation for the solution is
\[
\begin{equation}
u_N(x,t) = \sum_{k=-K}^K \hat{u}_k(t) \phi_k(x) = \sum_{k=-K}^K \hat{u}_k(t) e^{\mathrm{i}kx} \ ,
\end{equation}
\]
with \(N=2K+1\) unknown coefficients. Inserting \(u_N\) into the Burger equation (6.1) gives
\[
\begin{equation}
\sum_{k=-K}^K \frac{\mathrm{d}\hat{u}_k}{\mathrm{d}t} \phi_k(x) + \sum_{k=-K}^K \hat{u}_k\phi_k(x) \sum_{m=-K}^K \hat{u}_m\frac{\mathrm{d}\phi_m(x)}{\mathrm{d}x} = 0 \ .
\end{equation}
\]
The next step is to multiply with the \(N\) test functions \(\phi_l\) together with the integration over the domain obtaining a system of equations with \(l=0,\ldots,N-1\)
\[
\begin{equation}
\int_0^{2\pi} \phi_l(x) \sum_{k=-K}^K \frac{\mathrm{d}\hat{u}_k}{\mathrm{d}t} \phi_k(x) \mathrm{d}x + \int_0^{2\pi} \phi_l(x)\sum_{k=-K}^K \hat{u}_k\phi_k(x) \sum_{m=-K}^K \hat{u}_m\frac{\mathrm{d}\phi_m(x)}{\mathrm{d}x}\mathrm{d}x = 0 \ ,
\end{equation}
\]
which can be rewritten as follows using the product and derivative properties of the \(\phi_i\)
\[
\begin{equation}
\sum_{k=-K}^K \frac{\mathrm{d}\hat{u}_k}{\mathrm{d}t} \int_0^{2\pi}\phi_k(x) \phi_l(x)\mathrm{d}x + \sum_{k=-K}^K\sum_{m=-K}^K \mathrm{i} m \hat{u}_k \hat{u}_m\int_0^{2\pi} \phi_l(x)\phi_{k+m}(x) \mathrm{d}x = 0 \ ,
\end{equation}
\]
Employing the orthogonality relations (3.1)
\[
\begin{equation}
\int_0^{2\pi} \phi_k(x) \phi_{-l}(x) \mathrm{d}x = 2\pi \delta_{kl}
\end{equation}
\]
gives first
\[
\begin{equation}
\sum_{k=-K}^K \frac{\mathrm{d}\hat{u}_k}{\mathrm{d}t} \delta_{-l,k} + \sum_{k=-K}^K\sum_{m=-K}^K \mathrm{i} m \hat{u}_k \hat{u}_m \delta_{-l,k+m} = 0 \ ,
\end{equation}
\]
which allows to get rid of one summation in each term yielding
\[\begin{split}
\begin{equation}
\frac{\mathrm{d}\hat{u}_{-l}}{\mathrm{d}t} + \sum_{\substack{k=-K\\ -l=k+m\\ |m|\leq K}}^K \mathrm{i} m \hat{u}_k \hat{u}_m = 0 \ ,
\end{equation}
\end{split}\]
and after replacing \(-l\) with \(l\) one obtains finally a system of nonlinear equations for the coefficients \(\hat{u}_l\)
(6.2)\[\begin{split}
\frac{\mathrm{d}\hat{u}_l}{\mathrm{d}t} + \sum_{\substack{k=-K\\ l=k+m\\ |m|\leq K}}^K \mathrm{i} m \hat{u}_k\hat{u}_m = 0 \ ,
\end{split}\]
where \(l\) goes from \(-K\) to \(K\).
The fairly complex summation in the second term resembling a convolution is a consequence of the nonlinearity. Since we need to evaluate a sum for all components \(l\) of the solution, the number of operations is \(\mathcal{O}(K^2)\) which makes this sum evaluation the computationally most expensive part.
