Editing Spectral method (section)

===Nonlinear example===

We wish to solve the forced, transient, nonlinear [[Burgers' equation]] using a spectral approach.

Given <math>u(x,0)</math> on the periodic domain
<math>x\in\left[0,2\pi\right)</math>, find <math>u \in \mathcal{U}</math> such that
:<math>\partial_{t} u + u \partial_{x} u = \rho \partial_{xx} u + f \quad \forall x\in\left[0,2\pi\right), \forall t>0</math>
where &rho; is the [[viscosity]] coefficient. In weak conservative form this becomes
:<math>\left\langle \partial_{t} u , v \right\rangle = \Bigl\langle  \partial_x \left(-\tfrac12 u^2 + \rho \partial_{x} u\right) , v \Bigr\rangle + \left\langle f, v \right\rangle \quad \forall v\in \mathcal{V}, \forall t>0</math>
where  following [[inner product space|inner product]] notation. [[integration by parts|Integrating by parts]] and using periodicity grants
:<math>\langle \partial_{t} u , v \rangle = \left\langle   \tfrac12 u^2 - \rho \partial_{x} u  ,  \partial_x v\right\rangle+\left\langle f, v \right\rangle \quad \forall v\in \mathcal{V}, \forall t>0.</math>

To apply the Fourier–[[Galerkin method]], choose both
:<math>\mathcal{U}^N := \biggl\{ u : u(x,t)=\sum_{k=-N/2}^{N/2-1} \hat{u}_{k}(t) e^{i k x}\biggr\}</math>
and
:<math>\mathcal{V}^N :=\operatorname{span}\left\{ e^{i k x} : k\in -\tfrac12 N,\dots,\tfrac12N - 1\right\}</math>
where <math>\hat{u}_k(t):=\frac{1}{2\pi}\langle u(x,t), e^{i k x} \rangle</math>. This reduces the problem to finding <math>u\in\mathcal{U}^N</math> such that
:<math>\langle \partial_{t} u , e^{i k x} \rangle = \left\langle  \tfrac12 u^2 - \rho \partial_{x} u  ,  \partial_x e^{i k x}  \right\rangle + \left\langle f, e^{i k x} \right\rangle \quad \forall k\in \left\{ -\tfrac12N,\dots,\tfrac12N-1 \right\}, \forall t>0.</math>

Using the [[orthogonality]] relation <math>\langle e^{i l x}, e^{i k x} \rangle = 2 \pi \delta_{lk}</math> where <math>\delta_{lk}</math> is the [[Kronecker delta]], we simplify the above three terms for each <math>k</math> to see
:<math>
\begin{align}
\left\langle \partial_{t} u , e^{i k x}\right\rangle &= \biggl\langle     \partial_{t} \sum_{l} \hat{u}_{l} e^{i l x}     ,      e^{i k x} \biggr\rangle  = \biggl\langle    \sum_{l} \partial_{t} \hat{u}_{l} e^{i l x}    ,     e^{i k x} \biggr\rangle = 2 \pi \partial_t \hat{u}_k,
\\
\left\langle f , e^{i k x} \right\rangle &= \biggl\langle    \sum_{l} \hat{f}_{l} e^{i l x}    ,     e^{i k x}\biggr\rangle= 2 \pi \hat{f}_k, \text{ and}
\\
\left\langle
  \tfrac12 u^2 - \rho \partial_{x} u
  ,
  \partial_x e^{i k x}
\right\rangle
&=
\biggl\langle
    \tfrac12
    \Bigl(\sum_{p} \hat{u}_p e^{i p x}\Bigr)
    \Bigl(\sum_{q} \hat{u}_q e^{i q x}\Bigr)
    - \rho \partial_x \sum_{l} \hat{u}_l e^{i l x}
    ,
    \partial_x e^{i k x}
\biggr\rangle
\\
&=
\biggl\langle
    \tfrac12
    \sum_{p} \sum_{q} \hat{u}_p \hat{u}_q e^{i \left(p+q\right) x}
    ,
    i k e^{i k x}
\biggr\rangle
-
\biggl\langle
    \rho i \sum_{l} l \hat{u}_l e^{i l x}
    ,
    i k e^{i k x}
\biggr\rangle
\\
&=
-\tfrac12 i k
\biggl\langle
    \sum_{p} \sum_{q} \hat{u}_p \hat{u}_q e^{i \left(p+q\right) x}
    ,
    e^{i k x}
\biggr\rangle
- \rho k
\biggl\langle
    \sum_{l} l \hat{u}_l e^{i l x}
    ,
    e^{i k x}
\biggr\rangle
\\
&=
- i \pi k \sum_{p+q=k} \hat{u}_p \hat{u}_q - 2\pi\rho{}k^2\hat{u}_k.
\end{align}
</math>

Assemble the three terms for each <math>k</math> to obtain
:<math>
2 \pi \partial_t \hat{u}_k
=
- i \pi k \sum_{p+q=k} \hat{u}_p \hat{u}_q 
- 2\pi\rho{}k^2\hat{u}_k
+ 2 \pi \hat{f}_k
\quad k\in\left\{ -\tfrac12N,\dots,\tfrac12N-1 \right\}, \forall t>0.
</math>
Dividing through by <math>2\pi</math>, we finally arrive at
:<math>
\partial_t \hat{u}_k
=
- \frac{i k}{2} \sum_{p+q=k} \hat{u}_p \hat{u}_q 
- \rho{}k^2\hat{u}_k
+ \hat{f}_k
\quad k\in\left\{ -\tfrac12N,\dots,\tfrac12N-1 \right\}, \forall t>0.
</math>
With Fourier transformed initial conditions <math>\hat{u}_{k}(0)</math> and forcing <math>\hat{f}_{k}(t)</math>, this coupled system of ordinary differential equations may be integrated in time (using, e.g., a [[Runge Kutta]] technique) to find a solution. The nonlinear term is a [[convolution]], and there are several transform-based techniques for evaluating it efficiently. See the references by Boyd and Canuto et al. for more details.