The following question is from an exam for the course Model Reduction. There are no answers and I don't really know what steps to take. Looking for an expert that can show me how it's done.

The viscous Burgers’ equation

$$\frac{\partial w}{\partial t}+w \frac{\partial w}{\partial x}=2\frac{\partial^2 w}{\partial x^2}+e(x)u(t) \qquad \qquad (1)$$

represents the velocity $w(x,t)$ of an incompressible fluid a t location $x \in [0, L]$ and time $t > 0$. Here, $e:[0, L] \rightarrow \mathbb{R}$ is an indicator function for the location at which an external force with input $u(t)$ is applied to the fluid flow. Solutions $w$ of $(1)$ are approximated by the finite expansions

$$w_r(x,t)=\sum_{k=1}^{r}a_k(t)\varphi_k(x)$$ where $\{\varphi_k \ | \ k=1,\dots,r\}$ is an orthonormal set of square integrable functions in the Hilbert space $L_2([0, L]).$

Derive expressions for the coefficients $a_k$(t) such that $w_r$ is a solution of the Galerkin projection of $(1)$ on the space spanned by {$\varphi_k \ | \ k = 1,\dots,r $}.

Thanks very much in advance.

Ok, so I know that the Galerkin projection of the model requires 2 projections. One for the signal $w$ that solves $(1)$ which is given by $$w_r(x,t)=\sum_{k=1}^{r}a_k(t)\varphi_k(x)$$ and one for the residual expression of the partial differrential equation which is given by: $$\left \langle \frac{\partial w_r}{\partial t}+w_r \frac{\partial w_r}{\partial x}-2\frac{\partial^2 w_r}{\partial x^2}-e(x)u(t), \varphi_n \right \rangle$$ where $n = 1,...,r$. Combining these projections means substituting $w_r$ in the latter expression, which leads to: $$\left \langle \frac{d}{dt}\sum_{k=1}^{r}a_k(t)\varphi_k(x)+\frac{d}{dx}\sum_{k=1}^{r}\big( a_k(t)\varphi_k(x)\big)^2 -2 \frac{d^2}{dx^2}\sum_{k=1}^{r}a_k(t)\varphi_k(x)-e(x)u(t),\varphi_n \right \rangle$$ Splitting this bigger inner product results in: $$\left \langle \frac{d}{dt}\sum_{k=1}^{r}a_k(t)\varphi_k(x),\varphi_n \right \rangle+ \left \langle \frac{d}{dx}\sum_{k=1}^{r}\big( a_k(t)\varphi_k(x)\big)^2,\varphi_n \right \rangle - \left \langle 2 \frac{d^2}{dx^2}\sum_{k=1}^{r}a_k(t)\varphi_k(x),\varphi_n \right \rangle - \left \langle e(x)u(t),\varphi_n \right \rangle$$ The summations can be taken out of the inner products because of linearity. The constants can also be taken out. Furthermore the time derivative can be written as a dot. This leads to: $$\sum_{k=1}^{r} \dot{a}_k(t) \langle \varphi_k(x), \varphi_n \rangle + \sum_{k=1}^{r} a_k(t) \left \langle \bigg( \frac{d}{dx} \varphi_k(x) \bigg)^2, \varphi_n \right \rangle - 2\sum_{k=1}^{r} a_k(t) \left \langle \frac{d^2 \varphi_k}{dx^2}, \varphi_n \right \rangle - \langle e(x)u(t), \varphi_n \rangle $$ The basis is orthonormal so $k = n = 1$ and $k \neq n = 0$ which means $\langle \varphi_n, \varphi_k \rangle = 1$ so the Galerkin projection becomes: $$0 =\sum_{k=1}^{r} \dot{a}_k(t)+\sum_{k=1}^{r} a_k(t) \left \langle \bigg( \frac{d}{dx} \varphi_k(x) \bigg)^2, \varphi_n \right \rangle - 2\sum_{k=1}^{r} a_k(t) \left \langle \frac{d^2 \varphi_k}{dx^2}, \varphi_n \right \rangle - \langle e(x)u(t), \varphi_n \rangle $$

which should be the GAlerkin projection i think


I compared your solution to the theory in "Numerical Methods in Scientific Computing" by J. Kan, A. Segal and F. Vermolen. Most of what you have done agrees with what has been written there over the Galerkin Method.

There are just 3 things that need notice.

First $$ \langle w\frac{\partial w}{\partial x}, \phi_n \rangle \neq \langle \sum_{k=1}^r(a_k(t)\phi_k(x))^2, \phi_n \rangle $$ but $$ \langle w\frac{\partial w}{\partial x}, \phi_n \rangle = \langle \sum_{k=1}^ra_k(t)\phi_k(x)\frac{\partial}{\partial x}\sum_{l=1}^ra_l(t)\phi_l(x), \phi_n \rangle \\ = \sum_{k=1}^r\sum_{l=1}^ra_k(t)a_l(t)\langle \phi_k(x)\frac{\partial}{\partial x}\phi_l(x), \phi_n \rangle \\ $$

Secondly, according to extended form of Holand and Bell(1969) theorem 20, page 84, for linear basis functions we have in $n=1$ dimensions $$ \langle \phi_{k\pm1}, \phi_k \rangle = \frac{1!1!}{(1+1+1)!}h = \frac{1}{6}h $$ and $$ \langle \phi_{k}, \phi_k \rangle = \frac{2!}{(2+1)!}h = \frac{2}{6}h = \frac{1}{3}h $$ You have however not specified your $\phi_k$, so this not necessarily holds, but the assumption you did about the inner product of $\phi_n$ and $\phi_k$ should be done with causion.

Finally, you demand that your functions need to be twice differentiable. The convention is however to write the test functions with as low as possible derivatives. This has the advantage that there are more valid functions to use as basis. Thus $$ -\langle \frac{d}{dx}\phi_n, \frac{d}{dx}\phi_k \rangle $$ is preferred over
$$ \langle \frac{d^2}{dx^2}\phi_n, \phi_k \rangle $$ These are equivalent, since $\phi_n$ and $\phi_k$ have compact support, i.e. $\phi_i(x_j)=\delta_{ij}$.

Combining this I think the correct answer should be $$ \frac{d}{dt}\sum_{k=1}^ra_k(t)\bigg\langle \phi_n, \phi_k \bigg\rangle = -\sum_{k=1}^r\sum_{l=1}^ra_k(t)a_l(t)\bigg\langle \phi_k(x)\frac{\partial}{\partial x}\phi_l(x), \phi_n\bigg\rangle-2\sum_{k=1}^ra_k(t)\bigg\langle \frac{d}{dx}\phi_n, \frac{d}{dx}\phi_k \bigg\rangle+\bigg\langle e(x)u(t), \phi_n(x) \bigg\rangle $$ or, when factoring out the sum over k instead over the time derivative, $$ \sum_{k=1}^r\Bigg(\frac{d}{dt}a_k(t)\bigg\langle \phi_n, \phi_k \bigg\rangle + 2a_k(t)\bigg\langle \frac{d}{dx}\phi_n, \frac{d}{dx}\phi_k \bigg\rangle+a_k(t)\sum_{l=1}^ra_l(t)\bigg\langle \phi_k(x)\frac{\partial}{\partial x}\phi_l(x), \phi_n\bigg\rangle\Bigg)=\bigg\langle e(x)u(t), \phi_n(x) \bigg\rangle $$

  • $\begingroup$ Did you recieve the bounty? $\endgroup$ – user463102 Mar 19 at 16:18
  • $\begingroup$ Yes, I did. Thank you. $\endgroup$ – ChocolateRain Mar 20 at 17:53

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