0
$\begingroup$

I am looking for a manufactured (or analytical if it exists) solution to the 2d boundary-value problem

$$\frac{\partial u}{\partial t} = \mathbf{a} \cdot \nabla u + D \nabla^2 u \quad \quad \mbox{in } \Omega $$ $$ u \ \mathbf{a} \cdot \mathbf{\hat{n}} + D \ \nabla u \cdot \mathbf{\hat{n}} = 0 \quad \quad \mbox{on } \partial \Omega $$

where $\Omega$ is a rectangular domain and $\mathbf{a}$ and $D$ are independent of space and time.

I would highly appreciate any help / suggestion / reference on the matter.

$\endgroup$
2
  • $\begingroup$ You mean like $u\equiv 0$? Otherwise replace $L[u]=0$ in $Ω$, $R[u]=0$ on $∂Ω$ with $L[u]=L[p]$, $R[u]=R[p]$ for any sufficiently smooth function $p$, obviously $u=p$ is the reference or manufactured solution. $\endgroup$ Oct 1, 2021 at 13:22
  • 3
    $\begingroup$ But you need a source term to balance your manufactured solution. $\endgroup$
    – nicoguaro
    Oct 2, 2021 at 14:06

1 Answer 1

2
$\begingroup$

Usually manufactured solutions are used to verify a solver. As stated in the comment section, you should consider a source term both in the domain $\Omega$ and on the boundary $\partial \Omega$

$$ \frac{\partial u}{\partial t} - {\bf a} \cdot \nabla u - D \nabla^2 u = f({\bf x},t) \quad \text{in $\Omega$} , $$ $$ u {\bf a} \cdot {\bf \hat{n}} + D\nabla u \cdot {\bf \hat{n}} = g({\bf x}, t) \quad \text{on $\partial \Omega$}. $$

Here, you simply plug any $u$ of your choice, then you compute the source terms $f$ and $g$ by hand, and then you can test the consistency of your solver by comparing the analytical $u$ you choose and the numerical $u_h$ you have computed. Once you do it, you are sure that you solver works also on the original equation, i.e., by setting $f=g=0$.

For instance, if you plug $u = e^{-t}\sin(x)\cos(y)$ you get (double check!)

$$ f({\bf x},t) = e^{-t} (\cos(y) \sin(x) - 2 D \cos(y) \sin(x) + a_x \cos(x) \cos(y) - a_y \sin(x) \sin(y)), $$

$$ g({\bf x},t) = e^{-t} (D n_x \cos(x) \cos(y) - D n_y \sin(x) \sin(y) + a_x n_x \cos(y) \sin(x) + a_y n_y \cos(y) \sin(x)). $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.