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.

  • $\begingroup$ 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. $\endgroup$ Commented Oct 1, 2021 at 13:22
  • 3
    $\begingroup$ But you need a source term to balance your manufactured solution. $\endgroup$
    – nicoguaro
    Commented Oct 2, 2021 at 14:06

1 Answer 1


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!)

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

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


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