I was reading a PDE tutorial on solving the nonlinear poisson equation. The author of the tutorial defines the problem and asserts the weak form, but does not provide the derivation. I tried my own derivation, but my derivation seems to have an extra term, so I must be missing a step or some insight. I was hoping that someone could tell me where the error is.
The problem is defined as:
$$
– \nabla \cdot (q(u)\nabla u) = f , \quad \text{ in } \Omega
$$
$$
u = u_D, \quad \text{ on } \partial \Omega
$$
In this problem, the existence of the $q(u)$ function makes the problem nonlinear. We can assume that $q(u)$ is non-constant and the $u(x, y)$ is a function of $x,y$.
The weak form that is provided is:
$$
\int_\Omega (q(u) (\nabla u)^T \cdot \nabla v – f^Tv) dV = 0
$$
So the integral is over the volume, hence the $dV$.
I tried to derive this weak from by multiplying by the test function $v$ and integrating.
$$
-\int_\Omega \nabla \cdot (q(u)\nabla u) v dV = \int_\Omega f^T v dV
$$
To simplify the divergence term on the left side I tried the product rule:
$$
\nabla \cdot (q(u)\nabla u) v = (\nabla q(u) \nabla u)v + q(u)\nabla^2 u v
$$
The second term on right hand side will simplify to the usual $q(u)(\nabla u)^T \nabla v$ term, that we saw in the asserted solution above. However that first term $(\nabla q(u) \nabla u)v$ does not appear in the weak form. So I was wondering why that term seemed to disappear?
Best Answer
Here is a solution that I came up with, based upon the suggestions in the comments. Please let me know if I made any errors.
The original problem is: $$ -\nabla \cdot (q(u)\nabla u) = f, \text{ in } \Omega \\ u = u_d, \text{ on } \partial \Omega $$
The usual approach is to multiply both sides by a test function $v \in V$ and integrate over the domain $\Omega$.
$$ -\int_\Omega \nabla \cdot (q(u)\nabla u )vdV = \int_\Omega fv dV $$
We can modify the divergence term on the left and apply the product rule.
$$ \nabla \cdot (q(u)\nabla uv) = (\nabla \cdot(q(u)\nabla u))v + q(u)\nabla u \nabla v $$
Next we rearrange the terms to match the left hand term in the original nonlinear poisson equation
$$ (\nabla \cdot(q(u)\nabla u))v = \nabla \cdot (q(u)\nabla uv) - q(u)\nabla u \nabla v $$
Notice that the term on the left is the same as the term on the left when we integrated and multiplied the original nonlinear poisson equation with the test function $v$. When we substitute these terms back into the original equation.
$$ -\int_\Omega \nabla \cdot (q(u)\nabla u )vdV = -\int_\Omega \nabla \cdot (q(u)\nabla uv) + \int_\Omega q(u)\nabla u \nabla v $$
Then we apply the Divergence theorem to the first term on the right hand side. Note that this will produce a boundary integral with a value of 0, because we have a Dirichlet condition over the boundary.
$$ -\int_\Omega \nabla \cdot (q(u)\nabla uv) = -\int_{\partial \Omega} q(u)\nabla uv \cdot \hat{n} ds = 0 $$
Note that $\hat{n}$ is the outward normal vector at the boundary. After removing this term from the equation above we are left with:
$$ -\int_\Omega \nabla \cdot (q(u)\nabla u )vdV = \int_\Omega q(u)\nabla u \nabla v = \int_\Omega fv dV $$
Finally by simplifying to residual form, this derivation produces the desired outcome.
$$ \int_\Omega q(u)\nabla u \nabla v - fv dV = 0 $$
Again, please let me know if I made any errors.