Weak Form of the Poisson Problem

  • Thread starter Thread starter Morberticus
  • Start date Start date
  • Tags Tags
    Form Poisson Weak
Morberticus
Messages
82
Reaction score
0
Hi,

I know the weak form of the Poisson problem

\nabla^2 \phi = -f

looks like

\int \nabla \phi \cdot \nabla v = \int f v

for all suitable v. Is there a similarly well-known form for the slightly more complicated poisson problem?

\nabla (\psi \nabla \phi ) = -f

I am writing some finite element code and variational/weak forms are very handy.

Thanks in advance
 
Last edited:
Physics news on Phys.org
Morberticus said:
Hi,

I know the weak form of the Poisson problem

\nabla^2 \phi = -f

looks like

\int \nabla \phi \cdot \nabla v = \int f v

for all suitable v. Is there a similarly well-known form for the slightly more complicated poisson problem?

\nabla (\psi \nabla \phi ) = -f

I am writing some finite element code and variational/weak forms are very handy.

Thanks in advance

By the product rule
<br /> \int_V v\nabla \cdot(\psi \nabla \phi)\,dV = \int_V\nabla\cdot(v\psi \nabla \phi) - \psi (\nabla \phi) \cdot (\nabla v)\,dV = \int_{\partial V} v\psi \nabla \phi \cdot dS - \int_V\psi (\nabla \phi) \cdot (\nabla v)\,dV
and hence the weak form of \nabla \cdot(\psi\nabla\phi) = - f is
<br /> \int_V \psi (\nabla \phi) \cdot (\nabla v)\,dV = \int_V fv\,dV<br />
 
  • Like
Likes 1 person
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
Back
Top