Bilinear Form & Linear Functional: Symmetric & Coercive?

[squenshl]

Homework Statement

The bilinear form are symmetric, i.e. a(u,v) = a(v,u) for all u and v. Find the bilinear form and the linear functional for the problem -\Deltau + b . \nablau + cu = f(x) in \Omega
u = 0 on the boundary.
Is this bilinear form for this problem symmteric? Is it coersive (assume C > 0)?

Homework Equations

The Attempt at a Solution

 

[HallsofIvy]

?? Why have you not shown any work? What is the "bilinear form" associated with this problem?

 

[squenshl]

a(u,v) = \int_D a₁\nablau . \nablav dx + \int_D a₀uv dx

 

[squenshl]

To start off for this problem do we multiply by v, integrate over D and use the divergence theorem?
So we get (-\Deltau + b . \nablau + cu - f)v dx = 0 in D which implies \int_D (-\Deltau + b . \nablau + cu)v dx = \int_D fv dx which implies some integral + some integral + \int_D cuv dx = \int_D fv dx, I just don't know how to find those integrals using the divergence theorem. Do i change that -\Deltau to -\nabla.\nablau and would this make it easier?

 

[squenshl]

I guess what I'm asking is how to find \int_D -\nabla . \nablau + b . \nablau dx, b and c are function of x.

 

[LPerrott]

In order to get the bilinear form you multiply the equation by a test function (in this case v), then you integrate over the domain Omega, then you integrate by parts (Note: you always integrate by parts when formulating a weak statement for a PDE .. the entire point is to transfer derivatives onto the test function, thus requiring less regularity), then you can apply divergence theorem to get rid of boundary terms. The coercivity can be tricky unless the function b is bounded below by a constant, but you can always use Poincare's inequality.

 

[squenshl]

Thanks.
Now all I need to do is just evaluate the integral \int_D -\nabla . \nablau + b . \nablau dx

 

[LPerrott]

You don't evaluate any integrals ... and secondly what you have listed is not the bilinear form. The bilinear form is an operator on two functions B(u,v)= integral (_____) ... you don't actually evaluate anything. All you need to show is that B(u,v)=B(v,u) and B is coercive .. try reading up a little on the topic , then work on the problem

 

[squenshl]

Sorry.
What I was meant to ask is how do I apply the divergence theorem to that integral above.

 

[LPerrott]

You don't use divergence theorem on the integral above. You need to first multiply by a test function, v, then integrate by parts. When you integrate by parts, the laplacian term will actually decompose into two terms. One will be gradient of u times gradient of v and the other will involve divergence. You can then use the divergence theorem on the divergence term to get an integral on the boundary, then you use boundary conditions

 

[squenshl]

In my lecture notes it says to use the divergence theorem on \int_D (-\nabla . \nablau + b . \nablau)v dx.
Do I get \int_D -\nablau . \nablav dx - \int_d v\partialu/\partial\nu d(sigma). (d is the boundary)