# Green's formula

1. Nov 8, 2008

### tom_rylex

1. The problem statement, all variables and given/known data
Consider the wave operator
$$\Box^2 = \frac{\partial^2}{\partial t^2} - \frac{\partial^2}{\partial x^2}$$
Write Green's forumula for an arbitrary domain $\Omega$ with boundary $\Gamma$

2. Relevant equations
define a vector q, where
$$q \cdot e_t = n \cdot e_t$$
$$q \cdot e_x = -n \cdot e_x$$
where e_x, e_t are unit vectors in x, t dimensions, n is the normal.
Show Green's formula takes the form
$$\int_\Omega (v\Box^2u - u\Box^2 v)dx dt = \int_\Gamma \left( v\frac{\partial u}{\partial q} -u\frac{\partial u}{\partial q} \right)dl$$
where dl is an element of arclength along the bounding curve $\Gamma$
(u and v are arbitrary functions in R1)

3. The attempt at a solution
I'm starting with Green's formula for an arbitrary linear operator:
$$\int_\Omega (vLu - L^*v) = \int_\Gamma div(J(u,v))$$
Where L is an ODE operator, and J is the conjunct of u and v. For the operator above, this takes the form of
$$\int_\Omega (vLu - L*v) = \int_\Gamma div \left[e_t \left( v\frac{\partial u}{\partial t} -u\frac{\partial u}{\partial t}\right) + u grad_x v - v grad_x u \right]$$
Where grad_x is the gradient along x. Now, the div operator should give us the same a the dot product of the normal to the line space:
$$= \int_\Gamma n\cdot\left[e_t \left( v\frac{\partial u}{\partial t} -u\frac{\partial u}{\partial t}\right) + u grad_x v - v grad_x u\right]$$

I believe that $n \cdot grad_x = \frac{\partial}{\partial n}$. So, substituting the form of q above, I have
$$= \int_\Gamma \left( q\cdot e_t \left( v\frac{\partial u}{\partial t} -u\frac{\partial u}{\partial t}\right) + \left(v\frac{\partial u}{\partial q} - u\frac{\partial u}{\partial q}\right)\right)$$

I'm almost there, I'm just not sure how $\left( q\cdot e_t \left( v\frac{\partial u}{\partial t} -u\frac{\partial u}{\partial t}\right)$ disappears.