Math Help - Adjoint of a differential operator

1. Adjoint of a differential operator

Hey there,
I need to prove the next statement:
Let $L = A\frac{\partial ^2 }{\partial{x^2}} + 2B \frac{\partial ^2 }{\partial{x} \partial{y}}+C\frac{\partial ^2 }{\partial{y^2}} + D \frac{\partial }{\partial{x}}+E \frac{\partial}{\partial{y}} +F$ be an operator that its coefficients A,B,C,D,E are continiously differentiable twice in the plane.

Show (directly from the definition of an adjoint of an operator) that $L^{**}=L$ .

The definition of an adjoint of an operator is+my problem:
Code:


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$L*v = \frac{\partial ^2 }{\partial{x^2}}(Av) + \frac{\partial ^2 }{\partial{x} \partial{y}}(2Bv)+\frac{\partial ^2 }{\partial{y^2}}(Cv) - \frac{\partial }{\partial{x}}(Dv)- \frac{\partial}{\partial{y}}(Ev) +Fv$.
My problem is , that when I substitute L* in the expression for the adjoint, I get partial derivatives of fourth order!!! how can I prove this equality using only this definition?
Hope you'll be able to help me !

2. Okay, exactly why is that definition a problem?

since what you want to prove, L**= L, does not have a "v", I recommend that you separate the v from the coefficients.

Use the product rule:
$\frac{\partial^2}{\partial x^2}(Av)= \frac{\partial}{\partial x}\left(A\frac{\partial v}{\partial x}+ \frac{\partial A}{\partial x}v\right)= A\frac{\partial^2 v}{\partial x^2}+ 2\frac{\partial A}{\partial x}\frac{\partial v}{\partial x}+ \frac{\partial^2 A}{\partial x^2}v$, etc.

3. Thanks !!