Finding eigenfunction to this operator I have the operator
$$
A : -e^{-2ax}
       \frac{\partial}{\partial x}
          \left(e^{2ax}\frac{\partial}{\partial x}\right)\\
D_A = \left( v \in C^2[0,L] \quad | \quad v(0) = v(L) = 0 \right)
$$
and I need to find the eigenfunction $p$. But first I have a question. How do I apply this operator to a function $u$? I see no $u$ in the formula. Is it a typo, should it be $\frac{\partial u }{\partial x}$ in the last partial derivative within the parenthesis? If this is the case, this is my attempt:
$$Ap = \lambda p$$
$$-2ap' - p'' - \lambda p = 0$$
$$p'' + 2ap' + \lambda p = 0$$
$$p(x) = C_1 e^{-a + \sqrt{a^2 - \lambda}} + C_2e^{-a - \sqrt{a^2 - \lambda}}.$$
But this doesn't at all look like the correct answer
$$e^{-ax}\sin{\frac{k\pi x}{L}}, k=1,2,3,...$$
What am I doing wrong and how do I find the right answer (without complicating things)?
 A: Hint: Your functions are
$$
p(x) = C_1 e^{(-a + \sqrt{a^2 - \lambda})x} + C_2e^{(-a - \sqrt{a^2 - \lambda})x}
$$
Consider how they behave when $\lambda > a^2$. Then consider what $\lambda$ and the $C_i$ need to be to satisfy the boundary conditions.
A: 
How do I apply this operator to a function $u$?

Your $D_A$ looks like the function space of all 2 times differentiable functions on $[0,L]$. So this seems like a function of one variable.
And therefore these seem ordinary differential operators, no partial ones, in the definition of $A$.
Thus I believe $A$ applies like this
$$
\begin{align}
A u 
&= -e^{-2ax} \partial_x \left(e^{2ax} \partial_x u \right) \\
&= -e^{-2ax}\left(e^{2ax} 2a \partial_x u + e^{2ax} \partial_x^2 u\right) \\
&= - (\partial_x^2 u + 2a \partial_x u)\\
\end{align}
$$
to $u \in D_A$.
A: How do we apply an operator such as
$A = -e^{-2ax} \dfrac{\partial}{\partial x} \left (e^{2ax} \dfrac{\partial}{\partial x} \right ) \tag 0$
to a function $u$?  Typically we work from right to left, taking the argument $u$ to occupy the first or right-most "empty slot"; bearing this in mind:
We start with the following calculation:
$Au = -e^{-2ax} \dfrac{\partial}{\partial x} \left (e^{2ax} \dfrac{\partial}{\partial x} \right )u = -e^{-2ax} \dfrac{\partial}{\partial x} \left (e^{2ax} \dfrac{\partial u}{\partial x} \right )$
$=-e^{-2ax} \left ( 2ae^{2ax} \dfrac{\partial u}{\partial x} + e^{2ax} \dfrac{\partial^2u}{\partial x_2} \right ) = -2a\dfrac{\partial u}{\partial x} - \dfrac{\partial u^2}{\partial x^2}; \tag 1$
if $\lambda$ is an eigenvalue of this operator, then
$-e^{-2ax} \dfrac{\partial}{\partial x} \left (e^{2ax} \dfrac{\partial}{\partial x} \right )u = \lambda u, \tag 2$
then, via (1),
$-2a\dfrac{\partial u}{\partial x} - \dfrac{\partial u^2}{\partial x^2} = \lambda u, \tag 3$
or
$\dfrac{\partial u^2}{\partial x^2} + 2a\dfrac{\partial u}{\partial x} + \lambda u = 0; \tag 4$
since this is a linear, constant coefficient, second order ordinary differential equation, we may find solutions of the general form $e^{\rho x}$ where $\rho$ is a root of the quadratic equation
$\rho^2 + 2a \rho + \lambda = 0; \tag 5$
we have
$\rho_\pm = \dfrac{-2a \pm \sqrt{4a^2 - 4 \lambda}}{2} = -a \pm \sqrt{a^2 - \lambda}; \tag 6$
thus we examine functions of the form
$u(x) = C_+ e^{\rho_+ x} + C_- e^{\rho_- x} \tag 7$
which satisfy
$u(0) = u(L) = 0; \tag 8$
we note that (7) and (8) yield
$C_+ + C_- = u(0) = 0, \tag 9$
so that
$C_+ = -C_-; \tag{10}$  
we can thus take
$C_+ = C = -C_-, \tag{11}$
and write
$u(x) = C(e^{\rho_+ x} - e^{\rho_- x}); \tag{12}$
we next note that for $\lambda < a^2$ in (6), we obtain two distinct real $\rho_\pm$, and if we apply the condition
$u(x) = C(e^{\rho_+ L} - e^{\rho_- L}) = u(L) = 0, \tag{13}$
we see that with $C \ne 0$ it leads to
$e^{\rho_+ L} = e^{\rho_- L}, \tag{14}$
which is impossible if $\rho_+ \ne \rho_-$; therefore (4), (5) have no solutions in $D_A$ with $\lambda < a^2$; since solutions to (4) are eigenfunctions of the operator $A$, we see it has no eigenvalues $\lambda < a^2$.  If $\lambda = a^2$, then we see via (6) that (5) has a repeated root $\rho = \rho_\pm = -a$; it is well-known the solutions in this case are $e^{\rho x} = e^{-ax}$ and $xe^{-ax}$; then
$u(x) = C_1 e^{-ax} + C_2 x e^{-ax} = (C_1 + C_2x) e^{-ax}, \tag{15}$
whence
$C_1 = u(0) = 0, \tag{16}$
and hence
$u(x) = C_2 x e^{-ax}, \tag{17}$
whence
$C_2 L e^{-aL} = u(L) = 0 \Longrightarrow C_2 = 0 \Longrightarrow u(x) = 0, \tag{18}$
eigenfunctions cannot be zero; thus we rule out the case $\lambda = a^2$.
We are left only with the possibility that
$\lambda > a^2; \tag{19}$
then we have
$\rho_\pm = -a \pm i \sqrt{\lambda - a^2}; \tag{20}$
formulas (7), (9) and (12) still hold so we still write
$u(x) = C(e^{\rho_+ x} - e^{\rho_- x}); \tag{21}$
with $\rho_\pm$ as in (20) we find
$e^{\rho_\pm x} = e^{-ax} e^{\pm \sqrt{\lambda - a^2} x} = e^{-ax}(\cos \sqrt{\lambda - a^2} x \pm i \sin \sqrt{\lambda - a^2} x), \tag{22}$
so that
$e^{\rho_+ x} - e^{\rho_- x} = 2i e^{-ax} \sin \sqrt{\lambda - a^2} x, \tag{23}$
and
$u(x) = 2C i e^{-ax} \sin \sqrt{\lambda - a^2} x; \tag{24}$
the criterion
$u(L) = 0 \tag{25}$
forces
$\sin \sqrt{\lambda - a^2} L = 0, \tag{26}$
or
$\sqrt{\lambda - a^2} L = k\pi, \; 0 \ne k \in \Bbb Z \tag{27}$
(we preclude $k = 0$ to enforce $u(x) \ne 0$); then
$\lambda = a^2 + \dfrac{k^2 \pi^2}{L^2}, \; 0 \ne k \in \Bbb Z, \tag{28}$
and the eigenfunctions may be taken to be
$e^{-ax} \sin \sqrt{\lambda - a^2} x = e^{-ax} \sin \dfrac{x k\pi}{L}, 0 \ne k \in \Bbb Z. \tag{29}$
