I need to solve following differential equation $y''(x) -k = y\delta(x-x_0)$
subject to boundary conditions
\begin{eqnarray} y(x=-a) = 0 \\ y(x=b) = p \end{eqnarray}
I am not sure if it is possible to solve at all. I got solution for a similar problem.
Can anyone suggest any way for moving ahead?
EDIT Physical background:
Non dimensional differential equation for Poiseuille flow, is $$y''(x) = k $$ Where $y(x)$ is transverse velocity and $k$ contains information of viscosity, gravity.
to account for additional friction between wall and fluid velocity a term proportional to velocity and in the vicinity of wall is added(in the limit a $\delta$ function) so that differential equation is $$y''(x) -k = \delta(x-x_0)y(x)$$
Solve the equation seperately for $x>x_0$ and $x<x_0$.
For $x<x_0$ we have $$y'' - k =0 \to y = y_1(x) = \frac{k}{2}x^2 + Ax + B$$
For $x>x_0$ we have $$y'' - k =0 \to y = y_2(x) = \frac{k}{2}x^2 + Cx + D$$
First apply the boundary conditions (I assume $x_0 \in (-a,b)$). This gives the equations
$$y_1(-a) = 0$$ $$y_2(b) = p$$
Now apply continuity for $y$ and $y'$ at $x=x_0$ to determine the unknown constants in the expressions above. This gives the two equations
$$y_1(x_0) = y_2(x_0)$$ $$y_1'(x_0) = y_2'(x_0)$$
The four equations above are just enough to specify the four constants $A,B,C,D$ in the solution.
${\bf EDIT}$: I don't think we can assume $y'$ being continious across the gap. To determine the final equation we can integrate the equation over $x=x_0$ to find
$$0 \equiv \lim_{\epsilon\to 0}\int_{x_0-\epsilon}^{x_0+\epsilon}y''(x) - k - y\delta(x-x_0) dx = \lim_{\epsilon\to 0}\left[y'(x) - kx\right]_{x-\epsilon}^{x+\epsilon} + y(x_0) = y_2'(x_0) - y_1'(x_0) - y(x_0)$$
Where $y(x_0) = y_1(x_0) = y_2(x_0)$. Thus $y_2'(x_0) - y_1'(x_0) - y(x_0) = 0$ should be the second equations you need.
To speed up the algebra it can be useful to write the solutions as $y_1(x) = \frac{k}{2}(x+a)^2 + A(x+a) + B$ and $y_2(x) = \frac{k}{2}(x-b)^2 + C(x-b) + D$ instead. Then applying the boundary conditions gives the simple equations $y_1(-a) = B = 0$ and $y_2(b) = D = p$.