This problem confused me for a long time. I have 2 ODEs which describe part of our domain. They are connected at middle: $$ \frac{d^2}{dx^2} u = -a, x<x_0 $$ $$ \frac{d^2}{dx^2} u - \frac{u}{b^2}= 0, x>x_0 $$ The whole domain is defined as $$ -W < x $$ With the following conditions $$ \lim_{x->-W} u = 0 $$ $$ \lim_{x->+\infty} u = c \;\;or\;\; \lim_{x->L} u = c $$
u, u' and u'' needs to be continous.
a, b, W and c are constants. If not possible with $\infty$ one can replace $+\infty$ with a constant L.
Update: As a hint, the following functions are solutions of individual equations
$$u_1(x) = \alpha x^2 + \beta x + \gamma $$ $$u_2(x) = c_1 exp(-x/b) + c_2 exp(x/b) $$ Where $\alpha$, $\beta$, $\gamma$, $c_1$ and $c_2$ are constants related to the constants in the ODEs.
Problem is that we usually put $c_2 = 0$ since the exp(x/b) diverges. If we do so, 2nd derivative of u would not be continous at $x_0$.
My numerical simulated solution looks like $$ \frac{1}{1+exp(x/b)}$$ But it does't match exactly. The form of numerical solution and 2 regions are clear in the following link: http://www.wolframalpha.com/input/?i=1%2F%281%2Bexp%28x%29%29+from+x%3D-10+to+x%3D10
Ok:
You have: $$u(x) \equiv \left\{ \begin{array}{ll} u_1(x) = \alpha e^{-x/b} + \beta e^{x/b} & \text{if } x > x_0 \\ u_2(x) = -a x^2 + \gamma x + \delta & \text{if } x < x_0 \end{array} \right. $$
Because of boundedness at $x \to \infty$, we must set $\beta = 0$.
Because of boundedness at $x \to -\infty$, the only solution satisfying this would be that whereas $a = 0$ (which is a given constant) and, of course, $\gamma = 0$, so we have: $$ \delta = c.$$ Otherwise, if $a \neq 0$, the solution is unbounded at $x \to \infty$.
Then we arrive at:
$$u(x) \equiv \left\{ \begin{array}{ll} u_1(x) = \alpha e^{-x/b} & \text{if } x > x_0 \\ u_2(x) = c & \text{if } x < x_0 \end{array} \right. $$
Hope this helps.
Cheers!
PD: Forgive me for changing your notation.