I look for the eigenfunction-expansion coefficients corresponding to a given initial ($t=0$) profile. I am struggling with formulating the correct orthogonality relations for the eigenfunctions.
Imagine a particle diffusing on an interval $[0,L]$. For $x\in (0,w)$ a constant force $F(x)=-f \equiv -2g$ is acting towards origin. The corresponding Smoluchowski equation for the probability is $$ \partial_t p(x,t) = -\partial_{x} J, \qquad -J = (\partial_x p(x,t) - F(x)p(x,t) ) $$ The eq. is separable as $p=X(x)T(t)$, where T'' = -c_n T, and $$ X'' + 2g X + c_n X = 0, \qquad x\in[0,w) \\ X'' + c_n X = 0, \qquad x\in(w,L] $$ If I impose boundary conditions $P(0)=0$ and $J(L)=0$ the possible solutions on each subinterval may be conveniently written as: $$ X^1_n = B_n \exp(-gx) \sin(k_n x) , \qquad x\in[0,w) \\ X^2_n = Q_n\frac{ \cos(c_n(L-x)) }{ \cos(c_n(L-w)) }, \qquad x\in(w,L] $$ where $c_n = g^2 + k_n^2$. The matching condition for $x=w$ should be the continuity continuous $X$ and of the current, expressed as $$ X^1_n(w) = X^2_n(w) \\ X^1_n\,{'}(w) + 2g X^1_n(w) = X^2_n\,{'}(w) $$ The second should be justified enough by integrating the ODE equation for $X$ over $\int_0^{w\pm\delta}$, and taking difference of the two terms (please correct me, were I wrong). The two equations (requiring zero determinant of the matrix corresponding to the homogeneus system for $B_n, Q_n$) yield $Q_n=B_n \exp(-gw)\sin(k_n w)$ and condition for $c_n$: $$ c_n \tan(c_n(L-w)) = +g + k_n \cot k_n w $$
Numerically solving for $c_n$ i am able to specify $X^1_n, X^2_n$.
Now, I am looking for a way to find a orthogonality relations between the functions $X_n$ on whole interval $(0,L)$, assuming the correct weight function is $w(x)=\exp(2gx)$ on $(0,w)$ and unity on the rest I get $$ const \times \delta_{nm} \stackrel{?}{=} \int_0^w \sin k_n x\sin k_m x + A_n A_m \int_w^L \frac{ \cos(c_n(L-x)) }{ \cos(c_n(L-w)) } \frac{ \cos(c_m(L-x)) }{ \cos(c_m(L-w)) } \\ \text{where} A_n = \exp(-gw)\sin(k_n w) $$
But this seems not to work (almost sure I have the numerical quadrature ok). I would appreciate all your insights :)