I'm trying to verify another result from Coddington-Levinson's ODE:
Given $$Lx=ix'=lx, \qquad x(0)-x(1)=0$$
We seek to find Green's function. The author gives it as:
$$G(t,s,l)=\begin{cases} \frac{ie^{il(s-t)}}{1-e^{il}} & 0\leq t\leq s \leq 1 \\ \frac{ie^{il(1+s-t)}}{1-e^{il}} & 0\leq s \leq t\leq 1 \end{cases}$$
I'm trying to follow the constructive proof given. We have that the fundamental solution is $$y_1(t)=e^{-ilt}$$ and thus the Wronskian $$W(y_1)(s)=e^{-ils}$$ and therefore $$K(t,s,l)=\begin{cases} \frac{e^{-ilt}}{ie^{-ils}} & 0\leq s\leq t \\ 0 & t<s\leq 1 \end{cases}$$ and we seek the Green's function of the form $$G(t,s,l)=K(t,s,l)+c(s)y_1(t)$$ Now I get stuck when I try to apply the boundary condition. We require that $$0=U_1(G)=U_1(K(t,s,l))+c(s)U_1(y_1)$$ where $$U_1(x)=x(0)-x(1)$$ and $$U_1(y_1)=y_1(0)-y_1(1)=1-e^{-il}$$ But I'm unsure of how to compute $U_1(K(t,s,l))$ since $K$ is defined piecewise and there don't seem to be any nice simplifications: $$K(0,s,l)-K(1,s,l)=\begin{cases} \frac{1}{ie^{-ils}} & 0\leq s\leq 0 \\ 0 & 0<s\leq 1 \end{cases}-\begin{cases} \frac{e^{-il}}{ie^{-ils}} & 0\leq s\leq 1 \\ 0 & 1<s\leq 1 \end{cases}=$$ $$\begin{cases} -i & s=0 \\ 0 & 1<s\leq 1 \end{cases}-\frac{e^{-il}}{ie^{-ils}}$$