Solving a BVP using Green's function

Question

Solve $\mathcal{L}u(x)=f(x)$ with the boundary conditions $u'(0)=a$, $u'(b)=\beta$ where $\mathcal{L}=\dfrac{d^2}{dx^2}+\dfrac{d}{dx}$ using Green's function.

$\mathbf{My \ attempt}$ Consider the homogeneous problem $\mathcal{L}u_H(x)=0$, $\mathcal{L}u_H^*(x)=0$ with homogeneous boundary conditions. Then we got $u_H(x)=C, u_H^*(x)=D$ where $C,D$ constants. Then I have the Modified and Adjoint Modified Green’s Function are given by the solution of $\mathcal{L}G=\delta(x-s)-1,\ \mathcal{L^*}G^*=\delta(x-s)-1 $ where $G^*$ denoted the adjoint modified Green's function, and 1 in the right hand side of $G,G^*$ came from $u_H(s)\frac{u_H(x)}{\int_0^1u_H(x)^2},\ u_H^*(s)\frac{u_H^*(x)}{\int_0^1(u_H(x)^*)^2}$ respectively. Now to find $u(x)$, we have $$\langle G^*,\mathcal{L}u\rangle-\langle u,\mathcal{L}^*G^*\rangle=J(u,G^*)=G(1)(a+u(1))+G(0)(-b-u(0))$$ i.e., \begin{align*} \int_0^1G^*f-u(\delta(x-s)-1)\,dx &=\int_0^1G^*f\,dx-u(s)-\int_0^1u\,dx\\ &=G^*(1)(a+u(1))+G^*(0)(-b-u(0)) \end{align*} by changing the variable of $s,x$ we got $$u(x)=\int_0^1G^*f-\int_0^1u\,ds-(G^*(1)(a+u(1))+G^*(0)(-b-u(0)))$$ since $\int_0^1u\,ds$ does not depend on $x$ it is a constant; so my final solution for $u$ is given by $$u(x)=\int_0^1G^*f\,ds-(G^*(1)(a+u(1))+G^*(0)(-b-u(0)))-A.$$ Now I am asked to find the solution of $u$ when $f=0$ and $a,b$ are nonzero, I could not find that with my final form of the solution of $u$, I do not know if my solution seems in the right form.

Answer 1

I'm not following how you're using homogeneous BCs to find the Green's function. The operator $L$ is not just defined by the differential operator, but also by the boundary/initial conditions. Here's what I have.

1. Assume $0<s<b.$ For $x<s,$ apply the BC on the left to obtain the solution to $Lu_1(x)=0, u'(0)=a$ as $$u_1(x)=C_1-ae^{-x}.$$
2. Then for $x>s,$ apply the BC on the right to obtain the solution to $Lu_2(x)=0, u'(b)=\beta$ as $$u_2(x)=C_2-\beta e^{-(x-b)}.$$
3. Next we apply the continuity/discontinuity conditions \begin{align*} \lim_{x\to s^+}u_2(x)&=\lim_{x\to s^-}u_1(x)\\ \lim_{x\to s^+}u_2'(x)-\lim_{x\to s^-}u_1'(x)&=1,\qquad\text{or}\\ C_1-ae^{-s}&=C_2-\beta e^{-(s-b)}\\ ae^{-s}&=1+\beta e^{-(s-b)},\qquad\text{or}\\ a&=e^s+\beta e^b \end{align*} Now this is an interesting set of equations for two reasons. One is that we cannot actually solve for $C_1$ and $C_2$ exactly, because both of them only show up in one equation. The second is that the discontinuity relation imposes a constraint on the BCs! But, given the form of the solutions, this actually makes some sense, because with the inflexibility of these solutions, if the BCs are too "extreme", you're not going to be able to solve them with the given Green's function constraints (particularly continuity). We can simplify a tad: \begin{align*} C_1-1-\beta e^{-(s-b)}&=C_2-\beta e^{-(s-b)}\\ C_1&=C_2+1=:C. \end{align*} So our Green's function we can write as $$G(x,s)= \begin{cases} C-(e^s+\beta e^b)e^{-x},&x<s\\ C-1-\beta e^{-(x-b)},&x>s. \end{cases} $$
4. Finally, you would write $$u=\int_0^x\left[C-1-\beta e^{-(x-b)}\right]f(s)\,ds +\int_x^b \left[C-(e^s+\beta e^b)e^{-x}\right]f(s)\,ds.$$