# Converting between Solution forms using Green's Functions in Linear Differential Equation

EDIT: Bounty is over tomorrow so I tried to clean up the question a bit, and put the additional work below as optional to read. I summarized the current results and the solution form I am trying to convert to.

I know the solution from the book, but I am unsure how it was reached. I am nearly there, but am having trouble getting the correct form, since the final form looks so much different from the work I've done leading up to it.

After a lot of hunting, I realized it's using Green's functions to show the answer, but I couldn't find anything helpful enough that was in 3 variables like my problem. So I need help formulating the solution in the proper form instead of the basic linear form.

The basic idea of the system here is that there is a variable $$x$$, and it changes, up and down, with $$m$$ the running minimum and $$M$$ the running maximum. Once the variable $$x$$ reaches a distance $$L$$ above $$m$$ OR a distance $$L$$ below $$M$$, (for the first time) the value of the function $$u$$ is known. This basic idea is already imposed in the differential equation and its boundaries below.

I mostly need help on figuring out how to formulate the Green's functions/integrals so the solution form looks the way it does below. Most of the work I have shown below is for solving the general system, so it can be skipped if it's not relevant for finding the solution form. I basically just solve for $$a(m,M)$$ and $$b(m,M)$$ in the linear solution form $$u(x,m,M) = a(m,M)x + b(m,M)$$ which may be helpful for the Green's functions but I am unsure.

Thanks!

Problem: \begin{align} \partial_x^2u(x,m,M) = 0, m < x < M \tag{1} \\ \partial_Mu(x,m,M)|_{x=M} = 0 \tag{2} \\ \partial_mu(x,m,M)|_{x=m} = 0 \tag{3} \\ u(m,m,L+m) = \Phi(m) \tag{4} \\u(M,M-L,M) = \Phi(M) \tag{5} \end{align}

Solution:

$$u(x,m,M) = L^{-1}((M-x)\Phi(M-L) + (x-m)\Phi(m+L)) + L^{-2}{\huge(}\int_{M-L}^m(x-y)\Phi(y)dy + \int_M^{m+L}(y-x)\Phi(y)dy{\huge)}$$

So far I have the equation $$u(x,m,M) = \frac{\Phi(L+m) - \Phi(m)}{L} * x + \frac{(L+m)\Phi(m) - m\Phi(L+m)}{L}$$ but have no idea how to get this to the solution form. I am pretty sure this is a correct equation, but if you have doubts you can check my work below.

Equating the two answers, I need to find a way to show the following equality: $$(m+L - x)\Phi(m) = \\(M-x)\Phi(M-L) + L^{-1}{\huge(}\int_{M-L}^m(x-y)\Phi(y)dy + \int_M^{m+L}(y-x)\Phi(y)dy{\huge)}$$

but I am unsure if it is just better to work with the original equations.

### ADDITIONAL WORK BUT NOT NECESSARY

Work so far (mine plus some steps in the book):

Using $$(1)$$, the solution form is linear in $$x$$, so: $$u(x,m,M) = a(m,M)x + b(m,M) \tag{6}$$

Using $$(6)$$ to rewrite $$(4)$$ and $$(5)$$: $$u(m,m,L+m) = \Phi(m) = a(m,L+m)m + b(m,L+m) \tag{7}$$ $$u(M,M-L,M) = u(L+m,m,L+m) = \Phi(L+m) =\\ a(m,L+m)(L+m) + b(m,L+m) \tag{8}$$

Equation $$(8)$$ comes from the fact that if $$x$$ reaches a distance $$L$$ above $$m$$ (for the first time), then $$M = L + m$$, and so $$x = M = L +m$$.

From $$(7)$$ and $$(8)$$ we calculate $$a(m,L+m)$$ and $$b(m,L+m)$$, by subtracting $$(7)$$ from $$(8)$$ to solve for $$a(m,L+m)$$ and then substituting $$a(m,L+m)$$ into $$(7)$$ to solve for $$b(m,L+m)$$: $$a(m,L+m) = \frac{\Phi(L+m) - \Phi(m)}{L} \tag{9}$$ $$b(m,L+m) = \frac{(L+m)\Phi(m) - m\Phi(L+m)}{L} \tag{10}$$

The book says that equations $$(2)$$ and $$(3)$$ can be integrated (from $$L+m$$ to $$M$$) and it can be shown that $$\partial_{mM}b(m,M) = 0$$. And therefore the following equation is true $$\partial_mb(m,M) = \partial_mb(m,L+m) = \partial_m{\huge(}\frac{(L+m)\Phi(m) - m\Phi(L+m)}{L}{\huge)} \tag{11}$$

And finally, (I think many steps are skipped), the book says that by integrating $$(11)$$, the solution above is reached.

So I think we are supposed to use $$(11)$$ to calculate $$b(m,M)$$, so I did the following: $$\int_{M-L}^m\partial_{m}b(m,M)dm = b(m,M) - b(M-L,M) = \int_{M-L}^m\partial_m{\huge(}\frac{(L+m)\Phi(m) - m\Phi(L+m)}{L}{\huge)} = \\ {\huge(}\frac{(L+m)\Phi(m) - m\Phi(L+m)}{L}{\huge)} - {\huge(}\frac{(M)\Phi(M-L) - (M-L)\Phi(M)}{L}{\huge)}$$

And so solving for $$b(m,M)$$, using (from $$(5)$$) that $$b(M-L,M) = \Phi(M) - a(M-L,M)*M$$ and from $$(4)$$ and $$(5)$$ that $$a(M-L,M) = \frac{\Phi(M) - \Phi(M-L)}{L}$$, we have:

$$b(m,M) = b(M-L,M) + {\huge(}\frac{(L+m)\Phi(m) - m\Phi(L+m)}{L}{\huge)} - {\huge(}\frac{(M)\Phi(M-L) - (M-L)\Phi(M)}{L}{\huge)} = \frac{(L+m)\Phi(m) - m\Phi(L+m)}{L}$$

So I have $$b(m,M)$$ solved for, but have no idea how to get from there to the correct solution for $$u(x,m,M)$$ since it seems like the solution uses the boundary conditions somehow instead of directly plugging in $$b(m,M)$$.

Any help would be greatly appreciated, apologies in advance for the ginormous question. Thanks!