I am trying to find out the form for solutions to this equation:
$\partial_x^2u(x,M) = 0$, where x and M are not independent (M is the running maximum of process x, but I don't think that's necessary for this question I am asking)
So I am trying to find an equation for $u(x,M)$ and was doing this by integrating twice. I am unsure if I am doing this correctly, and wanted some help checking the steps.
$\int \partial_x^2u(x,M)dx = \int0 dx$
$ \partial_xu(x,M) + g(M) = f(M)$
then I combine the functions $g$ and $f$ together to just $g$ since they are arbitrary
$\partial_xu(x,M) + g(M) = 0$
then integrate again:
$\int \partial_xu(x,M)dx + \int g(M)dx = \int0dx$
$u(x,M) + j(M) + g(M)(x + k(M)) = l(M)$
So here to calculate $\int g(M)dx$ I am unsure if we can hold $g(M)$ constant since $M$ is a function of $x$, but since the partial derivative would keep $M$ fixed, I assumed it was and calculated $g(M) * \int1dx = g(M) * (x + k(M))$
So the final solution seems to be (combining $j$ and $l$ to $j$):
$u(x,M) + j(M) + g(M)(x + k(M)) = 0$
$u(x,M) + j(M) + g(M)x + g(M)k(M) = 0$
Can this be further simplified? Is $g(M)k(M)$ able to be made into some other function $r(M)$ or is it supposed to be kept separate?
Any help would be appreciated. Thanks!
I found out from a bit of research/textbooks that the general form for the solution here is $u(x,M) = a(M)x + b(M)$, which confirms the process used above.
Since $g(M)k(M)$ is just a function of $M$, then it can be written as $r(M)$ and then you can add it to $j(M)$ to get the new function $-b(M)$. Then rewrite $g(M)$ as $-a(M)$, and then we get the form $u(x,M) = a(M)x + b(M)$.
Using the negative sign in front of the functions can be done since the functions are just arbitrary.