hutchphd said:
Man I don't understand this at all...Can you flesh it out for the slow kids? Sorry
That's good you don't, because it was full of mistakes
Ok, consider the change of variables
##u=x_1+x_2##, ##v=x_1-x_2## (not the change of variables I said, I flipped it)
Then by the chain rule, ##\frac{\partial f}{\partial x_1}= \frac{\partial f}{\partial u} \frac{\partial u}{\partial x_1} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial x_1}##
The partial derivatives of ##u## and ##v## are just 1 of course.
Which I will rewrite in the notation of the original post as ##\partial_{x_1} = (\partial_u+\partial_v)##
Similarly, ##\partial_{x_2} = (\partial_u-\partial_v)##
Notice that these partial derivative operators just multiply like ##\partial_x \partial_y## is the partial derivative with respect to y, then with respect to x (and these commute).
So ##\partial_{x_1} \partial_{x_2} = (\partial_u+\partial_v)(\partial_u-\partial_v)##.
Expanding the right hand side gives ##\partial^2_u+\partial_v\partial_u - \partial_u \partial_v - \partial^2_v##. The middle two terms cancel, leaving us with ##\partial^2_u - \partial^2_v,## (not the plus sign I promised)
Then I made a mistake, since I forgot ##u## showed up on the right hand side. If you had ##\partial_x u = f(x)## and you had some solution ##u_f##, then every given any solution of ##u_x =0##, say ##u_0##, then ##u_f+u_0## is a solution to ##\partial_x u =f(x)##. But this differential equation isn't in that form so that's not a useful observation.
My apologies for the mistakes.Edit to add:
I realize there are some simple solutions to this of the form ##f(u)+g(v)##. Then we get ##f''(u)=f(u)## and ##g''(v)=-g(v)## so anything like ##Ae^u + B e^-{u} + C \cos(v) + D\sin(v)## is a solution. To translate back to x coordinates, ##Ae^{x_1+x_2}+Be^{-(x_1+x_2)}+ C \cos(x_1-x_2)+D\sin(x_1-x_2)## are all solutions of the two dimensional equation (I'm still ignoring ##t##n from the o.p., since you can throw in functions of ##t## wherever you would have an arbitrary constant)