I have some questions about the process involved when integrating higher order partial derivatives. I was going through a textbook on engineering mathematics on PDEs.
If $~\dfrac{\partial^2 u}{\partial x \partial y} = \sin(x+y)~$, it states that $~\dfrac{\partial u}{\partial x} = -\cos(x+y) + \phi(x)~$.
My understanding was that $~\dfrac{\partial^2 u}{\partial x \partial y}= \dfrac{\partial }{\partial x}\left(\dfrac{\partial u}{\partial y}\right) = \sin(x+y)~$.
My question is this: Can you really integrate $~\dfrac{\partial }{\partial x}\left(\dfrac{\partial u}{\partial y}\right)$ with respect to $~y~$? I thought this had to be at least integrated first wrt $~x~$ and then wrt $~y~$. In essence, since the derivative was first wrt $~y~$ and then wrt $~x~$, shouldn't the integral be first wrt $~x~$ and then wrt $~y~$?
I hope this makes sense.
Thanks
Edit: Here's the full question-Solve the equation $~\dfrac{\partial^2 u}{\partial x \partial y} = \sin(x+y)~$, given that at $~y = 0~$, $~\dfrac{\partial u}{\partial x} = 1~$ and at $~x = 0~$, $~u = (y-1)^2 ~$. Judging by the initial conditions, one way we can solve this equation is by first integrating wrt $~y~$ in order to obtain an equation in $~\dfrac{\partial u}{\partial x}~$ and hence making use of the initial condition.
Best Answer
In this case, the function $\sin(x + y)$ is infinitely differentiable with respect to both partial derivatives. Therefore, by Clairaut's theorem, the order of partial derivatives isn't pertinent. Therefore, we may rewrite $\frac{\partial^2u}{\partial x\partial y} = \frac{\partial}{\partial x}\frac{\partial u}{\partial y} = \frac{\partial}{\partial y}\frac{\partial u}{\partial x} = \sin(x + y)$. From this, we can see that integrating this with respect to $y$ is legal, and we can apply the fundamental theorem of calculus to simplify the integration.