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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.

nmasanta
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John_dydx
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    Schwarz theorem... Look here http://en.wikipedia.org/wiki/Symmetry_of_second_derivatives – DiegoMath Nov 17 '14 at 02:56
  • Schwarz's theorem isn't necessarily applicable. Continuity of the crossed second order partial derivatives is required. As it is we only have continuity of one of those derivatives. Does the the continuity of the other one follow? – Git Gud Nov 17 '14 at 02:58
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    As I see it, the answer is yes, you can really integrate with respect to $y$ (it's an integrable function, after all), but it won't be necessarily true that $\frac{\partial u}{\partial x}(x,y) = -\cos(x+y) + \phi(x)$. I think you're correct and the integration should be done with respect to $x$ first. – Git Gud Nov 17 '14 at 03:02
  • @DiegoMath, thanks. So does this mean they are assuming that $f_{xy} == f_{yx}$? I know this doesn't always hold but I'll read the theorem in more details. I love PDEs! – John_dydx Nov 17 '14 at 03:03
  • @GitGud, Thanks for your comments. I'll add the full question to my post so you can see the full picture. – John_dydx Nov 17 '14 at 03:05

2 Answers2

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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.

Alex Ortiz
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$$\dfrac{\partial^2 u}{\partial x \partial y} = \sin(x+y)$$

Since $\quad\dfrac{\partial^2 u}{\partial x \partial y}=\dfrac{\partial^2 u}{\partial y \partial x}\quad$ it doesn't matter if you first integrate with respect to $x$ or if you first integrate with respect to $y$.

For example let $\quad v=\dfrac{\partial u}{\partial x}.\quad$ The eqution becomes : $$\dfrac{\partial v}{ \partial y}= \sin(x+y)$$

$$v=-\cos(x+y)+\text{constant with respect to } y$$ Any function of $x$ but not of $y$ is constant with respect to $y$. Thus $$v=-\cos(x+y) + \phi(x)$$ $$\dfrac{\partial u}{\partial x} = -\cos(x+y) + \phi(x)$$ Then we integrate with respect to $x$ $$u=-\sin(x+y)+\int\phi(x)dx+\text{constant with respect to } x$$ Any function of $y$ but not of $x$ is constant with respect to $x$. Thus $$u=-\sin(x+y) +\int\phi(x)dx+ F(y)$$ $$u=-\sin(x+y) +\Phi(x)+ F(y)$$ $\Phi$ and $F$ are arbitrary functions.

In the above example of calculus we first integrate with respect to $y$. Instead of, try to first integrate with respect to $x$ . You will see that the final result will be the same.

JJacquelin
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