Does it exist and is it continuous? (exact)

[nonequilibrium]

Hello,

Let M be a function from R^2 to R with image M(x,y).
Given is that M has continuous partial derivatives \frac{\partial M(x,y)}{\partial y} & \frac{\partial M(x,y)}{\partial x}.

Question:
Does
\frac{\partial^2}{\partial x \partial y} \int M(x,y) \mathrm d x
exist and is it continuous?

It is used in my DE course, but I don't find it self-evident. The professor's argument was that since M was already differentiable with respect to x and y, and integrating makes it more continuous, it will definitely be differentiable. Okay I find it somewhat self-evident (with this reasoning) that
\frac{\partial}{\partial y}\int M(x,y) \mathrm d x
exists, but how does one convince himself of this last expression being differentiable with respect to x? And even if it is differentiable with respect to x, the derivative might have an essential discontinuity (of course I'm not saying it can in this case: I believe my professor; but I don't believe in his hand-waving reasoning and I'm looking for a more rigorous/insightful argument)

 

[JG89]

Are you talking about the first set of lecture notes that the professor posted? If so, then
<br /> \frac{\partial^2}{\partial x \partial y} \int M(x,y) \mathrm d x = \frac{\partial}{\partial y} M(x,y)<br />

and it was said in the notes to assume that this function is continuous.

EDIT: I'm assuming you are in MAT267

EDIT 2: You do go to U of T, right?

 

[nonequilibrium]

Are you talking about the first set of lecture notes that the professor posted? If so, then
<br /> \frac{\partial^2}{\partial x \partial y} \int M(x,y) \mathrm d x = \frac{\partial}{\partial y} M(x,y)<br />

and it was said in the notes to assume that this function is continuous.

EDIT: I'm assuming you are in MAT267

EDIT 2: You do go to U of T, right?

Hello. Nope, I'm from Belgium, but I suppose the theorem of exact differential equations are popular in all DE courses :p

As for your comment: that equality you give is only true if you can "mix/switch" partial derivatives, which you can only do if you already know \frac{\partial^2}{\partial x \partial y} \int M(x,y) \mathrm d x is continuous (and existing, of course), the latter being exactly my question.

 

[JG89]

I didn't switch anything. \frac{\partial}{\partial x \partial y} \int M(x,y) dx means that you first differentiate \int M(x,y) dx with respect to x, then differentiate the resulting expression with respect to y. By the fundamental theorem of calculus, differentiating \int M(x,y) dx with respect to x gives M(x,y), and differentiation with respect to y can be written \frac{\partial}{\partial y} M(x,y). But I guess your question is whether \frac{\partial}{\partial y} M(x,y) is continuous. In my set of lecture notes, I am to assume it is continuous, so I can't help you there. Sorry!

 

[JG89]

As a counter example, what about the function M(x,y) = y^2sin(\frac{1}{y}) + 0x

\frac{\partial^2}{\partial x \partial y} \int M(x,y) dx = \frac{\partial^2}{\partial x \partial y} \int y^2sin(\frac{1}{y}) + 0x dx = \frac{\partial}{\partial y} y^2 sin(\frac{1}{y} )

But the partial derivative with respect to y of y^2 sin(\frac{1}{y}) is not continuous when y = 0.

 

[nonequilibrium]

JG89: with regard to your first post: you seem to be a bit confused, the expression means you first differentiate with respect to y and then with respect to x, not the other way around.

as for your second post: your M(x,y) does not have continuous partial derivatives (because dM/dy is not continuous in zero), which I stated as a given.

 

[nonequilibrium]

JG89: you seem to be a bit confused, the expression means you first differentiate with respect to y and then with respect to x, not the other way around.