# Does it exist and is it continuous? (exact)

• nonequilibrium
In summary, the conversation discusses the existence and continuity of the expression \frac{\partial^2}{\partial x \partial y} \int M(x,y) \mathrm d x in a differential equations course. The professor argues that since M has continuous partial derivatives with respect to x and y, and integrating makes it more continuous, the expression will definitely be differentiable. The student is unsure about this reasoning and is looking for a more rigorous argument. The conversation also mentions a counter example where the function M(x,y) does not have continuous partial derivatives.
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)

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

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?

Last edited:
JG89 said:
Are you talking about the first set of lecture notes that the professor posted? If so, then
$$\frac{\partial^2}{\partial x \partial y} \int M(x,y) \mathrm d x = \frac{\partial}{\partial y} M(x,y)$$

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.

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!

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.

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.

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.

## 1. What does it mean for something to exist?

The concept of existence is a philosophical question that has been debated by scientists and philosophers for centuries. In simple terms, something exists if it has a tangible form or can be perceived by our senses. However, the definition of existence can vary depending on one's beliefs and perspectives.

## 2. How do scientists determine if something exists?

Scientists use the scientific method to determine the existence of something. This involves making observations, forming a hypothesis, conducting experiments, and analyzing the results. If the results consistently support the hypothesis, the existence of the phenomenon can be confirmed.

## 3. What is meant by "continuous" in terms of existence?

In science, continuity refers to the idea that something exists without interruption or gaps. In other words, it is a state of being unbroken and connected. For example, the flow of time is continuous as it never stops or has any gaps.

## 4. Can something exist without being continuous?

Yes, it is possible for something to exist without being continuous. For instance, some phenomena, such as volcanic eruptions or earthquakes, occur in sporadic intervals rather than continuously. These events still exist, but they are not continuous.

## 5. How does the concept of continuity relate to the concept of infinity?

Continuity and infinity are closely related concepts in mathematics. In simple terms, something is continuous if it has no gaps or interruptions, while something is infinite if it has no end or limit. However, not all continuous phenomena are infinite, and not all infinite phenomena are continuous. For example, the set of all real numbers is infinite but not continuous, as there are gaps between each number.

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