Symbol for partial derivative not used for partial integrals?

[MostlyHarmless]

$${\frac{∂(xy)}{∂x}=x}$$ Going backwards. If we took,
$$∫x dy$$ we get $$xy+f(x)$$

Now, the only way that
$$∫x dy$$
is a valid operation, is if we know that we came from a partial derivative. Why, when taking a partial integral, do we not denote like this:
$$∫x ∂y$$
It seems like it would be less confusing.

 

[micromass]

The $d$ in the integral is actually the symbol for the so-called "exterior derivative". You should read up on differential forms, I think that will clear up some confusions. A book like Spivak's "calculus on manifolds" is a nice place to start!

 

[MostlyHarmless]

The first equation was supposed to be with respect to y not x. >.<

But regardless, thanks for the info. I'll look into it when I get some time. I should really be doing homework right now too.

What has my life become that my idea of procrastination is participating in a math forum??

 

[micromass]

What has my life become that my idea of procrastination is participating in a math forum??

A few years ago, I had to study for an exam. I said I was going to check the internet for 5 minutes and I found PF. Now I'm a mentor...

 

[MostlyHarmless]

This IS a great place. Very intellectually stimulating. I could read threads for hours and hours here.

 

[lurflurf]

https://www.physicsforums.com/showthread.php?t=648467&highlight=handbook

It is used sometimes to make clear that partial integration is being used.

 

[Curious3141]

Now I'm a mentor...

You're also mental. :tongue:

 

[WannabeNewton]

You're also mental. :tongue:

Ain't that the truth (did you know he threatened to mail me 100s of copies of stewart if I didn't do Reed's Functional Analysis text?)

 

[Curious3141]

Ain't that the truth (did you know he threatened to mail me 100s of copies of stewart if I didn't do Reed's Functional Analysis text?)

So call his bluff and sell them on ebay. PROFIT!

 

[micromass]

So call his bluff and sell them on ebay. PROFIT!

I would think that is is very immoral to actually demand money for a book like Stewart...

 

[joeblow]

The expression $$\partial x$$ is precisely equal to dx if all other "variables" are kept constant. Thus, in iterated integrals, where you *do* keep all variables constant, $$dx = \partial x.$$

 

[Fredrik]

Now, the only way that
$$∫x dy$$
is a valid operation, is if we know that we came from a partial derivative.

That's not true. The y after the d let's us know that the function we're dealing with is the map that takes y to x (as opposed to e.g. the map that takes x to x), and also that we are supposed to evaluate the primitive function of this map at y.

Define $g:\mathbb R\to\mathbb R$ by $g(t)=x$ for all $t\in\mathbb R$.

The result of the "integration" would typically be stated as "xy+C, where C is a constant". But this is just a strange way to say that for all real numbers C, the function $G_C:\mathbb R\to\mathbb R$ defined by $G_C(t)=xt+C$ for all $t\in\mathbb R$ is a primitive function of g. If $f:\mathbb R\to\mathbb R$, then f(x) is a real number. So $G_{f(x)}$ is a primitive function of G. That's why you can write
$$\int x dy = G_{f(x)}(y) =xy+f(x).$$

 

[Fredrik]

It's hard to explain this stuff because of how weird the indefinite integral notation really is. Consider e.g. $\int x^2 dx$. What exactly does this notation represent? Is it a number, a function, or a set of functions? People often write things like
$$\int x^2 dx=\frac{x^3}{3}+C$$ and then say something like "...where C is a constant". But there are several strange things going on here. First of all, x is a dummy variable on the left, but not on the right. Second, C is an arbitrary constant.

The second observation suggests that $\int x^2dx$ should really be thought of as a set. The first observation suggests that this is a set of functions, not numbers. So if we want our notation to make sense, we should write something like this instead:
$$\int x^2 dx =\left\{t\mapsto \frac{t^3}{3}+C\,\bigg|\,C\in\mathbb R\right\}.$$ If you don't like the "mapsto" notation ($\mapsto$), then we'd have to write something like this instead:
$$\int x^2 dx =\left\{F_C\,\big|\,C\in\mathbb R\right\},$$ where for each real number C, $F_C:\mathbb R\to\mathbb R$ is defined by $F_C(t)=t^3/3+C$ for all $t\in\mathbb R$.

 

[MostlyHarmless]

That's not true. The y after the d let's us know that the function we're dealing with is the map that takes y to x (as opposed to e.g. the map that takes x to x), and also that we are supposed to evaluate the primitive function of this map at y.

Define $g:\mathbb R\to\mathbb R$ by $g(t)=x$ for all $t\in\mathbb R$.

The result of the "integration" would typically be stated as "xy+C, where C is a constant". But this is just a strange way to say that for all real numbers C, the function $G_C:\mathbb R\to\mathbb R$ defined by $G_C(t)=xt+C$ for all $t\in\mathbb R$ is a primitive function of g. If $f:\mathbb R\to\mathbb R$, then f(x) is a real number. So $G_{f(x)}$ is a primitive function of G. That's why you can write
$$\int x dy = G_{f(x)}(y) =xy+f(x).$$

If you could take $$∫xdy=xy+c$$ Withoutt knowing we are working backwards from a partial derivative. Then there would be no need to use separation of variables when solving a differential equation. For example. $$xdy-ydx=0$$ The solution to this is not $$xy-yx=c$$ Which would be 0=c. The solution is $$lny-lnx=c$$ I could be misguided in this, but that is my understanding.

 

[Fredrik]

If you could take $$∫xdy=xy+c$$ Withoutt knowing we are working backwards from a partial derivative. Then there would be no need to use separation of variables when solving a differential equation. For example. $$xdy-ydx=0$$ The solution to this is not $$xy-yx=c$$ Which would be 0=c. The solution is $$lny-lnx=c$$ I could be misguided in this, but that is my understanding.

I think the problem here is that to obtain xy-yx=c, you simply inserted an integral sign in front of each term. Why should the equality still hold after that operation?

I don't see what your example has to do with separation of variables. That phrase usually refers to a rewrite like $\phi(x,y)=X(x)Y(y)$ that gives us two ODE's instead of one PDE.

You wrote $x dy - y dx = 0$. I interpret this as a weird way of writing dy/dx = y/x, which can also be written as f'(x)=f(x)/x. This implies
$$\frac 1 x =\frac{f'(x)}{f(x)} =\frac{d}{dx}\log f(x).$$ Now this is something we can integrate, to get $\log x=\log f(x)+C$, and if we set y=f(x), this is equivalent to your result.

I suppose we can obtain this result more quickly by rewriting the original equation as
$$\frac{dx}{x}=\frac{dy}{y}.$$ Now we actually get the right result if we insert an integral sign in front of both sides. But unless you know a theorem that explains the exact conditions that must be satisfied for this sort of trick to work, you're going to have to treat the result as a guess, and verify the result by other means.

 

[MostlyHarmless]

I was just trying to think of some example of a differential equation, and that is what I came up with.

I'm not sure if we are on the same page. I failed to mention that my calculus background is half way through elementary differential equations, and oversight on my part. When I made the OP I had Differential equations in mind. We have several methods of solving ODE, generally all of these methods attempt to get "only x's in front of dx" and "only y's in front of dy." If we are allowed to take $$∫xdy$$ without the distinction of "xdy" coming from a partial derivative. Then differential equations, would be a bucket of rainbows and lollipops, no? So that is my issue.

To restate it, my understanding is that $$∫xdy$$ is not a valid operation, in the context of differential equations at least, unless you know that xdy came from taking a partial derivative.

 

[Fredrik]

I disagree, for the reasons stated in post #12. It seems to me that you're confusing the validity of $\int x dy$ with the validity of inserting integral signs in front of each term in any equation involving dx and dy separately. The latter operation seems to be valid when each term involves only one variable, but I don't immediately see why.

It's been a long time since I studied differential equations, so I don't remember any of those methods to solve them. If you're half-way through the course, you already know more of them than I do. Separation of variables (the kind I described) and the rewrite $f'(x)/f(x)=d/dx \log f(x)$ are pretty much the only tricks I remember.

 

[pwsnafu]

To restate it, my understanding is that $$∫xdy$$ is not a valid operation, in the context of differential equations at least, unless you know that xdy came from taking a partial derivative.

Can you come up with an example where you don't know the number of dependent variables?

 

[Fredrik]

Can you come up with an example where you don't know the number of dependent variables?

We only have to think about dependent/independent variables when the notation is bad to begin with, e.g. when y is a function and someone has written y instead of y(x) out of laziness. If we assume that the notation contains all the information we need, then expressions like $\int x dy$ are unambiguous.

 

[pwsnafu]

We only have to think about dependent/independent variables when the notation is bad to begin with, e.g. when y is a function and someone has written y instead of y(x) out of laziness. If we assume that the notation contains all the information we need, then expressions like $\int x dy$ are unambiguous.

That's side stepping my point. This thread is framed in the context of DEs. You always know what your variables are when you set up your DE in the first place. That's why I'm asking the OP for an example where variables are unknown. I can't envision a case where OP's problem crops up.

 

[lurflurf]

You do not always know what your variables are when you set up your DE in the first place. Who told you that? It is common to have variables that are related. Many methods depend upon transformations, changes of variables, or relations that simplify the equation(s). This question is more about notation, use whichever notation is more clear at a particular time. A common use of the partial notation (and a case where it is a helpful reminder) is when there is an exact equation like

$$0=\mathrm{d}F=F_x \, \mathrm{dx}+F_y \, \mathrm{dy}$$
an obvious solution is
$$F=\int \! F_x \, \partial x$$
but to avoid extraneous solution we must enforce a condition like
$$F_y=\dfrac{\partial}{\partial y}\int \! F_x \, \partial x$$

 

[Fredrik]

$$0=\mathrm{d}F=F_x \, \mathrm{dx}+F_y \, \mathrm{dy}$$
an obvious solution is
$$F=\int \! F_x \, \partial x$$

I don't see why we would want to write $\partial x$ instead of $\mathrm{d}x$.

I don't see in what sense $F=\int F_x\partial x$ is an obvious solution. The only way I can see that it is, is to first realize that your equation says that the partial derivatives of F are zero everywhere, so the only solutions are constant functions. Then it's obvious that $F=\int 0 \mathrm{d}x$ is a solution.

 

[lurflurf]

Suppose the particular example
0=dF=F_x+F_ydy=2(x-y)(dx-dy)=(2x-2y)dx+(2y-2x)dy
with solution
F=(x-y)²+Constant

It does not matter if we use ∂ or d inside our integral so long as the variables are clearly understood. I believe that using ∂ is a helpful reminder that we must have

$$F_y=\dfrac{\partial}{\partial y}\int \! F_x \, \partial x$$

preventing an error such as

F=x(x-2y)+Constant

 

[Fredrik]

I think the "0=" part of that string of equalities shouldn't be there. 0=dF implies that F is constant.

I still don't understand the rest. It's probably because I'm terrible at differential equations. The problem you're considering is to find all functions F such that
\begin{align}
F_x=2x-2y\\
F_y=2y-2x.
\end{align} I don't even know how to go about to solve this sort of thing systematically, but I see of course that all maps of the form $(x,y)\mapsto (x-y)^2+C$, where C is a real number, satisfy the equations above.

What you're doing looks really odd to me. You seem to be saying that the $\partial$ symbol in $F=\int F_x\partial x$ tells us to consider both equations above, rather than just the first one. Then why would we use that integral notation at all, instead of simply writing down the two equations above? What you're doing looks like a very strange statement of the problem, and not at all like a solution to the problem.

 

[lurflurf]

http://www.cliffsnotes.com/study_guide/Exact-Equations.topicArticleId-19736,articleId-19710.html

0=dF does not require F=0 only that the gradient is perpendicular to displacement which can be written in vector form as

$$0=\mathrm{d}\vec{r} \cdot \vec{\nabla}\mathrm{F}(\vec{r}) $$

which tells us we have a conservative field and integrals are path independent and can be written as differences of the potential function

$$\int_{(x_0,y_0)}^{(x,y)}\mathrm{d}\vec{r} \cdot \vec{\nabla}\mathrm{F}(\vec{r})=\int_{(x_0,y_0)}^{(x,y)}(F_x \mathrm{dx}+F_y \mathrm{dy})=\int_{(x_0,y_0)}^{(x,y_0)} F_x \partial \mathrm{x}+\int_{(x,y_0)}^{(x,y)} F_y \partial \mathrm{y}=F(x,y)-F(x_0,y_0)$$

The integral is the systematic way of solving as F can be found by integrating F_x or F_y we just need to make sure they match up.

 

[WannabeNewton]

Micromass answered this in literally the first response post. The symbol $d$ denotes the exterior derivative and when doing things like path integrals or surface integrals or w\e you are taking advantage of differential forms in order to actually perform the integration. It is not as if the notation is completely arbitrary. You can't just go ahead and replace $\int_{(x_{0},y_{0})}^{(x,y)}F_xdx$ with $\int_{(x_{0},y_{0})}^{(x,y)}F_x\partial x$. The second expression has no meaning whereas the first expression codifies the fact that you need differential forms in order to perform such integrals and the $d$ is an actual map not just a choice of notation.

 

[Fredrik]

Micromass answered this in literally the first response post. The symbol $d$ denotes the exterior derivative and when doing things like path integrals or surface integrals or w\e you are taking advantage of differential forms in order to actually perform the integration.

But when we're talking about an ordinary Riemann integral of a function from ℝ or ℝ² into ℝ, there's no need to mention differential forms.

and the $d$ is an actual map not just a choice of notation.

In the simple case I described above, which is also the case the OP was concerned with, it is just a choice of notation, and doesn't require any knowledge of differential forms.

 

[Fredrik]

http://www.cliffsnotes.com/study_guide/Exact-Equations.topicArticleId-19736,articleId-19710.html

0=dF does not require F=0 only that the gradient is perpendicular to displacement which can be written in vector form as

$$0=\mathrm{d}\vec{r} \cdot \vec{\nabla}\mathrm{F}(\vec{r}) $$

Thanks for the link. That page was good reminder of some things I had forgotten. The "integrate and merge" trick still looks to me like a clever way to guess the solution. I haven't really thought it through though. Maybe I will find it obvious that it must work if I think about it.

I still don't like the $\partial$ notation in the integrals, and I don't even see how it's supposed to be useful. If it's just a reminder that "integrate" is only the first step in "integrate and merge", then I don't think it's useful.

I didn't say that dF=0 implies that F=0. I said that it implies that the partial derivatives of F are 0, and equivalently, that F is constant. If dF=0, then the right-hand side of the equality in your quote is zero for all $\mathrm{d}\vec{r}\in\mathbb R^3$, and this forces $\vec{\nabla}\mathrm F$ to be zero.

Note that the simplest meaningful definition of dF is as a function from ℝ⁴ into ℝ:
$$dF(x,y,h,k)=D_1f(x,y)h+D_2f(x,y)k.$$ If we use this definition, then dF=0 means that dF(x,y,h,k)=0 for all x,y,h,k.