Why the integral of a differential does not give the function back in 2D?

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  • #1
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Let f be a 2 variables function.

1) ##f(x,y)=g(x)+h(y)\Rightarrow df=g'(x)dx+h'(y)dy\Rightarrow\int df=g(x)+k(y)+h(y)+l(x)=f(x,y),\textrm{ if } k=l=0##

2) ##f(x,y)=xy\Rightarrow df=ydx+xdy\Rightarrow\int df=2xy+k(y)+l(x)\neq f(x,y)##

Why in the second case the function cannot be recovered ?
 

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  • #2
mjc123
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You are assuming that y is constant when integrating over x, and vice versa, which is not necessarily true. You need to know how y varies with x over the integration path.
For example, if you integrate from x1 to x2 along a line y = constant, then df = ydx (as dy =0) and the integral = xy + k.
If you integrate along the line y = x, y(x) = x and x(y) = y, so df = xdx + ydy and integral = ½x2 + ½y2 + k = xy + k.
 
  • #3
PeroK
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Let f be a 2 variables function.

1) ##f(x,y)=g(x)+h(y)\Rightarrow df=g'(x)dx+h'(y)dy\Rightarrow\int df=g(x)+k(y)+h(y)+l(x)=f(x,y),\textrm{ if } k=l=0##

2) ##f(x,y)=xy\Rightarrow df=ydx+xdy\Rightarrow\int df=2xy+k(y)+l(x)\neq f(x,y)##

Why in the second case the function cannot be recovered ?
What you have done there bears some resemblance to mathematics; but, it's essentially your own personal symbolic manipulation.
 
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  • #4
jk22
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Can we rename the variable and write : ##\int df=\int d(\underbrace{xy}_{u})=\int du=u+C=xy+C## ?

Also I thought another way like computing to higher order with McLaurin :

##f(dx,dy)=f(0,0)+D_xf(0,0)dx+D_yf(0,0)dy+D_{xx}f(0,0)dx^2/2+D_{yy}f(0,0)dy^2/2+D_{xy}f(0,0)dxdy##

##\Rightarrow f(dx,dy)=dxdy\Rightarrow f(x,y)=\iint dxdy=\int x+D(y) dy=xy+k(x)+l(y)##

But anyhow this would imply that : ##d(xy)=dxdy## which is wrong.
 
  • #5
jk22
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You are assuming that y is constant when integrating over x, and vice versa, which is not necessarily true. You need to know how y varies with x over the integration path.
For example, if you integrate from x1 to x2 along a line y = constant, then df = ydx (as dy =0) and the integral = xy + k.
If you integrate along the line y = x, y(x) = x and x(y) = y, so df = xdx + ydy and integral = ½x2 + ½y2 + k = xy + k.
I rather see as x and y are independent. It's clear that if one assumes a common cause parameter that determines the variables x and y it could change, (like in Bell's theorem)
 
  • #6
mjc123
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They may be independent, in the sense that you can arbitrarily choose your path in the xy plane, but you can't integrate over x while holding y constant, and integrate over y while holding x constant simultaneously, which was what you were trying to do above. However, you can do it sequentially. If f only depends on x and y, then f(x2,y2) - f(x1,y1) depends only on the start and finish points, not on the path taken between them. So we can do the integral in two steps: first integrate from x1 to x2 with y = y1 constant, then from y1 to y2 with x = x2 constant. Then
∫df = ∫y1dx + ∫x2dy = x2y1 - x1y1 + x2y2 - x2y1 = x2y2 - x1y1
Note that the answer is the sum of two terms both of which have the form Δ(xy), but they are not the same, and neither is equal to x2y2 - x1y1; it is their sum that has this value.
 
  • #7
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What about if we say x depends on a parameter fw1 and y on fw2 where fwi represented free will of person i. Could freewill be linked to a single parameter time ?
 
  • #8
hutchphd
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yes
 
  • #9
Delta2
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According to the gradient theorem of multivariable calculus:
https://en.wikipedia.org/wiki/Gradient_theorem
and doing a bit of abuse of notation we can say that ##\int df=f## as long as we define $$df=\nabla f\cdot d\mathbf{r}$$ where ##d\mathbf{r}=dx\hat x+dy\hat y+dz\hat z+...##.

So what you doing wrong and you can't recover f in the case of f(x,y)=xy?

First of all the notation ##\int df## is a bit "naked" in the case that f is a function of many variables, because we have to include a path of integration (which is not necessary in the case of one variable because there the path of integration is alaways a segment on the x-axis). Of course the gradient theorem tells us that the integral does not depend on the path of integration but only on the start and end points of the path, however we need the path of integration when we going to calculate ##\int df## the way you trying to do it.

The important thing here is that you have to keep the path of integration the same when you break up the integral to ##\int xdy## plus ##\int y dx## and that's where you go wrong. In the first integral you consider the path of integration to be along the y-axis but in the second integral you consider it to be along the x-axis. It has to be the same, either along the y-axis for both or along the x-axis for both.

So if we consider a path of integration P that is a segment of the y-axis it ##\int_P x dy=xy## but ##\int_P ydx=0## because since that path is on y-axis, the x coordinate doesn't change so it is dx=0 along that path and hence the integral is zero as well. Similarly if we choose as P a segment on the x-axis and apply it to both the integrals it will be ##\int_P x dy=0## but ##\int_P y dx=xy##. In any case it will be $$\int_P df=\int_P ydx+\int_P x dy=xy+0 (or 0+xy)=xy$$.
 
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  • #10
jk22
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In fact to have two path we would need ##\int x_1dy_1+\int y_2dx_2##.
Classically

What If we suppose the path were probabilists ? Then like in quantum mechanics, it could follow both path, or a superposition of them ? It gives back xy because of the sum of the weights.
 
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  • #11
mjc123
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What do you mean by "follow both paths"? It could follow two different paths between the same start and finish points, but that's not what you were doing above. You were treating ∫xdy as an integral along the y-axis and ∫ydx as an integral along the x axis. From the same starting point, these have different end points. It's not the same integral.
 
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  • #12
WWGD
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Can we rename the variable and write : ##\int df=\int d(\underbrace{xy}_{u})=\int du=u+C=xy+C## ?

Also I thought another way like computing to higher order with McLaurin :

##f(dx,dy)=f(0,0)+D_xf(0,0)dx+D_yf(0,0)dy+D_{xx}f(0,0)dx^2/2+D_{yy}f(0,0)dy^2/2+D_{xy}f(0,0)dxdy##

##\Rightarrow f(dx,dy)=dxdy\Rightarrow f(x,y)=\iint dxdy=\int x+D(y) dy=xy+k(x)+l(y)##

But anyhow this would imply that : ##d(xy)=dxdy## which is wrong.
This is kind of a mess, evaluating a function defined on ##\mathbb R^2## on pairs of differential forms and doing the same for its Taylor Series. There is something for motivating expressions but this seems too loose.
 
  • #13
romsofia
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so, first off, and I'm not sure why no one pointed it out (EDIT: I lied someone did whoops!), if you have ##\int df## and ##f = xy## then saying ##u = xy## doesn't do anything, you first have to take the differential of ##xy##, so no you can't just say ##\int df = \int du = u + C## if ##f = xy##. ##d## is an operator, you can't just ignore it willynilly.

What you *can* write is: $$f = xy \rightarrow df = d(xy) = ydx+xdy \rightarrow \int df = \int ydx + \int xdy$$

Secondly, I think the point you're wanting to get into is: Different paths taken between two points, and in that case I'd get a book on calculus of variations and go from there.
 
  • #14
jk22
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What if in fact x and y would represent the position of relativistic particles with their own proper time ##x(\tau_1)y(\tau_2)##
 
  • #15
Frabjous
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What if in fact x and y would represent the position of relativistic particles with their own proper time ##x(\tau_1)y(\tau_2)##
You have been told multiple times that you need to define a path. Giving different interpretations of the functions is not going to change that mathematical fact.

You have brought up Bell’s theorem, freewill, probability, quantum mechanics and relativity.
 
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