Proof of function existence

In summary: Suppose all second partial derivatives of F = F (x, y) are continuous and F_{xx} + F_{yy} = 0 on an open rectangle R. Show that F_ydx - F_xdy = 0 is exact on R, and therefore there’s a function G such thatG_x = −F_y and Gy = F_x in R.≈≈≈≈≈≈≈To prove that F_ydx + F_xdy = 0 is exact on R,ibilityI have F_{xx} + F_{yy} = 0which is F_{xx}=-F_{yy}Integrating both sides
  • #1
Medicol
223
54
Suppose all second partial derivatives of [itex]F = F (x, y)[/itex] are continuous and [itex]F_{xx} + F_{yy} = 0[/itex] on an open rectangle [itex]R[/itex].
Show that [tex]F_ydx - F_xdy = 0[/tex] is exact on [itex]R[/itex], and therefore there’s a function [itex]G[/itex] such that
[tex]G_x = −F_y[/tex] and [tex]Gy = F_x[/tex] in [itex]R[/itex].
≈≈≈≈≈≈≈≈
To prove that [itex]F_ydx + F_xdy = 0[/itex] is exact on [itex]R[/itex],
I have [tex]F_{xx} + F_{yy} = 0[/tex]
which is [tex]F_{xx}=-F_{yy}[/tex]
Integrating both sides and cancel out the constants I obtain
[tex]F_x=-F_y[/tex]
This proves the function [itex]F_ydx - F_xdy = 0[/itex] is exact on [itex]R[/itex]​
Could you help me prove the existence of [itex]G[/itex] ? Thank you...
 
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  • #2
Medicol said:
Suppose all second partial derivatives of [itex]F = F (x, y)[/itex] are continuous and [itex]F_{xx} + F_{yy} = 0[/itex] on an open rectangle [itex]R[/itex].
Show that [tex]F_ydx - F_xdy = 0[/tex] is exact on [itex]R[/itex], and therefore there’s a function [itex]G[/itex] such that
[tex]G_x = −F_y[/tex] and [tex]Gy = F_x[/tex] in [itex]R[/itex].
≈≈≈≈≈≈≈≈
To prove that [itex]F_ydx + F_xdy = 0[/itex] is exact on [itex]R[/itex],
I have [tex]F_{xx} + F_{yy} = 0[/tex]
which is [tex]F_{xx}=-F_{yy}[/tex]
Integrating both sides and cancel out the constants I obtain
[tex]F_x=-F_y[/tex]
This proves the function [itex]F_ydx - F_xdy = 0[/itex] is exact on [itex]R[/itex]​
Could you help me prove the existence of [itex]G[/itex] ? Thank you...

If your rectangle lives in the plane, then just show the form [tex]F_ydx - F_xdy = 0[/tex] is closed, since in a rectangle, every closed form is exact. If you haven't seen this, what results can you use?
 
  • #3
Medicol said:
[tex]F_{xx}=-F_{yy}[/tex]
Integrating both sides and cancel out the constants

How do you know constants are identical?
 
  • #4
I answered so, I meant to choose 2 exact constants to give both a go.
I'm thinking because I already answered the first which is also the main part of the problem, I may continue to put "the given function has become exact on R, so there must be a function G that satisfies both of the given conditions in R"

Is this a correct solution ?
 
  • #5


Sure, to prove the existence of G, we can use the definition of an exact differential form. A differential form is exact if it is the differential of some function. In this case, we want to show that F_ydx - F_xdy is the differential of some function G.

First, let's rewrite the given equation as F_ydx + F_xdy = 0. Using the definition of a differential form, we can rewrite this as dG = F_ydx + F_xdy, where G is the unknown function we are trying to find.

Next, we can use the fact that all second partial derivatives of F are continuous to apply Clairaut's theorem. This theorem states that if the second partial derivatives of a function are continuous, then the order of differentiation does not matter. In other words, F_{xy} = F_{yx}.

Using this, we can rewrite the equation as dG = F_ydx + F_xdy = F_{yx}dx + F_{xy}dy.

Now we can use the definition of a partial derivative to rewrite this as dG = (F_xdx + F_ydy)dx + (F_ydx + F_xdy)dy.

Since F_{xx}=-F_{yy}, we can substitute this into the first term to get dG = (-F_{yy}dx + F_{yy}dy)dx + (F_ydx + F_xdy)dy.

Simplifying, we get dG = (-F_{yy}dx + F_ydy)dx + (F_xdy + F_xdy)dy.

Now we can group like terms and factor out the dx and dy to get dG = (-F_{yy} + F_x)dx + (F_y + F_x)dy.

We can see that this is equivalent to dG = (-F_ydx + F_xdy) + (F_xdx + F_ydy).

Since we know that these two terms cancel out, we are left with dG = 0, which means that G is a constant function.

Therefore, we have found a function G that satisfies G_x = -F_y and G_y = F_x, and we have shown that F_ydx - F_xdy is exact on R.
 

What is proof of function existence?

Proof of function existence is a scientific method used to demonstrate that a particular biological entity, such as a protein or gene, has a specific function in a living organism or system.

Why is proof of function existence important?

Proof of function existence is important because it provides evidence for the role of a biological entity in a living system. This information can help researchers understand the mechanisms of life and potentially lead to new medical treatments or technologies.

How is proof of function existence determined?

Proof of function existence can be determined through a variety of methods, including genetic manipulation, biochemical assays, and physiological experiments. These methods aim to isolate and manipulate the biological entity of interest to demonstrate its specific function.

What are some challenges in obtaining proof of function existence?

One challenge in obtaining proof of function existence is the complexity of biological systems. Many factors can influence the function of a biological entity, making it difficult to isolate and demonstrate its specific function. Additionally, the ethical considerations of conducting experiments on living organisms can also pose challenges.

Can proof of function existence be used in other fields besides biology?

Yes, the concept of proof of function existence can also be applied in other scientific fields, such as engineering and technology. In these fields, it may refer to demonstrating the function of a machine or system, rather than a biological entity.

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