Exact Differentials: Proving Existence of u(x,y) in Connected Open Region

[STEMucator]

Homework Statement

Let R be a connected open region ( in the plane ). Suppose that F = (M,N) is a vector function defined on R and is such that for any ( piecewise smooth ) curve C in R :

$\int_C Fdp$

depends on only the endpoints of C ( that is, any two curves from P₁ to P₂ in R give the same value for the integral).

Prove that there exists a function u(x,y) defined on R such that ∇u = F.

( i.e u_x = M and u_y = N )

Homework Equations

Err I think this may have to do with simply connected regions?

The Attempt at a Solution

I'm not quite sure where to start with this one? I'm having trouble seeing how the info provided leads to what I need.

I think it has to do with if R is a simply connected open region and Mdx + Ndy is such that M_y = N_x in R, then the differential is exact.

Any push in the right direction would be great.

 

[mighty2000]

You are correct in thinking that the information provided may have to do with simply connected regions. To prove the existence of a function u(x,y) such that ∇u = F, we can use the fact that R is a connected open region and F = (M,N) is a vector function defined on R. From this, we know that the integral of F along any curve C in R only depends on the endpoints of C. This property is known as path independence.

To prove the existence of u(x,y), we can use the fundamental theorem of line integrals, which states that if a vector function F = (M,N) is path independent, then there exists a function u(x,y) such that ∇u = F. This function u(x,y) is known as the potential function for F.

To show that u(x,y) exists, we can start by considering any two points P1 and P2 in R. Since R is a connected open region, we can choose a path C from P1 to P2 that lies entirely within R. Since the integral of F along any curve C in R only depends on the endpoints of C, we can say that the integral of F along this path C is equal to the difference in values of u at P2 and P1. This can be written as:

∫_C Fdp = u(P2) - u(P1)

Since this is true for any two points P1 and P2 in R, we can say that u(x,y) is a potential function for F, and therefore ∇u = F.

To summarize, we have shown that because R is a connected open region and F = (M,N) is path independent, there exists a function u(x,y) such that ∇u = F. This function u(x,y) is known as the potential function for F. We can also use the fact that R is a simply connected region to show that the differential is exact, which further supports our proof.