Proving that a function is gradient vector of another function

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Homework Help Overview

The discussion revolves around proving that the gradient of a scalar field is symmetric, particularly focusing on the relationship between partial derivatives of the gradient components. Participants are exploring the implications of this symmetry in the context of scalar functions and their gradients.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the symmetry of partial derivatives and the need to prove the reverse direction of the initial claim. Suggestions include constructing a scalar function by integrating the partial derivatives and exploring the existence of certain functions based on the conditions of the partial derivatives.

Discussion Status

The discussion is active, with participants offering various approaches to tackle the proof. Some guidance has been provided regarding integration of partials, but there is no explicit consensus on the best method to proceed. The original poster expresses difficulty with the proof's other direction, indicating ongoing exploration.

Contextual Notes

There is mention of a time constraint, as the original poster is on an extension, which may influence the urgency and focus of the discussion.

tewaris
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Trying to prove that the gradient of a scalar field is symmetric(?) Struggling with the formatting here. Please see the linked image. Thanks.

http://i.imgur.com/9ZelT.png
 
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so you have shown if g is the gradient of a scalar function then
[tex]\frac{\partial g_i}{\partial x_j} = \frac{\partial g_j}{\partial x_i}[/tex]

this is because the partial derivatives coummte, ie
[tex]\frac{\partial^2 V}{\partial x_j x_i} = \frac{\partial^2 V}{\partial x_i x_j}[/tex]

now you need to conisder the other direction of the proof
 
Unfortunately it's the other direction that has me baffled. Any help would be highly appreciated considering I am already on an extension. Thanks!
 
how about trying to construct the scalar function by integrating the partials you have? or you could maybe try a contradiction... though I'm not convinced on that one
 
Integrating the partials is a good idea.

In two dimensions for example, we want to find a function V(x,y) such that
[tex]\frac{\partial V}{\partial x} = g_1(x,y)[/tex]
and
[tex]\frac{\partial V}{\partial y} = g_2(x,y)[/tex]

We can integrate the first equation w.r.t. x and the second w.r.t. y and get by the fundamental theorem of calculus
[tex]V(x,y) = \int g_1(x,y)dx + F(y)[/tex]
[tex]V(x,y) = \int g_2(x,y)dy + G(x)[/tex]
The constant of integration when you integrate w.r.t. x is really an arbitrary function of y, and vice versa. (here the integration is really just any choice of antiderivative that you want, because we're writing out the constant of integration explicitly). So we're set as long as we can find functions G(x) and F(y) such that
[tex]\int g_1(x,y)dx + F(y)=\int g_2(x,y)dy + G(x)[/tex]

So the question boils down to why do these functions F and G exist given the condition on the partial derivatives of g1 and g2?
 

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