Proving that a function is gradient vector of another function

In summary: The answer is that the functions F and G exist because they're solutions to the equation \frac{\partial V}{\partial x} = g_1(x,y) \frac{\partial V}{\partial y} = g_2(x,y)This equation is called the Euler–Lagrange equation. It's a second order differential equation, and it has two solutions. The two solutions areF(y) = g_1(x,y)G(x) = g_2(x,y)In other words, F and G are solutions to the Euler–Lagrange equation if and only if the gradient
  • #1
tewaris
2
0
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
 
Last edited:
Physics news on Phys.org
  • #2
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
 
  • #3
Unfortunately it's the other direction that has me baffled. Any help would be highly appreciated considering I am already on an extension. Thanks!
 
  • #4
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
 
  • #5
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?
 

FAQ: Proving that a function is gradient vector of another function

What does it mean for a function to be a gradient vector of another function?

When a function is the gradient vector of another function, it means that the first function is the rate of change of the second function in a particular direction. In other words, the gradient vector represents the direction of steepest ascent of the second function.

How do I prove that a given function is the gradient vector of another function?

To prove that a function is a gradient vector of another function, you must show that the partial derivatives of the first function with respect to each variable are equal to the components of the gradient vector of the second function in the same order. This can be done using the gradient operator.

Can a function be the gradient vector of more than one function?

Yes, a function can be the gradient vector of more than one function. This is because the gradient vector represents the direction of steepest ascent, and different functions may have the same direction of steepest ascent.

Are there any properties of gradient vectors that can help prove a function is a gradient vector?

Yes, there are several properties of gradient vectors that can be useful in proving that a function is a gradient vector. These include the fact that the gradient vector is perpendicular to level curves of the function, and that the magnitude of the gradient vector represents the rate of change of the function in the direction of steepest ascent.

Is it possible for a function to have a gradient vector that is not continuous?

No, it is not possible for a function to have a gradient vector that is not continuous. The gradient vector is a vector field, and as such, its components must be continuous in order for the vector field to be well-defined. If the components of the gradient vector are not continuous, then the function cannot be the gradient vector of another function.

Similar threads

Replies
8
Views
1K
Replies
9
Views
1K
Replies
6
Views
3K
Replies
4
Views
1K
Replies
20
Views
4K
Replies
4
Views
2K
Replies
7
Views
2K
Back
Top