When the gradient of a vector field is symmetric?

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SUMMARY

The gradient of a vector field is symmetric if and only if the vector field is the gradient of a function, as established by Marek L. Szwabowicz in his 2008 paper on Pure Strain Deformations of Surfaces. For the function f=5x^3+3xy-15y^3, the gradient results in a vector field, and the gradient of this gradient yields a symmetric tensor. This conclusion is supported by the mixed partial derivative property, which states that if the derivatives are continuous, the order of differentiation does not affect the outcome.

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Homework Statement



"A gradient of a vector field is symmetric if and only if this vector field is a gradient of a function"
Pure Strain Deformations of Surfaces
Marek L. Szwabowicz
J Elasticity (2008) 92:255–275
DOI 10.1007/s10659-008-9161-5

f=5x^3+3xy-15y^3
So the gradient of this function is a vector field, right? Now the grad of this grad is a tensor which is symmetric and according to Marek it's always like that.
Can you guys think of any reason or proof for it?

Homework Equations


The Attempt at a Solution


Maybe it has something to do with double differentiation...but I can't figure out why...
 
Last edited:
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It's based on the "mixed partial derivative property":
As long as the derivatives are continuous,
[tex]\frac{\partial f}{\partial x\partial y}= \frac{\partial f}{\partial y\partial x}[/tex]
 

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