Suppose that we have a vector valued function $D(x)$ with derivative $H(x)$ and that both of these are smooth. Under what conditions does there exist a function $f(x)$ such that $\nabla f(x) = D(x)$? Is there a functional form for it? I was looking for the multivariate equivalent of the second fundamental theorem of calculus, but was coming up empty.

In my particular case, $H$ is symmetric and semi-definite.

  • $\begingroup$ Helmholtz's theorem is known as the fundamental theorem of vector calculus. As usual in higher dimensions, the statement and application is not as straightforward as in the one-dimensional case. Consider for example change-of-variables in integration. $\endgroup$ – RRL Apr 20 at 0:51

By Helmholtz's theorem -- the fundamental theorem of vector calculus -- any continuously differentiable vector field that vanishes at infinity sufficiently fast can be decomposed into irrotational and solenoidal parts of the form

$$\mathbf{D} = \nabla f + \nabla \times \mathbf{a}$$

We can find $f$ and $\mathbf{a}$ given the divergence and curl of the vector field since

$$\nabla \cdot \mathbf{D} = \nabla \cdot \nabla f + \nabla \cdot \nabla \times \mathbf{a} = \nabla^2f$$


$$\nabla \times \mathbf{D} = \nabla \times \nabla f + \nabla \times \nabla \times \mathbf{a} = \nabla \times \nabla \times \mathbf{a},$$

leading to linear partial differential equations that can be solved with suitable boundary conditions.

If $\mathbf{D}$ is curl-free (irrotational), then $\mathbf{D} = \nabla f$.

  • $\begingroup$ Thank you for your response. I'm still a bit fuzzy as to what the conditions are for the existence of $f$. Is it the vector field vanishing fast enough? Obviously, a non-symmetric $H$ would preclude $D$ being a gradient. $\endgroup$ – rasta Apr 20 at 1:46
  • 1
    $\begingroup$ Do you have a specific example? As you say if $H = \nabla f$ then the matrix of $DH$ would have to have a an obvious symmetry. What you have may just not reduce to a simple solenoidal or irrotational form. The trace of $DH$ is the divergence so if that is not $0$ then we can’t have $H = \nabla \times a$. $\endgroup$ – RRL Apr 20 at 2:17
  • $\begingroup$ Sorry, a little confused by your comment. The way I had is set up was $D=\nabla f$ and $H = \nabla^2 f$. H is positive semi-definite. The actual problem is related to generalized method of moments and would require too much background to post here. $\endgroup$ – rasta Apr 20 at 2:27
  • $\begingroup$ I meant $D$ but wrote $H$ instead in the comment. I also wanted to use $D$ as the derivative operator, so DD is even more confusing. I can refresh my memory about GMM. If I come up with anything I’ll get back to you. $\endgroup$ – RRL Apr 20 at 2:56
  • $\begingroup$ It seems that $D$ is a gradient of a function iff $H$ is symmetric. See (in 2D): steiner.math.nthu.edu.tw/disk3/cal01/reconst1.pdf . Related: math.stackexchange.com/questions/834607/… $\endgroup$ – rasta Apr 21 at 23:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.