# Conditions under which a known vector valued function the gradient of some function

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.

• 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. – RRL Apr 20 at 0:51

## 1 Answer

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

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

and

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

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

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

• 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. – rasta Apr 20 at 1:46
• 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 $H = \nabla \times a$. – RRL Apr 20 at 2:17
• 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. – rasta Apr 20 at 2:27
• 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. – RRL Apr 20 at 2:56
• 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/… – rasta Apr 21 at 23:21