Proving h(u,v) = U(u) + V(v) for huv=0 ∀ u,v

[bobbarker]

Homework Statement

"Let h_uv = 0 \forall u, v. Show that h is of the form h(u,v) = U(u) + V(v)."

Homework Equations

n/a

The Attempt at a Solution

The problem doesn't really seem that complex to me, in fact, from PDEs a few years ago I remember this quite readily. However, the proof/demonstration is giving me some trouble. I started off supposing that h_uv = 0 \forall u,v but h were of the form h(u,v) = U(u)*V(v). Then
h_u(u,v) = U'(u)*V(v)
h_uv(u,v) = U'(u)*V'(v) = 0
But that just means that either/both U,V are constants, which isn't prohibited in the initial setup of the problem.

Any ideas? Thanks! :)

 

[LCKurtz]

If f'(x) = 0 then you know f is independent of x (f(x) = c).

So if you have this function h_u whose partial with respect to v is zero, what does that say about h_u's dependence on v?

 

[bobbarker]

So since h_u is independent of v and h_v is independent of u, and disregarding the possibility that the functions U,V are constants, then we must have h(u,v) = U(u) + V(v)? Since h_u is not necessarily 0 but h_uv=0, and similarly h_v is not necessarily 0 but h_vu=0?

 

[LCKurtz]

So since h_u is independent of v and h_v is independent of u, and disregarding the possibility that the functions U,V are constants, then we must have h(u,v) = U(u) + V(v)? Since h_u is not necessarily 0 but h_uv=0, and similarly h_v is not necessarily 0 but h_vu=0?

No, that isn't very precise and is confusing. Think about integrating both sides of (h_u)_v = 0 with respect to v. If this were a function of one variable and an ordinary antiderivative, you would get a constant. What would the "constant" of integration look like in this case when, in effect, we are taking an anti-partial derivative with respect to v?