# Showing scalar functions u(x,y,z) and v(x,y,z) are related

1.a. Show that ∇F[u(x,y,z),v(x,y,z)] = Fu∇v + Fv∇u
1.b. Show that a necessary and sufficient condition that u and v are functionally related by the equation F(u,v) = 0 is ∇u x ∇v = 0

## Homework Equations

∇ = $\frac{\partial}{\partial x}$$\widehat{i}$ + $\frac{\partial}{\partial y}$$\widehat{j}$ + $\frac{\partial}{\partial z}$$\widehat{k}$

3. The Attempt at a Solution 1.a
∇F[u(x,y,z),v(x,y,z)] = (Fuux + Fvvx)$\widehat{i}$ + (Fuuy + Fvvy)$\widehat{j}$ + (Fuuz + Fvvz)$\widehat{k}$ = Fu∇u + Fv∇v

4. The attempt at solution 1.b
I'm honestly stuck. The necessary and sufficient condition throws me. If I work from the assumption that F[u,v] = 0, I can get:

∇F x ∇v = Fu∇u x ∇v = 0
∇F x ∇u = Fv∇v x ∇u = 0

Either of which lead to ∇u x ∇v = 0

But this seems to show neither necessity nor sufficiency.

I know that this leads to developing the Jacobian and I have an inkling that the delta function may help, but can't get anywhere with that. Any pointers would be greatly appreciated.

On the contrary, your work so far demonstrates necessity. You proved $F(u,v) = 0 \implies \nabla u \times \nabla v = 0$. In other words, you proved $\nabla u \times \nabla v = 0$ is necessary follows from $F(u,v) = 0$.

Now prove $\nabla u \times \nabla v = 0 \implies F(u,v) = 0$ to demonstrate sufficiency.

Last edited:
On the contrary, your work so far demonstrates necessity. You proved $\nabla u \times \nabla v = 0 \implies F(u,v) = 0$. In other words, you proved $F(u,v) = 0$ necessarily follows from $\nabla u \times \nabla v = 0$.

Now prove $F(u,v) = 0 \implies \nabla u \times \nabla v = 0$ to demonstrate sufficiency.

Let F[u,v] = uv

∇F = u∇v + v∇u

Repeat the cross products? I guess I'm lost on the starting point to demonstrate sufficiency.

My previous post was in error, which I will proceed to correct with a detailed explanation.

Assume $F(u,v) = 0$.

$$F(u,v) = 0$$

$$\implies \begin{cases} \frac{dF}{dx} = 0 \\\\ \frac{dF}{dy} = 0 \\\\ \frac{dF}{dz} = 0 \end{cases}$$

$$\implies \begin{cases} \frac{\partial F}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial F}{\partial v}\frac{\partial v}{\partial x} = 0 \\\\ \frac{\partial F}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial F}{\partial v}\frac{\partial v}{\partial y} = 0 \\\\ \frac{\partial F}{\partial u}\frac{\partial u}{\partial z} + \frac{\partial F}{\partial v}\frac{\partial v}{\partial z} = 0 \end{cases}$$

$$\implies \begin{cases} \frac{\partial F}{\partial u}\frac{\partial u}{\partial x} = -\frac{\partial F}{\partial v}\frac{\partial v}{\partial x} \\\\ \frac{\partial F}{\partial u}\frac{\partial u}{\partial y} = -\frac{\partial F}{\partial v}\frac{\partial v}{\partial y} \\\\ \frac{\partial F}{\partial u}\frac{\partial u}{\partial z} = -\frac{\partial F}{\partial v}\frac{\partial v}{\partial z} \end{cases}$$

$$\implies \begin{cases} \frac{\partial u}{\partial x} \frac{\partial v}{\partial y} = \frac{\partial u}{\partial y}\frac{\partial v}{\partial x} \\\\ \frac{\partial u}{\partial x} \frac{\partial v}{\partial z} = \frac{\partial u}{\partial z}\frac{\partial v}{\partial x} \\\\ \frac{\partial u}{\partial y} \frac{\partial v}{\partial z} = \frac{\partial u}{\partial z}\frac{\partial v}{\partial y} \end{cases}$$

Now, compute $\nabla u \times \nabla v$.

$$\nabla u \times \nabla v = (\frac{\partial u}{\partial y}\frac{\partial v}{\partial z} - \frac{\partial u}{\partial z}\frac{\partial v}{\partial y})\mathbf{i} - (\frac{\partial u}{\partial x}\frac{\partial v}{\partial z} - \frac{\partial u}{\partial z}\frac{\partial v}{\partial x})\mathbf{j} + (\frac{\partial u}{\partial x}\frac{\partial v}{\partial y} - \frac{\partial u}{\partial y}\frac{\partial v}{\partial x})\mathbf{k} = 0\mathbf{i} + 0\mathbf{j} + 0\mathbf{k} = \mathbf{0}$$

We have proved $\nabla u \times \nabla v = \mathbf{0}$ necessarily follows from $F(u,v) = 0$. In other words, $\nabla u \times \nabla v = \mathbf{0}$ is necessary for $F(u,v) = 0$.

Reversing the direction of the proof shows $\nabla u \times \nabla v = \mathbf{0} \implies F(u,v) = 0$. Therefore, $\nabla u \times \nabla v = \mathbf{0}$ is sufficient for $F(u,v) = 0$.

You will notice we did not directly use the result from 1.a. in our answer to 1.b. I will investigate further and report any connection I discover. Cheers!