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

[bla1089]

1.a. Show that ∇F[u(x,y,z),v(x,y,z)] = F_u∇v + F_v∇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)] = (F_uu_x + F_vv_x)$\widehat{i}$ + (F_uu_y + F_vv_y)$\widehat{j}$ + (F_uu_z + F_vv_z)$\widehat{k}$ = F_u∇u + F_v∇v4. 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 = F_u∇u x ∇v = 0
∇F x ∇u = F_v∇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.

 

[QED Andrew]

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.

 

[bla1089]

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.

 

[QED Andrew]

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!