Getting a solution from a system of PDEs

[CrosisBH]

Homework Statement

From Griffith's E&M 1.50b
Show that $$\vec{F}_3=yz\hat{x}+zx\hat{y}+xy\hat{z}$$ can be written both as the gradient of a scalar and as the curl of a vector. Find scalar and vector potentials for this function.

Relevant Equations

None specifically (I think), all I need is the definition of the gradient of a scalar function, and the curl of a vector function.

I want to start off here saying I took the problem has finding a potential function, and not a general solution, so I worked to only find one function that works.

I already confirmed that this function can be written as a curl of a vector function and the gradient of a scalar function.

Since it can:

$$\vec{F}_3 = \nabla v$$
and so
$$yz = \frac{\partial v}{\partial x}$$
$$zx = \frac{\partial v}{\partial y}$$
$$xy = \frac{\partial v}{\partial z}$$

Here's where the physicist rigor comes in. I only took the first function and integrated it with respect with x (treating the partial derivitive as a normal derivitive) and got that v = xyz + C(y,z), and I just set the constant = 0 and got a solution of v = xyz.

The second part is what's tripping me. Where I have to write this as a curl of a vector function.
$$\vec{F}_3 = \nabla \times \vec{C}$$

I expanded the right hand side out to vector components and got this system:
$$yz = \frac{\partial C_z}{\partial y} - \frac{\partial C_y}{\partial z}$$
$$xz = \frac{\partial C_x}{\partial z} - \frac{\partial C_z}{\partial x}$$
$$xy = \frac{\partial C_y}{\partial x} - \frac{\partial C_x}{\partial y}$$

I have no idea how to force a solution out of this, and I don't have the mathematical knowledge to solve this system of PDEs for a general solution. I'm kinda stuck. Any help is appreciated.

 

[Orodruin]

I only took the first function and integrated it with respect with x (treating the partial derivitive as a normal derivitive) and got that v = xyz + C(y,z), and I just set the constant = 0 and got a solution of v = xyz.

The first part is fine, but you cannot do the second part without additional argumentation, i.e., inserting your v into the remaining differential equations and finding out that the partial derivatives of C(x,y) are both zero, actually making C a constant. It is otherwise somewhat confusing to call a function (which C(x.y) is) a constant.

For your remaining question, note that the vector potential is not going to be unique so you are going to need additional constraints in order to find it unambiguously. A common approach is to introduce an additional constraint, for example $C_z = 0$, try it out and see where it gets you.

 

[CrosisBH]

The first part is fine, but you cannot do the second part without additional argumentation, i.e., inserting your v into the remaining differential equations and finding out that the partial derivatives of C(x,y) are both zero, actually making C a constant. It is otherwise somewhat confusing to call a function (which C(x.y) is) a constant.

For your remaining question, note that the vector potential is not going to be unique so you are going to need additional constraints in order to find it unambiguously. A common approach is to introduce an additional constraint, for example $C_z = 0$, try it out and see where it gets you.

I did your suggestion, and worked out the system and got a vector of $\vec{C} = \left(\frac{1}{2}xz^2 - \frac{1}{2}xy^2\right)\hat{x} - \frac{1}{2} yz^2 \hat{y}$ as a potential function and worked it out and the curl of the function is the original function. So for these types of problems that demand a solution but not a specific one, I can keep imposing reasonable constrants until a solution pops up? (By the way, thanks a bunch)

 

[Orodruin]

So for these types of problems that demand a solution but not a specific one, I can keep imposing reasonable constrants until a solution pops up?

If by ”reasonable” you mean that they are not so stringent that they rule out all solutions. Otherwise there is nothing ”physically reasonable” in a particular choice of constraint other than that it might make it easier to find a solution.

 

[Dr Transport]

For a vector $\vec{B} = \vec{\nabla} \times \vec{A}$, isn't the solution $\vec{A} = \frac{1}{2} \vec{B} \times \vec{r}$?

 

[Orodruin]

For a vector $\vec{B} = \vec{\nabla} \times \vec{A}$, isn't the solution $\vec{A} = \frac{1}{2} \vec{B} \times \vec{r}$?

No, this is true if $\vec B$ is constant, which is not the case here. Furthermore the solution is not unique neither for the scalar nor for the vector potential.