Equation related to the wave equation

[fluidistic]

Homework Statement

Consider the following system of equations: $$\frac{\partial \vec H}{\partial t} -i \vec \nabla \times \vec H =0$$ where $$\vec H$$ is a vector field.
1)Show that $$\vec Y =\partial _t \vec H$$ satisfies the wave equation.
2)Demonstrate that if $$\vec \nabla \cdot \vec H=0$$ initially, then it remains true for all time.

Homework Equations

The wave equation. Namely I must show that $$\frac{\partial ^2 \vec Y}{\partial t^2}- \triangle \vec Y =0$$.

The Attempt at a Solution

1)I must show that $$\frac{\partial ^2}{\partial t^2}(i \vec \nabla \times \vec H)=(i \vec \nabla \time \vec H )$$.
I have that $$\frac{\partial \vec Y}{\partial t}i \vec \nabla \times \vec H=i \frac{\partial}{\partial t}(\vec \nabla \times \vec H )$$.
I'm not sure how to proceed... I'm really lost.
I'd like a tip.


2)None yet, will do after 1).


Thanks for any help.

 

[fzero]

Start with

$$\frac{\partial \vec H}{\partial t} = i \vec \nabla \times \vec H ~~(*)$$

and compute $$\partial/\partial t$$ of both sides. You can use (*) once more to relate the 2nd time derivative of $$\vec{H}$$ to the curl of its curl. This can be related by a standard identity to the sum of $$\nabla^2 \vec{H}$$ and another term that is proportional to the gradient of $$\nabla \cdot \vec{H}$$. So $$\vec{H}$$ almost satisfies the wave equation.

If we take another time derivative, we can use (*) to get rid of the term that involved the divergence. I have a feeling that this part is relevant to part 2.

 

[fluidistic]

Thank you fzero.
I just don't know how to compute $$\frac{\partial }{\partial t}i \vec \nabla \times \vec H$$.
I think I could try to workout the rest (I've noticed the identity you talk about as $$\vec \nabla \times (\vec \nabla \times \vec H )=\vec \nabla (\vec \nabla \cdot \vec H ) - \triangle \vec H$$.)

 

[fzero]

Thank you fzero.
I just don't know how to compute $$\frac{\partial }{\partial t}i \vec \nabla \times \vec H$$.
I think I could try to workout the rest (I've noticed the identity you talk about as $$\vec \nabla \times (\vec \nabla \times \vec H )=\vec \nabla (\vec \nabla \cdot \vec H ) - \triangle \vec H$$.)

It's simple because the derivatives commute:

$$\frac{\partial }{\partial t}i \vec \nabla \times \vec H = i \vec \nabla \times \frac{\partial \vec H}{\partial t} = i \vec \nabla \times ( i \vec \nabla \times \vec H ),$$

where in the last step, we used (*).

 

[fluidistic]

Thanks a lot fzero. I solved part 1), though I'd like a clarification about when you said "So H almost satisfies the wave equation".
I've showed (and you probably did too) that $$\frac{\partial \vec H}{\partial t}$$ satisfies the wave equation. As it is linear, any derivative is also a solution, but I'm not sure -though I strongly believe- that $$\vec H$$ also satisfies it.
So why did you use to word "almost"?

I'll try part 2) now.

 

[fzero]

$$\vec{H}$$ satisfies

$$\frac{\partial^2 \vec{H}}{\partial t^2} - \Delta \vec{H} = - \nabla (\nabla \cdot \vec{H}).$$

That's what I mean by "almost." Part 2 of the question addresses a condition under which $$\vec{H}$$ actually does satisfy the wave equation.