From the image you uploaded, the problem asks to prove the following relationships based on Maxwell's equations in a vacuum:

$\nabla^2 \vec{E} = \frac{\partial^2 \vec{E}}{\partial t^2} \quad \text{and} \quad \nabla^2 \vec{B} = \frac{\partial^2 \vec{B}}{\partial t^2}$

Given Maxwell's equations in a vacuum:

1. $\nabla \cdot \vec{E} = 0$
2. $\nabla \cdot \vec{B} = 0$
3. $\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$
4. $\nabla \times \vec{B} = \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}$

Proof for $\nabla^2 \vec{E} = \frac{\partial^2 \vec{E}}{\partial t^2}$:

1. Start with $\nabla \times \vec{B} = \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}$.
2. Taking the curl of both sides: $\nabla \times (\nabla \times \vec{B}) = \nabla \times \left(\mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}\right)$
3. Using the vector identity $\nabla \times (\nabla \times \vec{B}) = \nabla(\nabla \cdot \vec{B}) - \nabla^2 \vec{B}$, and since $\nabla \cdot \vec{B} = 0$, this reduces to: $-\nabla^2 \vec{B} = \mu_0 \epsilon_0 \frac{\partial}{\partial t} (\nabla \times \vec{E})$
4. Substitute $\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$ from Maxwell's equations: $-\nabla^2 \vec{B} = \mu_0 \epsilon_0 \frac{\partial}{\partial t} \left(-\frac{\partial \vec{B}}{\partial t}\right)$
5. Simplifying gives: $\nabla^2 \vec{B} = \mu_0 \epsilon_0 \frac{\partial^2 \vec{B}}{\partial t^2}$

This shows that the wave equation for $\vec{B}$ holds.

Proof for $\nabla^2 \vec{B} = \frac{\partial^2 \vec{B}}{\partial t^2}$:

The proof for the electric field $\vec{E}$ follows a similar process:

1. Start with $\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$.
2. Take the curl of both sides: $\nabla \times (\nabla \times \vec{E}) = -\nabla \times \frac{\partial \vec{B}}{\partial t}$
3. Again, use the vector identity $\nabla \times (\nabla \times \vec{E}) = \nabla(\nabla \cdot \vec{E}) - \nabla^2 \vec{E}$, and since $\nabla \cdot \vec{E} = 0$: $-\nabla^2 \vec{E} = -\frac{\partial}{\partial t} (\nabla \times \vec{B})$
4. Substitute $\nabla \times \vec{B} = \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}$: $-\nabla^2 \vec{E} = -\frac{\partial}{\partial t} \left(\mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}\right)$
5. Simplifying gives: $\nabla^2 \vec{E} = \mu_0 \epsilon_0 \frac{\partial^2 \vec{E}}{\partial t^2}$

Thus, we have shown the wave equation for $\vec{E}$ holds as well.

Both $\vec{E}$ and $\vec{B}$ satisfy wave equations where the wave speed $v = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$, which is the speed of light in a vacuum.

Let me know if you'd like more details or if you have questions.

Related Questions:

1. What are the physical meanings of Maxwell’s equations?
2. How is the speed of light derived from Maxwell’s equations?
3. Why does the wave equation apply to both electric and magnetic fields?
4. What are the implications of the wave equation for electromagnetic fields in a vacuum?
5. How do boundary conditions affect Maxwell’s equations in different media?

Tip:

Maxwell's equations not only unify electric and magnetic fields but also predict the existence of electromagnetic waves, traveling at the speed of light.