The image shows two mathematical identities to be proven:

(a) $\delta \mu = \frac{1}{2}(\Delta + \nabla)$

(b) $\Delta - \nabla^2 = \Delta \nabla^2$

(a) Proving $\delta \mu = \frac{1}{2}(\Delta + \nabla)$

We need to investigate the left-hand side, $\delta \mu$, and express it in a form similar to $\frac{1}{2}(\Delta + \nabla)$, where $\Delta$ typically refers to the Laplacian operator, and $\nabla$ is the gradient operator. The $\frac{1}{2}$ suggests averaging these operators. However, the exact details depend on the definitions of $\mu$ and $\delta \mu$ in this context, which likely involve some vector calculus identities or manipulations.

(b) Proving $\Delta - \nabla^2 = \Delta \nabla^2$

This is likely another identity involving operators, where $\Delta$ is the Laplace operator and $\nabla^2$ might refer to the square of the gradient. To prove this, we would typically apply known properties of these operators.

Could you clarify the variables involved (e.g., $\mu$, $\delta \mu$) and confirm if they follow vector calculus definitions? Alternatively, we could use assumptions based on common operator identities to proceed.

Let me know how you'd like to approach this! Additionally, here are some related questions:

1. What are the definitions of $\delta \mu$ and $\mu$ in this context?
2. Are there any additional conditions or fields the operators are acting on?
3. Is this problem related to a specific field of physics (e.g., fluid mechanics, electromagnetism)?
4. Should the Laplacian $\Delta$ be interpreted as the divergence of the gradient ($\nabla \cdot \nabla$)?
5. Would you like me to proceed with assumptions based on general vector calculus?

Tip: In vector calculus, identities involving $\nabla$ (gradient), $\nabla \cdot$ (divergence), and $\nabla \times$ (curl) are key for proving operator relations.