If 𝑣 = 𝑟𝑚, where 𝑟2 = 𝑥2 + 𝑦2 + 𝑧2, then prove that 𝑣𝑥𝑥 + 𝑣𝑦𝑦 + 𝑣𝑧𝑧 = 𝑚(𝑚 + 1)𝑟𝑚−2.

1. Break Down the Problem

We need to prove that: $v_{xx} + v_{yy} + v_{zz} = m(m + 1) r^{m - 2}$ where $v = r^m$ and $r^2 = x^2 + y^2 + z^2$.

2. Relevant Concepts

1. Partial Derivatives: We will need to compute the second partial derivatives $v_{xx}$, $v_{yy}$, and $v_{zz}$.
2. Chain Rule: Since $v$ is expressed in terms of $r$, we will use the chain rule for differentiation.

3. Analysis and Detail

1. Calculate $v$ and its first derivatives: $v = r^m = (x^2 + y^2 + z^2)^{m/2}$ Now compute the first partial derivatives using the chain rule: $v_x = \frac{\partial v}{\partial x} = m r^{m - 2} x$ $v_y = \frac{\partial v}{\partial y} = m r^{m - 2} y$ $v_z = \frac{\partial v}{\partial z} = m r^{m - 2} z$

2. Calculate the second partial derivatives: Using the product and chain rule: $v_{xx} = \frac{\partial}{\partial x}\left(m r^{m - 2} x\right) = m \left( (m - 2) r^{m - 4} x^2 + r^{m - 2} \right)$ $v_{yy} = \frac{\partial}{\partial y}\left(m r^{m - 2} y\right) = m \left( (m - 2) r^{m - 4} y^2 + r^{m - 2} \right)$ $v_{zz} = \frac{\partial}{\partial z}\left(m r^{m - 2} z\right) = m \left( (m - 2) r^{m - 4} z^2 + r^{m - 2} \right)$

3. Add the second derivatives: $v_{xx} + v_{yy} + v_{zz} = m\left( (m - 2) r^{m - 4} (x^2 + y^2 + z^2) + 3r^{m - 2} \right)$ Since $r^2 = x^2 + y^2 + z^2$: $v_{xx} + v_{yy} + v_{zz} = m\left( (m - 2) r^{m - 4} r^2 + 3r^{m - 2} \right)$ Simplifying: $= m\left( (m - 2) r^{m - 2} + 3r^{m - 2} \right) = m\left( m + 1 \right) r^{m - 2}$

4. Verify and Summarize

We have computed $v_{xx} + v_{yy} + v_{zz}$ and shown that: $v_{xx} + v_{yy} + v_{zz} = m(m + 1) r^{m - 2}$ This completes the proof.

Final Answer

$v_{xx} + v_{yy} + v_{zz} = m(m + 1) r^{m - 2}$