The problem in the image is:

Given $u = ax^2 + by^2 + cz^2$, and $u_{xx} + u_{yy} + u_{zz} = 0$, prove that $a + b + c = 0$.

Solution:

We need to find the second partial derivatives of $u$ with respect to $x$, $y$, and $z$.

1. Expression for $u$:

   $u = ax^2 + by^2 + cz^2$

2. Second partial derivatives:

   • For $u_x$ and $u_{xx}$: $u_x = \frac{\partial u}{\partial x} = 2ax$ $u_{xx} = \frac{\partial^2 u}{\partial x^2} = 2a$

   • For $u_y$ and $u_{yy}$: $u_y = \frac{\partial u}{\partial y} = 2by$ $u_{yy} = \frac{\partial^2 u}{\partial y^2} = 2b$

   • For $u_z$ and $u_{zz}$: $u_z = \frac{\partial u}{\partial z} = 2cz$ $u_{zz} = \frac{\partial^2 u}{\partial z^2} = 2c$

3. Summing the second derivatives: According to the condition $u_{xx} + u_{yy} + u_{zz} = 0$: $2a + 2b + 2c = 0$

4. Simplifying: $a + b + c = 0$

Thus, we have proved that $a + b + c = 0$.

Would you like more details on this process or any related concepts? Here are some follow-up questions that might deepen your understanding:

1. What are the first-order partial derivatives of $u$ with respect to $x$, $y$, and $z$?
2. How do partial differential equations (PDEs) relate to this problem?
3. Can you find another function $u$ that satisfies the condition $u_{xx} + u_{yy} + u_{zz} = 0$?
4. What does this result mean geometrically in terms of the surface described by $u$?
5. How can this technique be applied to other quadratic forms?

Tip: Understanding the role of partial derivatives is crucial for solving PDEs and other problems in multivariable calculus.