To solve the first-order partial differential equation (PDE) using the method of characteristics, let's first outline the problem:

We are given the PDE: $u_x + u_y + u_z = 0$ and we want to find the general solution $u(x, y, z)$.

Step 1: Set up the characteristic equations

The method of characteristics involves converting the PDE into a system of ordinary differential equations (ODEs) along characteristic curves. The PDE can be written as: $\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} = 0.$ This gives the following system of characteristic equations: $\frac{dx}{ds} = 1, \quad \frac{dy}{ds} = 1, \quad \frac{dz}{ds} = 1.$

Step 2: Solve the characteristic system

From the equations $\frac{dx}{ds} = 1$, $\frac{dy}{ds} = 1$, and $\frac{dz}{ds} = 1$, we can integrate to find: $x = s + C_1, \quad y = s + C_2, \quad z = s + C_3,$ where $C_1, C_2, C_3$ are constants of integration.

Now, subtracting these equations from each other to eliminate $s$, we get: $x - y = C_1 - C_2 = C, \quad x - z = C_1 - C_3 = D,$ where $C$ and $D$ are constants. Therefore, the general solution must be a function of these two constants: $u(x, y, z) = f(x - y, x - z).$ This is the general solution for the PDE.

Step 3: Apply the auxiliary condition

We are given the auxiliary condition: $u(x, y, 0) = x^2 + y^2.$ Substitute $z = 0$ into the general solution $u(x, y, z) = f(x - y, x - z)$: $u(x, y, 0) = f(x - y, x) = x^2 + y^2.$ Now we need to find $f(\xi_1, \xi_2)$ such that: $f(x - y, x) = x^2 + y^2.$ Let’s assume $f(\xi_1, \xi_2) = g(\xi_1) + h(\xi_2)$. This gives: $g(x - y) + h(x) = x^2 + y^2.$ We can now separate variables. Since $g(x - y)$ is a function of $x - y$ and $h(x)$ is a function of $x$, the only way to satisfy this equation is: $g(x - y) = y^2, \quad h(x) = x^2.$ Thus, $g(\xi_1) = \xi_1^2$ and $h(\xi_2) = \xi_2^2$, giving: $f(\xi_1, \xi_2) = \xi_1^2 + \xi_2^2.$

Step 4: Write the particular solution

Substitute $\xi_1 = x - y$ and $\xi_2 = x - z$ into the expression for $f(\xi_1, \xi_2)$: $u(x, y, z) = (x - y)^2 + (x - z)^2.$ This is the particular solution to the PDE that satisfies the auxiliary condition $u(x, y, 0) = x^2 + y^2$.

Final Answer

The particular solution to the PDE $u_x + u_y + u_z = 0$ that satisfies $u(x, y, 0) = x^2 + y^2$ is: $u(x, y, z) = (x - y)^2 + (x - z)^2.$

Would you like more details or explanations on any step?

Here are 5 related questions to explore:

1. What is the geometric interpretation of characteristic curves in this context?
2. How do you handle non-linear PDEs with the method of characteristics?
3. Can the method of characteristics be used for second-order PDEs?
4. How does the auxiliary condition affect the general solution?
5. What happens if we change the auxiliary condition to something more complex?

Tip: Always double-check if the auxiliary condition is correctly applied to ensure consistency in solving PDEs using methods like characteristics.