Use the method of characteristics to find the general solution for 𝑢(𝑥, 𝑦, 𝑧) of 𝑢𝑥 + 𝑢𝑦 + 𝑢𝑧 = 0. Find a particular solution which satisfies the auxiliary condition 𝑢(𝑥,𝑦,0)=𝑥2 +𝑦2.

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.

Use the method of characteristics to find the general solution for u(x, y, z) of u_x + u_y + u_z = 0. Find a particular solution which satisfies the auxiliary condition u(x,y,0) = x^2 + y^2.

Learn how to solve the first-order PDE u_x + u_y + u_z = 0 using the method of characteristics. This step-by-step solution includes finding the general and particular solutions, satisfying the auxiliary condition u(x, y, 0) = x^2 + y^2. Suitable for undergraduate students.

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