(7) Consider the Cauchy problem
u
∂u
∂x +
∂u
∂y = 1, (x, y) ∈ R × (0,∞),
u(x, 0) = kx, x ∈ R,
with a given real parameter k. For which of the following values of k does the
above problem have a solution defined in R × (0,∞)?
(a) k = 0
(b) k = −2
(c) k = 4
(d) k = 1

We are given the following Cauchy problem:

\[
u \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = 1, \quad (x, y) \in \mathbb{R} \times (0,\infty)
\]
\[
u(x, 0) = kx, \quad x \in \mathbb{R}
\]

where \( k \) is a real parameter. We need to determine for which value of \( k \) the problem has a solution defined on \( \mathbb{R} \times (0,\infty) \).

### Step 1: Characteristics of the PDE

The given PDE is a first-order nonlinear partial differential equation. To solve it, we can use the method of characteristics.

The characteristic equations are:

\[
\frac{dx}{ds} = u, \quad \frac{dy}{ds} = 1, \quad \frac{du}{ds} = 1
\]

### Step 2: Solving the Characteristic Equations

From the second equation, we have:

\[
\frac{dy}{ds} = 1 \implies y = s + C_1
\]

Given that at \( s = 0 \), \( y = 0 \), we get \( C_1 = 0 \). Therefore, \( y = s \), or \( s = y \).

From the third equation:

\[
\frac{du}{ds} = 1 \implies u = s + C_2 = y + C_2
\]

From the first equation:

\[
\frac{dx}{ds} = u = y + C_2 \implies x = \frac{y^2}{2} + C_2y + C_3
\]

### Step 3: Using the Initial Condition

The initial condition is \( u(x, 0) = kx \). For \( y = 0 \):

\[
u(x, 0) = C_2 = kx
\]

Since \( C_2 \) should be independent of \( s \), it must be a constant. Therefore, \( C_2 = kx \), which implies that \( k = 0 \) or \( x = \frac{1}{k} \), provided \( k \neq 0 \).

### Step 4: Determining the Validity of the Solution

For \( k \neq 0 \), the solution \( u(x, y) \) involves \( x = \frac{1}{k} \) which suggests that the solution is only valid for specific values of \( x \), not for all \( x \). Hence, the solution is not valid on \( \mathbb{R} \times (0,\infty) \) when \( k \neq 0 \).

For \( k = 0 \), the solution \( u(x, y) \) does not depend on \( x \), and hence a solution defined on \( \mathbb{R} \times (0,\infty) \) exists.

### Conclusion

The solution is valid over \( \mathbb{R} \times (0,\infty) \) **only when \( k = 0 \)**.

Thus, the correct answer is:
**(a) \( k = 0 \)**.

Would you like a more detailed explanation or have any other questions?

---

Here are some related questions to consider:
1. How would the solution change if the initial condition were different?
2. What happens to the solution if the nonlinearity in the PDE is changed?
3. How does the method of characteristics apply to higher-dimensional problems?
4. Can we solve this problem using any numerical methods?
5. What is the physical interpretation of this PDE in a real-world scenario?

**Tip:** Always check the domain of the solution to ensure it fits the initial and boundary conditions.

Explore the solution to the Cauchy problem \( u \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = 1 \) with initial condition \( u(x, 0) = kx \). Learn which values of the parameter k allow for a solution defined in \( \mathbb{R} \times (0, \infty) \). Suitable for graduate-level study.

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