(5) Let u : R
2 → R be the solution to the Cauchy problem:
(
∂xu + 2∂yu = 0 for (x, y) ∈ R
2
,
u(x, y) = sin(x) for y = 3x + 1, x ∈ R.
Let v : R
2 → R be the solution to the Cauchy problem:
(
∂xv + 2∂yv = 0 for (x, y) ∈ R
2
,
v(x, 0) = sin(x) for x ∈ R.
Let S = [0, 1] × [0, 1]. Which of the following statements are true?
(a) u changes sign in the interior of S.
(b) u(x, y) = v(x, y) along a line in S.
(c) v changes sign in the interior of S.
(d) v vanishes along a line in S.

To solve this problem, we need to analyze the solutions \( u(x, y) \) and \( v(x, y) \) to the given partial differential equations (PDEs) and the Cauchy conditions provided.

### 1. Solving the PDE for \( u(x, y) \)

The PDE for \( u(x, y) \) is:
\[
\partial_x u + 2\partial_y u = 0
\]
This is a first-order linear PDE. The general method to solve such an equation involves finding the characteristics, which are given by the equation:
\[
\frac{dx}{1} = \frac{dy}{2} = \frac{du}{0}
\]
From \( \frac{dx}{1} = \frac{dy}{2} \), we integrate to find the characteristic curves:
\[
y - 2x = C_1
\]
where \( C_1 \) is a constant. Thus, the general solution is of the form:
\[
u(x, y) = f(y - 2x)
\]
where \( f \) is an arbitrary function determined by the initial condition.

The initial condition given is:
\[
u(x, y) = \sin(x) \quad \text{for} \quad y = 3x + 1
\]
Substitute \( y = 3x + 1 \) into the general solution:
\[
u(x, 3x + 1) = f((3x + 1) - 2x) = f(x + 1)
\]
So, we have:
\[
f(x + 1) = \sin(x)
\]
Thus, the function \( f \) can be written as:
\[
f(z) = \sin(z - 1)
\]
Therefore, the solution \( u(x, y) \) is:
\[
u(x, y) = \sin(y - 2x - 1)
\]

### 2. Solving the PDE for \( v(x, y) \)

The PDE for \( v(x, y) \) is the same:
\[
\partial_x v + 2\partial_y v = 0
\]
The solution follows a similar procedure:
\[
v(x, y) = g(y - 2x)
\]
where \( g \) is determined by the initial condition:
\[
v(x, 0) = \sin(x)
\]
Substitute \( y = 0 \) into the general solution:
\[
v(x, 0) = g(-2x) = \sin(x)
\]
Thus, the function \( g \) can be written as:
\[
g(z) = \sin\left(-\frac{z}{2}\right)
\]
Therefore, the solution \( v(x, y) \) is:
\[
v(x, y) = \sin\left(-\frac{y - 2x}{2}\right) = \sin\left(\frac{2x - y}{2}\right)
\]

### 3. Analyzing the Statements

- **(a) \( u \) changes sign in the interior of \( S \):**

  \( u(x, y) = \sin(y - 2x - 1) \). To check if it changes sign in \( S = [0, 1] \times [0, 1] \), consider the function within this domain. Since \( \sin(z) \) oscillates between \(-1\) and \(1\), and \( y - 2x - 1 \) takes different values within the region, \( u \) can indeed change sign. This statement is **true**.

- **(b) \( u(x, y) = v(x, y) \) along a line in \( S \):**

  \( u(x, y) = \sin(y - 2x - 1) \) and \( v(x, y) = \sin\left(\frac{2x - y}{2}\right) \). These two functions can be equal for specific values of \( (x, y) \), potentially along a line. To find such a line, equate:
  \[
  y - 2x - 1 = \frac{2x - y}{2}
  \]
  Solving this gives a line equation in \( (x, y) \). This statement is **true**.

- **(c) \( v \) changes sign in the interior of \( S \):**

  Since \( v(x, y) = \sin\left(\frac{2x - y}{2}\right) \) and the sine function oscillates, \( v \) can change sign in \( S \). This statement is **true**.

- **(d) \( v \) vanishes along a line in \( S \):**

  \( v(x, y) = 0 \) when \( \frac{2x - y}{2} = n\pi \) for integer \( n \). This defines lines in \( (x, y) \) space where \( v \) vanishes. This statement is **true**.

### Conclusion

All the statements (a), (b), (c), and (d) are true.

Would you like more details on any part of this explanation?

---

### Related Questions:
1. How are characteristic equations used to solve first-order PDEs?
2. What is the geometric interpretation of the solutions \( u(x, y) \) and \( v(x, y) \)?
3. Can we generalize this solution method to higher-dimensional PDEs?
4. What happens if the initial condition in the Cauchy problem is non-linear?
5. How would the solution change if the PDE had a different coefficient for \( \partial_y u \)?

**Tip:** When solving PDEs with initial conditions, always verify the consistency of the characteristic curves with the given conditions.

Let u : R^2 → R be the solution to the Cauchy problem: (∂xu + 2∂yu = 0 for (x, y) ∈ R^2, u(x, y) = sin(x) for y = 3x + 1, x ∈ R). Let v : R^2 → R be the solution to the Cauchy problem: (∂xv + 2∂yv = 0 for (x, y) ∈ R^2, v(x, 0) = sin(x) for x ∈ R). Let S = [0, 1] × [0, 1]. Which of the following statements are true? (a) u changes sign in the interior of S. (b) u(x, y) = v(x, y) along a line in S. (c) v changes sign in the interior of S. (d) v vanishes along a line in S.

Explore the solution to the Cauchy problem involving partial differential equations. Learn how to solve for u(x, y) and v(x, y) using the method of characteristics, and determine the validity of statements about sign changes and equality along lines within the region [0, 1] × [0, 1]. Suitable for undergraduate students.

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