(2) Solve using method of characteristics



ut +

u
2
2

x
= 0 in R × (0,∞),
u = g on Γ = R × {t = 0}
where g(x) = 1 − x.

The given partial differential equation (PDE) is:

\[
\begin{aligned}
&u_t + \left(\frac{u^2}{2}\right)_x = 0, \quad \text{in } \mathbb{R} \times (0, \infty), \\
&u(x, 0) = g(x) = 1 - x.
\end{aligned}
\]

We will solve this PDE using the method of characteristics.

### Step 1: Rewrite the PDE

The PDE can be rewritten as:

\[
u_t + u \cdot u_x = 0,
\]

where \( u_t = \frac{\partial u}{\partial t} \) and \( u_x = \frac{\partial u}{\partial x} \).

### Step 2: Identify the Characteristic Equations

The method of characteristics transforms the PDE into a system of ordinary differential equations (ODEs). The characteristic equations are given by:

\[
\frac{dx}{dt} = u, \quad \frac{du}{dt} = 0.
\]

### Step 3: Solve the Characteristic Equations

1. **Solve for \( u \):**
   \[
   \frac{du}{dt} = 0 \implies u = \text{constant along the characteristic curves.}
   \]

   Let's denote this constant by \( u_0 \). Thus, \( u = u_0 \) along each characteristic.

2. **Solve for \( x(t) \):**
   \[
   \frac{dx}{dt} = u_0.
   \]
   Integrating with respect to \( t \), we get:
   \[
   x(t) = u_0 t + x_0,
   \]
   where \( x_0 \) is the initial position of the characteristic curve at \( t = 0 \).

### Step 4: Apply the Initial Condition

The initial condition is \( u(x, 0) = g(x) = 1 - x \). At \( t = 0 \), \( x(0) = x_0 \), and:

\[
u_0 = u(x_0, 0) = 1 - x_0.
\]

So, \( u_0 = 1 - x_0 \).

### Step 5: Express the General Solution

The characteristic equation \( x(t) = u_0 t + x_0 \) can be rewritten as:

\[
x(t) = (1 - x_0) t + x_0.
\]

Solving for \( x_0 \), we have:

\[
x_0 = \frac{x - t}{1 - t}.
\]

Thus, the solution along the characteristic curves is:

\[
u(x, t) = u_0 = 1 - x_0 = 1 - \left(\frac{x - t}{1 - t}\right).
\]

Simplifying this expression:

\[
u(x, t) = \frac{1 - t - (x - t)}{1 - t} = \frac{1 - x}{1 - t}.
\]

### Final Solution

The solution to the PDE is:

\[
u(x, t) = 1 - \frac{x - t}{1 - t}, \quad \text{valid for } t < 1.
\]

For \( t \geq 1 \), the solution might develop shocks, requiring additional analysis (e.g., using the method of weak solutions or Rankine-Hugoniot conditions).

Would you like to go into more details or have any questions?

### Related Questions:
1. How do shocks form in the method of characteristics?
2. What are weak solutions in the context of hyperbolic PDEs?
3. How does the initial condition influence the characteristic curves?
4. Can you explain the Rankine-Hugoniot condition for solving shocks?
5. How would the solution change if the initial condition \( g(x) \) were different?

### Tip:
When dealing with quasilinear PDEs, always check for the potential development of shocks or discontinuities, especially when the solution depends on the initial condition in a non-linear manner.

Learn how to solve the partial differential equation ut + (u^2/2)x = 0 using the method of characteristics. This problem involves applying characteristic equations and initial condition methods. Suitable for advanced undergraduate students and researchers.

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