tut + xux = −2u, x ∈ R, t> 1; u(x, 1) = f(x), x ∈ R.

This problem presents a **partial differential equation (PDE)** in the variables \( x \), \( u \), and \( t \). The equation is:

\[
t u_t + x u_x = -2u,
\]

with the initial condition:

\[
u(x, 1) = f(x),
\]

where \( x \in \mathbb{R} \) (the set of real numbers), \( t > 1 \), and \( u = u(x,t) \) is the unknown function of both \( x \) and \( t \). 

### Approach: Method of Characteristics
This is a **first-order linear PDE**, and we can solve it using the **method of characteristics**. 

1. **Characteristic equations**: Start by rewriting the PDE as:

\[
t \frac{\partial u}{\partial t} + x \frac{\partial u}{\partial x} = -2u.
\]

This gives the system of ordinary differential equations (ODEs) for the characteristic curves:

\[
\frac{dt}{t} = \frac{dx}{x} = \frac{du}{-2u}.
\]

We can now solve each equation separately.

#### 1. Solving \(\frac{dt}{t} = \frac{dx}{x}\):

\[
\frac{dt}{t} = \frac{dx}{x} \implies \frac{d}{dt} \ln(t) = \frac{d}{dx} \ln(x).
\]

Integrating, we get:

\[
\ln(t) - \ln(x) = \text{constant} \implies \frac{t}{x} = C_1,
\]

where \( C_1 \) is a constant along the characteristics. This gives us a relationship between \( t \) and \( x \):

\[
x = C_1 t.
\]

#### 2. Solving \(\frac{dt}{t} = \frac{du}{-2u}\):

\[
\frac{dt}{t} = \frac{du}{-2u} \implies \frac{d}{dt} \ln(t) = -\frac{d}{du} \ln(u^2).
\]

By integrating, we get:

\[
u = \frac{C_2}{t^2},
\]

where \( C_2 \) is another constant.

#### 3. General solution:

Since \( x = C_1 t \), \( C_1 = \frac{x}{t} \), and substituting this into the expression for \( u \), we get the general solution for \( u(x, t) \):

\[
u(x, t) = \frac{1}{t^2} f\left( \frac{x}{t} \right),
\]

where \( f \) is an arbitrary function determined by the initial condition.

### 4. Applying the initial condition:

The initial condition is \( u(x, 1) = f(x) \), so substitute \( t = 1 \) into the general solution:

\[
u(x, 1) = \frac{1}{1^2} f\left( \frac{x}{1} \right) = f(x).
\]

Thus, the general solution satisfies the initial condition, and the final solution is:

\[
u(x, t) = \frac{1}{t^2} f\left( \frac{x}{t} \right).
\]

### Summary:

The solution to the PDE is:

\[
u(x, t) = \frac{1}{t^2} f\left( \frac{x}{t} \right),
\]

where \( f(x) \) is the function given by the initial condition \( u(x, 1) = f(x) \).

Would you like further details or explanations? Here are some related questions for deeper exploration:

1. What are characteristic curves, and how are they used in solving PDEs?
2. How does the initial condition affect the general solution in this method?
3. Can this method of characteristics be applied to non-linear PDEs?
4. What happens if we change the initial condition to \( u(x, t_0) = g(x) \)?
5. How does the behavior of the solution change as \( t \to \infty \)?

**Tip**: The method of characteristics is powerful for solving first-order PDEs, as it transforms the PDE into simpler ODEs along certain paths called characteristic curves.

Explore the solution to the partial differential equation tut + xux = −2u using the method of characteristics. This solution involves solving a system of ordinary differential equations to derive the general solution u(x, t) = (1/t^2) * f(x/t). Suitable for advanced calculus or introductory PDE courses.

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