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### Problem 1:
Consider the PDE for \( u(t, x) \):

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

with initial conditions \( u(1, x) = 0 \), \( x \in \mathbb{R} \).

#### Part (1a):
Let \( t = t(\tau; s) \), \( x = x(\tau; s) \), and \( U(\tau; s) = u(t(\tau; s), x(\tau; s)) \). For constant \( s \) parameterizing the Cauchy data, we need to determine the ordinary differential equations (ODEs) for \( t(\tau; s) \) and \( x(\tau; s) \), along with the initial conditions describing the characteristic curves.

To find these ODEs, we substitute the given PDE into the method of characteristics. The characteristic equations are:

\[
\frac{dt}{d\tau} = -t, \quad \frac{dx}{d\tau} = x, \quad \frac{dU}{d\tau} = 2x - U.
\]

The initial conditions are \( t(0; s) = 1 \) and \( x(0; s) = s \).

#### Part (1b):
Solve the ODEs for \( t(\tau; s) \) and \( x(\tau; s) \).

- For \( t(\tau; s) \):
  
\[
\frac{dt}{d\tau} = -t \quad \Rightarrow \quad t(\tau; s) = e^{-\tau}.
\]

- For \( x(\tau; s) \):

\[
\frac{dx}{d\tau} = x \quad \Rightarrow \quad x(\tau; s) = se^{\tau}.
\]

#### Part (1c):
On these characteristic curves, determine the ODE for \( U(\tau; s) \), along with the initial condition. This equation must not contain \( x \) and \( t \).

Given that \( U(\tau; s) = u(t(\tau; s), x(\tau; s)) \), the ODE for \( U(\tau; s) \) is:

\[
\frac{dU}{d\tau} = 2se^{\tau} - U,
\]

with the initial condition \( U(0; s) = 0 \).

#### Part (1d):
Solve the ODE for \( U(\tau; s) \). We use the integrating factor method.

The integrating factor \( \mu(\tau) = e^{\tau} \), so:

\[
\frac{d}{d\tau}\left(e^{\tau}U\right) = 2se^{2\tau} \quad \Rightarrow \quad U(\tau; s) = se^{\tau}(e^{\tau} - 1).
\]

#### Part (1e):
Determine \( u(t, x) \) by expressing \( \tau \) and \( s \) in terms of \( t \) and \( x \).

Since \( t = e^{-\tau} \) and \( x = se^{\tau} \), we have:

\[
\tau = -\ln(t), \quad s = xt.
\]

Thus:

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

#### Part (1f):
Sketch the characteristic curves for \( t > 0 \).

The characteristic curves in the \( (t, x) \) plane are given by \( x = se^{\tau} = xt \), so \( x \) and \( t \) are proportional. The curves are straight lines emanating from the origin.

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### Next Steps

Would you like more details on any of these parts, or have any specific questions?

### Relative Questions
1. How would the solution change if the initial condition were different?
2. What is the physical interpretation of the characteristic curves in this context?
3. How does the solution behave as \( t \) approaches 0 or infinity?
4. What would happen if the PDE had an additional term, say \( \frac{\partial^2 u}{\partial x^2} \)?
5. How would you numerically approximate the solution to this PDE?

### Tip
Always check your initial conditions after solving the differential equations to ensure the solution is correct.

Explore the solution process for a partial differential equation (PDE) with initial conditions using the method of characteristics. Learn about characteristic equations, integrating factor method, and solving PDEs step-by-step.

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