Solve completely: (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 a first-order nonlinear PDE. To solve it using the method of characteristics, we proceed as follows:

### 1. Problem Statement
We are given the PDE:
\[
u_t + \frac{u^2}{2} u_x = 0 \quad \text{in} \; \mathbb{R} \times (0, \infty),
\]
with the initial condition:
\[
u(x, 0) = g(x) = 1 - x \quad \text{on} \; \Gamma = \mathbb{R} \times \{t = 0\}.
\]

### 2. Characteristics Equations
The method of characteristics involves converting the PDE into a set of ordinary differential equations (ODEs) along characteristic curves. The characteristic equations for the given PDE are:

\[
\frac{dx}{dt} = \frac{u^2}{2}, \quad \frac{du}{dt} = 0.
\]

### 3. Solving the ODEs

#### a) Solve \(\frac{du}{dt} = 0\)
This implies that \( u \) is constant along the characteristic curves, i.e.,
\[
u(x(t), t) = u_0,
\]
where \( u_0 \) is a constant determined by the initial condition.

#### b) Solve \(\frac{dx}{dt} = \frac{u^2}{2}\)
Given that \( u \) is constant along a characteristic, we have:
\[
\frac{dx}{dt} = \frac{u_0^2}{2}.
\]
Integrating this with respect to \( t \), we obtain:
\[
x(t) = \frac{u_0^2}{2} t + x_0,
\]
where \( x_0 \) is the initial position of the characteristic curve at \( t = 0 \).

### 4. Determine \( u_0 \) from the Initial Condition
The initial condition is \( u(x, 0) = 1 - x \). At \( t = 0 \), we have:
\[
u_0 = 1 - x_0.
\]
Substituting \( u_0 = 1 - x_0 \) into the characteristic equation for \( x(t) \):
\[
x(t) = \frac{(1 - x_0)^2}{2} t + x_0.
\]
This equation describes the characteristic curves in the \( x \)-\( t \) plane.

### 5. Express the Solution \( u(x, t) \)
To find the solution \( u(x, t) \), we solve for \( x_0 \) in terms of \( x \) and \( t \):
\[
x - x_0 = \frac{(1 - x_0)^2}{2} t.
\]
Given \( u = 1 - x_0 \), the solution \( u(x, t) \) is implicitly given by the equation:
\[
x = x_0 + \frac{(1 - x_0)^2}{2} t,
\]
where \( x_0 = 1 - u \).

Thus, the solution \( u(x, t) \) satisfies the implicit equation:
\[
x = 1 - u + \frac{u^2}{2} t.
\]

This is the implicit form of the solution, and in general, it may not be possible to explicitly solve for \( u \) in terms of \( x \) and \( t \) without further simplifications or numerical methods.

### Summary
The solution \( u(x, t) \) is determined implicitly by the characteristic equations:
\[
x = 1 - u + \frac{u^2}{2} t,
\]
where \( u(x, t) \) is the value of the function \( u \) at time \( t \) and position \( x \).

---

Do you want more details on any step, or have any questions?

Here are some related questions to explore further:
1. How do we interpret the characteristic curves in the \( x \)-\( t \) plane?
2. What is the significance of the initial condition \( g(x) = 1 - x \) in the solution?
3. Can the solution \( u(x, t) \) be simplified or expressed in an alternative form?
4. How would the solution change if a different initial condition \( g(x) \) were given?
5. What are some numerical methods that could be used to solve this PDE?

**Tip:** When solving nonlinear PDEs using the method of characteristics, always pay attention to the possibility of shocks or discontinuities forming in the solution.

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

Learn how to solve a first-order nonlinear partial differential equation using the method of characteristics. This problem covers the key concepts of characteristic equations and initial conditions with a step-by-step solution, suitable for undergraduate mathematics or advanced high school students.

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