This problem involves solving a first-order partial differential equation (PDE) using the method of characteristics. The equation is:

$\partial_t u + u \partial_x u = 0, \quad u(0, x) = |x|$

Step 1: Setting up the characteristic equations

The general form of a first-order PDE is:

$a(x,t,u) \partial_x u + b(x,t,u) \partial_t u = c(x,t,u)$

In this case, comparing with the equation given:

$\partial_t u + u \partial_x u = 0$

We can identify:

• $a(x,t,u) = u$
• $b(x,t,u) = 1$
• $c(x,t,u) = 0$

Now, the characteristic equations are derived from the following system: $\frac{dt}{ds} = 1, \quad \frac{dx}{ds} = u, \quad \frac{du}{ds} = 0$

Step 2: Solving the characteristic equations

From $\frac{du}{ds} = 0$, we deduce that $u$ is constant along a characteristic curve. Thus, $u = u_0$, where $u_0$ is a constant along each characteristic.

Next, from $\frac{dt}{ds} = 1$, we get $t = s + t_0$. Setting $t_0 = 0$, we find $t = s$, so the characteristic curves are parameterized by $s = t$.

From $\frac{dx}{ds} = u = u_0$, we obtain $x = u_0 s + x_0$. Since $s = t$, this gives:

$x = u_0 t + x_0$

Step 3: Applying the initial condition

The initial condition is $u(0, x) = |x|$. When $t = 0$, we have $u = u_0$ and $x = x_0$, so the initial condition tells us that:

$u(0, x_0) = |x_0|$

Thus, $u_0 = |x_0|$. Therefore, the solution along the characteristic curve is:

$u(t, x) = |x_0| = |x - u_0 t|$

Step 4: Final solution

The solution is implicitly given by the characteristics, and we know that $u$ remains constant along each curve. Using the initial condition $u(0, x) = |x|$, the solution to the PDE is:

$u(t, x) = |x|$

This is the solution of the initial value problem for $t > 0$.

Would you like a more detailed explanation of any specific part of this solution? Here are some related questions to deepen your understanding:

1. How do the characteristic curves help in solving first-order PDEs?
2. Why is $u$ constant along each characteristic line in this problem?
3. What is the geometric interpretation of the characteristic equations?
4. How can the method of characteristics be applied to nonlinear PDEs in general?
5. How would the solution change if the initial condition was a different function of $x$?

Tip: The method of characteristics is a powerful tool for solving first-order PDEs, especially when the equation is nonlinear. Understanding the geometric properties of characteristic curves is key to mastering this technique.