Questions: Method of Characteristics

Question 1

For the transport equation u_t + cu_x = 0, what are the characteristic curves?

• A. Lines x = ct + x₀
• B. Lines x = -ct + x₀
• C. Parabolas x = ct² + x₀
• D. Lines t = cx + t₀

Answer: (A)

The characteristics satisfy dx/dt = c, giving x = ct + x₀. Along these lines, the solution is constant: u(x,t) = u₀(x - ct), showing that the initial profile translates to the right at speed c.

Question 2

The method of characteristics can be applied directly to elliptic PDEs.

• T. True
• F. False

Answer: False

Elliptic PDEs have no real characteristic curves—their discriminant B² - AC is negative, so the characteristic equation has no real solutions. The method of characteristics applies to first-order PDEs and to hyperbolic (and sometimes parabolic) second-order PDEs.

Question 3

What happens when characteristics cross in a quasilinear first-order PDE?

Model answer: Shock waves form and the classical solution breaks down

When characteristics intersect, the solution would need to be multi-valued, which is physically impossible. Instead, a discontinuity called a shock wave develops, and one must work with weak solutions and entropy conditions to select the physically relevant solution.

Question 4

For the equation u_t + u·u_x = 0 (Burgers' equation) with u(x,0) = -x, at what time do characteristics first cross?

• A. t = 0
• B. t = 1
• C. t = 2
• D. They never cross

Answer: (B)

The characteristics are x = -x₀·t + x₀ = x₀(1-t). All characteristics pass through x = 0 when t = 1, so they first cross at t = 1. This is when the solution develops a shock.