Let u(x, y) be the solution of the Cauchy problem u·uₓ + uy = 0 for x ∈ ℝ, y > 0, with u(x, 0) = x for x ∈ ℝ. Which of the following is the value of u(2, 3)?

C. Characteristics: dx/dy = u with u constant along them, so x = x₀ + u·y and u = x₀. Hence u = x/(1 + y), and u(2, 3) = 1/2.

Part BCSIR NET June 2023burgers-characteristics

Burgers characteristics

Let u(x, y) be the solution of the Cauchy problem $u\cdot{}u_{x} + u_y = 0$ for $x \in \mathbb{R}, y > 0$ , with u(x, 0) = x for $x \in \mathbb{R}$ . Which of the following is the value of u(2, 3)?

A.2
B.3
C.1/2✓ Correct
D.1/3

The idea, in plain words

Follow the curves along which the solution stays constant. Each starting point carries its own value along a straight line whose slope is that very value, so tracing back from the query point to the initial line gives the answer directly.

Solution

Characteristics: dx/dy = u with u constant along them

so $x = x_{0} + u\cdot{}y$ and $u = x_{0}$ .

Hence u = x/(1 + y), and u(2, 3) = 1/2.

Why each option is right or wrong

Checked against 2 symbolic computations

A.WrongExecution slip
False — 2 is the x-coordinate, that is the initial value u(2, 0).

The solution changes as y increases; reading the answer straight off the initial data ignores the evolution entirely.
B.Wrong
False — 3 is the y-coordinate.

The characteristics carry the initial value forward, and the answer depends on both coordinates through the combination x/(1 + y).
C.Correct
True.

Along characteristics, u is constant and dx/dy = u

so $x = x_{0} + u_{0}y$ with $u_{0} = x_{0}$ from the initial data.

Hence $x = x_{0}(1 + y)$

giving $x_{0} = x/(1 + y)$ and u = x/(1 + y).

At (2, 3): u = 2/4 = 1/2 (verified).

The trigger: for a quasilinear equation of this form, solve for $x_{0}$ in terms of (x, y) and substitute into the initial data.
D.WrongExecution slip
False — 1/3 is what you get from x/(1 + y) with the roles of x and y swapped, or from 1/y.

Note that this solution never forms a shock for y > 0

since the characteristics fan out rather than converge: the denominator 1 + y stays positive.

2 of these 3 wrong options correspond to a named reasoning error — the same errors recur across subjects.

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

Burgers characteristics

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