Skip to content
Part BCSIR NET June 2024transport-only-translates

Transport only translates

Let u = u(x, t) be the solution of the following initial value problem: ut+2024ux=0u_{t} + 2024u_{x} = 0 for x∈R,t>0x \in \mathbb{R}, t > 0, and u(x,0)=u0(x)u(x, 0) = u_{0}(x) for x∈Rx \in \mathbb{R}, where u0:R→Ru_{0} : \mathbb{R} \to \mathbb{R} is an arbitrary C1C^{1} function. Consider the following statements: S1S_{1}: If At:=A_{t} := {x∈R:u(x,t)<1x \in \mathbb{R} : u(x, t) < 1} and |AtA_{t}| denotes the Lebesgue measure of AtA_{t} for every t ≥ 0, then |AtA_{t}| = |A0A_{0}|, ∀t>0.S2\forall{}t > 0. S_{2}: If u0u_{0} is Lebesgue integrable, then for every t > 0, the function x ↦ u(x, t) is Lebesgue integrable. Then

  1. A.both S1S_{1} and S2S_{2} are true✓ Correct
  2. B.S1S_{1} is true but S2S_{2} is false
  3. C.S2S_{2} is true but S1S_{1} is false
  4. D.both S1S_{1} and S2S_{2} are false

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests execution slip.

See pricing

50 are analysed free — try those first.

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

Related counterexample: A first-order quasilinear Cauchy problem with smooth data has a global smooth solution

More on this topic

The chapter behind this: First-order PDE: characteristics, Lagrange and Charpit — free to read

From Partial Differential Equations › First-order PDE: Lagrange, Charpit, characteristics

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

ShareWhatsAppTelegram