Skip to content
Part CCSIR NET December 2025differentiate-e-and-substitute-the-pde-the-boundary-term-vanishes-because-fixed-endpoints-force-u-t-to-vanish-there-too

Differentiate e and substitute the pde the boundary term vanishes because fixed endpoints force u t to vanish there too

Let u(x,t) be the solution of the initial-boundary value problem utt−uxx+ut=0,0<x<π,t>0,u(0,t)=0,u(π,t)=0,t>0,u(x,0)=sin⁡x,ut(x,0)=0,0<x<πu_{tt} - u_{xx} + u_{t} = 0, 0<x<\pi, t>0, u(0,t)=0, u(\pi,t)=0, t>0, u(x,0)=\sin x, u_{t}(x,0)=0, 0<x<\pi. For t≥0, define E(t)=∫0π(ut2+ux2)E(t) = \int_{0}^\pi (u_{t}^{2}+u_{x}^{2}) dx. Then which of the following statements are true?

  1. A.E(t)≥πE(t) \ge \pi for all t>0
  2. B.E(t)≤πE(t) \le \pi for all t>0✓
  3. C.E(t)≥π/2E(t) \ge \pi/2 for all t>0
  4. D.E(t)≤π/2E(t) \le \pi/2 for all t>0✓

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.

See pricing

50 are analysed free — try those first.

Related counterexample: All the classical equations satisfy a maximum principle

More on this topic

The chapter behind this: The three classical equations — free to read

From Partial Differential Equations › Laplace, heat and wave equations: separation of variables

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

ShareWhatsAppTelegram