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Part CCSIR NET June 2025on-the-unit-circle-x-squared-times-y-squared-peaks-at-one-quarter

On the unit circle x squared times y squared peaks at one quarter

Suppose u = u(x, y) is the solution of the boundary value problem ∂2u/∂x2+∂2u/∂y2+∂u/∂x=0\partial^{2}u/\partial{}x^{2} + \partial^{2}u/\partial{}y^{2} + \partial{}u/\partial{}x = 0 in {(x,y)∈R2:x2+y2<1(x, y) \in \mathbb{R}^{2} : x^{2} + y^{2} < 1}, u(x,y)=1+2x2y2u(x, y) = 1 + 2x^{2}y^{2} on {(x,y)∈R2:x2+y2=1(x, y) \in \mathbb{R}^{2} : x^{2} + y^{2} = 1}. Then which of the following statements are true?

  1. A.The minimum value of u is 1✓ Correct
  2. B.The maximum value of u is 3
  3. C.The minimum value of u is 2
  4. D.The maximum value of u is 3/2✓ Correct

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: 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

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