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Part BCSIR NET December 2023

Classification and canonical forms

The partial differential equation x2uxx−2x^{2} u_{xx} - 2xy uxu_{x}ᵧ −3y2uyy+ux−uy=0- 3y^{2} u_yy + u_{x} - u_y = 0 is

  1. A.elliptic in {(x, y) : y > 0}
  2. B.parabolic in {(x, y) : x > 0, y > 0}
  3. C.hyperbolic in {(x, y) : xy ≠ 0}✓ Correct
  4. D.parabolic in {(x, y) : xy ≠ 0}

The idea, in plain words

Classifying a second-order equation means testing one combination of its three leading coefficients. Here that combination comes out positive almost everywhere, which is the hyperbolic case — the type of equation that carries waves.

Solution

B2−4B^{2} - 4AC =4x2y2+12x2y2=16x2y2>0= 4x^{2}y^{2} + 12x^{2}y^{2} = 16x^{2}y^{2} > 0 wherever xy ≠ 0: hyperbolic.

Why each option is right or wrong

Checked against 1 symbolic computation
  • A.WrongExecution slip

    False.

    With A=x2,B=−2A = x^{2}, B = -2xy, C=−3y2C = -3y^{2}, SymPy expands the discriminant B2−4B^{2} - 4AC to 16x2y216x^{2}y^{2}

    which is a square and therefore never negative.

    Elliptic requires B2−4B^{2} - 4AC < 0

    so the equation is elliptic nowhere — least of all on a half-plane where x is unrestricted.

  • B.WrongExecution slip

    False.

    Parabolic requires B2−4B^{2} - 4AC = 0

    but on {x > 0, y > 0} the discriminant 16x2y216x^{2}y^{2} is strictly positive.

    The sign error that leads here is dropping the minus in C=−3y2C = -3y^{2}: with C=+3y2C = +3y^{2} the discriminant would be 4x2y2−12x2y2=−8x2y2<04x^{2}y^{2} - 12x^{2}y^{2} = -8x^{2}y^{2} < 0, and the classification would flip entirely.

  • C.Correct

    True.

    B2−4B^{2} - 4AC =4x2y2−4⋅x2⋅(−3y2)=4x2y2+12x2y2=16x2y2(= 4x^{2}y^{2} - 4\cdot{}x^{2}\cdot(-3y^{2}) = 4x^{2}y^{2} + 12x^{2}y^{2} = 16x^{2}y^{2} (verified).

    This is strictly positive exactly when xy ≠ 0

    which is the hyperbolic condition.

    The trigger: write down A, B, C with their signs before anything else — the two minus signs here (in B and in C) are what the question is testing, and they combine to make the discriminant larger, not smaller.

    The first-order terms ux−uyu_{x} - u_y never affect classification.

  • D.WrongBoundary and endpoint

    False.

    The region xy ≠ 0 is exactly where the discriminant is non-zero

    so it is precisely where the equation is not parabolic.

    Parabolic behaviour occurs only on the degenerate set xy = 0, where all three second-order coefficients collapse.

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

Related counterexample: A PDE has a single type throughout its domain

More on this topic

The chapter behind this: Classification of second-order PDE — free to read

From Partial Differential Equations › Classification and canonical forms

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

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