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Classify a second order PDE

Short answer

Coefficients: A = x^2, B = 3, C = x; discriminant B^2 − 4AC = 9 − 4x^3. The PDE is hyperbolic for x^3 < 9/4, parabolic on x^3 = 9/4, and elliptic for x^3 > 9/4.

Classify the PDE: x² ∂²u/∂x² + 3 ∂²u/∂x∂y + x ∂²u/∂y² + 17 ∂u/∂t = 100u.

For A u_xx + B u_xy + C u_yy + (lower-order terms) = 0, the sign of B² − 4AC decides the type: negative is elliptic, zero is parabolic and positive is hyperbolic.

In the given equation the second order terms are x² u_xx + 3 u_xy + x u_yy. The terms 17 ∂u/∂t and −100u are of lower order (first order and zero order), so they do not affect the classification; the principal part involves x and y.

A = x², B = 3, C = x

B² − 4AC = 9 − 4x³

  • 9 − 4x³ > 0, that is x³ < 9/4 (x < 1.31): hyperbolic
  • 9 − 4x³ = 0, that is x³ = 9/4 (x ≈ 1.31): parabolic
  • 9 − 4x³ < 0, that is x³ > 9/4 (x > 1.31): elliptic

So the type changes with the region: hyperbolic for x < (9/4)^(1/3), parabolic on x = (9/4)^(1/3), elliptic for x > (9/4)^(1/3).

✓ Answer checked by Claude Opus 5.5 (high effort)

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