Core Definition
Let z=f(x,y) have continuous second-order partial derivatives on an open disk centered at a critical point (a,b) where βf(a,b)=0. The Hessian determinant (or discriminant) D is defined by:
D(x,y)=det[fxxβfyxββfxyβfyyββ]=fxxβfyyββ(fxyβ)2
Evaluating D(a,b) and fxxβ(a,b) yields four distinct geometric classifications:
- Local Minimum: If D(a,b)>0 and fxxβ(a,b)>0 (or fyyβ(a,b)>0).
- Local Maximum: If D(a,b)>0 and fxxβ(a,b)<0 (or fyyβ(a,b)<0).
- Saddle Point: If D(a,b)<0.
- Inconclusive Test: If D(a,b)=0 (the critical point may be a local minimum, local maximum, saddle point, or flat trough; higher-order analysis or geometric inspection is required).
Concept Explanation
Geometric and Curvature Foundations
The Second Derivative Test analyzes the quadratic curvature of the Taylor expansion of f(x,y) near the critical point (a,b):
f(a+h,b+k)βf(a,b)+21β[fxxβh2+2fxyβhk+fyyβk2]
The bracketed term is a quadratic form Q(h,k)=Ah2+2Bhk+Ck2, with matrix:
H=[fxxβfxyββfxyβfyyββ]
(where fxyβ=fyxβ by Clairaut's Theorem).
Completing the square in the quadratic form reveals why the discriminant D=ACβB2 dictates surface geometry:
D>0 (Definite Curvature β Bowl Shape):
- Since fxxβfyyβ>(fxyβ)2β₯0, both pure second derivatives fxxβ and fyyβ must share the same non-zero sign.
- If fxxβ>0, the surface is concave upward in all vertical cross-sections, forming a bowl. The point is a local minimum.
- If fxxβ<0, the surface is concave downward in all vertical cross-sections, forming an inverted dome. The point is a local maximum.
D<0 (Indefinite Curvature β Saddle Shape):
- The quadratic form takes on both positive and negative values depending on the direction of travel β¨h,kβ©.
- Along one trajectory, the cross-section curves upward (f increases); along a transverse trajectory, the cross-section curves downward (f decreases).
- The point is a saddle point (resembling a mountain pass or a horse saddle).
Local Minimum (D > 0, f_xx > 0) Saddle Point (D < 0)
\ / .-' \ / '-.
\ * / / \ / \
'---' ( * x * )
Bowl Shape \ / \ /
'-. / \ .-'
Mountain Pass
Why fxxβ Determines the Sign When D>0
When D=fxxβfyyββ(fxyβ)2>0, we have fxxβfyyβ>(fxyβ)2β₯0. This forces fxxβ and fyyβ to have the exact same algebraic sign. Hence, checking fxxβ>0 is mathematically equivalent to checking fyyβ>0.
Step-by-Step Procedure
- Find all critical points (a,b) satisfying fxβ=0 and fyβ=0 (using Critical Points in Multivariable Functions).
- Compute second-order partial derivatives: fxxβ, fyyβ, and fxyβ.
- Formulate the general discriminant equation:
D(x,y)=fxxβ(x,y)fyyβ(x,y)β[fxyβ(x,y)]2
- For each critical point (a,b):
- Calculate the numerical value of D(a,b).
- If D(a,b)>0, evaluate the sign of fxxβ(a,b).
- Classify according to the discriminant rules.
Engineering Applications
- Structural Mechanics: In finite element analysis (FEA), elastic potential energy fields must possess positive-definite Hessians (D>0,fxxβ>0) at loaded equilibrium points to ensure dynamic stability; negative discriminants indicate buckling or snapping modes.
- Aerospace Engineering: In aircraft wing design, the lift-to-drag aerodynamic efficiency surface E(Ξ±,Ξ²) is optimized. Engineers seek a true local maximum (D>0,EΞ±Ξ±β<0) to guarantee robust cruise performance against perturbations in angle of attack Ξ± or flap deflection Ξ².
- Electrical & Audio Engineering: Audio dynamic amplifiers use automatic gain control loops where stable operating quiescent points require positive definite matrices, preventing harmonic distortion or self-sustaining oscillations.
Example Problems
Example 1: Multi-Point Classification (Local Minimum and Saddle Point)
Find and classify all critical points of:
f(x,y)=2x3β6x+2y2β4y
Solution
First-order partial derivatives:
fxβ=6x2β6,fyβ=4yβ4
Critical points:
6x2β6=0βΉx2=1βΉx=Β±1
4yβ4=0βΉy=1
Critical points: (1,1) and (β1,1).
Second-order partial derivatives:
fxxβ=12x,fyyβ=4,fxyβ=0
Discriminant formula:
D(x,y)=(12x)(4)β(0)2=48x
Classification:
- At (1,1):
D(1,1)=48(1)=48>0,fxxβ(1,1)=12(1)=12>0
Since D>0 and fxxβ>0, (1,1) is a local minimum with value:
f(1,1)=2(1)3β6(1)+2(1)2β4(1)=2β6+2β4=β6
- At (β1,1):
D(β1,1)=48(β1)=β48<0
Since D<0, (β1,1) is a saddle point with value:
f(β1,1)=2(β1)3β6(β1)+2(1)2β4(1)=β2+6+2β4=2
Example 2: Non-Zero Cross-Derivative & Local Maximum
Find and classify all critical points of:
f(x,y)=4xyβx4βy4
Solution
First-order partial derivatives:
fxβ=4yβ4x3=4(yβx3)
fyβ=4xβ4y3=4(xβy3)
Locate critical points:
From fxβ=0, y=x3. Substitute into fyβ=0:
xβ(x3)3=0βΉxβx9=0βΉx(1βx8)=0
This gives three real solutions for x: x=0,1,β1.
- If x=0βΉy=03=0βΉ(0,0)
- If x=1βΉy=13=1βΉ(1,1)
- If x=β1βΉy=(β1)3=β1βΉ(β1,β1)
Second-order partial derivatives:
fxxβ=β12x2,fyyβ=β12y2,fxyβ=4
Discriminant:
D(x,y)=(β12x2)(β12y2)β(4)2=144x2y2β16
Classification:
- At (0,0):
D(0,0)=144(0)β16=β16<0βΉSaddleΒ point
Value: f(0,0)=0.
- At (1,1):
D(1,1)=144(1)2(1)2β16=128>0,fxxβ(1,1)=β12(1)2=β12<0βΉLocalΒ maximum
Value: f(1,1)=4(1)(1)β14β14=2.
- At (β1,β1):
D(β1,β1)=144(β1)2(β1)2β16=128>0,fxxβ(β1,β1)=β12(β1)2=β12<0βΉLocalΒ maximum
Value: f(β1,β1)=4(β1)(β1)β(β1)4β(β1)4=2.
Connections & References
- Prerequisites:
- Direct Successors:
- Course Sessions: