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The two-variable Hessian determinant test

Statement

Let aa be a critical point of a C2C^2 function of two variables, and put A=fxx(a)A=f_{xx}(a), B=fxy(a)B=f_{xy}(a), C=fyy(a)C=f_{yy}(a) and Δ=ACB2\Delta=AC-B^2. If Δ>0\Delta>0, then A>0A>0 gives a strict local minimum and A<0A<0 a strict local maximum. If Δ<0\Delta<0, there is neither. If Δ=0\Delta=0, this test gives no conclusion.

Facts & Assumptions

Given: aa is a critical point of a C2C^2 scalar field ff on an open subset of R2\mathbb R^2.

[L2]

The second-derivative test classifies a critical point from definiteness or indefiniteness of its Hessian (The multivariable second-derivative test by definiteness of the Hessian).

Proof

technique · calculation
1.1

By [L1], the Hessian quadratic form is Q(s,t)=As2+2Bst+Ct2Q(s,t)=As^2+2Bst+Ct^2. If A0A\ne0, completing the square gives Q=A(s+(B/A)t)2+(Δ/A)t2Q=A(s+(B/A)t)^2+(\Delta/A)t^2.

L1algebra
2.1

If Δ>0\Delta>0, the two coefficients in step 1.1 have the sign of AA, so QQ is positive definite for A>0A>0 and negative definite for A<0A<0.

step 1.1algebra
2.2

If Δ<0\Delta<0 and A0A\ne0, then Q(1,0)=AQ(1,0)=A while Q(B/A,1)=Δ/AQ(-B/A,1)=\Delta/A, which have opposite signs; hence QQ is indefinite. If A=0A=0, then B2=Δ<0-B^2=\Delta<0, so B0B\ne0, and Q(1,t)=2Bt+Ct2Q(1,t)=2Bt+Ct^2 has both signs for sufficiently small positive and negative tt.

step 1.1algebra
3.1

Apply [L2] to steps 2.1 and 2.2. When Δ=0\Delta=0 and A0A\ne0, step 1.1 makes Q=A(s+(B/A)t)2Q=A(s+(B/A)t)^2, so it is semidefinite but not definite; when A=0A=0, then B=0B=0 and Q=Ct2Q=Ct^2, with the same conclusion (including Q=0Q=0). Thus this is the inconclusive case of [L2].

L2step 1.1step 2.1step 2.2algebra

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