Why separability simplifies classification
A general function of three variables can couple its coordinates: changing one coordinate may affect how the function responds to another. A separable function has a simpler structure. It is a sum of three one-variable functions, so its change can be understood coordinate by coordinate.
This structure does not make every stationary point easy to classify. It does, however, make the Hessian especially transparent: its diagonal entries are the second derivatives of the one-variable summands, and its off-diagonal entries vanish. The signs of those diagonal entries give a direct classification when they are all strictly positive, all strictly negative, or include both signs.
Assume throughout that the one-variable functions are twice continuously differentiable near the point under consideration. This guarantees that the first and second partial derivatives used below exist and are continuous.
Stationary points and the diagonal Hessian
For a separable function, each partial derivative involves only one coordinate:
Consequently, a point \((a,b,c)\) is stationary precisely when all three one-variable derivative equations hold: \(g'(a)=0\), \(h'(b)=0\), and \(k'(c)=0\). The Hessian at that point is diagonal.
Write \(A=g''(a)\), \(B=h''(b)\), and \(C=k''(c)\). For a displacement \(\mathbf{u}=(u_1,u_2,u_3)^\top\), the Hessian quadratic form is \(\mathbf{u}^{\top}\mathbf{H}_f(a,b,c)\mathbf{u}=Au_1^2+Bu_2^2+Cu_3^2\). Thus its signs can be read directly from the three one-variable second derivatives.
Why the sign rules work
The reason is visible in the second-order expansion at a stationary point. For sufficiently small displacements, the change in \(f\) is its Hessian quadratic form plus a remainder that is smaller than the squared distance to the point. In this separable setting, the quadratic part is a sum of three independent squared-coordinate terms.
Here the notation \(o(u_1^2+u_2^2+u_3^2)\) means that the remainder, divided by \(u_1^2+u_2^2+u_3^2\), tends to zero as the displacement tends to zero. If all three coefficients are positive, the quadratic part is positive for every nonzero displacement; sufficiently close to the point, the smaller remainder cannot change that sign. The function therefore increases away from the point, giving a strict local minimum. If all three coefficients are negative, the same reasoning gives a strict local maximum.
For the saddle conclusion, suppose for example that \(A>0\) and \(B<0\). Restrict attention first to the \(x\)-axis through the point, so the displacement is \((t,0,0)\). The change in the function is \(\frac12At^2+o(t^2)\), which is positive for all sufficiently small nonzero \(t\). Along the \(y\)-axis, the change is \(\frac12Bt^2+o(t^2)\), which is negative for all sufficiently small nonzero \(t\). Thus arbitrarily close to the stationary point there are function values both above and below its value. That is exactly the saddle-point behavior. A zero coefficient in the third coordinate does not remove either of these two directions.
Worked Example: A strict local minimum
Worked Example: A strict local minimum
Classify the stationary point of \(f(x,y,z)=(x-1)^2+2(y+2)^2+3(z-1)^2\).
Differentiate each one-variable term:
Set each derivative equal to zero. The equations give \(x=1\), \(y=-2\), and \(z=1\). Therefore the only stationary point is \((1,-2,1)\). The second derivatives are \(f_{xx}=2\), \(f_{yy}=4\), and \(f_{zz}=6\), with all mixed partials equal to zero. Every diagonal entry is positive, so the theorem classifies the point as a strict local minimum.
To check the function value and the classification directly, let \(u=x-1\), \(v=y+2\), and \(w=z-1\). Then \(f(x,y,z)-f(1,-2,1)=u^2+2v^2+3w^2\). The value at the stationary point is \(0\), and the change is positive whenever \((u,v,w)\ne(0,0,0)\). This confirms the strict local minimum.
Worked Example: Mixed coordinate curvatures give a saddle
Worked Example: Mixed coordinate curvatures give a saddle
Classify the origin for \(f(x,y,z)=x^2-2y^2+3z^2\).
The first partial derivatives are \(f_x=2x\), \(f_y=-4y\), and \(f_z=6z\). They all vanish at \((0,0,0)\), so the origin is stationary. The diagonal Hessian entries are \(2,-4,6\): there are positive and negative entries, so the Hessian is indefinite and the origin is a saddle.
The function value at the origin is \(0\). Along the \(x\)-axis, \(f(t,0,0)=t^2>0\) for \(t\ne0\). Along the \(y\)-axis, \(f(0,t,0)=-2t^2<0\) for \(t\ne0\). These points approach the origin as \(t\) approaches zero, verifying that values both above and below \(f(0,0,0)\) occur arbitrarily close to it.
Zero second derivatives: what the test can and cannot say
If one or more diagonal entries are zero, inspect the signs of the others before deciding whether the second-derivative test is inconclusive. When there is at least one positive entry and at least one negative entry, the Hessian is indefinite and the point is a saddle, even if another entry is zero. In contrast, if all entries are nonnegative, or all are nonpositive, and at least one is zero, the Hessian is semidefinite. In that situation the second-derivative test may be inconclusive: the zero-curvature coordinate has no quadratic term to determine how the function changes in that direction.
For example, both functions in the next example have the same Hessian at the origin. Their classifications differ because the second derivatives do not capture the first nonzero change in the \(z\)-direction.
Worked Example: The same semidefinite Hessian, different behavior
Compare \(f_1(x,y,z)=x^2+y^2+z^4\) and \(f_2(x,y,z)=x^2+y^2-z^4\) at the origin.
For \(f_1\), the first partial derivatives are \(2x\), \(2y\), and \(4z^3\), so the origin is stationary. The second partial derivatives on the diagonal are \(2,2,12z^2\), and therefore the Hessian at the origin is \(\operatorname{diag}(2,2,0)\). Its entries are nonnegative, with one zero; the second-derivative test alone does not classify the point. Directly, \(f_1(x,y,z)=x^2+y^2+z^4\ge0=f_1(0,0,0)\), with equality only at the origin. Hence the origin is a strict local minimum.
For \(f_2\), the first partial derivatives are \(2x\), \(2y\), and \(-4z^3\), so the origin is also stationary. Its Hessian at the origin is again \(\operatorname{diag}(2,2,0)\). But \(f_2(t,0,0)=t^2>0\) and \(f_2(0,0,t)=-t^4<0\) for every nonzero \(t\). These values approach \(f_2(0,0,0)=0\) from above and below, so the origin is a saddle. The direct checks show why the shared semidefinite Hessian cannot settle either classification.
Combining one-variable information
Separability also gives a useful way to reason when one-dimensional behavior is known beyond the second derivatives. Define the coordinate changes \(\Delta_g=g(a+u)-g(a)\), \(\Delta_h=h(b+v)-h(b)\), and \(\Delta_k=k(c+w)-k(c)\). Then the total change is exactly their sum:
If each one-variable function has a local minimum at its corresponding coordinate, then each change is nonnegative for sufficiently small coordinate changes, so their sum is nonnegative. The point is a local minimum of \(f\). If each has a local maximum, the sum is nonpositive and the point is a local maximum. If one coordinate has arbitrarily close values strictly above its value at the stationary coordinate, and another has arbitrarily close values strictly below its value, varying those coordinates one at a time proves that the three-variable point is a saddle. These observations can resolve some cases where a second derivative is zero, but they require information about the one-variable functions beyond that second derivative.
A practical classification procedure
For \(f(x,y,z)=g(x)+h(y)+k(z)\), solve \(g'(x)=0\), \(h'(y)=0\), and \(k'(z)=0\) independently.
At each stationary point \((a,b,c)\), compute \(g''(a)\), \(h''(b)\), and \(k''(c)\).
All strictly positive means a strict local minimum; all strictly negative means a strict local maximum; both a positive and a negative entry mean a saddle.
If the entries have only one weak sign and at least one is zero, the second-derivative test may be inconclusive. Examine the one-variable changes or higher-order terms if a classification is needed.
Common pitfall: treating every zero entry as inconclusive
A zero diagonal entry does not, by itself, make the second-derivative test inconclusive. For example, a diagonal Hessian with entries \(2,-2,0\) has both positive and negative entries, so it is indefinite. The directions associated with \(2\) and \(-2\) already give function changes of opposite signs near the stationary point; the zero entry is irrelevant to the saddle classification.
The correct order is to inspect all entries first. Conclude “saddle” if the Hessian has both signs. Only when the Hessian is semidefinite with a zero entry should you say that the second-derivative test is inconclusive. Then look beyond the quadratic terms rather than guessing the classification from the zero alone.
Check Your Understanding
Use the coordinatewise structure and the signs of the second derivatives to answer each question. When a zero entry appears, distinguish an indefinite Hessian from a semidefinite one.
- For \(f(x,y,z)=x^2+4y^2+5z^2\), find the stationary point and classify it.
- At a stationary point, the diagonal Hessian entries are \(-3,0,-1\). Is the Hessian indefinite or semidefinite, and what can the second-derivative test conclude?
- At a stationary point, the diagonal Hessian entries are \(2,-2,0\). Why does the zero entry not prevent a saddle classification?
- Give two separable functions with the same Hessian \(\operatorname{diag}(2,2,0)\) at the origin but different classifications there.
- Explain why the sum of three one-variable functions has zero mixed partial derivatives.