Examples from Seen Questions

A composite function and a partial differential equation

Let $f:\mathbb R\to\mathbb R$ be an odd differentiable function. Define

(a) Differentiating $F$ with respect to $y$ gives

Find $g(x,y)$.

(b) $F$ satisfies

Find a polynomial $h(x,y)$ that makes this identity hold for every such $f$.

(c) Given that $F(x,x)=-6x^2$ for all $x\in\mathbb R$, find the rule for $f$ and then the rule for $F$.

Solution

Set

The function $q$ is a polynomial, so it is differentiable everywhere. Since $f$ is differentiable everywhere, the multivariable chain rule applies throughout $\mathbb R^2$:

(a) Finding $g$

Differentiate the inner function with respect to $y$, holding $x$ fixed:

Therefore

so

(b) Finding $h$

The other partial derivative is

Substituting both derivatives into the proposed equation gives

Choosing

makes the two terms identical, so their difference is zero:

For a particular $f$, the equation need not determine $h$: if $f$ is constant, both partial derivatives of $F$ vanish and any $h$ works. The displayed polynomial works for every differentiable $f$.

(c) Recovering $f$ and $F$

Putting $y=x$ in the definition of $F$ gives

The supplied condition therefore becomes

The quantity $4x^2$ ranges over $[0,\infty)$. If $t\ge0$, choose $x=\frac{\sqrt t}{2}$ to obtain

The hypothesis that $f$ is odd extends this rule to negative inputs. If $t<0$, then $-t>0$, so

The same formula also gives $f(0)=0$, so altogether

Substitution into $F(x,y)=f(q(x,y))$ gives

As a check,

and the partial derivatives of this final expression satisfy the equation by the cancellation in part (b).

The diagonal condition samples $f$ only at non-negative arguments, since $x^2+xy+2y^2=(x+y/2)^2+7y^2/4\ge0$. Oddness supplies the missing negative half-line.

The key course idea is the partial derivative of a composite function: differentiate the outer function and multiply by the partial derivative of the inner function.

Critical points depending on a parameter

Consider

(a)

Find $a$ such that $f$ has at least two critical points. Give the answer as a set or an interval.

(b)

For the case where $f$ has multiple critical points, classify the saddle points, local minimum points and local maximum points and give their coordinates.

Solution

Because $f$ is a polynomial, its gradient exists everywhere, so the critical points are exactly the solutions of $f_x=f_y=0$. These equations also give the condition on $a$. The Hessian then classifies the non-zero candidates; at the origin the Hessian test is inconclusive, so we compare values along two lines.

The first partial derivatives are

At a stationary point,

Adding these equations eliminates the parameter term:

Over the reals,

and

with equality only at $(x,y)=(0,0)$. Hence every real stationary point satisfies $y=-x$. Substituting this into the first stationary equation gives

Thus $(0,0)$ is always a critical point. The other possibilities satisfy

Since $x^2\ge0$ and $a\ne0$, these additional real points exist exactly when $a>0$. Therefore the answer to part (a) is

For every such $a$, put $s=\sqrt a$. The complete list of critical points is

To classify the two non-zero points, compute the second partial derivatives:

At either point with $x^2=y^2=a$, the Hessian is

Also, $f_{xx}=10a>0$. The second-order partial derivative test therefore classifies both points as local minima:

At the origin,

so the Hessian test does not decide the classification. Compare the function along two lines through the origin. On $y=x$,

whereas on $y=-x$,

whenever $0<|t|<\sqrt{2a}$. Since $f(0,0)=0$, every neighbourhood of the origin contains points where $f$ is larger and points where $f$ is smaller than its value at the origin. Thus the origin is a saddle point.

There are no other critical points, so the requested classification for $a>0$ is

and

This follows the local extrema workflow: the stationary equations supply every candidate and the Hessian test classifies the two non-zero points. Where the determinant is zero, a direct comparison along lines exhibits the opposite signs that define a saddle. The parameter $a$ decides whether the two minima exist.

Finding and classifying a critical point

Let $f$ be the function

(a) Finding the single critical point

Find the single critical point of $f$.

Because $f$ is a polynomial, its first partial derivatives exist everywhere in $\mathbb{R}^2$. Compute them by treating the other variable as constant:

A stationary point and hence a critical point, must satisfy

Solving the two equations separately gives

and

Therefore the single critical point is

(b) Classifying the critical point

The possible choices are:

• local minimum;
• local maximum;
• inconclusive;
• saddle point.

To classify the point, use the second-order partial derivative test. The Hessian matrix is constant:

Its determinant is

Since $D<0$, the Hessian is indefinite and the second-order test classifies the critical point as a saddle point. Thus the correct option is saddle point.

The classification can be checked directly by completing the square:

At the critical point,

If we move in the $x$-direction by setting $x=\frac18+t$ and $y=-\frac12$, then

If instead we move in the $y$-direction by setting $x=\frac18$ and $y=-\frac12+t$, then

Arbitrarily close to the critical point there are therefore both larger and smaller function values, which is the defining behaviour of a saddle point.

The local extrema method locates the candidate with the stationary-point equations and classifies it with the Hessian determinant. The completed-square form shows the same thing directly: the surface curves down in one coordinate direction and up in the other.