3 calculus 3 examples

The function {f(x,y)=\dfrac{xy}{x^2+y^2}} might be the world’s most popular example demonstrating that the existence of partial derivatives does not imply differentiability.

xy/(x^2+y^2)
xy/(x^2+y^2)

But in my opinion, it is somewhat extreme and potentially confusing, with discontinuity added to the mix. I prefer

\displaystyle  f(x,y)=\frac{xy}{\sqrt{x^2+y^2}}

pictured below.

xy/sqrt(x^2+y^2)
xy/sqrt(x^2+y^2)

This one is continuous. In fact, it is Lipschitz continuous because the first-order partials {f_x} and {f_y} are bounded. The restriction of {f} to the line {y=x} is {f(x,y)=x^2/\sqrt{2x^2} = |x|/\sqrt{2}}, which is a familiar single-variable example of a nondifferentiable function.

To unify the analysis of such examples, let {f(x,y)=xy\,g(x^2+y^2)}. Then

\displaystyle    f_x = y g+ 2x^2yg'

With {g(t)=t^{-1/2}}, where {t=x^2+y^2}, we get

\displaystyle    f_x = O(t^{1/2}) t^{-1/2} + O(t^{3/2})t^{-3/2} = O(1),\quad t\rightarrow 0

By symmetry, {f_y} is bounded as well.

My favorite example from this family is more subtle, with a deceptively smooth graph:

Looks like xy
Looks like xy

The formula is

\displaystyle    f(x,y)=xy\sqrt{-\log(x^2+y^2)}

Since {f} decays almost quadratically near the origin, it is differentiable at {(0,0)}. Indeed, the first order derivatives {f_x} and {f_y} are continuous, as one may observe using {g(t)=\sqrt{-\log t}} above.

And the second-order partials {f_{xx}} and {f_{yy}} are also continuous, if just barely. Indeed,

\displaystyle    f_{xx} = 6xy g'+ 4x^3yg''

Since the growth of {g} is sub-logarithmic, it follows that {g'(t)=o(t^{-1})} and {g''(t)=o(t^{-2})}. Hence,

\displaystyle    f_{xx} = O(t) o(t^{-1}) + O(t^{2}) o(t^{-2}) = o(1),\quad t\rightarrow 0

So, {f_{xx}(x,y)\rightarrow 0 = f_{xx}(0,0)} as {(x,y)\rightarrow (0,0)}. Even though the graph of {f_{xx}} looks quite similar to the first example in this post, this one is continuous. Can’t trust these plots.

Despite its appearance, f_{xx} is continuous
Despite its appearance, f_{xx} is continuous

By symmetry, {f_{yy}} is continuous as well.

But the mixed partial {f_{xy}} does not exist at {(0,0)}, and tends to {+\infty} as {(x,y)\rightarrow (0,0)}. The first claim is obvious once we notice that {f_x(0,y)= y\, g(y^2)} and {g} blows up at {0}. The second one follows from

\displaystyle    f_{xy} = g + 2(x^2+y^2) g' + 4x^2y^2 g''

where {g\rightarrow\infty} while the other two terms tend to zero, as in the estimate for {f_{xx}}. Here is the graph of {f_{xy}}.

Up, up and away
Up, up and away

This example is significant for the theory of partial differential equations, because it shows that a solution of the Poisson equation {f_{xx}+f_{yy} = h } with continuous {h} may fail to be in {C^2} (twice differentiable, with continuous derivatives). The expected gain of two derivatives does not materialize here.

The situation is rectified by upgrading the continuity condition to Hölder continuity. Then {f} indeed gains two derivatives: if {h\in C^\alpha} for some {\alpha\in (0,1)}, then {f\in C^{2,\alpha}}. In particular, the Hölder continuity of {f_{xx} } and {f_{yy} } implies the Hölder continuity of {f_{xy} }.

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