# Counterintuitive PDE

After thinking about it for a while and consulting other students, no one seems to be able to find an example of the following:

Given the PDE

$\dfrac{\partial f}{\partial x} = 0 \quad$ on $U = { (x,y) \in \mathbb R^2 ; y>0, 1 < x^2 + y^2 < 4}$

I am looking for a solution $f$ that does not only depend on $y$.

How can this be?!

The exercise is taken form Lee's "Introduction to smooth manifolds", p. 517 at the end of the chapter on the Frobenius theorem.

(Note: According to the errata, the condition on $U$ is $y > 0$, not $x > 0$ as your copy of the book might state).

S. L.

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Does a constant $f$ depend only on $y$? It doesn't depend on $x$. –  Ross Millikan Nov 7 '10 at 4:56

I will use the corrected version mentioned by Douglas, i.e. $U$ will be the domain defined by $y>0$ and $1< x^2 +y^2< 4$. Consider a $C^\infty$ function $\phi (y)$ which is equal to $1$ for negative $y$, $0$ for $y\geq 1/2$ and strictly decreasing for $0 < y < 1/2$.

The required function $f$ is then defined by:

$f(x,y) =-\phi (y)$ if $(x,y)\in U$ and $x\leq 0$

$f(x,y) =+\phi (y)$ if $(x,y)\in U$ and $x\geq 0$.

It does not only depend on $y$ since $f(-3/2,1/4)<0$ and $f(+3/2,1/4)>0$. Nevertheless we do have $\dfrac {\partial f}{\partial x}=0$

PS The Mean Value Theorem doesn't apply because the segment joining the two points $(-3/2,1/4)$ and $(+3/2,1/4)$ (for example) is not entirely included in $U$.

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Is it only me? The segment of the second sentence "strictly decreasing for $0<y<1/2$." doesn't render . Could someone please have a look at the source and correct it if the segment is indeed not rendered for all viewers. Thanks a lot. –  Georges Elencwajg Nov 7 '10 at 13:55
Moron, thank you very, very much for your perfect edit. Could you please explain what mistake I made? –  Georges Elencwajg Nov 7 '10 at 14:09
I think we need to put spaces around < and >. –  Aryabhata Nov 7 '10 at 14:20
Thanks, Moron, I wouldn't have thought of that! –  Georges Elencwajg Nov 7 '10 at 14:23
Alternatively, there's \lt and \gt . –  Ｊ. Ｍ. Nov 7 '10 at 16:03

The Mean Value Theorem implies that any function f differentiable on U must be constant w.r.t. x for any fixed y.

There is an error in the text -- it should say y>0 instead of x>0. (see http://www.math.washington.edu/~lee/Books/Smooth/errata.pdf)

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Thanks for the errata link. The mean value theorem only implies that $f$ is constant if the horizontal cross-sections are intervals. For U whose intersections with horizontal lines are not single intervals, the function can be locally but not globally constant. This distinction is the content of the exercise. –  T.. Nov 7 '10 at 15:34
Thanks for the link! This explains a lot... –  Sam Nov 7 '10 at 15:56
$f$ can be locally but not globally independent of $x$.
For $|y| < 1$ the function can have different values on the left and right sides of the annulus. The restriction of the function to the left or the right side depends only on $y$. For example, $f = {\rm sgn(x)}(1-y^2)$ for $|y| \leq 1$ and $0$ in the rest of the region.