Analytic hypoellipticity of linear ordinary differential operators - MathOverflow

Analytic hypoellipticity of linear ordinary differential operators

2010-09-23T23:39:36Z

Let $P = a_n(x) D_x^n + a_{n-1}(x) D_x^{n-1} + \ldots + a_0(x)$ be a linear ordinary differential operator with polynomial (or real analytic) coefficients $a_j(x)$. Suppose that $a_n(x)$ doesn't vanish on an interval $(a,b)$ and that $u$ is a weak solution of $P u = 0$ on $(a,b)$. It can be concluded from the elliptic regularity theorem that $u$ is in fact real analytic and a classical solution to the differential equation.

In other words, the operator $P$ is analytically hypoelliptic.

However, appealing to the elliptic regularity theorem seems a bit much for the above case of an ordinary differential operator. Does anyone know a (canonical) reference for this (probably much more classical) case? I browsed through several books on differential equations but the closest I could find was in G. Folland's Fourier analysis and its applications where he mentions this fact (with just $C^\infty$ smoothness) on the top of page 344 without reference or proof.

Update: Thanks for the answers so far! While very helpful, I think I didn't make it clear what I was looking for. For an audience which may not be familiar with the theory of elliptic operators, what would be the proper way to (in one sentence plus a reference) justify that a weak solution in the case at hand is automatically real analytic? Currently, I have to make reference to a textbook on PDEs which treats elliptic regularity even though the problem for linear ODEs seems so much simpler.

Thank you!

Answer by Andreas Minne for Analytic hypoellipticity of linear ordinary differential operators

2010-09-24T00:16:06Z

This is probably not what you are looking for, but the book An Introduction to Ordinary Differential Equations by Earl A. Coddington considers existence and uniqueness of classical solutions to $Pu=0$ (given initial conditions). Maybe the uniqueness part can be altered to include weak solutions as well.

Edit (since I don't have enough points to comment, I'll put my response here): Yes I am aware that it is about regularity, but since we have existence of a classical solution to this problem, uniqueness is enough, or have I missed something?

Answer by Dick Palais for Analytic hypoellipticity of linear ordinary differential operators

2010-09-24T00:26:41Z

It is easy to generalize the proof tha a differentiable real ODE has differentiable solutions (using a fxed point argument) from the real to the complex case, and deduce that a holomorphic ODE has local holomorphic solutions. Then given a real analytic ODE, use the same power series to look at it as complex holomorphic. Finally go back to the real case by taking real parts. The details of this argument can be read here (see page 32).

http://ode-math.com/PDF_Files/ChapterFirstPages/Chap1Frst35Pages.pdf

Answer by Deane Yang for Analytic hypoellipticity of linear ordinary differential operators

2010-09-24T01:26:29Z

(EDITED in tone)

I do not know any reference for this, but the proof is almost exactly the same (but easier) than the one for higher dimensions. What's nice about the $1$-dimensional case is that you really can work with $C^k$, where $k$ is any integer, and don't need to mess with Sobolev or Holder spaces as you do in higher dimensions.

So here's an outline of a proof that does not rely on any results from elliptic PDE's and uses only freshman calculus, basic results from ODE's, and the notion of a weak solution:

Use the fundamental theorem of calculus to show that $D$ is an elliptic operator in the sense that if $u$ is a weak solution to $Du = f$, then $u$ has one more order differentiability than $f$.
Use #1 to show that $D^n$ is elliptic in the sense that if $u$ is a weak solution to $D^nu = f$, then $u$ has $n$ more orders of differentiability than $f$.
Now use the standard bootstrapping argument to show that $a_n D^n + \cdots + a_0$ is an elliptic DO.
Now use the results above along with the uniqueness theorems for smooth and real analytic ODE's to conclude that if $f$ is real analytic, so is $u$.

Answer by Denis Serre for Analytic hypoellipticity of linear ordinary differential operators

2010-09-24T05:55:44Z

There are two parts in the statement. First, weak solutions are classical solutions. Second, that they are real analytic.

Weak = classical. This is very simple for linear ODEs, by duality. We know that the classical solutions form an $n$-dimensional space. Then it is enough to prove that the image of ${\mathcal D}(a,b)$ under the adjoint operator $P^*$ is of codimension $n$ at most; this will ensure that the weak solutions of $Pu=0$ form a linear space of dimension $\le n$. Since it contains the space of classical solutions, they are equal.

The codimension $\le n$ is true, because if you take a $\phi\in{\mathcal D}(a,b)$ and solve the Cauchy problem $P^*z=\phi$ with $z(a)=z'(a)=\cdots=z^{n-1}(a)=0$, then $z$ is in ${\mathcal D}(a,b)$ if and only if it vanishes at $b$ together with its derivatives up to order $n-1$. Because these are $n$ linear conditions, we get the codimension $\le n$.

Real analyticity follows from the theory of differential equations with holomorphic coefficients. It has been known since the XIXth century, probably by Cauchy himself. The important point, as in the real case, is that $a_n$ does not vanish. Fuchs went much further by studying regular singularities in holomorphic ODEs, when $a_n$ has simple zeroes. Even the case of irregular singularities (higher order zeroes of $a_n$) is interesting. This yields monodromy groups, which are subgroups of $GL_n({\mathbb C})$, defined up to conjugacy. It is an important open question whether given points $z_1,\ldots,z_m$ and monodromy groups $G_1,\ldots,G_m$, there exists a complex ODE with singularities $z_j$ and monodromy group $G_j$ at $z_j$.

Answer by Piero D'Ancona for Analytic hypoellipticity of linear ordinary differential operators

2010-09-24T09:46:30Z

The answers already given are quite complete and say it all, but maybe the OP was in search of a very simple explanation of what is happening. Let me try.

You will agree that it is not necessary to work in the full generality of an n-th order equation; if we can do it for the first order case, it is easy to generalize by induction. Moreover, we can assume that the highest order coefficient is equal to 1 (just divide by it). So essentially the question is: let $u$ be a weak solution of the equation $$ u' + a(x) u = f(x) $$ on an open interval, with $ a(x) $ and $ f(x) $ analytic. Then we want to prove that $u$ is also analytic. A further reduction is possible: multiply both sides by $\exp (A(x)) $ where $A'=a$ and call $ v = e^A u $, $g=e^A f$. Then we are reduced to proving the same thing for the simple equation $$ v ' = g. $$ Is $v$ analytic if $g$ is analytic? A further reduction is possible! just call $ w=v-G $, where $G'=g$. Nice trick eh? we are reduced to the even simpler equation $$ w' = 0 $$ and our serach will be over if we prove the fundamental fact that any weak solution of $w'=0$ must be a constant function. Notice that the same chain of arguments applies if the coefficients are $C^\infty$ (you get that $u$ is also $C^\infty$) and if $f$ is just $C^k$.

Now, in the greatest possible generality, if $w$ is any distribution on an open interval, with vanishing derivative, then $w$ must be a constant. This is proved in the following way: by definition, we know that $ w(\phi')=0 $ for any test function $\phi$. Fix a test function $\chi$ with $\int\chi=1$. Let $\psi$ be an arbitrary test function, define $$ \psi_1 = \psi-\chi \cdot \int \psi $$ and notice that $\int\psi_1=0$. This means that $ \psi_1 $ can be written as the derivative of another test function (guess which one?) $\psi_1=\psi_2'$ and hence $$w(\psi_1) = w(\psi_2') = 0 $$ by assumption. This implies $$ w(\psi) = w(\chi) \cdot \int \psi $$ which in the language of distributions means precisely that $w$ is equal to the constant $w(\chi)$ (notice that $\chi$ is fixed once and for all). That's it.