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I hope somebody can help me to understand how to verify the sufficient condition for hypoellipticity for the equation with $m$ variables?

Hormander says: span of operators $X_k,[X_k,X_j],[X_k[X_j,X_l]]...$ spans the space, i.e. the rank of Lie algebra generated by the operators ${X_0,...,X_r}$ is equal to $m$. In his paper(1967) he provides an example of $$u_{xx}+xu_y-u_t=0$$ which satisfies that. I can't understand how to apply his sufficient condition in this case.

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Following Hörmander's notation [see (1.6) below], let us write $u_{xx}+xu_y-u_t=(X_1^2+X_0)u$, where $X_1=\frac{\partial}{\partial x}$ and $X_0=x\frac{\partial}{\partial y}-\frac{\partial}{\partial t}$. We must check that at every point of $\mathbb R^3$ the vector fields $X_0,X_1$, and their commutators, have 3-dimensional span. This is immediate from $$[X_1,X_0]=\frac{\partial}{\partial x}\left(x\frac{\partial}{\partial y}-\frac{\partial}{\partial t}\right)-\left(x\frac{\partial}{\partial y}-\frac{\partial}{\partial t}\right)\frac{\partial}{\partial x}=\frac{\partial}{\partial y}$$

(Added) Let's check the spanning property. A set of vectors spans the ambient space (in this case 3 dimensional) iff the matrix of their components has maximal rank (ie, 3). Here we have $$\begin{pmatrix} 1&0&0\\0&x&-1\\0&1&0 \end{pmatrix} $$ which has rank 3 no matter what x is. Note that (a) both the original fields and their commutators are included; (B) in general the number of fields may be greater than the dimension, so the matrix is not necessarily square.

(Remark) Hörmander does not distinguish the time derivative in his Theorem 1.1, and I followed his approach by putting it along with a space derivative into $X_0$. Noticing that the time derivative does not contribute to the commutators anyway, one may want to reformulate the Hörmander condition for parabolic equation so that the time derivative is left out of it completely. That is, we could write the equation $u_t=(X_1^2+\widetilde X_0)u$ where $X_1=\frac{\partial}{\partial x}$ and $\widetilde X_0=x\frac{\partial}{\partial y}$. Then we want $X_1$, together with all commutators of $\widetilde X_0$ and $X_1$, to span $\mathbb R^2$ at every point. I believe this is the correct parabolic form of the Hörmander condition, and the one given on Wikipedia is stated incorrectly (it appears to exclude $\widetilde X_0$ from participating in commutators, not just from the spanning set).

For ease of reference, here is the main result of Hörmander's 1967 paper Hypoelliptic second order differential equations.

Hypoelliptic second order differential equations, page 149

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    Leonid, thanks for the answer, that's helpful. I can see the way you calculate $[X_0,X_1]$, however when you obtain $\partial/\partial y$, what exactly that implies? I mean how it follows from the fact what I see on RHS, what if I got $0$, or $\partial/\partial x$ or $\partial/\partial t$?2012-07-18
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    I have followed the calculations you have provided but for the case of the pde $x^2u_{xx}+xu_x+\frac{x-y}{T-t}u_y-u_t=0$, then following the same notation $X_1=x\frac{\partial}{\partial x}$ and $X_0=x\frac{\partial}{\partial x}+\frac{x-y}{T-t}\frac{\partial}{\partial y}-\frac{\partial}{\partial t}$. Then $[X_0,X_1]=\frac{x}{T-t}\frac{\partial}{\partial y}$, can I conclude it is hypoelliptic with an analogy above?2012-07-18
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    By the way, I don't think there is an issue of putting the time derivative together with a space derivative since it is a particular case of a degenerate elliptic equation anyway.2012-07-18
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    @Medan For every point (x,y,t) in the domain of the operator, the vectors that you got must. span 3-d space with the basis $\partial/\partial x$, .. $\partial/\partial t$. The way to check this is to write each of these basis elements in terms of your field. Easy to do in my case, but not in yours. Indeed, what do you get when x=0?2012-07-18
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    ok, but if check hypo ellipticity inside of a bounded domain $D=(0,T)\times(0,x_m)\times(y_1,y_2)$? Then there is no $0$ in there and I have no problem and have it hypoelliptic inside which technically implies the super regularity inside, but not including the boundary, which probably just continuous there.2012-07-18
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    Leonid, when you say they should span $3-d$ space, this is due to the fact that there are $t,x,y$ 3 variables. However, I am still confused: for any $x,y,t$ my $[X_0,X_1]$ should not be zero to satisfy that, correct? This, is what it means to span? Because this is just a computation of a commutator of two operators of a Lie algebra...or commutator is what represents a linear combination?2012-07-18
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    @Medan I added the verification of spanning property to my answer.2012-07-18
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    this is very hekpful, so in my case of the second equation $$\begin{pmatrix} x&0&0\\x&\frac{x-y}{T-t}&-1\\0&\frac{x}{T-t}&0 \end{pmatrix} $$ Then assuming that I am considering only a domain $x>0, t I am fine and the rank is 3. Note, that this equation lacks diffusion in $y$ dimension and one might think that I should not expect any smooth properties along $y$ dimension as it is just transport equation and it carries all the irregularity in the boundary condition. However, if the coefficient in front of $u_y$ depends on $x$, it infinitely smooth for whatever bad bdry condition!2012-07-18
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    @Medan Exactly: and the reason is that the diffusion in the $x$ direction spills over to the other directions via the commutators. Of course it does not do it as quickly as in other directions, which comes into play when one considers equations where the right hand side is not smooth.2012-07-18
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    I wonder if it is "bad" to have such a singularity for $t=T$ in order to make a conclusion? Because if $t \rightarrow T$ the entries in the matrix that contain that in the denominator blow up, but it is still ok to conclude that the rank is 3 no matter what the magnitude of the elements?2012-09-23
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    @Medan As long as we are talking about interior regularity, the bad behavior at the boundary is not an issue. The solution is smooth inside of the domain. Smoothness is a local property, so when verifying it you can restrict to subdomain where everything is nice. The singularity as $t\to T$ indicates we don't know anything about the smoothness up to the boundary, i.e., whether the solution and its derivatives have continuous extensions to $t=T$.2012-09-23
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    thank you very much for these comments. They are helpful.2012-09-24