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I have attempted to get an answer on the math.stackexchange but have not got any answer for a while. Thus, I am posting the question here. I am analyzing the following problem in order to establish that it is well-posed and state the regularity of the solution.

$$u_t=u_{xx}+\frac{1}{T-t}(x-y)u_y,\;u(0,x,y)=u_0(x,y),\; x,y\in R, \;t \in [0,T]$$

I think I have found a reference to show existence of that equation. However, for the uniqueness and continuous dependance I have difficulty due to "bad" behavior of the coefficient $\frac{1}{T-t}(x-y)$ as $t$ approaches $T$. Moreover, application of the Hormander theorem gives the regularity only inside of the domain. In fact, I can have this problem well-posed on any finite interval $[0,t_1]$ for $t_1 < T$, but not on the whole interval $[0,T]$.

I can do change of variables $z:=y(T-t)$ and $u(t,x,y)=v(t,x,z)$. Then, after application of the chain rule, the problem becomes much simpler to analyze:

$$v_t=v_{xx}+xv_z,\;v(0,x,z)=u_0(0,x,zT),\; x,z\in R, \;t \in [0,T]$$

and one might notice that $v(T,x,0)=u(T,x,y)$. This is where I am having some difficulty to continue. First, the second formulation satisfies the conditions for existence and uniqueness, does that imply the original problem has a unique solution as well? Second, the energy estimates are used to show continuous dependence on the initial data, do they imply the original problem is also continuously dependent on the initial data? Third, application of the Hormander condition gives the regularity on the entire time domain $[0,T]$, does that imply the regularity of the original problem? Basically, I want to know if it is equivalent to consider the alternative formulation in order to establish existence, uniqueness, continuous dependance and regularity of the original problem. Would like to hear any suggestions or answers to that question.

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  • $\begingroup$ Possibly, you should say in what spaces you are looking for uniqueness/regularity/continuous dependence $\endgroup$
    – Carlo Mantegazza
    Commented Feb 5, 2013 at 0:21
  • $\begingroup$ I would like to have estimates in $L^2$ space. $\endgroup$
    – Kamil
    Commented Feb 5, 2013 at 1:43

1 Answer 1

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Your last equation $$ \mathcal K=v_t-xv_z-v_{xx}=0 \tag 1$$ is indeed a particular case of Hörmander's $X_0-\sum_{1\le j\le r}X_j^2$ with $$ X_0=\partial_t-x\partial_z, X_1=\partial_x, r=1, [X_0,X_1]=\partial_z. $$ However, it is also exactly Kolmogorov equation, as studied by Andrei Kolmogorov in his 1937 Annals paper. This article was in fact the starting point of Lars Hörmander's work on this topic. It turns out that there is an explicit parametric construction for (1): a change of variables straightening the vector field $X_0$ is $$ \begin{cases} s=t,\ \\ x_1=x,\ \\ x_2=z+xt, \end{cases} $$ so that $ \mathcal K=\partial_s-(\partial_{x_1}+s\partial_{x_2})^2 $ and the latter can be Fourier transformed to the ODE $$ \partial_s+(\xi_1+s\xi_2)^2. $$ The latter is of course explicitly solvable: we have an explicit integral expression $$ v(t,x,z)=v(s,x_1,x_2-sx_1)=w(s,x_1,x_2)=\iint e^{i(x_1\xi_1+x_2\xi_2)}e^{-\int_0^t(\xi_1+s\xi_2)^2ds} \hat w_0(\xi_1,\xi_2) d\xi_1d\xi_2, $$ $$ v(t,x,z)=\iint e^{i(x\xi_1+(z+xt)\xi_2)}e^{-\int_0^t(\xi_1+s\xi_2)^2ds} \hat v_0(\xi_1,\xi_2) d\xi_1d\xi_2,\quad v_0(x,z)=v(0,x,z). $$

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  • $\begingroup$ yes, I have gotten to that point too, I must agree the transformed equation has a lot of nice properties, but I need to know if I can utilize those to state the properties of the original equation as this is the one I need to analyze. $\endgroup$
    – Kamil
    Commented Feb 3, 2013 at 22:50
  • $\begingroup$ It seems to me that the formulas above give an explicit integral solution to your problem. From this explicit expression, it is for instance easy to prove hypoellipticity and regularization properties, even to find the exact fractional amount of $z$ derivative that you gain. $\endgroup$
    – Bazin
    Commented Feb 7, 2013 at 20:14
  • $\begingroup$ yes, I can find the solution however, the regularity is harder to observe. Even though the tranformed function $v$ is perfectly regular due to satisfaction of the Hormander condition, I can't transfer those properties to the original function, can I? I agree I can conclude that I have a solution up to a change of variables, but I don't think regularity follows the same way. I might be wrong, but if you could please add a few lines to your answer with respect to regularity of the original question that would be very helpful as this is the main answer I am looking for! $\endgroup$
    – Kamil
    Commented Feb 14, 2013 at 3:28
  • $\begingroup$ I have made an explicit computation and you can find as well an explicit solution with my formula above by plugging the values of $v$ in terms of your $u$. The regularity business, say for the function $v$ follows from the explicit integral expression: you get easily that the $L^2$ norm of $\mathcal K v$ controls the $H^1$ norm in the $x$ variable of $v$. The expression of $w$ shows that you control 2/3 of derivatives for the $z$ variable: if you want an isotropic control then you cannot do better than $2/3$. To see that is not completely obvious: just compute exactly the integral in the phase $\endgroup$
    – Bazin
    Commented Feb 16, 2013 at 18:14
  • $\begingroup$ @Bazin: there is something important is missing that I do not understand in your answer. Change of variables works great but we should track the change in the domain along with change of variables and as you can see the domain in $z$ is shrinking. It does it to such an extent that at $t=T$ it becomes simply a point. So, you can't define that the solution there as it is not a domain anymore. How would you argue that? $\endgroup$
    – Kamil
    Commented Mar 5, 2013 at 1:40

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