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The equation \begin{gather} \frac{\partial u}{\partial t} (t,x) = \frac{1}{2} \text{Trace}[\sigma(x) \sigma(x) (\text{Hessian}_x u)(x,t)] + \langle \mu (x) , (\nabla_x u) (t,x) \rangle, \\ u(0,x) = \varphi(x), \end{gather} is called Kolmogorov partial differential equation and is associated to a particular type of SDE I care about.

I know about a result which states, that this equation has a unique, at most polynomially growing viscosity solution under some assumptions including continuity for $\varphi$ and local Lipschitz continuity for $\sigma, \mu$.

I have seen several solution-existence theorems of this type associated with PDEs and I am confused by the emphasis placed on "at most polynomial growth" of solutions seemingly required for uniqueness in several cases. I am fairly new to the theory of partial differential equations and I wonder why we only seem to care about solutions which grow at most polynomially.

What about other solutions which might exist and do not fulfill this growth requirement?

Are at most polynomially growing solutions the only ones which are interesting in applications?

I see no reason to single out at most polynomially growing solutions and give them special attention if there might be many other solutions; it seems like a rather arbitrary property to me. So why are we satisfied with existence theorems like the one about Kolmogorov PDEs above?

If someone could shed some light on what makes solutions of this type particularly interesting, that would be fantastic!

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    $\begingroup$ This is a crosspost: math.stackexchange.com/questions/3167453/… $\endgroup$ – Peter McNamara Apr 1 '19 at 5:56
  • $\begingroup$ The usual assumption for existence and uniqueness to linear PDEs is exponential growth and not polynomial. In probability theory sometimes subquadratic growth pops up, if you intend to plug the solution in an exponential in order to use Girsanov's theorem. If you are looking for a counterexample in the PDE theory you might consider mathoverflow.net/questions/82408/…. $\endgroup$ – Kore-N Apr 1 '19 at 14:21
  • $\begingroup$ Thank you. But this just shifts the question to why we only seem to care about solutions which grow exponentially then? $\endgroup$ – Joker123 Apr 3 '19 at 13:28
  • $\begingroup$ ... with some delay: Because gaussian processes have at most moments of order $e^{c|x|^2}$. So if you just plug in something worse you will obtain something with infinite average (which is not good for a Radon-Nikodym derivative for example). $\endgroup$ – Kore-N Aug 20 '19 at 7:48

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