a question about first-order hyperbolic equations - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T21:19:18Z http://mathoverflow.net/feeds/question/111915 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/111915/a-question-about-first-order-hyperbolic-equations a question about first-order hyperbolic equations Uday 2012-11-09T16:37:44Z 2012-11-09T19:33:47Z <p>Performing certain manipulations on pseudo-differential equations I have come across the following first order equation: $$D_{t}u-\lambda(t,x,D_{t},D_{x})u=0, \ \ (*)$$ where $\lambda$ is a scalar pseudo-differential operator with the principal symbol being real-valued and independent of $D_{t}.$ But, the lower-order terms of $\lambda$ depend on $D_{t}$ (or $\tau$ at the symbol level). </p> <p>I was expecting a hyperbolic equation. But, I find that standard text books(like M.Taylor's 'Pseudo-differential operators') treat only equations in which $\lambda$ term is independent of $D_{t}$ (or $\tau$ at the symbol level). </p> <p>For the equation $(*)$ to be hyperbolic, is it necessary that $\lambda$ to be independent of $D_{t}$? Are there any references which discuss these issues? </p> http://mathoverflow.net/questions/111915/a-question-about-first-order-hyperbolic-equations/111924#111924 Answer by Bazin for a question about first-order hyperbolic equations Bazin 2012-11-09T18:53:36Z 2012-11-09T18:53:36Z <p>Your principal symbol is $\tau-\lambda_1 (t,x,\xi)$, with a real-valued $\lambda_1$, but you have lower order terms which may depend on $\tau$. You can get rid of that dependency as follows: take for instance $\lambda_0$ of order $0$. You can find an operator $M$ of order $0$ such that $$Op\bigl(\tau-\lambda_1 (t,x,\xi)-\lambda_0(t,x,\tau,\xi)\bigr)= e^{-iM}Op\bigl(\tau-\lambda_1 (t,x,\xi)\bigr)e^{iM}+Op(S^{-1}).$$ In fact, since $e^{iM}$ is a pseudodifferential operator of order 0, the composition formula gives $$e^{-iM}Op\bigl(\tau-\lambda_1 (t,x,\xi)\bigr)e^{iM}=D_t+\frac{\partial M}{\partial t} -e^{-iM}[Op(\lambda_1),e^{iM}]-Op(\lambda_1),$$ and since $e^{-iM}[Op(\lambda_1),e^{iM}]=Op(${$\lambda_1,m$}$)+Op(S^{-1})$ where $m$ is of order 0 and {} is the Poisson bracket. To get rid of $\lambda_0$, you have only to solve $$\frac{\partial m}{\partial t}-\text{ {\lambda_1,m } }=-\lambda_0,$$ which is a linear transport equation of real principal type in $m$. You can of course iterate this business to get a remainder of order as negative as you like. This is explained in the Chapter 23 on hyperbolic equations in Hörmander third volume of ALPDO.</p> http://mathoverflow.net/questions/111915/a-question-about-first-order-hyperbolic-equations/111931#111931 Answer by Deane Yang for a question about first-order hyperbolic equations Deane Yang 2012-11-09T19:33:47Z 2012-11-09T19:33:47Z <p>If I understand correctly, you have a PDE of the form $$(D_t - \lambda_1(t, x, D_x) + \lambda_0(t, x, D_t, D_x))u = f,$$ where $\lambda_1$ is a first order pseudodifferential operator and $\lambda_0$ is a zero-th order pseudodifferential operator. It seems to me that the proofs of many if not all estimates, including energy integral estimates, and theorems about regularity, uniqueness, and existence for the equation $$(D_t - a(t,x)D_x + b(t,x))u = f,$$ as presented in books like Taylor can be extended to your PDE.</p>