# Wave equation v.s.Schrödinger equation

The motivation of comparison of this two kind of operators is that,$$\partial_{tt}-\Delta=(\partial_{t}-i\sqrt{-\Delta})(\partial_{t}+i\sqrt{-\Delta})$$ From the above that a wave operator can be seen as the product of two (general type) schr$\ddot{o}$dinger operators.Indeed,compared with other kinds of differential operators(elliptic,parabolic),these two operators seem to share more in common.Such as they don't have the usual global regularity property,and the use of harmonic analysis has made a great success in these two areas.

So I'm very curious to know the intersting links or fundemental differences between this two operator in order to understand them better.

*Edit:*In the above,I'm refering to a general dispersive equation:$u_{t}-i\phi({D})u=0$.When $\phi=-|\xi|^{2}$,it's the free schr$\ddot{o}$dinger equation.

• That is not a Schrodinger equation... – Chris Gerig Aug 3 '12 at 4:14
• well, I think you can look it as a fractional schrodinger operator. – user23078 Aug 3 '12 at 4:56

There is no direct link in the way you write it down. In order to decompose the wave equation you will need a Clifford algebra of matrices $\gamma$, such that $\gamma_i\gamma_j + \gamma_j\gamma_i = 2\eta_{ij}$, and define the operator $D=i\gamma_k\partial^k$. Then you will have $D^2=\partial_{tt}-\Delta$. But note that this operator will not apply on ordinary scalar functions but on spinors. Finally, when you consider the "massive" operator $D-I$, you are able to recover in some limit the Schroedinger equation.
• Well,I found a more direct link (in Terry Tao's book)between this two operators,which says that let $u\in C(R\times R^{d})$,and defines $v\in C(R\times R^{d+1})$ by $v(t,x_{1},\dots,x_{d},x_{d+1})=e^{-i(t+x_{d+1})}u(\frac{t-x_{d+1}}{2},x_{1},\dots,x_{d})$.Then v solves the d+1 dimensional wave equation if and only if u solves the d dimensioanal Schrodinger equation. – user23078 Aug 5 '12 at 8:47