2
$\begingroup$

I have been reading the paper on wave equations, and I have some confusion in notations.

Consider the initial value problem(IVP)(Wave equation):

$\frac{\partial ^2 u } {\partial t^2}(x,t) = \Delta u (x,t), (x,t) \in \mathbb R^{d} \times \mathbb R,$ $u(x,0) =f(x), \frac{\partial }{ \partial t} u(x, 0) =g(x).$

The solution to the above Cauchy Problem is $$u(x,t)=\cos (t \sqrt{- \Delta}f) (x) + \left( \frac{\sin (t \sqrt{- \Delta})}{\sqrt{- \Delta}}g \right) (x)$$

My Question is: What does the square root sign tells us?

Is it true that to study wave equation is equivalent to study the multiplier $e^{it |\xi|}$ ($f\mapsto (e^{i t |\xi|}\hat{f})^{\vee}$)? ( I don't see why this multiplier cover both the term in the solution. Can we deduce some thing about the multiplier $\cos t |\xi|$ and $\frac{\sin t |\xi|}{|\xi|}$

Background: I am familiar with the following solution to the wave equation: $u(x,t)= f\ast \partial_t W_t(x) + g\ast W_t (x),$ where $W_t = [ \frac{ \sin 2\pi t |\xi|}{ 2\pi |\xi|}]^{\vee}.$

(But here I do not see any square root sign; I guess, there must some thing to do with square root sign ....)

Improve this question
$\endgroup$
3
  • 3
    $\begingroup$ $x\mapsto\cos x$ and $x\mapsto\frac{\sin x}x$ are both power series in $x^2$, so in effect, the formula does not really involve any square root. $\endgroup$
    – Sebastian Goette
    Apr 11, 2016 at 14:42
  • $\begingroup$ $\sqrt{x^2} = |x|$. $\endgroup$
    – Willie Wong
    Apr 11, 2016 at 15:11
  • 1
    $\begingroup$ Also, you probably want both the forward wave $\exp i t |\xi|$ and the reverse wave $\exp -it |\xi|$ propagators to construct the full solution. Note also that $\partial_t \sin(t |\xi|) / |\xi| = \cos (t |\xi|)$. $\endgroup$
    – Willie Wong
    Apr 11, 2016 at 15:16

0

Reset to default