- #1
KFC
- 488
- 4
Hi there,
I am solving a second-order wave equation of the following form:
##v^2\dfrac{\partial^2 y}{\partial x^2} = \dfrac{\partial^2y}{\partial t^2}##
where ##t## is time; ##y## is the wave function. I learn the wave function by assuming the system is linear and has a linear solution. The book tells to try a solution of sinusoidal wave so I get the following solution
##
y = A\sin(kx - \omega t)
##
M question is why the wave equation has to be a second-order PDE but not a first-order PDE. I saw some explanation online, it is stated that for a physical system, we need to know both initial position and initial speed so to learn the time evolution of the system. This makes sense to me. However, besides this points, is it anything to do with the time reversibility to explain why we need a second-order PDE instead?
I am solving a second-order wave equation of the following form:
##v^2\dfrac{\partial^2 y}{\partial x^2} = \dfrac{\partial^2y}{\partial t^2}##
where ##t## is time; ##y## is the wave function. I learn the wave function by assuming the system is linear and has a linear solution. The book tells to try a solution of sinusoidal wave so I get the following solution
##
y = A\sin(kx - \omega t)
##
M question is why the wave equation has to be a second-order PDE but not a first-order PDE. I saw some explanation online, it is stated that for a physical system, we need to know both initial position and initial speed so to learn the time evolution of the system. This makes sense to me. However, besides this points, is it anything to do with the time reversibility to explain why we need a second-order PDE instead?
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