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Change of variables for 2nd order differential

  1. Aug 7, 2014 #1
    Hi there, I'm having some difficulty in understanding how the change of variables by considering a retarded time frame can be obtained for this particular eqn I have.

    Say I have this original equation,
    \frac{\partial A}{\partial z} + \beta_1 \frac{\partial A}{\partial t}+ \frac{i \beta_2}{2} \frac{\partial^2 A}{\partial t^2}+\frac{\alpha}{2} A = i \gamma |A|^2 A

    Considering a frame of reference moving with speed [itex]\nu_g[/itex] (i.e. group velocity), a new variable T can be defined as follows
    [itex]T=t-\frac{z}{\nu_g} = t-\beta_1 z[/itex]
    where [itex]\beta_1=\frac{1}{\nu_g}[/itex]

    Apparently, the differential equation can be reduced to
    \frac{\partial A}{\partial z} -\frac{i \beta_2}{2} \frac{\partial^2 A}{\partial T^2}+\frac{\alpha}{2} A = i \gamma |A|^2 A

    From my understanding, the first order differential term [itex]\beta_1 \frac{\partial A}{\partial t} [/itex] disappears because
    \frac{\partial A}{\partial t} = \frac{\partial A}{\partial T} \times \frac{\partial T}{\partial t}

    T =t-\frac{z}{\nu_g} = t-\beta_1 z \\
    \therefore \frac{\partial T}{\partial t} = 1-\frac{\partial}{\partial t} (\beta_1 z) \\
    = 1-\beta_1 \nu_g = 0 \phantom{0} (\because \beta_1=\frac{1}{\nu_g})

    \frac{\partial A}{\partial t} = 0

    However, I can't wrap my head around how to transform the second derivative [itex] \frac{\partial^2 A}{\partial t^2}[/itex]?

    \frac{\partial^2 A}{\partial t^2} = \frac{\partial}{\partial t}(\frac{\partial A}{\partial t})
    = \frac{\partial}{\partial t}(\frac{\partial A}{\partial T} \times \frac{\partial T}{\partial t})

    Which seems to give me zero following from the above argument for the first order differential term [itex]\frac{\partial A}{\partial t}[/itex]...
    Am I missing something here? I would really appreciate if anyone could help. Thanks in advance!
  2. jcsd
  3. Aug 7, 2014 #2


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    Gold Member

    To me it seems that the change of variable (z,t) => (z,T) leads to


    z is an independent variable and therefore


    The wave packet represented by this equation might well have an average position.
    But don't name it "z", name it something else than the independent variable z of the wave equation.
  4. Aug 7, 2014 #3
    Thanks maajdl for the quick reply. Hmm... if ∂T/∂t=1, then the first order derivative of ∂A/∂T will still exist...

    Based on my understanding, z is the direction of propagation of the pulse. So the pulse travels with a group velocity [itex]\nu_g[/itex].
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