# Change of variables for 2nd order differential

1. Aug 7, 2014

### test1234

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 $\nu_g$ (i.e. group velocity), a new variable T can be defined as follows
$T=t-\frac{z}{\nu_g} = t-\beta_1 z$
where $\beta_1=\frac{1}{\nu_g}$

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

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

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

Hence,
$\frac{\partial A}{\partial t} = 0$

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

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

Which seems to give me zero following from the above argument for the first order differential term $\frac{\partial A}{\partial t}$...
Am I missing something here? I would really appreciate if anyone could help. Thanks in advance!

2. Aug 7, 2014

### maajdl

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

∂T/∂t=1

z is an independent variable and therefore

∂z/∂t=0

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.

3. Aug 7, 2014

### test1234

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 $\nu_g$.