# I Change of variable (2nd order ODE)

1. Jul 7, 2016

Hi there, I was just trying to perform a change of variable on a differential equation as shown below. The original second order ODE is written in terms of a dependent variable $\theta$ and independent variable $q$. I have used the expression $q = \sqrt{\frac{z\tau'}{LT_R}}$ and $\tau' = T'(1-e^{\frac{-\tau}{T'}})$ (where $z$ and $\tau$ are also independent variables, while $L, T',$ and $T_R$ are known constants). I am only considering the case of $z = 0.01L$ and am thus only solving this equation for a varying $\tau$. (In this case, $\tau$ is a purely real and positive number.) I have posted my work below to transform the second order ODE into a function of $t$ instead of $q$. (I may have used $t$ and $\tau$ interchangeably out of bad habit--my apologies if this causes confusion.) If you could possibly check it to see if there are any errors, that would be greatly appreciated as I'm still a little uncertain if I've applied the chain rule correctly in certain parts.

(My apologies if this is the wrong section to post such question, although this is not homework.)

The final solution I get (it is partially cut in the last image) is:

$$100e^{\frac{2t}{T''}}(1-e^{\frac{-t}{T''}})[(1 + \frac {e^{\frac{-t}{T''}}}{2(1-e^{\frac{-t}{T''}})}) \frac {d\theta}{dt} + T'' \frac {d^2\theta}{dt^2} ] + 50e^{\frac{t}{T''}}\frac {d\theta}{dt} = sin(\theta)$$

which can also be expressed as:

$$100T''(e^{\frac{2t}{T''}}-e^{\frac{t}{T''}}) \frac {d^2\theta}{dt^2} + 50e^{\frac{t}{T''}}[(e^{\frac{t}{T''}}-1)(2+ \frac {1}{e^{\frac{t}{T''}}-1}) + 1]\frac {d\theta}{dt} = sin(\theta)$$

which is equivalent to:

$$100T''(e^{\frac{2t}{T''}}-e^{\frac{t}{T''}}) \frac {d^2\theta}{dt^2} + 100e^{\frac{2t}{T''}}\frac {d\theta}{dt} = sin(\theta)$$

Last edited: Jul 7, 2016
2. Jul 7, 2016

### Irene Kaminkowa

Hi,
It looks like some oscillation with attenuation.
Can you set the problem? And what is your purpose?

3. Jul 7, 2016

I am trying to solve the ODE:

$$\frac {d^2 \theta}{dq^2} + \frac {1}{q} \frac {d \theta}{dq} = sin(\theta)$$

at $z = 0.01L$ for all $\tau$. I know $q = \sqrt{\frac{z\tau'}{LT_R}}$ and $\tau' = T'(1-e^{\frac{-\tau}{T'}})$ and so have applied the above change of variables to the initial ODE. This is all being done to find $\theta$ at $z = 0.01L$ for all $\tau$ so that I can use this as a boundary condition for another problem I am working on that involves modelling an electric field propagating in a homogenous medium. For more context, page 32 from "Super-radiance multi atomic coherent emission" by Benedict et al. may help. But it is all once again being done essentially to find $\theta$ at an early $z$ for the entire $\tau$ domain to use as a boundary condition for further equations which are found in the aforementioned textbook. I couldn't use $z = 0$ since this makes $q = 0$ for all time. I am just skeptical if my solution is exactly correct since if I rearrange the above equation with just $\frac {d^2 \theta}{d\tau^2}$ isolated, then I have to divide by 0 in the case $\tau = 0$, but this is a physical situation and I wouldn't expect this to happen.

4. Jul 7, 2016

### Irene Kaminkowa

If T' = const then it is not a good choice of the name. T0 could be better.
I'd like to specify, whether
$$\tau' = T'\left ( 1-e^{\frac{\tau}{T'}} \right )$$
means something like
$$\frac{d \tau}{dt} = T_0\left ( 1-e^{\frac{\tau}{T_0}} \right )$$
?

5. Jul 7, 2016

Ahh sorry for the confusion. I was following the same notation from where I got the equations; $T'$ is constant and $\tau'$ is not the derivative but rather just another symbol used to denote it is different from $\tau$

6. Jul 7, 2016

### Irene Kaminkowa

Ah, now I see. You do want to change the variables in your equation )
May I ask, what are you going to do next? Find analitical solutions? Or numerical ones?

7. Jul 7, 2016

Numerical solutions

8. Jul 7, 2016

### Irene Kaminkowa

Then, it is not necessary to change the variables. Your can try RK4 right on these data.

9. Jul 8, 2016

Fair enough, I will certainly implement that. Although I am still interested in obtaining the expression for the second order DE in terms of $\tau$.