Splitting 2nd order DE into two 1st order DE's

[markov4]

Homework Statement

1.a) Taking the Schrödinger second order differential equation given, split it into two first order differential equations for numerical solving.
We're given the relevant constants in a table, such as \hbar, \alpha, etc. 'z' is taken to be an indepedent variable which we incorporate into our initial conditions (which we must state).

The Attempt at a Solution

Have i done this right? Is it right to say that \Psi(z)=vz ? Since the \Psi(z) that is in the equation, the one multiplied by ..+[U(z)-E_n]\Psi(z) has to be set to something too (?).

 

[gabbagabbahey]

Is it right to say that \Psi(z)=vz ?

No, just leave \Psi(z) as \Psi(z)

It's true that d\Psi=vdz, but until you solve for v(z) you have no way of integrating vdz.

 

[markov4]

^^But for the second part of the question, we're required to "Write a program that uses an inbuilt numerical solver (such as Matlab's ode45) to numerically solve for \Psi(z) for z₁<z<z₂..." so we have to choose our own initial conditions to incorporate into our program, along with using ode45 in matlab.

So then does that mean \Psi(z)=vz still? I don't understand why you would integrate it though? Unless I've completely missed something in the numerical solving part of this course.

 

[Gigasoft]

v depends on z, so you can't integrate \frac{d\Psi}{dz}=v to get \Psi=vz.

 

[markov4]

^^So when i put these coupled equations into matlab, what do i define \Psi(z) to be?

 

[gabbagabbahey]

^^So when i put these coupled equations into matlab, what do i define \Psi(z) to be?

It's one of the functions you are trying to solve your system of equations for...you simply leave it as \Psi(z)

 

[markov4]

^^I see then. So I'm assuming that ODE45 in MATLAB will accommodate for leaving \Psi(z) in the equation? I'm reading up on it now (ODE45) but i don't think i understand it.

 

[markov4]

Ok after doing some reading on ode45, i think i get it. So yes i'll leave Psi(z) as it is, and implementing it into my MATLAB code, it will be solved according to the initial conditions i set. Just need to try it out in MATLAB now!