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

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Homework Help Overview

The discussion revolves around splitting the Schrödinger second order differential equation into two first order differential equations for numerical solving. The context involves using constants provided in a table and defining initial conditions for the variable 'z'.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants explore the correct formulation of the wave function \Psi(z) and its relationship to the variable v. There are questions about the integration of the differential equation and how to set initial conditions for numerical solving in MATLAB.

Discussion Status

Some participants have provided guidance on leaving \Psi(z) as it is in the equations, while others express uncertainty about how to implement this in MATLAB. The conversation reflects a mix of understanding and confusion regarding the numerical solving process.

Contextual Notes

Participants are required to define initial conditions for their numerical solver and are discussing the implications of their choices on the formulation of the equations.

markov4
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Homework Statement


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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 [tex]\hbar[/tex], [tex]\alpha[/tex], 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


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

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markov4 said:
Is it right to say that [tex]\Psi(z)[/tex]=vz ?

No, just leave [itex]\Psi(z)[/itex] as [itex]\Psi(z)[/itex]

It's true that [itex]d\Psi=vdz[/itex], but until you solve for [itex]v(z)[/itex] you have no way of integrating [itex]vdz[/itex].
 
^^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 [tex]\Psi(z)[/tex] 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 [tex]\Psi(z)[/tex]=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.
 
v depends on z, so you can't integrate [tex]\frac{d\Psi}{dz}=v[/tex] to get [tex]\Psi=vz[/tex].
 
^^So when i put these coupled equations into matlab, what do i define [tex]\Psi(z)[/tex] to be?
 
markov4 said:
^^So when i put these coupled equations into matlab, what do i define [tex]\Psi(z)[/tex] to be?

It's one of the functions you are trying to solve your system of equations for...you simply leave it as [itex]\Psi(z)[/itex]
 
^^I see then. So I'm assuming that ODE45 in MATLAB will accommodate for leaving [tex]\Psi(z)[/tex] in the equation? I'm reading up on it now (ODE45) but i don't think i understand it.
 
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!
 

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