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

  1. Apr 25, 2010 #1
    1. The problem statement, all variables and given/known data
    15r1sm0.jpg

    1.a) Taking the Schrodinger 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).
    3. The attempt at a solution
    e6oprr.jpg

    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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  3. Apr 25, 2010 #2

    gabbagabbahey

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    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].
     
  4. Apr 25, 2010 #3
    ^^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.
     
  5. Apr 25, 2010 #4
    v depends on z, so you can't integrate [tex]\frac{d\Psi}{dz}=v[/tex] to get [tex]\Psi=vz[/tex].
     
  6. Apr 25, 2010 #5
    ^^So when i put these coupled equations into matlab, what do i define [tex]\Psi(z)[/tex] to be?
     
  7. Apr 25, 2010 #6

    gabbagabbahey

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    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]
     
  8. Apr 25, 2010 #7
    ^^I see then. So i'm assuming that ODE45 in matlab will accomodate 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.
     
  9. Apr 26, 2010 #8
    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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