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

In Summary,The student attempted to solve a system of differential equations using an inbuilt numerical solver (such as Matlab's ode45). However, they did not understand why integration was necessary. After reading up on ode45, they determined that the solution would be solved according to the initial conditions they set.
  • #1
markov4
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0

Homework Statement


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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).

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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  • #2
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].
 
  • #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.
 
  • #4
v depends on z, so you can't integrate [tex]\frac{d\Psi}{dz}=v[/tex] to get [tex]\Psi=vz[/tex].
 
  • #5
^^So when i put these coupled equations into matlab, what do i define [tex]\Psi(z)[/tex] to be?
 
  • #6
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]
 
  • #7
^^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.
 
  • #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!
 

1. What is the purpose of splitting a 2nd order differential equation into two 1st order differential equations?

Splitting a 2nd order differential equation into two 1st order differential equations allows for easier solution methods and can provide more insight into the behavior of the system being modeled.

2. How do you split a 2nd order differential equation into two 1st order differential equations?

To split a 2nd order differential equation, you can introduce a new variable, typically denoted as y, and express the 2nd order equation as two 1st order equations in terms of this new variable. This can be done using substitution or by rewriting the equation in a specific form.

3. Can any 2nd order differential equation be split into two 1st order differential equations?

Yes, any 2nd order differential equation can be split into two 1st order differential equations. However, the resulting equations may not always be simpler or easier to solve.

4. What are the benefits of splitting a 2nd order differential equation into two 1st order differential equations?

Splitting a 2nd order differential equation can make it easier to apply numerical methods or analytical techniques to solve the equations. It can also provide a deeper understanding of the behavior of the system being modeled.

5. Are there any limitations to splitting a 2nd order differential equation into two 1st order differential equations?

One limitation is that the resulting equations may be more complex or difficult to solve than the original 2nd order equation. Additionally, some physical systems may not lend themselves well to being split into two 1st order equations.

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