Boundary value problem

In summary, a boundary value problem is a mathematical problem that involves finding a solution to a differential equation subject to certain constraints or boundary conditions. There are two main types of boundary value problems: Dirichlet and Neumann. These problems are typically solved using numerical methods, but analytical solutions are also possible for simpler cases. Boundary value problems have many real-world applications and the proper definition of boundary conditions is crucial for obtaining accurate solutions.
  • #1
John O' Meara
330
0
Schodinger's equation for one-dimensional motion of a particle whose potential energy is zero is
[tex]\frac{d^2}{dx^2}\psi +(2mE/h^2)^\frac{1}{2}\psi = 0 [/tex]
where [tex] \psi [/tex] is the wave function, m the mass of the particle, E its kinetic energy and h is Planck's constant. Show that

[tex]\psi = Asin(kx) + Bcos(kx)[/tex] ( where A and B are constants) and [tex] k =(2mE/h^2)^\frac{1}{2}[/tex] is a solution of the equation.
Using the boundary conditions [tex]\psi=0[/tex] when x=0 and when x=a, show that
(i) the kinetic energy [tex]E=h^2n^2/8ma^2[/tex]
(ii) the wave function [tex]\psi = A sin(n\pi\times x/a)[/tex] where n is any integer. (Note if [tex] sin(\theta) = 0 then \theta=n\pi[/tex])

My attempt:

A*sin(0) + B*cos(0) = 0, => 0 +B =0 => B = 0.
Therefore
A*sin(k*a)=0, Therefore [tex] (2mE/h^2)^\frac{1}{2}a = n\pi => E=n^2\pi^2h^2/2ma^2[/tex]

Homework Statement


Homework Equations


The Attempt at a Solution

 
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  • #2
(ii) should read A*sin(n*pi*x/a), I'm just notgood at boundary value problems.
 
  • #3
John O' Meara said:
Schodinger's equation for one-dimensional motion of a particle whose potential energy is zero is
[tex]\frac{d^2}{dx^2}\psi +(2mE/h^2)^\frac{1}{2}\psi = 0 [/tex]
where [tex] \psi [/tex] is the wave function, m the mass of the particle, E its kinetic energy and h is Planck's constant. Show that

[tex]\psi = Asin(kx) + Bcos(kx)[/tex] ( where A and B are constants) and [tex] k =(2mE/h^2)^\frac{1}{2}[/tex] is a solution of the equation.
Using the boundary conditions [tex]\psi=0[/tex] when x=0 and when x=a, show that
(i) the kinetic energy [tex]E=h^2n^2/8ma^2[/tex]
(ii) the wave function [tex]\psi = A sin(n\pi\times x/a)[/tex] where n is any integer. (Note if [tex] sin(\theta) = 0 then \theta=n\pi[/tex])

My attempt:

A*sin(0) + B*cos(0) = 0, => 0 +B =0 => B = 0.
Therefore
A*sin(k*a)=0, Therefore [tex] (2mE/h^2)^\frac{1}{2}a = n\pi => E=n^2\pi^2h^2/2ma^2[/tex]
Well, first, you haven't shown that that [itex]\psi[/itex] does, in fact, satisfy the differential equation. After that, yes, B= 0. Now, assuming A is not 0, that is that [itex]\psi[/itex] is not itself identically 0, then yes, we must have sin(ka)= 0 so that [itex]ka= (2mE/h^2)^{\frac{1}{2}}= n\pi[/itex]. E follows eactly as you say.
[/QUOTE]
 
  • #4
I was able to show that [tex]\psi[/tex] is a solution of the equation, it is (i) and (ii) that I had trouble doing espically (ii) i.e., [tex]\psi = A sin(n\pi\times x/a)[/tex]. Where did he get the argument "[tex]n\pi\times x/a[/tex]", or more importantly how does he expect me to get that argument of the sine.
Also remember in (i) the answer he has for E [tex]=h^2n^2/8ma^2[/tex], not what I got for E. Thanks for the help.
 
  • #5
I hope someone can tell me how [tex]\psi[/tex] can go from [tex] Asin((2mE/h^2)^\frac{1}{2}x) to Asin(n\pi\times x/a)[/tex]. Thanks for the help.
 
  • #6
John O' Meara said:
I was able to show that [tex]\psi[/tex] is a solution of the equation, it is (i) and (ii) that I had trouble doing espically (ii) i.e., [tex]\psi = A sin(n\pi\times x/a)[/tex]. Where did he get the argument "[tex]n\pi\times x/a[/tex]", or more importantly how does he expect me to get that argument of the sine.
Well, when x= a, the argument is [itex]n\pi[/itex] what is [itex]sin(n\pi)[/tex]? Remember that you were told that [itex]\psi(0)= 0[/itex] and [itex]\psi(a)= 0[/itex]. Knowing that cos(0)= 1 tells us that the second constant, B, must be 0. That leaves Asin(kx). We must have Asin(ka)= 0 and we don't want A= 0 (that would mean our function is always 0) so we must have sin(ka)= 0. For what x is sin(x)= 0? Multiples of [itex]\pi[/itex] of course:
[itex]ka= n\pi[/itex]. For that to be true, k must be equal to [itex]n\pi/a[/itex]

Also remember in (i) the answer he has for E [tex]=h^2n^2/8ma^2[/tex], not what I got for E. Thanks for the help.
You were also told that [itex]k= \sqrt{2ME/h^2}[/itex] and you now know [itex]k= n\pi/a[\itex] so [itex]n\pi/a= \sqrt{2ME/h^2}[/itex]. Solve that for E.
 

1. What is a boundary value problem?

A boundary value problem is a mathematical problem that involves finding a solution to a differential equation subject to certain constraints or boundary conditions at the endpoints of the domain.

2. What are the types of boundary value problems?

The two main types of boundary value problems are Dirichlet boundary value problems, which specify the value of the solution at the boundary, and Neumann boundary value problems, which specify the value of the derivative of the solution at the boundary.

3. How are boundary value problems solved?

Boundary value problems are typically solved using numerical methods such as finite difference, finite element, or boundary element methods. Analytical solutions are also possible for some simple boundary value problems.

4. What are some real-world applications of boundary value problems?

Boundary value problems have a wide range of applications in many fields such as engineering, physics, and biology. Some examples include heat transfer, fluid mechanics, and population dynamics.

5. What is the importance of boundary conditions in solving boundary value problems?

Boundary conditions are crucial in solving boundary value problems as they define the behavior of the solution at the boundaries of the domain. Without proper boundary conditions, the solution may not be unique or may not accurately reflect the real-world situation being modeled.

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