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What's the difference between initial conditions and boundary conditions? 
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#1
Feb1005, 06:16 PM

P: 65

As the thread title says: What's the difference between initial conditions and boundary conditions?
Thanks in advance for helpful replies. :) 


#2
Feb1005, 06:26 PM

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It all depends on what the variables of the unknown function in the PDE are.And what the domain in which u search for the solution is.
Boundary conditions are basically independent of who the variables are and are concerned only with the values either the function (Dirichlet problem) or its normal derivative (Neumann problem) take on the boundary of the domain one searches for the solution. As for initial conditions,well they are bit restrictive since are applied only to a restrained field of PDEs,namely the ones in which "who are the variables" really matter.Namely,that,besides the space variables,time (with the significance from physics) joins in.It is wrt time that the "initial conditions" refer to.Namely the values of the unknown function (and/or its time derivative) take/s at a certain moment of time. Daniel. 


#3
Feb1005, 06:31 PM

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There's no difference from a mathematical point of view (As far as I recall).
In physical systems depending on time, it's usually customary to state to values of the function at time t=0 (or t=t0). When dealing with a timeindependent problem, you want to give the values at some point x=x0. For the first, initial value is used, for the second, the term 'boundary condition' is more customary. Boundary conditions are used often in PDE's, such as the Laplace equation for f(x,y,z), since there is no time dependence. In QM, we can state [itex]\Psi(t_0,x)[/itex] which is a initial value for the Schrodinger equation. Or we can state for example [itex]\Psi_(t,0)=0[/itex], which is a boundary condition. 


#4
Feb1005, 06:34 PM

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What's the difference between initial conditions and boundary conditions?
i would also say no difference at all. as to your aside about appreciating only helpful replies, there is an old saying that beggars should not be choosers. you always get your money's worth here.



#5
Feb1105, 07:06 AM

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PF Gold
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It's not neccessary to be dealing with partial differential equations to have "initial values" and "boundary values".
An initial value problem is a differential equations problem in which you are given the the value of the function and sufficient of its derivatives at ONE VALUE OF X. Typically, if you have a second order equation, you are given the value of the function and its first derivative at some value of x. (That value of x being the "initial" time often it is x= 0 but doesn't have to be.) A boundary value problem is a differential equations problem in which you are given the value of the function at several different values of x. Typically, if you have a second order equation, you are given the value of the function at the endpoints of an interval. (Those endpoints being the "boundary".) There is a critical theoretical difference between the two: There is a theorem (the "fundamental existence and uniqueness theorem for initial value problems" that say if f(x,y) is well behaved (continuous in both x and y and Lipschitz in y) then the initial value problem dy/dx= f(x,y) with y(x_{0})= y_{0} has a unique solution. There is no such theorem for boundary value problems. For example, the differential equation d^{2}y/dx^{2}= y With initial conditions y(0)= A, y'(0)= B has the unique solution y(x)= A cos(x)+ B sin(x) no matter what A and B are. The differential equation d^{2}y/dx^{2}= y with the boundary value conditions y(0)= 0, y([itex]\pi[/itex])= 0 has an infinite number of solutions (y= A sin(x) for any value of A). The differential equation d^{2}y/dx^{2}= y with the boundary value conditions y(0)= 0, y([itex]\pi[/itex])= 1 has NO solution. 


#6
Apr1708, 06:00 PM

P: 1

A more mathematical way to picture the difference between an initial value problem and a boundary value problem is that an initial value problem has all of the conditions specified at the same value of the independent variable in the equation (and that value is at the lower boundary of the domain, thus the term "initial" value). On the other hand, a boundary value problem has conditions specified at the extremes of the independent variable. For example, if the independent variable is time over the domain [0,1], an initial value problem would specify a value of y(t) and y'(t) at time t = 0, while a boundary value problem would specify values for y(t) at both t = 0 and t = 1.



#7
Mar3109, 01:16 PM

P: 89

can anyone tell me what one means by these bracket [tex]
\langle\psi(r)\psi^*(r) \rangle [/tex] , is it integral or what?????and is there any idea how to calculate this integral [tex] \int\int{\mid\langle\psi(r)\psi^*(s)\rangle\mid}^ 2 dr ds [/tex] 


#8
Mar3109, 02:26 PM

P: 490

I believe you'll find that the angle brackets mean "expected value", the bars   mean "modulus", and this is some sort of QM problem.
Or maybe it's just making me have false memories... And the star means "complex conjugate", by the way. What ends up being a good way to do these depends on what the wave functions look like. 


#9
Mar3109, 06:40 PM

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PF Gold
P: 39,497

For example, the initial value problem y"+ y= 0, y(0)= A, y'(0)= B has unique solution y(t)= A cos(t)+ B sin(t) for all A and B. But y"+ y= 0, y(0)= 0, y(\pi)= B has an infinite number of solution if B= 0, no solutions if B is not 0. 


#10
Mar3109, 06:43 PM

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PF Gold
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Anglea, PLEASE do not "hijack" someone else's thread to ask question unrelated to the topic!



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