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Extremely difficult System of Differentil Equations

by Noone1982
Tags: differentil, difficult, equations, extremely
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Noone1982
#1
Dec11-06, 12:13 PM
P: 83
The force on an electron is,

[tex]F\; =\; qV\times B[/tex]

where F is the force, V is velocity and B is the magnetic field. All are vectors except the charge of the electron q.

Expressing this as a cross product,

[tex]F\; =\; q\left[ \begin{array}{ccc} x^{\wedge } & y^{\wedge } & z^{\wedge } \\ V_{x} & V_{y} & V_{z} \\ B_{x}\left( t \right) & B_{y}\left( t \right) & B_{z}\left( t \right) \end{array} \right]\; =\; q\left( V_{y}B_{z}\left( t \right)\; -\; V_{z}By\left( t \right) \right)x^{\wedge }\; \; -\; q\left( V_{x}B_{z}\left( t \right)\; -\; V_{z}B_{x}\left( t \right) \right)y^{\wedge }\; +\; q\left( V_{x}B_{y}\left( t \right)\; -\; V_{y}B_{x}\left( t \right) \right)z^{\wedge }[/tex]

The x^ means a unit vector. Now we know that,

[tex]F\; =\; ma\; \; \; \; \; \; or\; \; \; \; \; \; a\; =\; \frac{F}{m}[/tex]

Lets define q/m = a

Now we have three differential equations,

[tex]\ddot{x}\; =\; a\left( \dot{y}B_{z}\left( t \right)\; -\; \dot{z}B_{y}\left( t \right) \right)[/tex]

[tex]\ddot{y}\; =\; a\left( \dot{x}B_{z}\left( t \right)\; -\; \dot{z}B_{x}\left( t \right) \right)[/tex]

[tex]\ddot{z}\; =\; a\left( \dot{x}B_{y}\left( t \right)\; -\; \dot{y}B_{x}\left( t \right) \right)[/tex]

There we have it. A system of three differential equations and I am stumped. Any takers?
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grant9076
#2
Dec12-06, 03:50 AM
P: n/a
Hi Noone,
Although there are many people here who know more about math than I ever will, I will do my best (hopefully one can jump in and help out). I believe that you can put these 3 differential equations in matrix form and find the characteristic equation (in polynomial form). You can then solve for the roots to find the eigenvalues which you can then use to find the eigenvectors. Using these (3) eigenvectors in a transpose matrix form and its inverse, you can diagonalize the original matrix to separate the variables. My lack of proficiency with using the special text prevents me from actually showing you how I would do it without creating confusion. I apologize for not being of more help.
dextercioby
#3
Dec12-06, 05:28 AM
Sci Advisor
HW Helper
P: 11,928
It's simple if you choose the system of coordinates as to ensure that in the system chosen, the magnetic field would have a single nonzero component. Why do you think physicists always choose a constant magnetic field along the Oz axis ? Simply because the system of ODE-s would simplify greatly.

Daniel.

CPL.Luke
#4
Dec12-06, 01:40 PM
P: 444
Extremely difficult System of Differentil Equations

hmm your substitution q/m=a is invalid, this would imply that f=q and we know that isn't true.
Noone1982
#5
Dec12-06, 02:25 PM
P: 83
Oops. I neglected to think people would confuse that a with acceleration. I should have chosen alpha or beta or gamma.

I know the problem is simple with the magnetic field as constants, but that is not my problem. I need to solve the DEs when the components of the magnetic field are functions.

Grant, a proffesor of mine suggested something similiar to your approach but unfortunately it would only work if the components of B are constants.
dextercioby
#6
Dec13-06, 12:55 AM
Sci Advisor
HW Helper
P: 11,928
The magnetic field can very well depend on time (it's called "nonconstant magnetic field"), it's just how the magnetic field looks spatially. THe simplifying assumption i suggested means the magnetic field is homogenous. That is the [itex] \vec{B} [/itex] is always parallel to itself in different points of space.

If the magnetic field is not homogenous, surely, the system of ODE-s retains the form you wrote initially modulo minor corrections suggested by CPL.Luke.

Daniel.
enricfemi
#7
Dec13-06, 01:41 AM
P: 192
hmm your substitution q/m=a is invalid, this would imply that f=q and we know that isn't true.
there is some misunderstanding.
noone didnot mean it.he means F=ma=qv*B
Noone1982
#8
Dec13-06, 09:53 AM
P: 83
Oh yes, it is homogeneous. What do you mean by modulo minor?
dextercioby
#9
Dec14-06, 01:42 AM
Sci Advisor
HW Helper
P: 11,928
modulo= without/less. I use it in my field of work. It comes from BRST. "minor"=small.

Daniel.
^_^physicist
#10
Dec14-06, 01:43 AM
P: 235
Hmm...I can say it would be much easier if we didn't have to worry about the z-componet, but being as that isn't the case.

Note that: Double dotted x = (dotted x)/dt. Also note that dotted y and dotted z = dy/dt and dz/dt. This will help you find the value for dotted x, and dotted y, which will be of use.

-----

The problem isn't too horrible, but has a lot of room for possible mistake, and can turn into a nest of trouble if you aren't careful.

Good Luck.
tehno
#11
Dec14-06, 01:00 PM
P: 363
Friendly suggestion to Noone1982:

Forget about electrons ,forcefields,and physical meaning of the equations here.
True mathematicians usually don't give a f*$k for the background ,origin and applications of (the system) of DE.
At first glance ,it seems the system of yours can be reduced to the system of algebraic equations with a proper operator method.
From that point on,things become trivial to think about.
CPL.Luke
#12
Dec14-06, 07:41 PM
P: 444
true, a laplace or fourier transform could probably take care of it. however you'd be left with inverse transforms of functions which is sometimes undesireable.
Noone1982
#13
Dec14-06, 08:20 PM
P: 83
Many words spoke but none so revealing. Speak with math and not words!
tehno
#14
Dec15-06, 01:22 PM
P: 363
Quote Quote by Noone1982 View Post
Many words spoke but none so revealing. Speak with math and not words!
You want me to solve your homework for you,what?
We have told you exactly how to simplify the problem.
Not probably ,but exactly -->The Laplace transform kills such sistems easily.



Quote Quote by CPL.Luke
true, a laplace or fourier transform could probably take care of it. however you'd be left with inverse transforms of functions which is sometimes undesireable
There are tables for a plethora of inverse transforms already made.
Therefore,one just need to know how to deal with system of 3 algebraic equations .
There is always,a direct method without transforms,but this is much longer.
Noone1982
#15
Dec15-06, 03:30 PM
P: 83
This is not homework. I do not know how to solve a system of DEs with Laplace transforms.
tehno
#16
Dec16-06, 01:54 PM
P: 363
Quote Quote by Noone1982 View Post
This is not homework. I do not know how to solve a system of DEs with Laplace transforms.
Well,this is a problem then.
I'm not a math teacher to guide you step by step how to learn ,and
,apply Laplace transform.
I suppose there are math proffessors here to that better ,only
the problem is if they are willing to (+ see my first post above ).
If the functions B(t) are constants,or very simple enough, than the system
like yours can be solved explicitly .If these functions are not simple than you may really need to use some numerical methods Matter of fact,you don't have to be expert on
systems of DEs to see that.This may be illustrated on example of single DE,without finite form solution like:
[tex]\frac{d^2 \theta}{dt^2}=sin\theta[/tex]
No Heaviside ,Laplace ,or Z-transformation in world will help you to find finite form solution there.

As for Laplace transform with constants case see:
http://www.sosmath.com/diffeq/laplac...plication.html
General sense of operator method is simplification and exchanging signs of integration and derivation by corresponding operators:
http://www.atp.ruhr-uni-bochum.de/rt...ol/node11.html

Summary on Laplace transform:
http://mathworld.wolfram.com/LaplaceTransform.html

Small table of functions transforms you can find here:
http://www.vibrationdata.com/Laplace.htm

Schaum's introduction for L. Transforms gives the table of over 300 transforms listed.

When you master basics of Laplace transforms and feel you are ready to
ask more questions let us know.Definitely,I will not do all for you from the begging to the end.
Noone1982
#17
Dec24-06, 09:40 AM
P: 83
Many, many thanks :)

I will try to use both methods, one using laplace to derive an exact expression for simpler constant B fields and the iterative one for B fields as a function of time. I got my iterative one to work, it can have some really funky motion although the drawback is that it's very slow to graph :(
tehno
#18
Dec26-06, 12:48 PM
P: 363
I think you got it.For start set in your system:
[tex]B_{x}(t)=A,B_{y}(t)=B,B_{z}(t)=C[/tex]

And solve it explicitly in Laplace domain.
Inverse transforms for time domain you can find in the tables.
Than observe what happens if the B functions are not constants.


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