Differential equation with cross product

In summary, the equation given is a system of differential equations that can be rewritten as a matrix equation and solved using eigen-vector decomposition if A has constant coefficients. Alternatively, you can also handle it by substituting for the second derivative of y and solving a simpler equation for x. If B is a constant, you can also use the trick mentioned by your friend to solve the equation.
  • #1
alsey42147
22
0
ok, i don't know what to do with something like this:

(d^2R/dt^2 ) + (dR/dt) x B = 0

where the capitals are vectors (sorry i suck at latex). R is a position vector in x-y plane and B is in the z-direction.

do i split this into equations for x and y directions separately and solve them? for x-direction i would get

d^2x/dt^2 + (dy/dt)|B| = 0

but how do i solve this? I'm pretty sure all the DEs I've ever solved had, for example, d^2x/dt^2 and dx/dt in them, but not d^2x/dt^2 and dy/dt.

or is there a quick way of doing it without having to expand the cross product?

any help would be well appreciated!
 
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  • #2
This is a SYSTEM of differential equations that you can rewrite as:
[tex]\frac{d\vec{U}}{dt}=A\vec{U}, \vec{U}=\frac{d\vec{R}}{dt}[/tex]
and A is a matrix.
If A has constant coefficients, then the system is readily solvable with eigen-vector decomposition.

Question:
Did you get this equation from a physical problem with a Coriolis term?
Just curious..

EDIT:
Insofar as B is constant, you may gain two decoupled 3.order diff.eq's in R (i.e 2.order in U).
 
Last edited:
  • #3
Maybe [itex] \vec{B} [/itex] is the magnetic field and he chose units such as q=1...
 
  • #4
damn, i wish i was more proficient with matrices and stuff. i might have to do a bit of quick revision.

the equation is just electron motion in a magnetic field without the constants.

my friend says that from the equation

d^2x/dt^2 + (dy/dt)|B| = 0

you just integrate to get

dx/dt + y|B| = constant

that doesn't seem quite right to me...is it? if it is that's all i need for now.
 
  • #5
Is B a constant, or a function of x and y?
 
  • #6
alsey42147 said:
ok, i don't know what to do with something like this:

(d^2R/dt^2 ) + (dR/dt) x B = 0

where the capitals are vectors (sorry i suck at latex). R is a position vector in x-y plane and B is in the z-direction.

do i split this into equations for x and y directions separately and solve them? for x-direction i would get

d^2x/dt^2 + (dy/dt)|B| = 0

but how do i solve this? I'm pretty sure all the DEs I've ever solved had, for example, d^2x/dt^2 and dx/dt in them, but not d^2x/dt^2 and dy/dt.

or is there a quick way of doing it without having to expand the cross product?

any help would be well appreciated!
Yes, writing R as <x, y, 0> you get
[tex]\frac{d^2x}{dt^2}+ B\frac{dy}{dt}= 0[/tex]
and
[tex]\frac{d^2y}{dt^2}- B\frac{dx}{dt}= 0[/itex]

You can, as Arildno suggested, introduce [itex]u= \frac{dx}{dt}[/itex] and [itex]v= \frac{dy}{dt}[/itex] and write this as a system of 4 first order differential equations.

Another way to handle it is this: differentiate the first equation again to get
[tex]\frac{d^3x}{dt^3}+ B\frac{d^2y}{dt^2}= 0[/itex]
and use the second equation to substitute for the second derivative of y
[tex]\frac{d^3x}{dt^3}+ B^2\frac{dx}{dt}= 0[/tex]
That's easy to solve.

Once you know x(t), you can use the first equation to solve for [itex]\frac{dy}{dt}[/itex] and integrate once more to find y(t).

(I just noticed J77's comment. I am assuming here that B is a constant.)
 
Last edited by a moderator:
  • #7
awesome, thanks. and yes B is constant.
 
  • #8
alsey42147 said:
damn, i wish i was more proficient with matrices and stuff. i might have to do a bit of quick revision.

the equation is just electron motion in a magnetic field without the constants.

my friend says that from the equation

d^2x/dt^2 + (dy/dt)|B| = 0

you just integrate to get

dx/dt + y|B| = constant

that doesn't seem quite right to me...is it? if it is that's all i need for now.

You can certainly use your friend's trick in the case of constant B. It doesn't help you an overly lot, though, compared to other techniques mentioned.
 

1. What is a cross product in the context of differential equations?

A cross product in differential equations is a mathematical operation that combines two vector fields to produce a third vector field. It is used to describe the relationship between two variables in a system, where the rate of change of one variable depends on the values of the other variable.

2. How is a cross product different from a dot product in differential equations?

In a cross product, the result is a vector, while in a dot product, the result is a scalar. This means that the cross product takes into account the direction of the vectors, while the dot product only looks at the magnitude. In differential equations, the cross product is used to describe relationships between vector fields, while the dot product is used to simplify equations and solve for unknown variables.

3. Can you give an example of a differential equation with a cross product?

Sure, a classic example is the Lorentz force equation, which describes the force on a charged particle in an electromagnetic field. It includes a cross product between the velocity of the particle and the magnetic field, which determines the direction of the force on the particle.

4. What are some real-world applications of differential equations with cross products?

Differential equations with cross products are used in many fields, including physics, engineering, and economics. For example, in physics, they are used to describe the motion of particles in a magnetic field, while in engineering, they can be used to model the behavior of mechanical systems. In economics, they are used to analyze the relationships between different economic variables.

5. Are there any special techniques for solving differential equations with cross products?

Yes, there are specific techniques for solving differential equations with cross products, such as the method of separation of variables or the method of variation of parameters. These techniques involve breaking down the equation into simpler parts and solving for each part separately before combining them to find the overall solution. However, the specific technique used will depend on the specific equation and its application.

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