General Solution of a Poisson Equation of a magnetic array

In summary: Your Name]In summary, the conversation discusses the Poisson equation and its application in a specific scenario involving a planar structure. The equation relates the scalar potential to the vector field, which in this case is represented by mok.cos(kx). The general solution of the Poisson equation includes a homogeneous and particular solution, with the latter being obtained using the method of undetermined coefficients. The choice of x=0 is arbitrary and does not affect the solution.
  • #1
tim9000
867
17
Hi,
I'll give some background, say you've got a planar structure of thickness 'd', lying on the z plane. Also say the upper and lower surfaces are y = 0 and y = -d, respectively.
The structure has scalar potentials inside it as so:
poissed off.PNG

As you can see the vector fields cancel out on one side, As it says below, there is a Poisson equation of:
poiss.PNG

BUT I HAVE NO IDEA WHY that is the poission equation, I get that Fi inside is a scalar potential, but why is mok.cos(kx) the vector field?, not like mx+my or something instead? It looks like they've just differentiated mx and that's the vector function, maybe just a coincidence?
I also have no idea how that is the general solution? Specifically the homogenious part.
I get that for the part of the particular you can solve the Poisson equation of using method of undetermined coefficients with a guess of
(Asin(kx) + Bcos(kx)) and just differentiate that twice for del2:
(Asin(kx) + Bcos(kx))'' = mok.Cos(kx)
therefore: -A.k2sin(kx)-B.k2cos(kx) = mok.Cos(kx)
therefore: -A.ksin(kx)-B.kcos(kx) = mo.Cos(kx) and equating coefficients yields:
B = - mo/k
A = 0
so Yp = - mo/k * Cos(kx)

But why does the homogenous part have exponentials and y in them? I thought they'd just be zero.
If someone could explain that or even just why the Poisson equation is what they say I'd be greatful.
By the way, I understand that a Poisson equation that = 0 is a Laplace equation, but as far as I know The General solution = homogenious + particular, solutions. As you can see in (3b) the particular solution I worked out is in there, but so is the 'homogenous' aspect which I can't figure out. As you can see in (2a) the Poisson is non-homogenous (not laplace) but it must use a homogenous part in the method of undetermined coefficients, I expect, to find (3a) and (3c) as well as the afformentioned aspect of (3b).

The boundary conditions are given later (below) so I didn't think they were used to find the general solution. It would seem weird that I can find half of the solution and not the rest. And that doesn't explain how they came to that value of the Poisson equation.
upload_2015-4-5_22-45-46.png


P.S I also wonder, which point is chosen as x = 0 on the diagram...?
THANKS!
 
Last edited:
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  • #2


Hi there,

Thank you for sharing your thoughts and questions about the Poisson equation and its application in this specific scenario. I would like to provide some insights and explanations that may help clarify your doubts.

Firstly, let's start with the Poisson equation itself. This equation is a fundamental equation in physics and mathematics that describes the relationship between the scalar potential and the vector field in a given region. In this case, the scalar potential is represented by Fi, while the vector field is represented by mok.cos(kx). The reason why mok.cos(kx) is used as the vector field is because it satisfies the boundary conditions of the problem, which are given by y=0 and y=-d. This means that at the upper and lower surfaces of the planar structure, the vector field must be equal to zero. By using mok.cos(kx), we can easily satisfy these boundary conditions.

Now, let's move on to the general solution of the Poisson equation. As you have correctly stated, the general solution is composed of the homogeneous and particular solutions. The homogeneous solution is the solution to the Laplace equation, which is the same as the Poisson equation but with a right-hand side of zero. This means that the homogeneous solution only depends on the boundary conditions and not on any external sources.

On the other hand, the particular solution is the solution to the Poisson equation with a non-zero right-hand side, which in this case is mok.cos(kx). As you have shown, this particular solution can be obtained by using the method of undetermined coefficients. However, it is important to note that the particular solution is not unique, as different choices of the trial function (in this case, Asin(kx)+Bcos(kx)) can lead to the same particular solution. This is why the homogeneous part is also present in the general solution, as it accounts for all the other possible choices of the trial function.

Regarding your question about the choice of x=0, this point is usually chosen arbitrarily and does not affect the solution as long as it is consistent throughout the problem.

I hope this helps to clarify your doubts and provides a better understanding of the Poisson equation and its application in this scenario. If you have any further questions, please do not hesitate to ask.
 

1. What is a Poisson equation of a magnetic array?

The Poisson equation of a magnetic array is a mathematical equation that describes the behavior of a magnetic field in a given space. It is based on the principles of electromagnetism and is commonly used in physics and engineering.

2. How is the general solution of a Poisson equation of a magnetic array determined?

The general solution of a Poisson equation of a magnetic array is determined by solving the equation using various mathematical methods, such as the finite difference method, the finite element method, or the boundary element method.

3. What factors affect the general solution of a Poisson equation of a magnetic array?

The general solution of a Poisson equation of a magnetic array is affected by several factors, including the geometry and properties of the magnetic array, the boundary conditions, and the presence of external sources such as electric currents.

4. Can the general solution of a Poisson equation of a magnetic array be applied to real-life situations?

Yes, the general solution of a Poisson equation of a magnetic array can be applied to real-life situations. It is commonly used in the design and analysis of magnetic devices, such as motors, generators, and sensors.

5. Are there any limitations to the general solution of a Poisson equation of a magnetic array?

Yes, there are limitations to the general solution of a Poisson equation of a magnetic array. It may not accurately describe the behavior of magnetic fields in highly complex systems or in the presence of non-linear materials. In these cases, more advanced numerical methods may be required to obtain accurate results.

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