What is the usage of Kelvin function?

In summary, the Kelvin function is a special case of the Bessel function, which has similar applications in solving differential equations. The Kelvin functions are the real and imaginary parts of the vth order of J(xe^{3πi/4}), and they can be represented by the formulas for Ber(x) and Bei(x). However, the method of separation of variables, which is used to solve PDEs, is not always applicable and depends on the equation and boundary conditions. Solutions to boundary value problems can take various forms and most complex ones do not have analytical solutions.
  • #1
Chuck88
37
0
When I was scanning the materials of Bessel function, I found that when the variable is:

[tex]
xe^{\frac{3\pi}{4}i}
[/tex]

the Bessel function is called Kelvin function. Then what is the practical appliance of this function?
 
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  • #3
Yes. But the problems is that the information on this website only tells me the application of Bessel functions, not the Kelvin functions. I do not quite understand the special case of Bessel functions. Can you provide me with further information?
 
  • #4
Well, let's see. A Bessel function of the first kind is defined as:

[tex]J_v(x)=\sum\frac{(-1)^m}{m!Γ(m+v+1)}(\frac{x}{2})^{2m+v}[/tex]

The Kelvin functions would be the real and imaginary parts of the vth order of [itex]J(xe^{3πi/4})[/itex], therefore:

[tex]J_v(x)=\sum_{m=0}^{m=\infty} \frac{(-1)^m}{m!Γ(m+v+1)}(\frac{xe^{3πi/4}}{2})^{2m+v}[/tex]

which ultimately wields the formulas for Ber(x) and Bei(x) that you can see here (although you have probably seen it already):

http://en.wikipedia.org/wiki/Kelvin_functions#Ker.28x.29

Now, since bessel functions are solutions to the bessel differential equation, it means that kelvin functions (that are derived from these solutions) would be applicable to the same problems.

Again, I'd never heard of Kelvin functions until you asked, so I'm just thinking logically. The theory suggests though that special cases of a differential equation are still solutions to that equation, which means that they describe a specific set of problems that this equation describes. For instance, we could state that Laplace's equation is a special case of Poisson's equation. The special case would still have similar applications, albeit more limited.
 
  • #5
meldraft said:
Well, let's see. A Bessel function of the first kind is defined as:

[tex]J_v(x)=\sum\frac{(-1)^m}{m!Γ(m+v+1)}(\frac{x}{2})^{2m+v}[/tex]

The Kelvin functions would be the real and imaginary parts of the vth order of [itex]J(xe^{3πi/4})[/itex], therefore:

[tex]J_v(x)=\sum_{m=0}^{m=\infty} \frac{(-1)^m}{m!Γ(m+v+1)}(\frac{xe^{3πi/4}}{2})^{2m+v}[/tex]

which ultimately wields the formulas for Ber(x) and Bei(x) that you can see here (although you have probably seen it already):

http://en.wikipedia.org/wiki/Kelvin_functions#Ker.28x.29

Now, since bessel functions are solutions to the bessel differential equation, it means that kelvin functions (that are derived from these solutions) would be applicable to the same problems.

Again, I'd never heard of Kelvin functions until you asked, so I'm just thinking logically. The theory suggests though that special cases of a differential equation are still solutions to that equation, which means that they describe a specific set of problems that this equation describes. For instance, we could state that Laplace's equation is a special case of Poisson's equation. The special case would still have similar applications, albeit more limited.

Thanks for your reply. I still have a question about the differential equation. Do you understand the "Seperation of Variables?" I am doubting the rightness of this method.
 
  • #6
Why? Can you formulate the question more explicitly?
 
  • #7
meldraft said:
Why? Can you formulate the question more explicitly?

The method of separation of variables is used to solve the problem of partial diffrential equation. For example, when the partial differential equation is:

[tex]
\frac{\partial u}{\partial t}-\alpha\frac{\partial^{2}u}{\partial x^{2}}=0
[/tex]

We could suppose that [tex]u(x,t)[/tex] is a solution concerning both x and t. It could also be represented as the product of two functions:

[tex]
u(x,t) = X(x)T(t)
[/tex]

My question is that the method of separation of variables is based on the assumption that our solution could be represented as the product of two functions, which are with respect to x and t repsectively. My question is that what if the solution u could not be represented as the product of two functions of x and t, like:

[tex]
u(x,t) = \frac{1}{xt+1}
[/tex]

We then could not use the method of separation of variables.

I know that the solution I provided above is NOT RIGHT. But is it possible that the solution is of the form of the solution I provided above? Which means that the solution could not be separated that easy?
 
  • #8
Separation of variables is just one out of many methods to solve PDEs. Whether or not an equation can be solved with that method depends on two things:

- The equation itself.
- The boundary conditions.

A very simple example is Laplace's equation. It can pretty much always be solved using SoV in a rectangular domain, but if the number and geometry of the boundary segments changes, then the variables can no longer be separated.

The reason why this method is used is that it reduces the PDE to a system of ODEs, and that is something we can easily solve. To answer your question though, no, this method is not always applicable, even for the same equation.

As for the question about the solution, just check whether it satisfied the equation. As long as it does, it can have any form. Solutions to boundary value problems are defined by the boundaries. Seeing as the boundaries can be just about anything, there are really a lot of forms that a solution can take. Keep in mind though that most complex boundary value problems do not even have an analytical solution.
 
  • #9
meldraft said:
Separation of variables is just one out of many methods to solve PDEs. Whether or not an equation can be solved with that method depends on two things:

- The equation itself.
- The boundary conditions.

A very simple example is Laplace's equation. It can pretty much always be solved using SoV in a rectangular domain, but if the number and geometry of the boundary segments changes, then the variables can no longer be separated.

The reason why this method is used is that it reduces the PDE to a system of ODEs, and that is something we can easily solve. To answer your question though, no, this method is not always applicable, even for the same equation.

As for the question about the solution, just check whether it satisfied the equation. As long as it does, it can have any form. Solutions to boundary value problems are defined by the boundaries. Seeing as the boundaries can be just about anything, there are really a lot of forms that a solution can take. Keep in mind though that most complex boundary value problems do not even have an analytical solution.

Thanks. Now I understand the constraints of "Seperation of Variables."
 

1. What is the Kelvin function?

The Kelvin function, also known as the Kelvin or modified Bessel function, is a special function in mathematics that is used to solve differential equations and describe physical phenomena in the fields of mathematics, physics, and engineering.

2. What is the usage of the Kelvin function?

The Kelvin function is used to describe and solve problems involving heat transfer, wave propagation, and diffusion processes. It is also commonly used in the study of fluid dynamics, quantum mechanics, and electromagnetism.

3. How is the Kelvin function different from other special functions?

The Kelvin function is unique in that it is a solution to a second-order linear differential equation that cannot be expressed in terms of elementary functions. It is also closely related to other special functions such as the Bessel function and the Hankel function.

4. What are the key properties of the Kelvin function?

The Kelvin function has several important properties, including asymptotic behavior, recurrence relations, and integral representations. It also has specific values at certain points known as zeros and poles, which are important in solving physical problems.

5. How is the Kelvin function used in real-world applications?

The Kelvin function is used in a wide range of real-world applications, including the design of heat exchangers, acoustic wave propagation, and modeling of electromagnetic fields. It is also used in the analysis of data from experiments and simulations in various scientific and engineering fields.

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