What is the usage of Kelvin function?

[Chuck88]

When I was scanning the materials of Bessel function, I found that when the variable is:

$$xe^{\frac{3\pi}{4}i}$$

the Bessel function is called Kelvin function. Then what is the practical appliance of this function?

 

[meldraft]

Without being familiar with Kelvin functions, since they are a special case of Bessel functions I would imagine they have similar applications, as described here:

http://en.wikipedia.org/wiki/Bessel_function#Applications_of_Bessel_function

 

[Chuck88]

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?

 

[meldraft]

Well, let's see. A Bessel function of the first kind is defined as:

$$J_v(x)=\sum\frac{(-1)^m}{m!Γ(m+v+1)}(\frac{x}{2})^{2m+v}$$

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

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

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.

 

[Chuck88]

Well, let's see. A Bessel function of the first kind is defined as:

$$J_v(x)=\sum\frac{(-1)^m}{m!Γ(m+v+1)}(\frac{x}{2})^{2m+v}$$

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

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

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 "separation of Variables?" I am doubting the rightness of this method.

 

[meldraft]

Why? Can you formulate the question more explicitly?

 

[Chuck88]

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:

$$\frac{\partial u}{\partial t}-\alpha\frac{\partial^{2}u}{\partial x^{2}}=0$$

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

$$u(x,t) = X(x)T(t)$$

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:

$$u(x,t) = \frac{1}{xt+1}$$

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?

 

[meldraft]

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.

 

[Chuck88]

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 "separation of Variables."