Using WKB(J) method to obtain asymptotics for an elastic string?

In summary: After the time and space coordinate separation you obtain the ordinary rather than partial derivatives. Your y(x) depends only on x. So my transformation applies well.
  • #1
dev00790
13
0
Hello,

The differential equation is question that models an elastic string is:
[tex]\frac {\partial^2 u}{\partial t^2} - \frac{\partial}{\partial x}\left( p(x) \frac{\partial u}{\partial x}\right)= 0[/tex]

Taking [tex]u(x,t) = e^{-i\lambda t} y(x) [/tex]
I simplify above diff.eqn. to: [tex]\lambda^2 y(x) + \frac{\partial}{\partial x}\left( p(x) y'(x)\right)= 0[/tex]

Then taking a WKB ansatz for [tex]y(x,\lambda)[/tex] as follows: [tex]y(x,\lambda) = e^{-i\lambda \xi(x)} \sum_{n=0}^{\infty} \frac{Y_{n}(x) }{\lambda^n}[/tex], I end up with;

[tex]\lambda^2 e^{-i\lambda \xi(x)} \sum_{n=0}^{\infty}\frac{Y_{n}(x) }{\lambda^n}
+ \frac{\partial}{\partial x}\left[p(x)e^{-i\lambda \xi(x)}\left(\sum_{n=0}^{\infty}\frac{Y_n'(x)}{\lambda^n} - i\lambda\xi(x)\sum_{n=0}^{\infty}\frac{Y_n(x)}{\lambda^n}\right)\right][/tex]

The Problem: I wonder if this last equation can be simplified? - I'm trying to obtain an recurrence relation for the eigenfunctions [tex]Y_n(x)[/tex] and their derivatives.

If it can't be simplified, what adjustments / changes are needed for this above method in order to obtain a recurrence relation?

Thanks, dev00790
 
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  • #2
The sign of the second equation is wrong, it should be read [itex]-\lambda^2 y(x)[/itex]. Appart from that, what is your question?
 
  • #3
Above post is edited, including further working and my question.

Replying to your question. I've checked my working, and i believe

[tex] \frac{\partial^2 u}{\partial t^2} = -\lambda^2 e^{-i\lambda t}y(x)[/tex]

[tex] \frac{\partial u}{\partial x} = e^{-i\lambda t}y'(x)[/tex]

Substituting into [tex] \frac {\partial^2 u}{\partial t^2} - \frac{\partial}{\partial x}\left( p(x) \frac{\partial u}{\partial x}\right)= 0 [/tex] then taking [tex]-e^{-i\lambda t}[/tex] outside bracket, and canceling by this factor this simplifies to
[tex] \lambda^2 y(x) + \frac{\partial}{\partial x}\left( p(x) y\'(x)\right)= 0 [/tex] ?
 
  • #4
dev00790 said:
[tex] \lambda^2 y(x) + \frac{\partial}{\partial x}\left( p(x) y\'(x)\right)= 0 [/tex] ?

Just curious. If the above linear DE can be solve for y(x) why do we need to consider that WKB ansatz (whatever it is) ?
 
  • #5
I've been using a WKB ansatz to deduce a recurrence relation in terms of [tex]Y_n[/tex] and [tex]Y_n{'}[/tex] and hence formulae for [tex]Y_n[/tex], as its the method I have been using to derive an asymptotic expansion of [tex]y(x,\lambda)[/tex] as n tends to infinity for the following equation:

[tex]y^{''}+ \lambda^2 p(x)y= 0[/tex].
This gave the following reccurence relation (with simplification): [tex]i\xi^{''}Y_{n+1}+2i\xi{'}Y'_{n+1}+Y_n{''}=0[/tex]

So I tried this method with [tex] \frac {\partial^2 u}{\partial t^2} - \frac{\partial}{\partial x}\left( p(x) \frac{\partial u}{\partial x}\right)= 0 [/tex] to see if method can still be used, if not how to change it / and or have to use a different method? - If so why?

Then hit the mentioned problem stated in my first post. Can anyone help with this please?
 
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  • #6
matematikawan said:
Just curious. If the above linear DE can be solve for y(x) why do we need to consider that WKB ansatz (whatever it is) ?

The equation d/dx[p(x)dy/dx]+λ2y(x)=0 can be transformed, by the variable changes dz=p(x)dx, w(z)=y(x(z)), to the form:

d2w/dz22p-1w=0

Now you can apply the WKB approximation which is rather accurate if p is not step-wise.

The latter equation can be also transformed in a Schroedinger-like form

u'' + Vu +λ2u = 0

and solved by the perturbation theory (see Mors, Feshbah book Methods of Theoretical Physics, V. 1). When n>>1, |λn2u|>>|Vu|, so the asymptotic solutions for un and λn are the solutions of the "non perturbed" equation:

u'' + λ2u = 0

Bob_for_short.
 
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  • #7
Bob_for_short, the following point you made;

The equation [tex]\frac{d}{dx}\left(p(x)\frac{dy}{dx}\right)+\lambda^2y(x)=0[/tex] can be transformed, by the variable changes dz=p(x)dx, w(z)=y(x(z))

but this is not the same equation as the one above: [tex]\frac{\partial}{\partial x}\left(p(x)\frac{dy}{dx}\right) + \lambda^2 y(x)= 0 [/tex]

Can one apply a transformation of variables to this equation as it involves partial differentiation? If so what would this transformation be? Thanks.
 
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  • #8
dev00790 said:
but this is not the same equation as the one above: [tex]\frac{\partial}{\partial x}\left(p(x)\frac{dy}{dx}\right) + \lambda^2 y(x)= 0 [/tex]

Can one apply a transformation of variables to this equation as it involves partial differentiation? If so what would this transformation be? Thanks.

After the time and space coordinate separation you obtain the ordinary rather than partial derivatives. Your y(x) depends only on x. So my transformation applies well.

Bob.
 
  • #9
The equation [tex]\frac{d}{dx}\left(p(x)\frac{dy}{dx}\right)+\lambda^2y(x)=0 [/tex] can be transformed, by the variable changes [tex]dz=p(x)dx,[/tex] [tex]w(z)=y(x(z)),[/tex] to the form:

[tex]\frac{d^2w}{dz^2}+\lambda^2 p^{-1}w=0[/tex]

Ok, so I try and do this transformation:

Using w(z)=y(x(z)), I get w'(z) = y'(x(z)).x'(z) = [y'(x(z))]/p(x) using [tex]\frac{dx(z)}{dx}=\frac{1}{p(x)}[/tex]

Thus w''(z) = y'(x(z)).x''(z) + x'(z).y''(x(z))

I cannot see how u arrive at [tex]\frac{d^2w}{dz^2}+\lambda^2 p^{-1}w=0[/tex] though?
Please explain this in detail. Thanks.
 
  • #10
I made a mistake, sorry. In fact dx=pdz or dz=p-1dx. Then

d/dx=z'd/dz=p-1d/dz; pd/dx=d/dz, OK?

y=w, OK?

So you directly obtain : p-1d2w/dz22w=0

or d2w/dz22pw=0

Bob.
 
  • #11
On Perturbation Theory for the Sturm-Liouville Problem with Variable Coefficients by Vladimir Kalitvianski: http://arxiv.org/abs/0906.3504. Lots of interesting results.
 

FAQ: Using WKB(J) method to obtain asymptotics for an elastic string?

1. What is the WKB(J) method?

The WKB(J) method is a mathematical technique used to obtain asymptotic solutions for differential equations, particularly in the study of wave propagation. It stands for Wentzel-Kramers-Brillouin-Jeffreys method, named after the scientists who developed it.

2. How does the WKB(J) method work?

The WKB(J) method involves finding an approximate solution to a differential equation by using a series expansion and then matching it to the exact solution. This is done by assuming the solution can be written as a product of two functions, one that varies rapidly and one that varies slowly. The rapidly varying function is then expanded in a power series and the slow function is determined by solving a simpler equation.

3. What is the application of the WKB(J) method in the study of elastic strings?

The WKB(J) method is commonly used to analyze the behavior of elastic strings, such as guitar strings or suspension bridge cables, under various conditions. It can be used to determine the resonant frequencies, amplitude of vibrations, and other characteristics of these strings.

4. What are the limitations of the WKB(J) method?

The WKB(J) method is most effective for linear equations and may not give accurate results for highly nonlinear systems. It also requires a good understanding of the equation and its boundary conditions, as well as careful matching of the approximate and exact solutions.

5. Can the WKB(J) method be used for other types of equations?

Yes, the WKB(J) method is a general technique that can be applied to a wide range of differential equations, including those in quantum mechanics and fluid dynamics. However, the specific form of the WKB(J) solution may vary depending on the equation being solved.

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