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

1. Jun 3, 2009

### dev00790

Hello,

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

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

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

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

The Problem: I wonder if this last equation can be simplified? - I'm trying to obtain an recurrence relation for the eigenfunctions $$Y_n(x)$$ 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

Last edited: Jun 3, 2009
2. Jun 3, 2009

### AiRAVATA

The sign of the second equation is wrong, it should be read $-\lambda^2 y(x)$. Appart from that, what is your question?

3. Jun 3, 2009

### dev00790

Above post is edited, including further working and my question.

$$\frac{\partial^2 u}{\partial t^2} = -\lambda^2 e^{-i\lambda t}y(x)$$

$$\frac{\partial u}{\partial x} = e^{-i\lambda t}y'(x)$$

Substituting into $$\frac {\partial^2 u}{\partial t^2} - \frac{\partial}{\partial x}\left( p(x) \frac{\partial u}{\partial x}\right)= 0$$ then taking $$-e^{-i\lambda t}$$ outside bracket, and canceling by this factor this simplifies to
$$\lambda^2 y(x) + \frac{\partial}{\partial x}\left( p(x) y\'(x)\right)= 0$$ ?

4. Jun 3, 2009

### matematikawan

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. Jun 3, 2009

### dev00790

I've been using a WKB ansatz to deduce a recurrence relation in terms of $$Y_n$$ and $$Y_n{'}$$ and hence formulae for $$Y_n$$, as its the method I have been using to derive an asymptotic expansion of $$y(x,\lambda)$$ as n tends to infinity for the following equation:

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

So I tried this method with $$\frac {\partial^2 u}{\partial t^2} - \frac{\partial}{\partial x}\left( p(x) \frac{\partial u}{\partial x}\right)= 0$$ 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?

Last edited: Jun 3, 2009
6. Jun 3, 2009

### Bob_for_short

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.

Last edited: Jun 3, 2009
7. Jun 3, 2009

### dev00790

Bob_for_short, the following point you made;

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

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

Last edited: Jun 3, 2009
8. Jun 3, 2009

### Bob_for_short

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. Jun 3, 2009

### dev00790

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 $$\frac{dx(z)}{dx}=\frac{1}{p(x)}$$

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

I cannot see how u arrive at $$\frac{d^2w}{dz^2}+\lambda^2 p^{-1}w=0$$ though?
Please explain this in detail. Thanks.

10. Jun 4, 2009

### Bob_for_short

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. Jun 18, 2009

### Bob_for_short

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