Wave equation with a slight alteration

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etotheipi
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I want to find the particular solution to the differential equation$$g(L-x) \frac{\partial^2 y}{\partial x^2} = \frac{\partial^2 y}{\partial t^2}$$with the boundary condition ##y(0) = 0## for all ##t##. If the coefficient of ##\frac{\partial^2 y}{\partial x^2}## were constant then it could be solved with ##y=A\sin{(kx -\omega t)}##, but how could I go about solving it in this case? Thanks
 
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  • #2
jasonRF
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Is ##g## a constant or a function? If I assume it is a constant, then this wave equation is applicable for an inhomogeneous media with a resonance at ##x=L##. What is the domain for this problem? Is it ##0\leq x## or is there another boundary? If there is no other boundary I'm assuming a radiation condition must apply. Also, what is the context of this problem? Finally, I see that this is listed as a "basic" thread, but in reality this problem is most suitable for advanced undergraduate or graduate students.

If you have learned about classification of partial differential equations, then you should notice that the variable coefficient changes sign at ##x=L## so the nature of the partial differential equation changes from hyperbolic for ##x<L## to elliptic for ##x>L## and therefore characteristics won't get you the solution you want.

Given that, I would either use separation of variables or simply assume harmonic time dependence. Either way you will get an ordinary differential equation with variable coefficients for the spatial variation of the solution. The solutions have different character on either side of the resonance (also called a turning point in this context) and you will need to match them. I expect there will be some subtleties here; the singularity will casue problems that are usually dealt with by adding a small amount of damping to the problem (which is partially why I asked about the context).

That should get you started, but again, this is not a basic problem.

jason
 
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  • #3
etotheipi
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Thanks for the reply! Here ##g## is just a constant, the gravitational field strength. Sorry I didn't specify! The domain is ##0\leq x \leq L## and there is no boundary condition on ##y(L)##. The context is a rope (fixed at one end) subjected to a body force of force density ##\rho \vec{g}##, with ##\vec{g}## constant, and subjected to small lateral oscillations. It is expected that various harmonics will be displayed.

I will have a look at your suggestions, specifically I'll try separating variables. I have essentially no prior background for methods to solve PDEs other than guessing, though, so I might need to go and read up about some theory beforehand!
 
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etotheipi
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How about this; the first harmonic at least looks like it could be solved with a function like$$y(x,t) = A(x)\sin{(\omega t)}$$We would have ##\frac{\partial ^2 y}{\partial t^2} = -\omega^2 A(x) \sin{(\omega t)}## and also ##\frac{\partial ^2 y}{\partial x^2} = A''(x) \sin{(\omega t)}##. That results in an ODE $$\omega^2 A(x) + g(L-x) A''(x) = 0$$
 
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jasonRF
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The restricted domain makes this a much easier problem than what I had assumed based on your original post. You are on the right track - now you need to solve this ODE. Have you learned about series solutions of ODEs?

jason
 
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  • #6
etotheipi
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The restricted domain makes this a much easier problem than what I had assumed based on your original post. You are on the right track - now you need to solve this ODE. Have you learned about series solutions of ODEs?
I can't remember exactly how to do this, but I'll try and refresh and will post however far I get up to!
 
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etotheipi
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I get these recurrence relations from inserting ##A(x) = \sum_{n=0}^{\infty} a_n x^n##:$$\omega^2 a_{n-1} + n(n+1)gLa_{n+1} -n(n-1)ga_n = 0$$ $$\omega^2 a_0 + 2gLa_2 = 0$$So I guess I just need to solve the recurrence relation... also ##a_0 = 0## because of the boundary conditions which means that ##a_2 = 0## also.
 
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jasonRF
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I won't check your algebra but it looks like you are on the right track.

By the way, for a unique solution you will need to specify initial conditions, as well as a boundary condition at ##x=L##. For example, for a freely swinging end ##\partial y(x=L,t)/\partial x = 0## is probably a reasonable choice. Your added boundary condition at ##x=L## will probably yield a set of discrete choices for ##\omega##, and it may require a summation of different modes to satisfy the initial conditions. This means that you would essentially be following the standard separation of variables technique (and you might need both cosine and sine terms). If you aren't familiar with the math a good free reference is
http://www.physics.miami.edu/~nearing/mathmethods/

If you just want to see what the different modes look like, then initial conditions might not be needed and you won't form a series to satisfy the initial conditions, but you will still need the extra boundary condition.

jason
 
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  • #9
pasmith
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There are certainly separable solutions of the form [itex]y(x,t)=X(x)\mathrm{e}^{\mathrm{i}\omega t}[/itex], where [itex]X[/itex] satisfies [tex]
g(L - x)X'' + \omega^2 X = 0.[/tex] Now the substitution [itex]x=L(1−z)[/itex] yields [tex]
z\frac{d^2X}{dz^2} + \lambda X = 0[/tex] where [itex]\lambda = (\omega^2L)/g[/itex].

Now that the singular point is at the origin, we can use Frobenius's Method directly: Seek a series of the form [itex]X(z)=\sum_{n=0}^\infty a_nz^{n+r}[/itex] where [itex]a_0 \neq 0[/itex]. This gives [tex]
a_0 r(r-1) + \sum_{n=0}^\infty \left(a_{n+1}(n + r + 1)(n + r) + \lambda a_n\right)z^{n+r} = 0[/tex] and taking [itex]r=1[/itex] yields the recurrence [tex]
a_{n + 1} = \frac{-\lambda}{(n+1)(n+2)}a_n[/tex] with solution [tex]
X(z) = a_0\sum_{n=0}^\infty \frac{(-1)^n (n+1)}{((n+1)!)^2}\lambda^n z^{n+1}.[/tex]

The linearly independent solution will have the form [tex]
CX(z) \ln z + \sum_{n=0}^\infty b_n z^n[/tex] (see https://en.wikipedia.org/wiki/Frobenius_method#Roots_separated_by_an_integer). In this case this solution is also nicely behaved at the origin as [itex]z^n \ln z \to 0[/itex].

For the special case [itex]\lambda = 0[/itex] we have of course [itex]X(z) = Az + B[/itex].
 
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  • #10
etotheipi
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Thanks @pasmith! I hadn't heard of Frobenius's method before so it was interesting to see how you handled the step with the recurrence relation. I will need to do some more reading around this area to get a little more confident with this!
 
  • #11
pasmith
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[tex]
X(z) = a_0\sum_{n=0}^\infty \frac{(-1)^n (n+1)}{((n+1)!)^2}\lambda^n z^{n+1}.[/tex]
If we set [itex]a_0 = A\lambda[/itex] then we can write this as [tex]
X(z) = AF(\lambda z)[/tex] where
[tex]
F(z) = \sum_{n=0}^\infty \frac{(-1)^n}{n!(n+1)!} z^{n+1}[/tex] satisfies [tex]
zF'' + F = 0.[/tex] This tells us that the eigenvalues [itex]\lambda[/itex] are, depending on the boundary condition at [itex]x = 0[/itex] ([itex]z = 1[/itex]), either zeroes of [itex]F[/itex] or zeroes of [itex]F'[/itex]. At [itex]x = L[/itex] ([itex]z = 0[/itex]) we have [itex]F(0) = 0[/itex] and [itex]F'(0) = 1[/itex].

The linearly independent solution will have the form [tex]
CX(z) \ln z + \sum_{n=0}^\infty b_n z^n[/tex] (see https://en.wikipedia.org/wiki/Frobenius_method#Roots_separated_by_an_integer). In this case this solution is also nicely behaved at the origin as [itex]z^n \ln z \to 0[/itex].
Note, however, that the derivative of [itex]z \ln z[/itex] does not exist at the origin and blows up as [itex]z \to 0^{+}[/itex], which suggets that you won't need it.
 
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