# Thread: Wave Equation with Gravity

1. ## Wave Equation with Gravity

I'm to find the vertical displacement of a vibrating string not neglecting gravity, with fixed ends.

The differential equation as given in Schuam's Outline of Advanced Mathematics (Problem 12.57) is

$\displaystyle $\frac{\partial^2 y}{\partial t^2}=\alpha ^{2}\frac{\partial^2 y}{\partial x^2}-g$$

With fixed ends at $\displaystyle x=0$ and $\displaystyle x=l$, and initial velocity and displacement, $\displaystyle f(x)$ and $\displaystyle g(x)$ respectively.

I am unsure of how to solve this differential equation. I tried assuming a solution of the type $\displaystyle y(x,t)=X(x)T(t)$ and then use seperation of variables. However this method does not work as I cannot seperate the variables due to the constant $\displaystyle g$. Any suggestions on how to start this problem would be greatly appreciated.

2. You could try using the Laplace Transform in the time domain, which would give you a second-order ODE in the x domain. You then solve that using the usual methods, at which time the fun begins: computing the inverse Laplace Transform. I have no idea how difficult it would be to compute the inverse LT. The one time I did this, it was the diffusion equation (essentially the heat equation). The inverse LT was not pretty. I had to go back to its complex line integral definition, and compute residues and such. But it might be easier for you here. I don't know.

3. Thanks for your reply Ackbeet. Although I have seen Laplace transforms before, we have not used them at all in this course. I think the instructor is expecting some other way of solving this problem. I will still have a look at the Laplace transform method, anyways, in hope that it will work out.

4. Well, the nice thing about the LT method is that you can easily incorporate initial conditions. Your f(x) and g(x) plug in nicely when you take the LT.

You might also try Green's functions. I, alas, know very little about them except the main idea. But they are supposed to be useful for inhomogeneous boundary value problems.

5. I was actually thinking about it in different way. Since the given DE is just the the non-homogeneous 1D wave equation, I could solve for the homogeneuos case first. Then since the diplacement due to gravity is a function of x only, I could assume a particular solution such as $\displaystyle y_{p}(x)$. Then the general solution would be a a combination of the homogeneous solution and the particular solution. Then from there I could apply the initial and boundary conditions. Or am I thinking about this the wrong way?

6. I suppose that approach might work. The one thing you'd have to be careful of, however, is that you have the ability to plug in the boundary conditions. By the way, do you know what f and g are? Or are they arbitrary? If they're arbitrary, then I think you're going to need some sort of infinite series solution, as in a Fourier series, so that you can create the f and g from appropriate selection of the constants in the series.

Those are my thoughts.

7. Yes, $\displaystyle f(x)$ and $\displaystyle g(x)$ are arbitrary. When I use this method I get the displacement due gravity, which is the particular solution, as a parabolic function ($\displaystyle y_{p}(x)=-\frac{1}{2}\frac{g}{\alpha ^{2}}x(l-x)$. Then for the homogeneous solution I get a Fourier series, where the coefficients depend on $\displaystyle f(x)$,$\displaystyle g(x)$ and $\displaystyle y_{p}(x)$.

8. Three thoughts:

1. The homogeneous solution Fourier Series coefficients don't depend on $\displaystyle y_{p},$ by definition.
2. Your choice of a $\displaystyle y_{p}$ will be annihilated by two spatial derivatives, as well as by the two time derivatives. Therefore, it cannot satisfy the original inhomogeneous pde.
3. I think you can tweak this method of solution just a little to get it to work. You're on the right track.

9. Thanks for your input Ackbeet.

$\displaystyle y_{p}(x)=-\frac{g}{\alpha ^{2}}x(l-x)$ will satisfy the inhomogeneous equation, because

$\displaystyle \frac{\partial^2 y_{p}}{\partial x^2}=\frac{g}{\alpha ^{2}}$ and $\displaystyle \frac{\partial^2 y_{p}}{\partial t^2}=0$

Also when I apply the initial condition $\displaystyle y(x,0)=f(x)$, I get

$\displaystyle y(x,0)=y_{h}(x,0)+y_{p}(x)=f(x)$

Thus, one of the coefficients of the fourier series of $\displaystyle y_{h}(x,t)$ will be dependent on $\displaystyle y_{p}(x)$, while the other will not since $\displaystyle y_{p}(x)$ will vanish when the second initial condition ($\displaystyle y'(x,0)=g(x)$) is applied. This is what I first came up with.

But I am also thinking that the coefficients of the fourier series cannot depend on $\displaystyle y_{p}(x)$, as you said, since it has to satisfy the homogeneous equation. I will just have to look over that, but otherwise I think I have got this problem.

10. Ah, good one. I was thinking two derivatives kills a quadratic, but it doesn't. You're right, and I'm wrong.

With your Fourier series, when you take a derivative, both sets of coefficients will remain, I think. One set multiplies the sines, and the other set multiplies the cosines. Taking a derivative does not annihilate either.

I'm not sure of the resolution of your coefficients depending on the particular solution, I'm afraid. It's been a while since I've done pde's.

Good work, though. I think you're almost there.

11. I'm pretty sure I have this problem solved. I just have to go through the small details to get the final answer.

Once again, thanks for your input Ackbeet.

12. You're welcome. Have a good one!