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Suppose we have this PDE problem $$\frac{\partial^2 \psi}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2 \psi}{\partial t^2}$$ $$\psi(0,t)=\psi(L,t)=0$$ It represents the vibrations of a string tightly stretched between two points. The standard technique for the solution is separation of variables $\psi(x,t)=T(t)y(x)$, giving the equations $y''=\frac{1}{c^2}\lambda y$ $T''=\lambda T$.

Every text I consulted assumes then that $\lambda<0$ and goes on. They say that the physics of the problem requires the solution to be a combination of sines and cosines. But is there a more rigorous mathematical way to see this? Are we losing possible solutions? What could happen if $\lambda>0$?

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Try it and see! What is the family of solutions of $y'' = \frac1{c^2}\lambda y$ for $\lambda > 0$? What happens when you apply the boundary conditions? –  Rahul Jun 13 '12 at 11:29
    
A bit more physics. This equation should satisfy energy conservation. And with $\lambda > 0$ the law of energy conservation (I think) is violated. Thus derive energy = const from this equation and show that it is violated with $\lambda > 0$. –  Yrogirg Jun 13 '12 at 11:34
    
@RahulNarain I have $y_\lambda = a_\lambda e^{\sqrt{\lambda/c}x} + b_\lambda e^{-\sqrt{\lambda/c}x}$, so introducing the boundary conditions and after a bit of algebra I obtained $1=e^{2\sqrt{\lambda/c}L}$, so $\lambda=0$. So it was only a problem of boundary conditions? –  Cauchy Jun 13 '12 at 11:44
    
Almost there. What do you get for $y(x)$? –  Rahul Jun 13 '12 at 11:48
    
@RahulNarain I think I understood. I get $y_{\lambda}(x) = 0$ for each $\lambda$ that is not a strictly positive integer, that is exactly the standard solution. Thanks –  Cauchy Jun 13 '12 at 11:53

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up vote 3 down vote accepted

It is true that the physics of the problem requires that $\lambda$ be $< 0 $, but for a more fundamental reason than the technical requirement that the answer involve trigonometeric functions.

The equation $T'' = \lambda T$ says that the acceleration of the string is proportional to the displacement. If the constant $\lambda$ were positive, it would say that the further the string were from the mean position of rest, the faster it would be accelerating, so the string would not be vibrating back and forth, but would rather be zooming away at a faster and faster rate.

The negative sign in $\lambda$ ensures that the string behave like a simple harmonic oscillator: as it is displaced from its mean position, there is a restoring force that pulls it back towards this mean position (rather than accelerating it further and further away from this position), so that it truly vibrates.

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Or with some algebra, the equation $T'' = \lambda T$ has the solution $T(t) = Aexp(\lambda t)$, now consider the solution of $y'' - \frac{\lambda}{c^2} y = 0$, it has the general form $y(x) = Bexp(\sqrt{\lambda/c^2} x)$. In the product $T(t)y(x)$, if $\lambda > 0$, then $T(t)$ would be monotonically growing, and regardless what constants we choose for the solution of $y(x)$, assuming $c^2 \ge 1$, we can not ''compensate'' this growing in the product $T(t)y(x)$, so the solution would be growing, but a string that bends infinitely long without never coming back isn't plausible. –  Stefan Jul 7 at 11:12

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