We now have all these equations that govern the dynamics of some simple continuous systems. We are in a rather simple framework, with systems that are of dimension one and that are unbound. And the equations are linear because we only considered small motions. The question now is : how do these systems move ? Let us consider the simplest one, the tensioned cable, here. The equation is so simple, with a second order derivative in time and a second order derivative in space. To make it even simpler, I can divide by the mass per unit length m and define a quantity c as sqrt(T /m). We will also use from now on a simple notation for the derivatives : a dot means d/dt and a prime means d/dx . The equation becomes yddot-c^2 y"=0 This is often called the wave equation, but it is a bit inappropriate to say this, because at this stage we don't see any waves it in. Moreover, there are plenty other equations that will give us waves! So, let us just call it the tensioned cable equation. What are the solution of this equation ? How does a tensioned cable move ? We have a very strong result which is that the solution y(x,t) will always be in the form of the sum of two functions y(x,t) = F(x-ct)+ G(x+ct). You can easily check this yourself by putting y into the equation, or by using x+ct and x-ct as new variables. This is much simpler because F and G are just funtions of one variable each. We shall call this the D'Alembert solution. D'Alembert was a French mathematician, physicist, philosopher in the 18th century. Thanks D'Alembert for that solution. A brilliant idea. Let us use this to see what happens after an initial deformation of the cable, say a shape y_0(x) here. This means that y(x,0) = y_0(x). I also need to give a condition on the velocity of the cable, for instance no velocity ydot(x,0)=0. Can I solve the equation? For this I need to find F and G Using the initial conditions I have y(x,0) which is F(x) + G(x) = y_0(x). And ydot(x,0) which is -cF'(x)+cG'(x)=0. From this you get that F(x)=G(x)=1/2 of the initial condition y_0(x). I know my functions F and G, which means that I know y for all space and time: y(x,t) =1/2 (F(x-ct) +G(x+ct)). What does this look like ? Here it is. The initial shape splits in two halves that go in opposite directions, corresponding to F and G . These two shapes stay undeformed. Why? Because they are always made with the same functions F and G and that changing time is just like changing the x coordinate. The shape at a given time t is found somewhere else at a later time. We have a wave propagation. What is the velocity of the propagation. It is the quantity c, the same in both directions. There is a special case of particular interest, that of a harmonic deformation. Imagine that the initial deformation is just sinusoidal in space, Ycos(kx). The quantity k here is called the wave number. Using the solution we just derived, we know what will happen: the displacement will be the sum of two waves. y(x,t) = (Y cosk(x_ct)+ Ycos k(x+ct)) But this is also equal to Y cos(kx) times cos(kct) This means that the initial shape oscillates. The two waves that go in opposite directions result in this oscillation. And there is a relation between the wave number k and the frequency of the oscillation omega = kc. In other terms there is a direct relation between the wave length of the sinus shape, lambda and the period of oscillation, T = lambda/c. I can also go the other way and ask: what is the shape phi(x) that oscillates at a frequency omega. By including such a displacement in the equation for the cable I have omga^2 phi(x) cos(omega t) + c^2 phi''(x) cos(omega t) = 0. The solution for phi(x) is a sinusoidal shape of wave number = omega/c. This the same oscillating solution of course, no surprise. More generally I can look for solutions y of the form of real part of e^i(k x-omega t). Using the cable equation I get omega^2- c^2 k^2 =0 This is the relation that connects the harmonic temporal variations (defined by the frequency omega) with the harmonic spatial variations (defined by the wave number k). This is called the dispersion relation D(omega,k) It is somehow the equivalent in the harmonic space (omega, k) of the original equation in the (x,t) physical space. In (x,t), the physical space, we had a dynamic equation with partial derivatives. In (omega, k), the harmonic space, we have a dispertion relation, which is quite simple. Both are equivalent The left one has the d'alembert solution with two waves a right and a left one. The right one, has two k for each omega, or two omega for each k, one positive and one negative, corresponding to the two directions Let us summarize We have very simple solutions for the tensioned cable: they are always the sum of two waves going in opposite directions. The velocity of the waves is c. The waves are undeformed. This quantity c is also the ratio between the frequencies and wave numbers, or equivalently the wavelength and the periods of oscillation. We have here what is called non-dispersive waves. There is no dispersion of the quantities. Actually, space and time are somehow equivalent in their role. They just differ by this constant, c, the wave velocity. This is not a surprise: the equation contains second order derivatives in time, and second order in space also. But we have other systems that have the same kind of equation, for fluids. What happens for them? Let us see that next.
