In simulation 29 we derived the linear wave equation, ∂2y/∂x2 = 1/v2 ∂2y/∂t2 for an elastic string by considering forces acting on a small section of the string. The right side of the equation is basically the vertical acceleration of a piece of the string and the left side is the force. Constants like the tension and mass per unit length appear in the speed, v which is constant for a linear system. But what happens if some other force acts on the string, friction for example?
In this simulation you can add two additional forces, friction and a nonlinear force. The addtion of these two forces changes the wave equation to be ∂2y/∂x2 - η ∂y/∂t + α y2 = 1/v2 ∂2y/∂t2. Here η is the friction coeficent so that the friction force, - η ∂y/∂t is proportional to velocity, ∂y/∂t. The simulation actually models a chain of masses coupled by springs as in simulation 30. If the masses are of unit size (equal to one) then v is determined by the spring constant between the masses. As mentioned earlier, in the limit as the distance between masses gets infinitely small the system reduces to an elastic string and the parameter v is determined by tension and mass per length. Although this system is actually discrete it will behave the same as an elastic string but with two additional forces.
What is the physical significance of the nonlinear term, αy2? If each of our masses in the chain of masses was attached to a spring whose other end was fixed there would be an additional force of F = -αy acting on the mass where α was the spring constant. In this case we have a special, nonlinear spring that provides a force proportional to displacement squared instead of just displacement as in a regular spring. A second difference is that in a normal spring the force is negative so that the spring force is pulls the mass back toward the equilibrium for positive and negative y. in this example our additional force is positive for all values of y so we have F = +αy2. The parameter α determines how large the nonlinear force is. In the continuum limit the nonlinear force acts like an extra nonlinear spring acting at each point along the string.
Note: There are still a few bugs in the simulation; energy should be conserved as long as friction is zero. And there is an aliasing problem that causes the string to look like it is bunched up (this is visualization problem only).
Set parameters before clicking play.
31.1. With friction and nonlinearity set to zero, try changing the amplitude and linear coupling parameters for the sine and Gaussian waves. What are the effects of amplitude and coupling on each type of intial condition?
31.2. Monitor the energy for a case without friction or nonlinearity. Is energy conserved?
31.3. Now turn on a small amount of friction and run the same case you did for the previous question. What happens? What happens to the energy? A physical example of this would be an elastic string in a liquid or dense gas; the energy of the string is gradually transfered to the liquid or gas as random thermal motion.
31.4. You may wonder what makes an equation nonlinear. In simulation 29 you showed that y(x,t) = y1(x,t) + y2(x,t) was a solution to the linear wave equation as long as y1(x,t) and y2(x,t) are also solutions. Try this with the equation ∂2y/∂x2 + α y2 = 1/v2 ∂2y/∂t2. Is the sum of two solutions also a solution?
31.5. It is also the case that trigonometric functions (sine, cosine and exponential) are generally not solutions to nonlinear equations. What happens to a sine wave initial condition over time with a small amount of nonlinearity?
32.6. As further evidence that trigonometric functions are not solutions, try A exp(i(k x - ω t)) as a possible solution to the nonlinear equation by substituting it into the equation in question 31.4. Is it a solution? Explain. Is it a solution to the equation with friction but no nonlinear term?
You should have been able to arrive at this expression in the previous question: ω2 = (k2 - A exp(i(k x - ω t)))/v2. We can get an approximate dispersion relation from this expression if we use the Taylor series expansion for the exponent: exp θ ≈ (1 + θ + θ2/2! + ... ) and keep only the first term. this gives us the aproximate dispersion relation for this equation as ω = (k2 + A)1/2/v.
31.7. The group velocity of a wave packet is given by vgroup = ∂ω(k) /∂k. Find an expression for the group velocity from the approximate expression of the dispersion relation for a nonlinear string. Is there dispersion for this string?
31.8. The group velocity is also dependent on the amplitude, A. Should taller waves travel faster or slower than short waves? For a small amount of nonlinearity (say 0.2) watch the Gaussian pulse for different amplitudes. Do taller waves travel faster? (Note: The Gaussian pulse is not a solution of this nonlinear wave equation so the pulse does not maintain its exact shape as it probagates.)
32.9. Because taller waves travel faster, a collection of waves made of several different frequencies will gradually pile up as the taller waves catch up with the slower, lower amplitude waves. You may be able to see the beginning of this effect with the Gaussian pulse. Try gradually steping up the nonlinearity. What do you see? (Note: The points on the string in the simulation are constrained to only move up and down so the simulation cannot represent at true breaking wave like at the beach. It is also the case that the simulation will fail for large amplitude waves. However you should be able to see a steepening of the leading edge with nonlinearity turned on. Waves break near the beach because as the move into more shallow water they begin to interact with the floor of the ocean, adding nonlinear forces to the wave.)