PythagorasD'AlembertDeriving the Wave EquationLeft EndRight EndPutting it together using F=maThe General SolutionSolving for X(x)Solving for T(t)Linear CombinationsThe Final SolutionSpecific Solutions with Initial ConditionsStanding WaveComputing the CoefficientsThe Solution at Various Times for One PeriodDisplacement of the string at L/2Computing the CoefficientsThe Solution at Various Times for One PeriodA More Interesting ExampleComputing the CoefficientsThe Solution at Various Times for One Period1/26 Using Differential Equations to Modela Vibrating StringBoden HegdalMichael Moore2/26 PythagorasThe science of waves and wave motion is essential to a wide rangeof applications. In it’s simplest form, a wave is a disturbance travelingthrough some medium.In 550 B.C. the Pythagorians observed that vibrating strings pro-duced sound and studied the mathematical relationship between thefrequency of the sound and the length of the string. In the seven-tenth century, the science of wave propagation received attention fromGalileo Galilei, Robert Boyle, and Isaac Newton.3/26 D’AlembertIt was not until the Eightenth Century that French mathematicianand scientist Jean Le Rond d’Alembert derived the wave equation.Jean le Rond D’Alembert’s “Reflexions sur la cause generale des vents”(“on the general theory of the winds”) is printed in 1747. It containedthe first general use of partial differential equations in mathematicalphysics. That same year he published his theory of vibrating stringswherein he described and solved the wave equation in two dimensions.4/26 Deriving the Wave Equation(x = 0) (x = L)String of length L lays on the x axisxx + ∆xTT• We will consider the segment from x to x + ∆x• Ignore all energy losses due to stretching and bending• Ignore all external forces5/26 Left EndθTTuTx(x, u(x, t))The Tension (T ) is always tangential to the string, where θ is theangular displacement from the horizontal. The slope of the string atany point x can be described as ∂u/∂x = tan(θ). Since u(x, t) is smallcompared to L, cos(θ) ≈ 1 so tan(θ) ≈ sin(θ). Thus,Tu= −T sin θ ≈ −T tan θ = −T∂u∂x(x, t)andTx= −T cos θ ≈ −T.6/26 Right EndθTTuTx(x + ∆x, u(x + ∆x, t))The Tension (T ) is always tangential to the string, where θ is theangular displacement from the horizontal. The slope of the string atany point x can be described as ∂u/∂x = tan(θ). Since u(x, t) is smallcompared to L, cos(θ) ≈ 1 so tan(θ) ≈ sin(θ). Thus,Tu= T sin θ ≈ T tan θ = T∂u∂x(x + ∆x, t)andTx= T cos θ ≈ T.7/26 Putting it together using F = maThe total force in the horizontal direc tion is Fx= −T + T = 0so there is no horizontal acceleration. The total force in the verticaldirection isFu≈ T∂u∂x(x + ∆x, t) −∂u∂x(x, t).The vertical acceleration is ∂2u/∂t2and the mass per unit length is ρ.Putting it together using F = ma we haveT∂u∂x(x + ∆x, t) −∂u∂x(x, t)= ρ∆x∂2u∂t2Now we can divide both sides by ∆x and take the the limit as ∆xapproaches zero.ρ∂2u∂t2= T lim∆x→01∆x∂u∂x(x + ∆x, t) −∂u∂x(x, t)(1)8/26 From the definition of a derivativelim∆x→0∂u∂x(x + ∆x, t) −∂u∂x(x, t)∆x=∂2u∂x2(2)Combining (2) and (3) and letting c2= T/ρ gives us the waveequation as defined by d’Alembert.∂2u∂x2=1c2∂2u∂t2(3)Now that we have derived the wave e quation a general solution foru(x, t) must be found.9/26 The General SolutionNow that we have the wave equation a general solution for u(x, t)can be obtained easily by s eparation of variables. First we will defineu(x, t) as a product of two independent functions.u(x, t) = X(x)T (t)Substituting this into (3) gives the following.[X(x)T (t)]xx=1c2[X(x)T (t)]ttTaking the second derivative and grouping variables givesX00(x)X(x)=T00(t)c2T (t)Setting both sides equal to a constant −λ gives us a system of twosecond order ordinary differential equations.X00(x) + λX(x) = 0 and T00(t) + λc2T (t) = 0 (4)10/26 Solving for X(x)We will start with the first equation and solve for X(x). There arethree cases depending on the value of λ. These cases will be λ < 0,λ = 0 and λ > 0. By using our boundary conditions we will identifythe appropriate case for the specific solutions.X(x):Case 1, λ < 0Let λ = −ω2, ω > 0. Substituting into (4) we haveX00− ω2X = 0Now we choose an integrating factor X(x) = erxgiving usr2erx− ω2erx= 0Diving both sides by erxand solving for r gives usr = ±ω.11/26 Now that we have two independent solutions they can be written as alinear combination.X(x) = C1eωx+ C2e−ωx(5)Using the first boundary condition X(0) = 0 gives usC1= −C2Substituting into (5) and using our second boundary condition X(L) =0 gives us e2ωL= 1. This is a false statement since neither L nor ω canbe zero. This means that λ is not less than zero.X(x):Case 2, λ = 0When λ = 0, X00= 0 andX(x) = C1x + C2(6)Using the first boundary condition X(0) = 0 gives usC2= 012/26 Substituting into (6) and using our second boundary condition X(L) =0 gives usC1= 0.If both C1and C2then our string does not ever move. This is trivialand therefore λ cannot be equal to zero.X(x):Case 3, λ > 0Let λ = ω2, ω > 0. Substituting into (4) we haveX00+ ω2X = 0Now we choose an integrating factor X(x) = erxgiving usr2erx+ ω2erx= 0Diving both sides by erxand solving for r gives us r2= −ω2orr = ±ωi.This means that X(x) = e±ωi. Using Euler’s identityeωxi= [cos ωx + i sin ωx]13/26 The general solution is a linear combination of the ReX(x) and ImX(x).X(x) = C1cos ωx + C2sin ωx (7)Using the first boundary condition X(0) = 0 gives usC1= 0Substituting into (7) and using our second boundary condition X(L) =0 gives us0 = C2sin ωLwhich is only satisfie d whenω =nπL, for n = 1, 2, 3, ...Therefore our general solution for X(x) isX(x) = sinnπxL14/26 Solving for T (t)We take (4) and let ω2= λc2.T00+ ω2T = 0This equation is the same form as case 3 of the previous section re-placing X(x) with T (t). Now we can see that the general solution ofT (t) will have the same form as the solution for X(x) in equation (7)except nowω =nπcLThus a fundamental set of solutions for T00+ ω2T