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Group 2 Bhadouria, Arjun Singh Glave, Theodore Dean Han, Zhe Chapter 5. Laplace Transform Chapter 19. Wave Equation Wave Equation Chapter 19 Overview • 19.1 – Introduction – Derivation – Examples • 19.2 – Separation of Variables / Vibrating String – 19.2.1 – Solution by Separation of Variables – 19.2.2 – Travelling Wave Interpretation • 19.3 – Separation of Variables/ Vibrating Membrane • 19.4 – Solution of wave equation – 19.4.1 – d’Alembert’s solution – 19.4.2 – Solution by integral transforms 19.1 - Introduction • Wave Equation – 𝑐2𝛻2𝑢 = 𝑢𝑡𝑡 – Uses: • • • • • Electromagnetic Waves Pulsatile blood flow Acoustic Waves in Solids Vibrating Strings Vibrating Membranes http://www.math.ubc.ca/~feldman/m267/separation.pdf Derivation u(x, t) = vertical displacement of the string from the x axis at position x and time t θ(x, t) = angle between the string and a horizontal line at position x and time t T(x, t) = tension in the string at position x and time t ρ(x) = mass density of the string at position x http://www.math.ubc.ca/~feldman/m267/separation.pdf Derivation • Forces: • Tension pulling to the right, which has a magnitude T(x+Δx, t) and acts at an angle θ(x+Δx, t) above horizontal • Tension pulling to the left, which has magnitude T(x, t) and acts at an angle θ(x, t) below horizontal • The net magnitude of the external forces acting vertically F(x, t)Δx • Mass Distribution: • 𝜌(𝑥) 𝛥𝑥 2 + 𝛥𝑢2 http://www.math.ubc.ca/~feldman/m267/separation.pdf Derivation Vertical Component of Motion Divide by Δx and taking the limit as Δx → 0. http://www.math.ubc.ca/~feldman/m267/separation.pdf Derivation http://www.math.ubc.ca/~feldman/m267/separation.pdf Derivation For small vibrations: Therefore, http://www.math.ubc.ca/~feldman/m267/separation.pdf Derivation Substitute into (2) into (1) http://www.math.ubc.ca/~feldman/m267/separation.pdf Derivation Horizontal Component of the Motion Divide by Δx and taking the limit as Δx → 0. http://www.math.ubc.ca/~feldman/m267/separation.pdf Derivation • For small vibrations: 𝑐𝑜𝑠θ~1 and 𝑑𝑇/𝑑𝑥(𝑥, 𝑡) ~ 0 Therefore, http://www.math.ubc.ca/~feldman/m267/separation.pdf Solution For a constant string density ρ, independent of x The string tension T(t) is a constant, and No external forces, F 𝑐 2 𝛻 2 𝑢 = 𝑢𝑡𝑡 𝑐 = √(𝑇/𝜌) http://www.math.ubc.ca/~feldman/m267/separation.pdf Separation of Variables; Vibrating String 19.2.1 - Solution by Separation of Variables Scenario u(x, t) = vertical displacement of a string from the x axis at position x and time t l = string length Recall: 𝑐 2 𝛻 2 𝑢 = 𝑢𝑡𝑡 (1) Boundry Conditions: u(0, t) = 0 for all t > 0 u(l, t) = 0 for all t > 0 (2) (3) Initial Conditions u(x, 0) = f(x) ut(x, 0) = g(x) for all 0 < x <l for all 0 < x <l (4) (5) http://logosfoundation.org/kursus/wave.pdf Procedure There are three steps to consider in order to solve this problem: Step 1: • Find all solutions of (1) that are of the special form 𝑢(𝑥, 𝑡) = 𝑋(𝑥)𝑇(𝑡)for some function 𝑋(𝑥) that depends on x but not t and some function 𝑇 (𝑡) that depends on t but not x. Step 2: • We impose the boundary conditions (2) and (3). Step 3: • We impose the initial conditions (4) and (5). http://logosfoundation.org/kursus/wave.pdf Step 1 – Finding Factorized Solutions 𝑢(𝑥, 𝑡) = 𝑋(𝑥)𝑇(𝑡) Let: 𝑋(𝑥)𝑇 ′′(𝑡) = 𝑐 2 𝑋′′(𝑥)𝑇 𝑋′′(𝑥) 1 𝑇′′(𝑡) (𝑡) ⇐⇒ = 2 𝑋(𝑥) 𝑐 𝑇(𝑡) Since the left hand side is independent of t the right hand side must also be independent of t. The same goes for the right hand side being independent of x. Therefore, both sides must be constant (σ). http://logosfoundation.org/kursus/wave.pdf Step 1 – Finding Factorized Solutions 𝑋 ′′ 𝑥 1 𝑇′′(𝑡) = 𝜎 2 = 𝜎 𝑋(𝑥) 𝑐 𝑇(𝑡) ⇐⇒ 𝑋 ′′ 𝑥 − 𝜎𝑋 𝑥 = 0 𝑇 ′′ 𝑡 − 𝑐 2 𝜎𝑇 𝑡 = 0 (6) http://logosfoundation.org/kursus/wave.pdf Step 1 – Finding Factorized Solutions Solve the differential equations in (6) 𝑋 𝑥 = 𝑒 𝑟𝑥 𝑇(𝑡) = 𝑒 𝑠𝑡 http://logosfoundation.org/kursus/wave.pdf Step 1 – Finding Factorized Solutions If 𝜎 ≠ 0, there are two independent solutions for (6) If 𝜎 = 0, http://logosfoundation.org/kursus/wave.pdf Step 1 – Finding Factorized Solutions Solutions to the Wave Equation For arbitrary 𝜎 ≠ 0 and arbitrary 𝑑1 , 𝑑2 , 𝑑3 , 𝑑4 For arbitrary 𝑑1 , 𝑑2 , 𝑑3 , 𝑑4 http://logosfoundation.org/kursus/wave.pdf Step 2 – Imposition of Boundaries 𝑋(0) = 𝑋(𝑙) = 0 For 𝜎 = 0 𝑑1 = 𝑑2 = 𝑋(𝑥) = 0 Thus, this solution is discarded. http://logosfoundation.org/kursus/wave.pdf Step 2 – Imposition of Boundaries 𝑋(0) = 𝑋(𝑙) = 0 For 𝜎 ≠ 0, when 𝑋 0 𝑑1 = −𝑑2 = 0 When 𝑋(𝑙) = 0 Therefore, http://logosfoundation.org/kursus/wave.pdf Step 2 – Imposition of Boundaries Since 𝜎 ≠ 0, in order to satisfy 𝑒 2√𝜎𝑙 = 1 An integer k must be introduced such that: Therefore, http://logosfoundation.org/kursus/wave.pdf Step 2 – Imposition of Boundaries Where, 𝛼𝑘 = 2𝚤𝑑1 𝑑3 + 𝑑4 and 𝛽𝑘 = −2𝑑1 (𝑑3 − 𝑑4 ) 𝑑1 , 𝑑3 , 𝑑4 are allowed to be any complex numbers 𝛼𝑘 and 𝛽𝑘 are allowed to be any complex numbers http://logosfoundation.org/kursus/wave.pdf Step 3 – Imposition of the Initial Condition From the preceding: which obeys the wave equation (1) and the boundary conditions (2) and (3), for any choice of 𝛼𝑘 and 𝛽𝑘 http://logosfoundation.org/kursus/wave.pdf Step 3 – Imposition of the Initial Condition The previous expression must also satisfy the initial conditions (4) and (5): (4’) (5’) http://logosfoundation.org/kursus/wave.pdf Step 3 – Imposition of the Initial Condition For any (reasonably smooth) function, h(x) defined on the interval 0<x<l, has a unique representation based on its Fourier Series: (7) Which can also be written as: http://logosfoundation.org/kursus/wave.pdf Step 3 – Imposition of the Initial Condition For the coefficients. We can make (7) match (4′) by choosing ℎ(𝑥) = 𝑓(𝑥) and 𝑏𝑘 = 𝛼𝑘 . Thus 𝛼𝑘 = 2 𝑙 𝑙 𝑘𝜋𝑥 𝑓(𝑥) sin 0 𝑙 𝑑𝑥. Similarly, we can make (7) match (5′) by choosing 𝑐𝑘𝜋 ℎ(𝑥) = 𝑔(𝑥) and 𝑏𝑘 = 𝛽𝑘 𝑙 Thus 𝑐𝑘𝜋 𝑙 𝛽𝑘 = 2 𝑙 𝑔(𝑥) 𝑙 0 𝑘𝜋𝑥 sin 𝑙 𝑑𝑥 http://logosfoundation.org/kursus/wave.pdf Step 3 – Imposition of the Initial Condition Therefore, (8) Where, http://logosfoundation.org/kursus/wave.pdf Step 3 – Imposition of the Initial Condition The sum (8) can be very complicated, each term, called a “mode”, is quite simple. For each fixed t, the mode 𝑘𝜋 is just a constant times sin( 𝑥) . As x runs from 0 to l, 𝑙 𝑘𝜋 the argument of sin( 𝑥) runs from 0 to 𝑘𝜋, which is k 𝑙 half–periods of sin. Here are graphs, at fixed t, of the first three modes, called the fundamental tone, the first harmonic and the second harmonic. http://logosfoundation.org/kursus/wave.pdf Step 3 – Imposition of the Initial Condition The first 3 modes at fixed t’s. http://logosfoundation.org/kursus/wave.pdf Step 3 – Imposition of the Initial Condition For each fixed x, the mode 𝑘𝑐𝜋 cos( 𝑡) 𝑙 𝑘𝑐𝜋 sin( 𝑡). 𝑘𝑐𝜋𝑙 is just a constant times plus a constant times As t 𝑘𝑐𝜋 increases by one second, the argument, 𝑡, of both cos( 𝑡) and 𝑙 𝑙 𝑘𝑐𝜋 𝑘𝑐 𝑘𝑐𝜋 sin( 𝑡) increases by 𝑙 , which is cycles (i.e. periods). So the 𝑙 2𝑙 𝑐 𝑐 fundamental oscillates at cps, the first harmonic oscillates at 2 cps, 2𝑙 2𝑙 𝑐 the second harmonic oscillates at 3 cps and so on. If the string has 2𝑙 𝑇 density ρ and tension T , then we have seen that 𝑐 = . So to 𝜌 increase the frequency of oscillation of a string you increase the tension and/or decrease the density and/or shorten the string. http://logosfoundation.org/kursus/wave.pdf Example Problem: Example Let l = 1, therefore, It is very inefficient to use the integral formulae to evaluate 𝛼𝑘 and 𝛽𝑘 . It is easier to observe directly, just by matching coefficients. Example Separation of Variables; Vibrating String 19.2.2 - Travelling Wave Interpretation Travelling Wave Start with the Transport Equation: where, u(t, x) – function c – non-zero constant (wave speed) x – spatial variable Initial Conditions http://www.math.umn.edu/~olver/pd_/pdw.pdf Travelling Wave Let x represents the position of an object in a fixed coordinate frame. The characteristic equation: Represents the object’s position relative to an observer who is uniformly moving with velocity c. Next, replace the stationary space-time coordinates (t, x) by the moving coordinates (t, ξ). http://www.math.umn.edu/~olver/pd_/pdw.pdf Travelling Wave Re-express the Transport Equation: Express the derivatives of u in terms of those of v: http://www.math.umn.edu/~olver/pd_/pdw.pdf Travelling Wave Using this coordinate system allows the conversion of a wave moving with velocity c to a stationary wave. That is, http://www.math.umn.edu/~olver/pd_/pdw.pdf Travelling Wave For simplicity, we assume that v(t, ξ) has an appropriate domain of definition, such that, Therefore, the transport equation must be a function of the characteristic variable only. http://www.math.umn.edu/~olver/pd_/pdw.pdf The Travelling Wave Interpretation http://www.math.umn.edu/~olver/pd_/pdw.pdf Travelling Wave Revisiting the transport equation, Also recall that: http://www.math.umn.edu/~olver/pd_/pdw.pdf Travelling Wave At t = 0, the wave has the initial profile • When c > 0, the wave translates to the right. • When c < 0, the wave translates to the left. • While c = 0 corresponds to a stationary wave form that remains fixed at its original location. http://www.math.umn.edu/~olver/pd_/pdw.pdf Travelling Wave As it only depends on the characteristic variable, every solution to the transport equation is constant on the characteristic lines of slope c, that is: where k is an arbitrary constant. At any given time t, the value of the solution at position x only depends on its original value on the characteristic line passing through (t, x). http://www.math.umn.edu/~olver/pd_/pdw.pdf Travelling Wave http://www.math.umn.edu/~olver/pd_/pdw.pdf 19.3 Separation of Variables Vibrating Membranes • Let us consider the motion of a stretched membrane • This is the two dimensional analog of the vibrating string problem • To solve this problem we have to make some assumptions Physical Assumptions 1. The mass of the membrane per unit area is constant. The membrane is perfectly flexible and offers no resistance to bending 2. The membrane is stretched and then fixed along its entire boundary in the xy plane. The tension per unit length T is the same at all points and does not change 3. The deflection u(x,y,t) of the membrane during the motion is small compared to the size of the membrane Vibrating Membrane Ref: Advanced Engineering Mathematics, 8th Edition, Erwin Kreyszig Derivation of differential equation We consider the forces acting on the membrane Tension T is force per unit length For a small portion ∆x, ∆y forces are approximately T∆x and T∆y Neglecting horizontal motion we have vertical components on right and left side as T ∆y sin β and -T ∆y sin α Hence resultant is T∆y(sin β – sin α) As angles are small sin can be replaced with tangents Fres = T∆y(tan β – tan α) Fres = TΔy[ux(x+ Δx,y1)-ux(x,y2)] Similarly Fres on other two sides is given by Fres = TΔx[uy(x1, y+ Δy)-uy(x2,y)] Using Newtons Second Law we get 2u xy 2 Tyu x x x, y1 u x x, y2 Tx u y x1 , y y u y x2 , y t Which gives us the wave equation: 2 2 2u u u 2 c 2 x 2 t 2 y …..(1) Vibrating Membrane: Use of double Fourier series • The two-dimensional wave equation satisfies the boundary condition (2) u = 0 for all t ≥ 0 (on the boundary of membrane) • And the two initial conditions (3) u(x,y,0) = f(x,y) (given initial displacement f(x,y) And (4) u g ( x, y) t t 0 Separation of Variables • Let u(x,y,t) = F(x,y)G(t) …..(5) • Using this in the wave equation we have F G c 2 FxxG Fyy G • Separating variables we get G 1 2 F F xx yy 2 cG F • This gives two equations: for the time function G(t) we have 2 …..(6) G G 0 And for the Amplitude function F(x,y) we have …..(7) Fxx Fyy 2 F 0 which is known as the Helmholtz equation • Separation of Helmholtz equation: F(x,y) = H(x)Q(y) • Substituting this into (7) gives d 2Q d 2H 2 Q H 2 HQ 2 dx dy …..(8) • Separating variables 1 d 2H 1 d 2Q 2 2 2 Q k 2 H dx Q dy • Giving two ODE’s 2 d H (9) k 2H 0 dx 2 And (10) d 2Q 2 p Q0 2 dy where p k 2 2 2 Satisfying boundary conditions • The general solution of (9) and (10) are H(x) = Acos(kx)+Bsin(kx) and Q(y) = Ccos(py)+Dsin(py) Using boundary condition we get H(0) = H(a) = Q(0) = Q(b) = 0 which in turn gives A = 0; k = mπ/a; C = 0; p = nπ/b m,n Ε integer • We thus obtain the solution Hm(x) = sin (mπx/a) and Qn(y) = sin(nπy/b) • Hence the functions (11)Fmn(x) = Hm(x)Qn(y) = sin(mπx/a)sin (nπy/b) Turning to time function As p2 = ν2-k2 and λ=cν we have λ = c(k2+p2)1/2 Hence λmn = cπ(m2/a2+n2/b2)1/2 Therefore …..(12) mx ny umn x, y, t Bmn cos mnt B sin mnt sin sin …(13) a b * mn Solution of the Entire Problem: Double Fourier Series …..(14) u ( x, y, t ) u mn ( x, y, t ) m 1 n 1 Bmn cos mnt B * mn m 1 n 1 mx ny sin mnt sin sin a b Using (3) mx ny u x, y,0 Bmn sin sin f ( x, y ) (15) a b m 1 n 1 • Using Fourier analysis we get the generalized Euler formula 4 mx ny f ( x, y) sin sin dxdy (16) ab 0 0 a b b a Bmn And using (4) we obtain * mn B 4 abmn mx ny 0 0 g ( x, y) sin a sin b dxdy (17) b a Example • Vibrations of a rectangular membrane Find the vibrations of a rectangular membrane of sides a = 4 ft and b = 2 ft if the Tension T is 12.5 lb/ft, the density is 2.5 slugs/ft2, the initial velocity is zero and the initial displacement is f x, y 0.1 4 x x 2 2 y y 2 ft Solution c 2 T / 12.5 / 2.5 5 [ft 2 / sec 2 ] * Bmn 0 as g(x, y) 0 4 mx ny 2 2 0.1 4 x x 2 y y sin sin dxdy 42 0 0 4 2 2 4 Bmn 0 m, n even 0.426 m, n odd 3 3 mn Which gives 1 5 ux, y, t 0.426 3 3 cos 4 m , n odd m n mx ny m 4n t sin sin 4 2 2 2 Ref: Advanced Engineering Mathematics, 8th Edition, Erwin Kreyszig 19.4 Vibrating String Solutions 19.4.1 d’Alembert’s Solution • Solution for the wave equation 2 2u u 2 c 2 t x 2 (1) can be obtained by transforming (1) by introducing independent variables v x ct , z x ct (2) • u becomes a function of v and z. • The derivatives in (1) can be expressed as derivatives with respect to v and z. u x uv v x u z z x uv u z u xx uv u z x uv u z x vx uv u z z z x uvv 2uvz u zz • We transform the other derivative in (1) similarly to get utt c uvv 2uvz u zz 2 • Inserting these two results in (1) we get 2u uvz 0 zv (3) which gives u (v ) ( z ) from (2) u ( x, t ) ( x ct ) ( x ct ) (4) • This is called the d’Alembert’s solution of the wave equation (1) D’Alembert’s solution satisfying initial conditions u x,0 f x ut x,0 g x (5) (6) u x,0 ( x) ( x) f x (7) ut x,0 c ' ( x) c ' ( x) g x (8) Dividing (8) by c and integrating we get x 1 ( x) ( x) k ( x0 ) g ( s) ds (9) c x0 where k ( x0 ) ( x0 ) ( x0 ) • Solving (9) with (7) gives x 1 1 1 ( x) f ( x) g ( s) ds k ( x0 ) 2 2c x0 2 x 1 1 1 ( x) f ( x) g ( s) ds k ( x0 ) 2 2c x0 2 • Replacing x by x+ct for φ and x by x-ct for ψ we get the solution x ct 1 1 u( x, t ) f x ct f x ct g ( s) ds 2 2c x ct 19.4.2 Solution by integral transforms Laplace Transform Semi Infinite string Find the displacement w(x,t) of an elastic string subject to: (i) The string is initially at rest on the x axis (ii) For time t>0 the left end of the string is sin t if 0 t 2 moved by w(0, t ) f (t ) 0 otherwise w( x, t ) 0 for t 0 (iii) lim x Solution 2 2w 2 w c 2 t x 2 • Wave equation: • With f as given and using initial conditions w( x , 0 ) 0 w 0 t t 0 • Taking the Laplace transform with respect to t 2 2w 2 w w 2 L 2 s Lw swx,0 c L 2 t t 0 t x 2 2 2 w st 2 w st L 2 e dt e w( x, t )dt 2 Lw( x, t ) 2 2 x x 0 x x 0 •We thus obtain 2 W 2 2 s W c x 2 2W s 2 thus 2W 0 2 x c •Which gives W ( x, s) A(s)e sx / c B(s)e sx / c • Using initial condition lim W ( x, s) lim e st w( x, t )dt e st lim w( x, t )dt 0 x x 0 0 x • This implies A(s) = 0 because c>0 so esx/c increases as x increases. • So we have W(0,s) = B(s)=F(s) • So W(x,s)=F(s)e-sx/c • Using inverse Laplace we get x x x w( x, t ) sin t if t 2 c c c and zero otherwise Travelling wave solution Ref: Advanced Engineering Mathematics, 8th Edition, Erwin Kreyszig References • H. Brezis. Functional Analysis, Sobolev Spaces and Partial Differential Equations. 1st Edition., 2011, XIV, 600 p. 9 illus. 10.3 • R. Baber. The Language of Mathematics: Utilizing Math in Practice. Appendix F • Poromechanics III - Biot Centennial (1905-2005) • http://www.math.ubc.ca/~feldman/m267/separa tion.pdf • http://logosfoundation.org/kursus/wave.pdf • http://www.math.umn.edu/~olver/pd_/pdw.pdf • Advanced engineering mathematics, 2nd edition, M. D. Greenberg • Advanced engineering mathematics, 8th edition, E. Kreyszig • Partial differential equations in Mechanics, 1st edition, A.P.S. Selvadurai • Partial differential equations, Graduate studies in mathematics, Volume 19, L. C. Evans • Advanced engineering mathematics, 2nd edition, A.C. Bajpai, L.R. Mustoe, D. Walker