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Second Order Linear Partial Differential Equations Part IV One▯dimensional undamped wave equation; D’Alembertsolution of the wave equation; damped wave equation and the general wave equation; two▯ dimensional Laplace equation The second type of second order linear partial differential equations in 2 independent variables is the one▯dimensional wave equation. Together with the heat conduction equation, they are sometimes referred to as the “evolution equations” because their solutions “evolve”, or change, with passing time. The simplest instance of the one▯dimensional wave equation problem can be illustrated by the equation that describes the standing wave exhibited by the motion of a piece of undamped vibrating elastic string. © 2008, 2012 Zachary S Tseng E▯4 ▯ 1 Undamped One-Dimensional Wave Equation: Vibrations of an Elastic String Consider a piece of thin flexible string of length L, of negligible weight. Suppose the two ends of the string are firmly secured (“clamped”) at some supports so they will not move. Assume the set▯up has no damping. Then, the vertical displacement of the string, 0 < x < L, and at any time t > 0, is given by the displacement function u(x,t). It satisfies the homogeneous one▯ dimensional undamped wave equation: 2 a u xxu tt Where the constant coefficient a is given by the formula a = T/ρ, such that a = horizontal propagation velocity of the wave motion, T = force of tension exerted on the string, ρ = mass density (mass per unit length). It is subjected to the homogeneous boundary conditions u(0,t) = 0, and u(L,t) = 0, t > 0. The two boundary conditions reflect that the two ends of the string are clamped in fixed positions. Therefore, they are held motionless at all time. The equation comes with 2 initial conditions, due to the fact that it contains the second partial derivative term utt The two initial conditions are the initial (vertical) displacement u(x,0), and the initial (vertical) velocity ut(x,0), both are arbitrary functions of x alone. (Note that the string is merely the medium for the wave, it does not itself move horizontally, it only vibrates, vertically, in place. The resulting wave form, or the wave▯like “shape” of the string, is what moves horizontally.) © 2008, 2012 Zachary S Tseng E▯4 ▯ 2 Hence, what we have is the following initial▯boundary value problem: (Wave equation) a u = u , 0 < x < L, t > 0 , xx tt (Boundary conditions) u(0,t) = 0 , and u(L,t) = 0 , (Initial conditions) u(x,0) = f (x) , and utx,0) = g(x) . We first let u(x,t) = X(x)T(t) and separate the wave equation into two ordinary differential equations. Substituting u = X″T and u = XT″ into xx tt the wave equation, it becomes 2 a X″T = XT″. © 2008, 2012 Zachary S Tseng E▯4 ▯ 3 Dividing both sides a X T : X ′′ T ′′ = 2 X a T A2 for the heat conduction equation, it is customary to consider the constant a as a function of t and group it with the rest of t▯terms. Insert the constant of separation and break apart the equation: X ′′= T ′′=−λ X a T X ′′ =−λ → X″ = −λX → X″ + λX = 0, X T′′ 2 =−λ 2 2 a T → T″ = −a λT → T″ + a λT = 0. The boundary conditions also separate: u(0,t) = 0 → X(0)T(t) = 0 → X(0) = 0 or T(t) = 0 u(L,t) = 0 → X(L)T(t) = 0 → X(L) = 0 or T(t) = 0 As usual, in order to obtain nontrivial solutions, we need to choose X(0) = 0 and X(L) = 0 as the new boundary conditions. The result, after separation of variables, is the following simultaneous system of ordinary differential equations, with a set of boundary conditions: X″ + λX = 0, X(0) = 0 and X(L) = 0, 2 T″ + a λT = 0 . © 2008, 2012 Zachary S Tseng E▯4 ▯ 4 The next step is to solve the eigenvalue problem X″ + λX = 0, X(0) = 0, X(L) = 0. We have already solved this eigenvalue problem, recall. The solutions are n π 2 Eigenvalues: λ = 2 , n = 1, 2, 3, … L nπ x Eigenfunctions: X nsin , n = 1, 2, 3, … L Next, substitute the eigenvalues found above into the second equation to find T(t). After putting eigenvalues λ into it, the equation of T becomes 2 2 2 n π T ′+ a 2 T =0 . L It is a second order homogeneous linear equation with constant coefficients. It’s characteristic have a pair of purely imaginarycomplex conjugate roots: anπ r = ± i. L Thus, the solutions are anπt anπt T nt) = A cns + B nin , n = 1, 2, 3, … L L Multiplying each pair of n and Tntogether and sum them up, we find the general solution of the one▯dimensional wave equation, with both ends fixed, to be © 2008, 2012 Zachary S Tseng E▯4 ▯ 5 ∞  an t anπt  nπ x u x t ) =∑  An cos L +B nsin L sin L . n=1  There are two sets of (infinitely many) arbitrary coefficients. We can solve for them using the two initial conditions. Set t = 0 and apply the first initial condition, the initial (vertical) displacement of the string u(x,0) = f(x), we have ∞ nπx u(x,0) =∑ (Ancos(0)+ B nin(0) sin ) n=1 L ∞ = A sin nπ x = f (x) n=1 n L Therefore, we see that the initial displacement f(x) needs to be a Fourier sine series. Since f(x) can be an arbitrary function, this usually means that we need to expand it into its odd periodic extension (of period 2L). The coefficientsnA are then found by the relanion n = b , wnere b are the corresponding Fourier sine coefficients of fhat is L A =b = 2 f (x)sinnπ x dx n n L ∫ L . 0 Notice that the entire sequence of the coeffinients A are determined exactly by the initial displacement. They are completely independent of the other sequence of coefficients B , which are determined solely by the second n initial condition, the initial (vertical) velocityof the string. To find B , we differentiate u(x,with respect to t and apply the initial velocity, u (x,0) = g(x). t © 2008, 2012 Zachary S Tseng E▯4 ▯ 6 ∞  anπ an t an π an t  n x utx t ) =∑ − An sin + Bn cos sin n 1 L L L L  L Set t = 0 and equate it with g(x): ∞ an π n x utx ,0) ∑ Bn sin = g(x). n=1 L L We see that g(x) needs also be a Fourier sine series. Expand it into its odd periodic extension (period 2L), if necessary. Once g(x) is written into a sine series, the previous equation becomes ∞ anπ nπ x ∞ nπ x u tx,0) =∑ B n sin = g(x)= ∑ bnsin n=1 L L n=1 L Compare the coefficients of theLlike sine terms, we see anπ 2 nπ x Bn =bn= ∫g(x)sin dx . L L 0 L Therefore, L B = L b = 2 g(x)sin nπ x dx n anπ n anπ ∫ L . 0 As we have seen, half of the particular solution is determined by the initial displacement, the other half by the initial velocity. The two halves are determined independent of each other. Hence, if the initial displacement f(x) = 0, then nll A = 0 and u(x,t) contains no sine▯terms of t. If the initial velocity g(x) = 0, then all B = 0 and u(x,t) contains no cosine▯terms of t. © 2008, 2012 Zachary S Tseng E▯4 ▯ 7 Let us take another look and summarize the result for these 2 easy special cases, when either f(x) or g(x) is zero. Special case I: Nonzero initial displacement, zero initial velocity: f(x) ≠ 0, g(x) = 0. Since g(x) = 0, then B = 0 for all n. 2 L nπ x A n= ∫ f ( )sin dx, n = 1, 2, 3, … L 0 L Therefore, ∞ u( , ) = A cosanπt sinnπ x ∑ n L L . n=1 © 2008, 2012 Zachary S Tseng E▯4 ▯ 8 The D’Alembert Solution In 1746, Jean D’Alembert produced an alternate formof solution to the wave equation. His solution takes on an especiallysimple form in the above case of zero initial velocity. Use the product formula sin(A)cos(B) = [sin(A − B) + sin(A + B)]/2, the solution above can be rewritten as ∞ 1  nπ (x at) nπ (x at)  u(x,t) = 2 ∑ A nn L +sin L  n 1   Therefore, the solution of the undamped one▯dimensional wave equation with zero initial velocity can be alternatively expressed as u(x,t) = [F(x − at) + F(x + at)]/. Such that F(x) is the odd periodic extension (period 2L) of the initial displacement f(x). An interesting aspect of the D’Alembert solution is that it readily shows that the starting waveform given by the initial displacement would keep its general shape, but it would also split exactly into two halves. The two halves of the wave form travel in the opposite directions at the same finite speed of a. (Notice that the two halves of the wave form are being translated/moved in the opposite direction at the rate of distance a per unit time.) © 2008, 2012 Zachary S Tseng E▯4 ▯ 9 Special case II: Zero initial displacement, nonzero initial velocity: f(x) = 0, g(x) ≠ 0. Since f(x) = 0, thnn A = 0 for all n. L 2 nπ x B n ∫ g(x)sin dx , n = 1, 2, 3, … anπ 0 L Therefore, ∞ u( , ) = B sinanπt sin nπ x ∑ n L L . n=1 © 2008, 2012 Zachary S Tseng E▯4 ▯ 10 Example: Solve the one▯dimensional wave problem 9u xxu tt , 0 < x < 5, t > 0, u(0,t) = 0, and u(5,t) = 0, u(x,0) = 4sin(πx) − sin(2πx) − 3sin(5πx), u (x,0) = 0. t 2 First note thata = 9 (so a = 3), and L = 5. The general solution is, therefore, ∞  3nπt 3nπt  nπ x u(x,t) = ∑  A nos +B snn sin . n 1 5 5  5 Since g(x) = 0, it must be that allnB = 0. We just need to find n . We also see that u(x,) = f(x) is already in the form of a Fourier sine series. Therefore, we just need to extract the corresponding Fourier sine coefficients: A 5 b 5 4, A 10b =101, A 25 =253, A n b n 0, for all other n, n ≠ 5, 10, or 25. Hence, the particular solution is u(x,t) = 4cos(3πt)sin(πx) − cos(6πt)sin(2πx) − 3cos(15πt)sin(5πx). © 2008, 2012 Zachary S Tseng E▯4 ▯ 11 We can also solve the previous example using D’Alembert’s solution. The problem has zero initial velocity and its initial displacement has already been expanded into the required Fourier sine series, u(x,0) = 4sin(πx) − sin(2πx) − 3sin(5πx) = F(x). Therefore, the solution can also be found by using the formula u(x, t) = [F(x − at) + F(x + at)]/2, where a = 3. Thus u(x,t) = [ [ 4sin(π(x + 3t)) + 4sin(π(x − 3t)) ] − [sin(2π(x + 3t)) + sin(2π(x + 3t)) ] − [3sin(5π(x + 3t)) + 3sin(5π(x + 3t)) ] ]/2 Indeed, you could easily verify (do this as an exercise) that the solution obtained this way is identical to our previous answer. Just apply the addition formula of sine function ( sin(α ± β) = sin(α)cos(β) ± cos(α)sin(β) ) to each term in the above solution and simplify. © 2008, 2012 Zachary S Tseng E▯4 ▯ 12 Example: Solve the one▯dimensional wave problem 9u = u , 0 < x < 5, t > 0, xx tt u(0,t) = 0, and u(5,t) = 0, u(x,0) = 0, ut(x,0) = 4. 2 As in the previous exampla = 9 (so a = 3), and L = 5. Therefore, the general solution remains ∞ u(x,t) =  A cos 3nπt +B sin 3nπt  sinnπ x n 1 n 5 n 5  5 . Now, f(x) = 0, consequently anl A = 0. We just need tn find B . The initial velocity g(x) = 4 is a constant function. It is not an odd periodic function. Therefore, we need to expand it into its odd periodic extension (period T = 10), then equate it with u (x,0). In short: L 5 B = 2 g x)sin n x dx = 2 4sin n x dx n anπ ∫ L 3nπ ∫ 5 0 0  80 =  2 2 , n = odd  3n π  0 , n = even Therefore, ∞ 80 3(2n −1)πt (2n −1)π x u(x,t) =∑ sin sin n=13(2n−1) 2π 2 5 5 . © 2008, 2012 Zachary S Tseng E▯4 ▯ 13 The Structure of the Solutions of the Wave Equation In addition to the fact that the constant a is the standing wave’s propagation velocity, several other observations can be readily made from the solution of the wave equation that give insights to the nature of the solution. To reduce the clutter, let us look at the form of the solution when there is no initial velocity (when g(x) = 0). The solution is ∞ anπt nπ x u(x,t) = ∑ A nos sin n 1 L L . The sine terms are functions of x. They described the spatial wave patterns (the wavy “shape” of the string that we could visually observe), called the normal modes, or natural modes. The frequencies of those sine waves that we could see, nπ/L, are called the spatial frequencies of the wave. Meanwhile, the cosine terms are functions of t, they give the vertical displacement of the string relative to its equilibrium position (which is just the horizontal, or the x▯axis). They describe the up▯and▯down vibrating motion of the string at each point of the string. These temporal frequencies (the frequencies of functions of t; in this case, the cosines’) are the actual frequencies of oscillating motion of vertical displacement. Since this is the undamped wave equation, the frequencies of the cosine terms, anπ/L (measured in radians per second), are called the natural frequencies of the string. In a string instrument, they are the frequencies of the sound that we could hear. The corresponding natural periods (= 2π/natural frequency) are therefore 2L/an. For n = 1, the observable spatial wave pattern is sin(πx/L). It is the string’s first natural mode. The first natural frequency of oscillation, aπ/L, is called the fundamental frequency of the string. It is also called, in acoustics, as the first harmonic of the string. © 2008, 2012 Zachary S Tseng E▯4 ▯ 14 For n = 2, the spatial wave pattern is sin(2πx/L) is the second natural mode. The second natural frequency of oscillation, 2aπ/L, is also called the second harmonic (or the first overtone) of the string. It is exactly twice of the string’s fundamental frequency. Acoustically, it produces a tone that is one octave higher than the first harmonic. For n = 3, the third natural frequency is also called the third harmonic (or the second overtone), and so forth. The motion of the string is the combination of all its natural modes, as indicated by the general solution. Lastly, notice that the “wavelike” behavior of the solution of the undamped wave equation, quite unlike the solution of the heat conduction equation discussed earlier, does not decrease in amplitude/intensity with time. It never reaches a steady state. This is a consequence of the fact that the undamped wave motion is a thermodynamically reversible process that needs not obey the second law of Thermodynamics. © 2008, 2012 Zachary S Tseng E▯4 ▯ 15 First natural mode (oscillates at the fundamental frequency / 1st harmonic): Second natural mode (oscillates at the 2nd natural frequency / 2nd harmonic): Third natural mode (oscillates at the 3rd natural frequency / 3rd harmonic): © 2008, 2012 Zachary S Tseng E▯4 ▯ 16 Summary of Wave Equation: Vibrating String Problems The vertical displacement of a vibrating string of length L, securely clamped at both ends, of negligible weight and without damping, is described by the homogeneous undamped wave equation initial▯boundaryvalue problem: a u xxu tt, 0 < x < L, t > 0, u(0,t) = 0, and u(L,t) = 0, u(x,0) = f(x), and t (x,0) = g(x). The general solution is ∞  anπt anπt  nπ x u(x,t) = ∑ A nos +B snn sin . n 1  L L  L The particular solution can be found by the formulas: L A = 2 f (x)sinnπ x dx n L ∫ L , and 0 2 L nπ x B n= g(x)sin dx anπ 0 L . The solution waveform has a (horizontal) propagation velocity of a. © 2008, 2012 Zachary S Tseng E▯4 ▯ 17 The General Wave Equation The most general form of the one▯dimensional wave equation is: 2 a u +xx(x,t) = u + γu ttku. t Where a = the propagation velocity of the wave, γ = the damping constant k = (external) restoration factor, such as when vibrations occur in an elastic medium. F(x, t) = arbitrary external forcing function (If F = 0 then the equation is homogeneous, else it is nonhomogeneous.) © 2008, 2012 Zachary S Tseng E▯4 ▯ 18 The Telegraph Equation The most well▯known example of (a homogeneous version of) the general wave equation is the telegraph equation. It describes the voltage u(x, t) inside a piece of telegraph / transmission wire, whose electrical properties per unit length are: resistance R, inductance L, capacitance C, and conductance of leakage current G: 2 a u =xx + γutt ku. t 2 Where a = 1/LC, γ = G/C + R/L, and k = GR/CL. © 2008, 2012 Zachary S Tseng E▯4 ▯ 19 Example: The One▯Dimensional Damped Wave Equation 2 a u xxu +ttu t, γ≠ 0. Suppose boundary conditions remain as the same (both ends fixed): (0,t) = 0, and u(L,t) = 0. The equation can be separated as follow. First rewrite it as: a X″T = XT″ + γXT′ , 2 Divide both sides ba XT, and insert a constant of separation: X ′′= T +γT ′= −λ X a T . Rewrite it into 2 equations: X″ = −λX → X″ + λX = 0, T″ + γT′ = − a λT → T″ + γT′ + a λT = 0. The boundary conditions also are separated, as usual: u(0,t) = 0 → X(0)T(t) = 0 → X(0) = 0 or T(t) = 0 u(L,t) = 0 → X(L)T(t) = 0 → X(L) = 0 or T(t) = 0 As before, setting T(t) = 0 would result in the constant zero solution only. Therefore, we must choose the two (nontrivial) conditions in terms of x: X(0) = 0, and X(L) = 0. © 2008, 2012 Zachary S Tseng E▯4 ▯ 20 After separation of variables, we have the system X″ + λX = 0, X(0) = 0 and X(L) = 0 , 2 T″ + γT′ + α λT = 0 . The next step is to find the eigenvalues and their corresponding eigenfunctions of the boundary value problem X″ + λX = 0, X(0) = 0 and X(L) = 0. This is a familiar problem that we have encountered more than once previously. The eigenvalues and eigenfunctions are, recall, n π 2 Eigenvalues: λ = 2 ,
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