0000001548 00000 n 0000066360 00000 n 500 1000 500 500 500 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 275 1000 666.7 666.7 888.9 888.9 0 0 555.6 555.6 666.7 500 722.2 722.2 777.8 777.8 The simplest solutions are plane waves in innite media, and we shall explore these now. (1) And we wish to solve the equation (1) given the conditions u(0,t) = u(L,t) = 0 for all t, (2) In particular, consider the initial-value problem 8 >< >: vtt c2 . endobj Consider a material in which B = H D = E J = = 0: (1) Then the Maxwell equations read >> ;&aJ7dIx!9mS@} lS6'-a0Lsz+?[50#`kA5j:zfAYOx)Iw*yjT/Ev}uhWm}43s x6S A58CUKsh%yk[hO. >> which is the 1D wave equation with solutions of propagating waves of permanent form. 255/dieresis] 593.8 500 562.5 1125 562.5 562.5 562.5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 unique. 0000049278 00000 n /FontDescriptor 9 0 R 812.5 875 562.5 1018.5 1143.5 875 312.5 562.5] /FontDescriptor 18 0 R 173/Omega/ff/fi/fl/ffi/ffl/dotlessi/dotlessj/grave/acute/caron/breve/macron/ring/cedilla/germandbls/ae/oe/oslash/AE/OE/Oslash/suppress/dieresis 756.4 705.8 763.6 708.3 708.3 708.3 708.3 708.3 649.3 649.3 472.2 472.2 472.2 472.2 1111.1 1511.1 1111.1 1511.1 1111.1 1511.1 1055.6 944.4 472.2 833.3 833.3 833.3 833.3 endobj The wave equation preserves the oddity of solutions. 0000045195 00000 n Rh%E(VxIo uD& _.XK7:;yRNcFE#U#4))%VL.1Gs(X 'L)A,n.~f|Q" /FirstChar 33 584.5 476.8 737.3 625 893.2 697.9 633.1 596.1 445.6 479.2 787.2 638.9 379.6 0 0 0 The solutions of the one wave equations will be discussed in the next section, using characteristic lines ct x = constant, ct+x = constant. /Subtype/Type1 %PDF-1.2 the construction of (unbounded) positive entire solutions for g = 2 , with certain estimates which are uniform for small parameter (0,0). 343.8 593.8 312.5 937.5 625 562.5 625 593.8 459.5 443.8 437.5 625 593.8 812.5 593.8 /Subtype/Type1 /Type/Font This paper obtains the traveling wave solution to a nonlinearly coupled wave equation. Wave equation The purpose of these lectures is to give a basic introduction to the study of linear wave equation. << 0000061245 00000 n Suppose that the function h(x,t) gives the the height of the wave at position x and time t. Then h satises the dierential equation: 2h t2 = c2 2h x2 (1) where c is the speed that . 907.4 999.5 951.6 736.1 833.3 781.2 0 0 946 804.5 698 652 566.2 523.3 571.8 644 590.3 /FirstChar 33 tions2; roughly speaking, the solutions of differential equations are themselves functions, while the solutions of normal algebraic equations are points within the domain of some equation-dependent function. 0000033856 00000 n Solution of the Wave Equation by Separation of Variables The Problem Let u(x,t) denote the vertical displacement of a string from the x axis at position x and time t. The string has length . In other words, a small change in either or results in a correspondingly small change in the solution . /BaseFont/ZZSIXJ+CMMI6 0000027337 00000 n A variety of ocean waves follow this 0000063293 00000 n )2J/srV;RvC) )F /#H@I%4,5eux . >> Substitution in equation (2.5c) shows that is a solution. /Type/Font /BaseFont/JBOAVI+MSBM10 0000068218 00000 n 0 0 0 0 0 0 0 0 0 0 777.8 277.8 777.8 500 777.8 500 777.8 777.8 777.8 777.8 0 0 777.8 x[n8}}SGS0LX 0}Pm;\~HQZNf^,x/:Ug>>{tF5atjVJ_SyI)yfYii\~Y/Y?V[vAo~zcGc3JI-xMvn\Lq S[G)?%/Cvfy*_mr3vi-Z#kf& Lh! 820.5 796.1 695.6 816.7 847.5 605.6 544.6 625.8 612.8 987.8 713.3 668.3 724.7 666.7 /Subtype/Type1 /BaseFont/FJDYGU+CMSY10 0000046578 00000 n << Solution of the One Dimensional Wave Equation The general solution of this equation can be written in the form of two independent variables, = V bt +x (10) = V bt x (11) By using these variables, the displacement, u, of the material is not only a function of time, t, and position, x; but also wave velocity, V b. /Name/F4 2 So the function is a solution to the wave equation. h]nR""lV'HT~( SR%dmt2{FxHM[.+/;5&LyB2 @sar7+x@2>_ vy7;"q(+;&N Y `W@iX-i W4oF. t c ux 0 For the well known wave equation utt the famous d'Alembert solution leads to c2 u xx 0 x)y= 0 solve the wave equation. Just as in the case of the wave equation, we argue from the inverse by assuming that there are two functions, u, and v, that both solve the inhomogeneous heat equation and satisfy the initial and Dirichlet boundary conditions of (4). /Widths[777.8 777.8 777.8 777.8 777.8 777.8 777.8 777.8 777.8 777.8 777.8 777.8 777.8 255/dieresis] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 642.9 885.4 806.2 736.8 35 0 obj /Subtype/Type1 777.8 777.8 777.8 777.8 777.8 777.8 777.8 777.8 777.8 777.8 777.8 777.8 777.8 777.8 The solution we were able to nd was u(x;t) := X1 n=1 g n cos n L ct + L nc h n sin n L ct sin n L x ; (2) by assuming the following sine Fourier series expansion of the initial data gand h: X1 n=1 g n sin n L x ; X1 n=1 h n sin n L cx : In order to prove that the function uabove is the solution of our problem, we cannot dif . /BaseFont/STCYUK+CMEX10 ~_|'iSQ"5TbIeHA`'eR&INLHS8r`+/1WCLgmhP'0grijrA"DhhG? /Subtype/Type1 ()MacCormack Method (1969) Predictor step : n+1 n n() j j j+1 t u=u-c u x n uj Correct step : 1111() 1 1 2 nnn nn jjj jj ct uuu . 43 0 obj Solution: The frequency fe n of the normal mode u n (x;t), given in (13), is fe n = 1 2 p c2 n k2 = 1 2 r c2n22 l2 k2 = cn 2l s 1 lk cn 2 = f n s 1 lk cn 2 (14) As the damping (k > 0) increases, the frequencies of the normal modes decrease. In addition, our argument works equally well for semilinear damped wave equations, when the coecient of the dissipation term is integrable (without sign condition) and space-independent . 687.5 312.5 581 312.5 562.5 312.5 312.5 546.9 625 500 625 513.3 343.8 562.5 625 312.5 depends . 531.3 826.4 826.4 826.4 826.4 0 0 826.4 826.4 826.4 1062.5 531.3 531.3 826.4 826.4 Moreover, any function y = f2(ct + x) will be a solution so that, generally, their superposition y = f1(ct x) + f2(ct + x) is the complete solution. 0000039327 00000 n 324.7 531.3 590.3 295.1 324.7 560.8 295.1 885.4 590.3 531.3 590.3 560.8 414.1 419.1 The general equation of wave motion is given by: y = Asin (kx - t) where A = amplitude, = angular frequency, k = angular wavenumber, t = time, and x is position. WATERWAVES 5 Wavetype Cause Period Velocity Sound Sealife,ships 10 110 5s 1.52km/s Capillaryripples Wind <101s 0.2-0.5m/s Gravitywaves Wind 1-25s 2-40m/s Sieches Earthquakes,storms minutestohours standingwaves Its left and right hand ends are held xed at height zero and we are told its initial . 826.4 826.4 826.4 826.4 826.4 826.4 826.4 826.4 826.4 826.4 1062.5 1062.5 826.4 826.4 589.1 483.8 427.7 555.4 505 556.5 425.2 527.8 579.5 613.4 636.6 272] 761.6 489.6 516.9 734 743.9 700.5 813 724.8 633.9 772.4 811.3 431.9 541.2 833 666.2 obeys the wave equation (1) and the boundary conditions (2 . 777.8 777.8 777.8 500 277.8 222.2 388.9 611.1 722.2 611.1 722.2 777.8 777.8 777.8 300 325 500 500 500 500 500 814.8 450 525 700 700 500 863.4 963.4 750 250 500] /Type/Font endobj More generally, using the fact that the wave equation is linear, we see that any nite linear combination of the functions un will also give us a solution of the wave equation on [0;l] satisfying our Dirichlet boundary conditions. 0000058123 00000 n Key Mathematics: More Fourier transform theory, especially as applied to solving the wave equation. We shall discuss the basic properties of solutions to the wave equation (1.2), as well as its multidimensional and non-linear variants. analytical solutions to the wave equation. 1062.5 826.4] 0000061940 00000 n I]nkuUU1^eM,6 /FirstChar 33 0000042001 00000 n 0000023978 00000 n /Widths[342.6 581 937.5 562.5 937.5 875 312.5 437.5 437.5 562.5 875 312.5 375 312.5 1000 1000 1055.6 1055.6 1055.6 777.8 666.7 666.7 450 450 450 450 777.8 777.8 0 0 13 0 obj Recall that the solution to the wave IVP on the whole line (u tt 2cu xx= 0; u(x;0 . 10 0 obj 0000041688 00000 n /Encoding 7 0 R 1444.4 555.6 1000 1444.4 472.2 472.2 527.8 527.8 527.8 527.8 666.7 666.7 1000 1000 This is the rst clue that wave solutions to the Schrodinger equation will not be identical to solutions of the classical wave equation. << *dGCfVaZ. 34 0 obj endobj << 0000046355 00000 n If t stands for time and ~x= fx igrepresent the observation point, such solutions of the wave equation, (@2 @t2 c2 o r 2)= 0; (1) 295.1 826.4 531.3 826.4 531.3 559.7 795.8 801.4 757.3 871.7 778.7 672.4 827.9 872.8 %PDF-1.5 /Name/F2 The wave equation is linear: The principle of "Superposition" holds. We call these travelling wave solutions and we can interpret these two functions as left and right . 761.6 679.6 652.8 734 707.2 761.6 707.2 761.6 0 0 707.2 571.2 544 544 816 816 272 D'Alembert's Formula - Initial Displacement Consider the wave equation . 0000063914 00000 n 295.1 826.4 501.7 501.7 826.4 795.8 752.1 767.4 811.1 722.6 693.1 833.5 795.8 382.6 666.7 722.2 722.2 1000 722.2 722.2 666.7 1888.9 2333.3 1888.9 2333.3 0 555.6 638.9 ?? Bf)h | CB2M%K*ZEJtzDMTUi%U6eQii65QmmH3D{9{5 _P6Jh/ gjq^%H;I: X_w);&FFi;Gzalx|[FDA\(i!:a'lOD7 fFG7=}o 2At,M-&pX8K1]M 761.6 272 489.6] 0000001603 00000 n 460.7 580.4 896 722.6 1020.4 843.3 806.2 673.6 835.7 800.2 646.2 618.6 718.8 618.8 Let 1141 A(x) Dx,x J(x)dx0 Then 221141A(x) Dx,x J x dx because 2acts only on xnot x1. endobj Solution. 6) Solve the heat equation ft = f xx on [0,] with the initial condition f(x,0) = |sin(3x)|. /Subtype/Type1 Suppose we have the wave equation u tt = a2u xx. 777.8 777.8 1000 500 500 777.8 777.8 777.8 777.8 777.8 777.8 777.8 777.8 777.8 777.8 /BaseFont/ZACWDP+CMMI8 Hb```f``sf`c`g`@ ;$AO=,Wx>333eE8f1Uo``9Pcn^u "[L}RFK{z2LSDIt-]5sWe9'/2Ov6.J'Z$ *wi Im=2"O/LHf6*XtrO//KsrGL0l59tT_\nWU\B;''sE=X]y+L0YGe466HkcYRu^iBw-]e^.w< a superposition)ofthe normal modes for the given boundary conditions. The . 0000046152 00000 n n\=bD12%F^Oy2#*1FppLZc"JAD; 545.5 825.4 663.6 972.9 795.8 826.4 722.6 826.4 781.6 590.3 767.4 795.8 795.8 1091 In particular the solution to the wave equation (14.1) for n=3is given by . >> Solution of the Wave Equation The solution of the wave equation will, of course, be a function of the variables x and t.We are going to show that any function of the form y= f1(ct - x) is a solution. 726.9 726.9 976.9 726.9 726.9 600 300 500 300 500 300 300 500 450 450 500 450 300 << endobj 0000003069 00000 n /FontDescriptor 30 0 R /LastChar 196 0000059410 00000 n The probability to /Type/Font We would like to summarize the properties of the obtained solutions, and compare the propagation of waves to conduction of heat. wave from the wave equation is invalid. (b) What is the expression for phase velocity of this wave and what issue is there with this expression? The derivatives of , , and now become: Figure 2.6b Spherical coordinates. << It means that light beams can pass through each other without altering each other. /\ /FontDescriptor 40 0 R In other words, we write u(x,y,t)= m=1 n=1 Cmn . The explicit solution (3) implies that for arbitrary continuous f;g;hone has a Prove that light obeys the wave equation . Moreover, the fact that there is a unique (up to a multiplicative constant) >> is a solution of the wave equation on the interval [0;l] which satises un(0;t) = 0 = un(l;t). << 31 0 obj 694.5 295.1] The solution is and . 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 458.3 458.3 416.7 416.7 iiHj(2rq+ `bU f14daf76q8+@ f,! /FirstChar 0 xY[o~`Bg"vCPhc}P$:&mo.Hi,;[8$9\_tL:c!LkKXQdWC[`"x)~ur?_O|ZVW#Rm,)yfMio1}& 0000067683 00000 n The wave equation governs a wide range of phenomena, including gravitational waves, light waves, sound waves, and even the oscillations of strings in string theory.Depending on the medium and type of wave, the velocity v v v can mean many different things, e.g. Edwards and Penney have a typo in the d'Alembert solution (equations (37) and (39) on page 639 in section 9.6). /Name/F11 Example 7.2. : Adding the three derivatives, we get. Summarizing these results, we have. /Name/F7 d'Alembert solution of the wave equation. 1277.8 811.1 811.1 875 875 666.7 666.7 666.7 666.7 666.7 666.7 888.9 888.9 888.9 656.3 625 625 937.5 937.5 312.5 343.8 562.5 562.5 562.5 562.5 562.5 849.5 500 574.1 7 0 obj 666.7 666.7 666.7 666.7 611.1 611.1 444.4 444.4 444.4 444.4 500 500 388.9 388.9 277.8 If the initial data . The 2D wave equation Separation of variables Superposition Examples Remarks: For the derivation of the wave equation from Newton's second law, see exercise 3.2.8. 720.1 807.4 730.7 1264.5 869.1 841.6 743.3 867.7 906.9 643.4 586.3 662.8 656.2 1054.6 CONCEPT:. 379.6 638.9 638.9 638.9 638.9 638.9 638.9 638.9 638.9 638.9 638.9 638.9 638.9 379.6 Hence, if u(x,t) is a solution, so is u(x,t). 675.9 1067.1 879.6 844.9 768.5 844.9 839.1 625 782.4 864.6 849.5 1162 849.5 849.5 734 761.6 666.2 761.6 720.6 544 707.2 734 734 1006 734 734 598.4 272 489.6 272 489.6 We now calculate the derivatives , etc. Question 1. Two-Dimensional Wave Equation The solution of the wave equation in two dimensions can be obtained by solving the three-dimensional wave equation in the case where the initial data depends only on xand y, but not z. /Type/Encoding 833.3 1444.4 1277.8 555.6 1111.1 1111.1 1111.1 1111.1 1111.1 944.4 1277.8 555.6 1000 << The wave equation in spherical coordinates is given in problem 2.6b. % 0000044674 00000 n 795.8 795.8 649.3 295.1 531.3 295.1 531.3 295.1 295.1 531.3 590.3 472.2 590.3 472.2 I. FT Change of Notation In the last lecture we introduced the FT of a function f (x) through the two equations () f x = f k . /Type/Font /Subtype/Type1 /Type/Font stream In general it can happen that the solution is di erentiable (u2C1), but that the derivatives v= u x and w= u t are not weak solutions of the above equations (6) and (5). 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 742.6 1027.8 934.1 859.3 Problems 2-4 belong together: 7) Verify that for any constants a,b the function h(x,t) = (ba)x/ +a is a solution to the heat . /Widths[791.7 583.3 583.3 638.9 638.9 638.9 638.9 805.6 805.6 805.6 805.6 1277.8 /Name/F8 41 0 obj Let d 1. In the previous section when we looked at the heat equation he had a number of boundary conditions however in this case we are only going to consider one type . Cauchy problem for the wave equation is . 444.4 611.1 777.8 777.8 777.8 777.8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Show that there is at most one solution to the Dirichlet problem (4). Wavelength (): Distance between two nearest particles vibrating in the same phase. /FontDescriptor 15 0 R SOLUTION OF THE WAVE EQUATION We need the solution of the equation 2 AJ0 We solve this by introducing the idea of a Green's function. 413.2 590.3 560.8 767.4 560.8 560.8 472.2 531.3 1062.5 531.3 531.3 531.3 0 0 0 0 << 0000011895 00000 n /Widths[272 489.6 816 489.6 816 761.6 272 380.8 380.8 489.6 761.6 272 326.4 272 489.6 One can see from the d'Alembert formula (see also the picture above) that the solution To see this, note that changing x into x leaves equation (92) unchanged, as does turning u into u. 334 405.1 509.3 291.7 856.5 584.5 470.7 491.4 434.1 441.3 461.2 353.6 557.3 473.4 and through a similar process as Equation (5.3), we can show that that uPI = uPI(x + ct). 0000062674 00000 n 0000034838 00000 n 16 0 obj 0000002831 00000 n It arises in different elds such as acoustics, electromagnetics, or uid dynamics. 611.1 777.8 777.8 388.9 500 777.8 666.7 944.4 722.2 777.8 611.1 777.8 722.2 555.6 In this method, a canonical form of the wave equation (3) is rst obtained using a suitable transformation. >> 0000024552 00000 n /Name/F1 At t= 0 the disturbance of the stringisgivenbyf(x) + g(x);astimeprogresses,twowaveformsmoveapart. /Widths[300 500 800 755.2 800 750 300 400 400 500 750 300 350 300 500 500 500 500 /LastChar 196 /Encoding 7 0 R Overview Wavesandvibrationsinmechanicalsystemsconstituteoneofthe As in the one dimensional situation, the constant c has the units of velocity. well posed. 3 0 obj 1001.4 726.4 837.7 509.3 509.3 509.3 1222.2 1222.2 518.5 674.9 547.7 559.1 642.5 /FirstChar 33 0000063707 00000 n 544 516.8 380.8 386.2 380.8 544 516.8 707.2 516.8 516.8 435.2 489.6 979.2 489.6 489.6 /FontDescriptor 27 0 R Writing , we proceed as in part (a). If f 1 (x,t) and f 2 (x,t) are solutions to the wave equation, then . We rst consider the IVP u tt= c2r2u; (x;y) 2R2; t>0; The Wave Equation In this chapter we investigate the wave equation (5.1) u tt u= 0 and the nonhomogeneous wave equation (5.2) u tt u= f(x;t) subject to appropriate initial and boundary conditions. /Type/Font /FirstChar 33 652.8 598 0 0 757.6 622.8 552.8 507.9 433.7 395.4 427.7 483.1 456.3 346.1 563.7 571.2 777.8 1000 1000 1000 1000 1000 1000 777.8 777.8 555.6 722.2 666.7 722.2 722.2 666.7 /LastChar 196 xZKsW[[eMuq%L 3.3 Particle Flux and Schr odinger Equation The solution of the Schr odinger equation is the wave function (~r;t) which describes the state of a particle moving in the potential U(~r;t). We can use an odd re ection to extend the initial condition, g odd(x) = 8 >< >: 1 x>0 0 x= 0 1 x<0; h odd(x) = 0: The particular solution to the extended PDE is u(x;t) = g odd(x+ 2t) + g odd(x 2t) 2: We now examine the cases depending on the sign . The one-dimensional wave equation Separation of variables The two-dimensional wave equation Rectangular membrane (continued) Since the wave equation is linear, the solution u can be written as a linear combination (i.e. 0000026832 00000 n We now have several constraints on the wave function : 1) It must obey Schrdinger's equation. These so called characteristic curves are the sound waves Starting with the right-hand side, we ignore for the time being and obtain. :TaAP|rj8\ALAc-8l3'1 ;Dt%j`.@"63=Qu8yK@+ZLsTvv00h`a`:` }%&1p6h2,g@74B63t^=LY,.,' u endstream endobj 154 0 obj 1140 endobj 97 0 obj << /Type /Page /Parent 91 0 R /Resources 98 0 R /Contents [ 113 0 R 133 0 R 138 0 R 140 0 R 142 0 R 147 0 R 149 0 R 151 0 R ] /MediaBox [ 0 0 612 792 ] /CropBox [ 0 0 612 792 ] /Rotate 0 >> endobj 98 0 obj << /ProcSet [ /PDF /Text ] /Font << /TT2 108 0 R /TT3 116 0 R /TT4 100 0 R /TT6 105 0 R /TT7 103 0 R /TT8 128 0 R /TT10 131 0 R /TT11 122 0 R /TT12 124 0 R /TT13 134 0 R /TT14 143 0 R >> /ExtGState << /GS1 152 0 R >> /ColorSpace << /Cs5 109 0 R >> >> endobj 99 0 obj << /Filter /FlateDecode /Length 8461 /Length1 12024 >> stream In its simp lest form, the wave . There are one way wave equations, and the general solution to the two way equation could be done by forming linear combinations of such solutions. HHC}*8jOnwM`t/')U ]Sg#A)B|U) Fs0!@.41E2Ak/STQ1*\WHaECP \wB$)_])_5}Ey'u;|MQ. The canonical form enable us to easily integrate the equation to obtain the general solution. If the data (f;g;h) are merely continuous then the solution uneed not be di eren-tiable. The Wave Equation Maxwell equations in terms of potentials in Lorenz gauge Both are wave equations with known source distribution f(x,t): If there are no boundaries, solution by Fourier transform and the Green function method is best. 875 531.3 531.3 875 849.5 799.8 812.5 862.3 738.4 707.2 884.3 879.6 419 581 880.8 In this work, the extended homogeneous balance method is used to derive exact solutions of nonlinear evolution equations. ryrN9y9KS,jQpt=K /Widths[1000 500 500 1000 1000 1000 777.8 1000 1000 611.1 611.1 1000 1000 1000 777.8 /LastChar 196 324.7 531.3 531.3 531.3 531.3 531.3 795.8 472.2 531.3 767.4 826.4 531.3 958.7 1076.8 /FontDescriptor 37 0 R bZ,|eRTX%1-rw* (iXjWo&x5+ k@v7(OEnd9.>F~D-TgiG7( ObBg(frB%9xs%huZ4i=I\Um&y5 )b@auZ5 ]SHM?GWY,f7 / ,h3 Time Period (T): Time taken by the wave to travel a distance . 489.6 489.6 489.6 489.6 489.6 489.6 489.6 489.6 489.6 489.6 272 272 272 761.6 462.4 0000059043 00000 n /Length 2922 4 Linear Surface Gravity Waves C, Dispersion, Group Velocity, and Energy Propagation30 4.1 Group Velocity . 379.6 963 638.9 963 638.9 658.7 924.1 926.6 883.7 998.3 899.8 775 952.9 999.5 547.7 DdUs[; Part A: Uniqueness of solution for one dimensional wave equation with nite length Theorem: The solution of the following problem, if it exists, is unique. endobj /LastChar 196 Remarks: The solution consists of the superposition of two traveling waves with speed c, but moving in opposite directions. 466.4 725.7 736.1 750 621.5 571.8 726.7 639 716.5 582.1 689.8 742.1 767.4 819.4 379.6] /Type/Font 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 663.6 885.4 826.4 736.8 450 500 300 300 450 250 800 550 500 500 450 412.5 400 325 525 450 650 450 475 400 Denoting the solutions for are ()x L un x t Fn x Gn t Bn nt Bn nt n ( , ) = ( ) ( ) = cos + *sin sin Solutions for the 1D Wave Equation are: As a result of solving for F, we have restricted These functions are the eigenfunctions of the vibrating string, and the values are called the eigenvalues. 0000027035 00000 n /Subtype/Type1 We will see this again when we examine conserved quantities (energy or wave action) in wave systems. We now consider the wave equation (14.1) utt4u=0with u(0,x)=f(x) and ut(0,x)=g(x) for xRn. >> 0000058356 00000 n 492.9 510.4 505.6 612.3 361.7 429.7 553.2 317.1 939.8 644.7 513.5 534.8 474.4 479.5 0000003344 00000 n 295.1 531.3 531.3 531.3 531.3 531.3 531.3 531.3 531.3 531.3 531.3 531.3 531.3 295.1 We shall discover that solutions to the wave equation behave quite di erently from solu- We want to solve the wave equation on the half line with Dirichlet boundary conditions. /Type/Encoding 0000045601 00000 n /Name/F10 /Widths[295.1 531.3 885.4 531.3 885.4 826.4 295.1 413.2 413.2 531.3 826.4 295.1 354.2 the nonlinear wave equation is the linear wave equation utt c2uxx = 0 under the assumption that the speed of sound cdepends on the solution uas well. 500 500 611.1 500 277.8 833.3 750 833.3 416.7 666.7 666.7 777.8 777.8 444.4 444.4 A href= '' https: //uclnatsci.github.io/Electromagnetism-Fluids-and-Waves/waves/waveequationsolns.html '' > 5 ( 14.1 ) for given Respect to and, the constant c has the units of velocity at which string displacements propagate c2u =. 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