Wave_Equation

Wave_Equation - P DE LECTURE NOTES, M ATH 237A-B 185 14....

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PDE LECTURE NOTES, MATH 237A-B 185 14. Wave Equation on R n ( Ref We now consider the wave equation (14.1) u tt 4 u =0 with u (0 ,x )= f ( x ) and u t (0 ,x )= g ( x ) for x R n . According to Section 13, the solution (in the L 2 — sense) is given by (14.2) u ( t, · )=(cos( t p 4 ) f + sin( t 4 ) 4 g. To work out the results in Eq. (14.2) we must diagonalize . This is of course done using the Fourier transform. Let F denote the Fourier transform in the x variables only. Then ¨ ˆ u ( t, k )+ | k | 2 ˆ u ( t, k )=0 with ˆ u (0 ,k )= ˆ f ( k ) and ˙ ˆ u ( t, k )=ˆ g ( k ) . Therefore ˆ u ( t, k )=cos( t | k | ) ˆ f ( k )+ sin( t | k | ) | k | ˆ g ( k ) . and so u ( t, x )= F 1 · cos( t | k | ) ˆ f ( k )+ sin( t | k | ) | k | ˆ g ( k ) ¸ ( x ) , i.e. sin( t 4 ) 4 g = F 1 · sin( t | k | ) | k | ˆ g ( k ) ¸ and (14.3) cos( t p 4 ) f = F 1 h cos( t | k | ) ˆ f ( k ) i = d dt F 1 · sin( t | k | ) | k | ˆ g ( k ) ¸ . (14.4) . Our next goal is to work out these expressions in x — space alone. 14.1. n =1 Case. As we see from Eq. (14.4) it su ces to compute: sin( t 4 ) 4 g = F 1 μ sin( t | ξ | ) | ξ | ˆ g ( ξ ) = lim M →∞ F 1 μ 1 | ξ | M sin( t | ξ | ) | ξ | ˆ g ( ξ ) = lim M →∞ F 1 μ 1 | ξ | M sin( t | ξ | ) | ξ | Bg. (14.5) This inverse Fourier transform will be computed in Proposition 14.2 below using the following lemma. Lemma 14.1. Let C M denote the contour shown in Figure 38, then for λ 6 =0 we have lim M →∞ Z C M e iλξ ξ =2 πi 1 λ> 0 . Proof. First assume that λ> 0 and let Γ M denote the contour shown in Figure 38. Then ¯ ¯ ¯ ¯ ¯ ¯ Z Γ M e iλξ ξ ¯ ¯ ¯ ¯ ¯ ¯ Z π 0 ¯ ¯ ¯ e iλMe ¯ ¯ ¯ =2 π Z π 0 dθe λM sin θ 0 as M →∞ .
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186 BRUCE K. DRIVER Therefore lim M →∞ Z C M e iλξ ξ = lim M →∞ Z C M + Γ M e iλξ ξ =2 πi res ξ =0 μ e iλξ ξ =2 πi. Figure 38. A couple of contours in C . If λ< 0 , thesameargumentshows lim M →∞ Z C M e iλξ ξ = lim M →∞ Z C M + ˜ Γ M e iλξ ξ and the later integral is 0 since the integrand is holomorphic inside the contour C M + ˜ Γ M . Proposition 14.2. lim M →∞ F 1 ³ 1 | ξ | M sin( t | ξ | ) | ξ | ´ ( x )=sgn( t ) π 2 1 | x | < | t | . Proof. Let I M = 2 π F 1 μ 1 | ξ | M sin( t | ξ | ) | ξ | ( x )= Z | ξ | M sin( ) ξ e · x dξ. Then by deforming the contour we may write I M = Z C M sin ξ e · x = 1 2 i Z C M e itξ e itξ ξ e · x = 1 2 i Z C M e i ( x + t ) ξ e i ( x t ) ξ ξ By Lemma 14.1 we conclude that lim M →∞ I M = 1 2 i 2 πi (1 ( x + t ) > 0 1 ( x t ) > 0 )= π sgn( t )1 | x | < | t | .
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Wave_Equation - P DE LECTURE NOTES, M ATH 237A-B 185 14....

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