To calculate a(x, y, t) and b(x, y, t), we must use the boundary conditions(A. 17)and (A. 18). To apply(A.17), we put z=0 into(A 28)to give a(r, y,t)+b(x, y, t)=f(x, y, t) (A.29) Using(A. 18)is a bit more complicated since we must compute ay/az, and z is a pa- rameter in the limits of the integral describing y. To compute the derivative we apply Leibnitz'rule for differentiatio f(x, a)dx= da )f(e(a), a) da)f((a),a)+ dx.(A.30) Using this on the integral term in(A 28)we have S(,y,s, t)dr ds=2 oz s Scx,y,s, t)dr ds. which is zero at z=0. Thus dy !(x,y,t)+-b'(x,y,t) where a'= da/at and b'= ab/at. Integration gives a(x, y, t)+b(, y, t)=c(x,y,I)dr 31) Equations(A 29)and(A 31) represent two algebraic equations in the two unknown functions a and b. The solutions are 2b(x,y, t)=f(x,y,t)+c/g(x,y,t)dr Finally, substitution of these into(A 28 ) gives us the solution to the inhomogeneous wave y(r, y ,z,t)= S(x,y,,τ)drd+ 3[f(x,,/~3 g(x, y, t)dt. This is known as the D' Alembert solution. The terms f(x, y, t z/c) contribute to y as waves propagating away from the plane z=0 in the +z-directions, respectively. The integral over the forcing term S is seen to accumulate values of s over a time interval The boundary conditions could have been applied while still in the temporal frequen domain(but not the spatial frequency domain, since the spatial position z is lost ). But to do this, we would need the boundary conditions to be in the temporal frequency domain This is easily accomplished by transforming them to give azv(y.2,O)=8(x,,0) 0 2001 by CRC Press LLC
To calculate a(x, y, t) and b(x, y, t), we must use the boundary conditions (A.17) and (A.18). To apply (A.17), we put z = 0 into (A.28) to give a(x, y, t) + b(x, y, t) = f (x, y, t). (A.29) Using (A.18) is a bit more complicated since we must compute ∂ψ/∂z, and z is a parameter in the limits of the integral describing ψ. To compute the derivative we apply Leibnitz’ rule for differentiation: d dα θ(α) φ(α) f (x,α) dx = dθ dα f (θ(α), α) − dφ dα f (φ(α), α) + θ(α) φ(α) ∂ f ∂α dx. (A.30) Using this on the integral term in (A.28) we have ∂ ∂z c 2 z 0 t+ z−ζ c t− z−ζ c S(x, y,ζ,τ) dτ dζ = c 2 z 0 ∂ ∂z t+ z−ζ c t− z−ζ c S(x, y,ζ,τ) dτ dζ, which is zero at z = 0. Thus ∂ψ ∂z z=0 = g(x, y, t) = −1 c a (x, y, t) + 1 c b (x, y, t) where a = ∂a/∂t and b = ∂b/∂t. Integration gives − a(x, y, t) + b(x, y, t) = c t −∞ g(x, y,τ) dτ. (A.31) Equations (A.29) and (A.31) represent two algebraic equations in the two unknown functions a and b. The solutions are 2a(x, y, t) = f (x, y, t) − c t −∞ g(x, y,τ) dτ, 2b(x, y, t) = f (x, y, t) + c t −∞ g(x, y,τ) dτ. Finally, substitution of these into (A.28) gives us the solution to the inhomogeneous wave equation ψ(x, y,z, t) = c 2 z 0 t+ z−ζ c t− z−ζ c S(x, y,ζ,τ) dτ dζ + 1 2 f x, y, t − z c + f x, y, t + z c + + c 2 t+ z c t− z c g(x, y,τ) dτ. (A.32) This is known as the D’Alembert solution. The terms f (x, y, t ∓ z/c) contribute to ψ as waves propagating away from the plane z = 0 in the ±z-directions, respectively. The integral over the forcing term S is seen to accumulate values of S over a time interval determined by z − ζ . The boundary conditions could have been applied while still in the temporal frequency domain (but not the spatial frequency domain, since the spatial position z is lost). But to do this, we would need the boundary conditions to be in the temporal frequency domain. This is easily accomplished by transforming them to give ψ( ˜ x, y,z,ω) z=0 = ˜f (x, y, ω), ∂ ∂z ψ( ˜ x, y,z,ω) z=0 = g˜(x, y, ω).
Applying these to(A. 25)(and again using Leibnitz rule) we have A(x, y, o)+B(x,y, o)=f(x, y, o), jkA(x, y, o)+jkB(x, y, o)=g(x, y, o), A(x, y, o)=f(x, y, o) (x 2B(x,y,o)=f(x,y,a)+c8(x,y, Finally, substituting these back into(A. 25)and expanding the sine function we obtain the frequency-domain solution that obeys the given boundary conditions ψ(x,y,z,o) c∫2「s(x,y,5,)e1=(=9)s(x,y,5,)e-( dc+ cg(x, y, o)e2z 8(x, y, o)e This is easily inverted using(A 27) to give(A. 32 Fourier transform solution of the 1-D homogeneous wave equation for dissipative media Wave propagation in dissipative media can be studied using the one-dimensional wave a2 23 a y(, y, z,t)=S(x, y, z, 1) This equation is nearly identical to the wave equation for lossless media studied in the previous section, except for the addition of the ay/at term. This extra term will lead to important physical consequences regarding the behavior of the wave solutions We shall solve(A33)using the Fourier transform approach of the previous section, but to keep the solution simple we shall only consider the homogeneous problem. We begin by writing y in terms of its inverse temporal Fourier transform y(x, y, z, t) ψ(x,y,z,o) ejor do Substituting this into the homogeneous version of(A 33 )and taking the time derivatives, we obtain 广[ jo)2+2()-n25|v(x The Fourier integral theorem leads to (A.34) 0 2001 by CRC Press LLC
Applying these to (A.25) (and again using Leibnitz’ rule) we have A˜ (x, y,ω) + B˜ (x, y,ω) = ˜f (x, y, ω), − jk A˜ (x, y,ω) + jkB˜ (x, y,ω) = g˜(x, y, ω), hence 2A˜ (x, y,ω) = ˜f (x, y,ω) − c g˜(x, y,ω) jω , 2B˜ (x, y,ω) = ˜f (x, y,ω) + c g˜(x, y,ω) jω . Finally, substituting these back into (A.25) and expanding the sine function we obtain the frequency-domain solution that obeys the given boundary conditions: ψ( ˜ x, y,z,ω) = c 2 z 0 S˜(x, y,ζ,ω)e j ω c (z−ζ) jω − S˜(x, y,ζ,ω)e− j ω c (z−ζ) jω dζ + + 1 2 ˜f (x, y,ω)e j ω c z + ˜f (x, y,ω)e− j ω c z + + c 2 g˜(x, y,ω)e j ω c z jω − g˜(x, y,ω)e− j ω c z jω . This is easily inverted using (A.27) to give (A.32). Fourier transform solution of the 1-D homogeneous wave equation for dissipative media Wave propagation in dissipative media can be studied using the one-dimensional wave equation ∂2 ∂z2 − 2 v2 ∂ ∂t − 1 v2 ∂2 ∂t 2 ψ(x, y,z, t) = S(x, y,z, t). (A.33) This equation is nearly identical to the wave equation for lossless media studied in the previous section, except for the addition of the ∂ψ/∂t term. This extra term will lead to important physical consequences regarding the behavior of the wave solutions. We shall solve (A.33) using the Fourier transform approach of the previous section, but to keep the solution simple we shall only consider the homogeneous problem. We begin by writing ψ in terms of its inverse temporal Fourier transform: ψ(x, y,z, t) = 1 2π ∞ −∞ ψ( ˜ x, y,z,ω)e jωt dω. Substituting this into the homogeneous version of (A.33) and taking the time derivatives, we obtain 1 2π ∞ −∞ (jω)2 + 2 (jω) − v2 ∂2 ∂z2 ψ( ˜ x, y,z,ω)e jωt dω = 0. The Fourier integral theorem leads to ∂2ψ( ˜ x, y,z,ω) ∂z2 − κ2 ψ( ˜ x, y,z,ω) = 0 (A.34)
where can solve the homogeneous ordinary differential equation(A.34)by inspection y(x,y, z, a)=A(x, y, o)e-k2+B(x, y, o)ek Here A and B are frequency-domain coefficients to be determined. We can either specify these coefficients directly, or solve for them by applying specific boundary conditions We examine each possibility below Solution to wave equation by direct application of boundary conditions. The solution to the wave equation(A 33)will be unique if we specify functions f(x, y, t) and g(x, y, t) such that ∫(x,y,t) g(r, y, t) Assuming the Fourier transform pairs f(r, y, t)+f(x, y, o)and g(x, y, t)+8(x, y, o) we can apply the boundary conditions(A36)in the frequency domain y(x,y,z,o)=f(x, y, co) g(r, y, o) From these we find A+B A+KB A B Substitution into(A 35) gives ψ(x,y,z,ω)=f(x,y,o) cosh xz+§(x,y,o) f(x,y,ω)-Q(x,y,z,ω)+g(x,y,o)Q(x,y,z,o) I(r, y, z, o)+y2(x, y, z, o) where Q= sinh Kz/K. Assuming that e(x, y, z, t)+ Q(x, y, z, o), we can employ the convolution theorem to immediately write down y(x, y, z, t) r. v.2 1)=f(x,y,1)*Q(x,y,x,1)+g(x,y,,D)*Q(x =ψ(x,y,z,1)+y2(x,y,z,t) (A.37) To find y we must first compute the inverse transform of o. Here we resort to a tabulated result 26 m+xP+-3+b(-)wa2-2 0 2001 by CRC Press LLC
where κ = 1 v p2 + 2 p with p = jω. We can solve the homogeneous ordinary differential equation (A.34) by inspection: ψ( ˜ x, y,z,ω) = A˜ (x, y,ω)e−κz + B˜ (x, y,ω)eκz . (A.35) Here A˜ and B˜ are frequency-domain coefficients to be determined. We can either specify these coefficients directly, or solve for them by applying specific boundary conditions. We examine each possibility below. Solution to wave equation by direct application of boundary conditions. The solution to the wave equation (A.33) will be unique if we specify functions f (x, y, t) and g(x, y, t) such that ψ(x, y,z, t) z=0 = f (x, y, t), ∂ ∂z ψ(x, y,z, t) z=0 = g(x, y, t). (A.36) Assuming the Fourier transform pairs f (x, y, t) ↔ ˜f (x, y,ω) and g(x, y, t) ↔ g˜(x, y,ω), we can apply the boundary conditions (A.36) in the frequency domain: ψ( ˜ x, y,z,ω) z=0 = ˜f (x, y, ω), ∂ ∂z ψ( ˜ x, y,z,ω) z=0 = g˜(x, y, ω). From these we find A˜ + B˜ = ˜f , −κ A˜ + κ B˜ = g˜, v or A˜ = 1 2 ˜f − g˜ κ , B˜ = 1 2 ˜f + g˜ κ . Substitution into (A.35) gives ψ( ˜ x, y,z,ω) = ˜f (x, y,ω) cosh κz + g˜(x, y,ω) sinh κz κ = ˜f (x, y,ω) ∂ ∂z Q˜ (x, y,z,ω) + g˜(x, y,ω)Q˜ (x, y,z,ω) = ψ˜ 1(x, y,z,ω) + ψ˜ 2(x, y,z,ω) where Q˜ = sinh κz/κ. Assuming that Q(x, y,z, t) ↔ Q˜ (x, y,z,ω), we can employ the convolution theorem to immediately write down ψ(x, y,z, t): ψ(x, y,z, t) = f (x, y, t) ∗ ∂ ∂z Q(x, y,z, t) + g(x, y,z, t) ∗ Q(x, y,z, t) = ψ1(x, y,z, t) + ψ2(x, y,z, t). (A.37) To find ψ we must first compute the inverse transform of Q˜ . Here we resort to a tabulated result [26]: sinh a √p + λ √p + µ √p + λ √p + µ ↔ 1 2 e− 1 2 (µ+λ)t J0 1 2 (λ − µ) a2 − t 2 , −a < t < a
Here a is a positive finite real quantity, and A and u are finite complex quantitie Outside the range [t a the time-domain function is zero Letting a=z/u, u=0, and 1=292 in the above expression, we find Q(x,D)=;h(2√a-p2)u+x/)-U0-x/0)(A here U(x)is the unit step function(A.5). From(A 37) we see that y2(r, y, z,t) g(r, y, t-t)Q(x, y, z, t)dt g(x,y,t-τ)Q(x,y,z,τ)dr Using the change of variables u=t-t and substituting(A. 38), we then have y2(x, y, z, t) g(x,y,u)euO(-=V22-(t-w)2-12)du (A.39) To find yI we must compute a0/az. Using the product rule we have a o(x, y, z, t) U-3t(32 [U(t +z/u)-U(t-z/u)]+ e-u(+x/)-U(-x/) Next, using dU(x)/dx=8(x)and remembering that Jo(x)=-J1(r) and Jo(0)=l,we can write de(, y,z, t) 1 26(+/)+8(-x 92041(B2-n) [U(t+z/v)-U(t-z/v) Convolving this expression with f(x, y, t) we obtain y1(x,y,z,1)= f(,y, t-)+sef(x,y,t+ 232e-g/f(x,y, me3V22-(-)2u2 力1 Finally, adding(A. 40)and(A 39 ), we obtain 收(xy,2D)=2gf(xy-)+ -f+2 h(v2-(-)n2 f(x, y, ue B√2-(-m2dn+ (t-l) A.41) fote that when h =0 this reduces to (,2D=(3-)+2(+)+28(yda hich matches(A. 32)for the homogeneous case where S=0 0 2001 by CRC Press LLC
Here a is a positive, finite real quantity, and λ and µ are finite complex quantities. Outside the range |t| < a the time-domain function is zero. Letting a = z/v, µ = 0, and λ = 2 in the above expression, we find Q(x, y,z, t) = v 2 e− t J0 v z2 − v2t 2 [U(t + z/v) − U(t − z/v)] (A.38) where U(x) is the unit step function (A.5). From (A.37) we see that ψ2(x, y,z, t) = ∞ −∞ g(x, y, t − τ)Q(x, y,z,τ) dτ = z/v −z/v g(x, y, t − τ)Q(x, y,z,τ) dτ. Using the change of variables u = t − τ and substituting (A.38), we then have ψ2(x, y,z, t) = v 2 e− t t+ z v t− z v g(x, y, u)e u J0 v z2 − (t − u)2v2 du. (A.39) To find ψ1 we must compute ∂ Q/∂z. Using the product rule we have ∂ Q(x, y,z, t) ∂z = v 2 e− t J0 v z2 − v2t 2 ∂ ∂z [U(t + z/v) − U(t − z/v)] + + v 2 e− t [U(t + z/v) − U(t − z/v)] ∂ ∂z J0 v z2 − v2t 2 . Next, using dU(x)/dx = δ(x) and remembering that J 0(x) = −J1(x) and J0(0) = 1, we can write ∂ Q(x, y,z, t) ∂z = 1 2 e− t [δ(t + z/v) + δ(t − z/v)] − − z 2 2v e− t J1 v √z2 − v2t 2 v √z2 − v2t 2 [U(t + z/v) − U(t − z/v)]. Convolving this expression with f (x, y, t) we obtain ψ1(x, y,z, t) = 1 2 e− v z f x, y, t − z v + 1 2 e v z f x, y, t + z v − − z 2 2v e− t t+ z v t− z v f (x, y, u)e u J1 v z2 − (t − u)2v2 v z2 − (t − u)2v2 du. (A.40) Finally, adding (A.40) and (A.39), we obtain ψ(x, y,z, t) = 1 2 e− v z f x, y, t − z v + 1 2 e v z f x, y, t + z v − − z 2 2v e− t t+ z v t− z v f (x, y, u)e u J1 v z2 − (t − u)2v2 v z2 − (t − u)2v2 du + + v 2 e− t t+ z v t− z v g(x, y, u)e u J0 v z2 − (t − u)2v2 du. (A.41) Note that when = 0 this reduces to ψ(x, y,z, t) = 1 2 f x, y, t − z v + 1 2 f x, y, t + z v + v 2 t+ z v t− z v g(x, y, u) du, which matches (A.32) for the homogeneous case where S = 0.
Solution to wave equation by specification of wave amplitudes. An alternative to direct specification of boundary conditions is specification of the amplitude functions A(x, y, a)and B(x, y, a)or their inverse transforms A(x, y, t)and B(x, y, t). If we specify the time-domain functions we can write y(x, y, z, t) as the inverse transform of (A 35) For example, a wave traveling in the +z-direction behaves as y(x,y,z,1)=A(x,y,1)*F+(x,y,z,t) F+(x,y,z,t)纱e We can find F+ using the following Fourier transform pair [26) e-÷v(p+p)2-a2 4e58(-x/)+x-p 1 Here x is real and positive and 11(x) is the modified Bessel function of the first kind and order 1. Outside the range x/u<t the time-domain function is zero. Letting p=S and σ= s we find F+(x,y,,)≈23-l1(9√P2-(x/u2) U(-z/)+e-=8(-z/).(A.4) Note that F+ is a real functions of time, as expected Substituting(A. 44) into(A42) and writing the convolution in integral form we have g2z9-1(2√r2-(x/u)2 y(x, y, z,t)=A(x, y, t-t) 2yr2-(x/1) dt+ (A.45) The 3-D Greens function for waves in dissipative media To understand the fields produced by bounded sources within a dissipative medium we ay wish to investigate solutions to the wave equation in three dimensions. The Greens function approach requires the solution to 9a1a2 n2a-pm)r;=-80)(r-r) -8(r)8(x-x')8(y-y)8(z-z) That is, we are interested in the impulse response of a point source located atr=r We begin by substituting the inverse temporal Fourier transform relations G(rr; t) G(rr; o)e/dc 8(t) btaining 2Q 0 2001 by CRC Press LLC
Solution to wave equation by specification of wave amplitudes. An alternative to direct specification of boundary conditions is specification of the amplitude functions A˜ (x, y,ω) and B˜ (x, y,ω) or their inverse transforms A(x, y, t) and B(x, y, t). If we specify the time-domain functions we can write ψ(x, y,z, t) as the inverse transform of (A.35). For example, a wave traveling in the +z-direction behaves as ψ(x, y,z, t) = A(x, y, t) ∗ F+(x, y,z, t) (A.42) where F+(x, y,z, t) ↔ e−κz = e− z v √p2+2 p. We can find F+ using the following Fourier transform pair [26]): e− x v √(p+ρ)2−σ2 ↔ e− ρ v x δ(t − x/v) + σ x v e−ρt I1 σ t 2 − (x/v)2 t 2 − (x/v)2 , x v < t. (A.43) Here x is real and positive and I1(x) is the modified Bessel function of the first kind and order 1. Outside the range x/v < t the time-domain function is zero. Letting ρ = and σ = we find F+(x, y,z, t) = 2z v e− t I1( t 2 − (z/v)2) t 2 − (z/v)2 U(t − z/v) + e− v z δ(t − z/v). (A.44) Note that F+ is a real functions of time, as expected. Substituting (A.44) into (A.42) and writing the convolution in integral form we have ψ(x, y,z, t) = ∞ z/v A(x, y, t − τ) 2z v e− τ I1( τ 2 − (z/v)2) τ 2 − (z/v)2 dτ + + e− v z A x, y, t − z v , z > 0. (A.45) The 3-D Green’s function for waves in dissipative media To understand the fields produced by bounded sources within a dissipative medium we may wish to investigate solutions to the wave equation in three dimensions. The Green’s function approach requires the solution to ∇2 − 2 v2 ∂ ∂t − 1 v2 ∂2 ∂t 2 G(r|r ;t) = −δ(t)δ(r − r ) = −δ(t)δ(x − x )δ(y − y )δ(z − z ). That is, we are interested in the impulse response of a point source located at r = r . We begin by substituting the inverse temporal Fourier transform relations G(r|r ;t) = 1 2π ∞ −∞ G˜ (r|r ; ω)e jωt dω, δ(t) = 1 2π ∞ −∞ e jωt dω, obtaining 1 2π ∞ −∞ ∇2 − jω2 v2 − 1 v2 (jω)2 G˜ (r|r ; ω) + δ(r − r ) e jωt dω = 0.