23.9 Hints and answers
23.9 Hints and answers23.1 Definey(−x)=y(x) and use the cosine Fourier transform inversion theorem;
y(x)=(2/π)^1 /^2 exp(−x^2 /2).
23.3 f′′(x)−f(x)=expx;α=3/4,β=1/2,γ=1/4.
23.5 (a)φ(x)=f(x)−(1+2n)Fnxn−(1− 2 n)F−nx−n. (b) There are no solutions for
λ=[1±(1− 4 n^2 )−^1 /^2 ]−^1 unlessF±n=0orFn/F−n=∓[(1− 2 n)/(1 + 2n)]^1 /^2.
23.7 (a)a(ni)=
∫b
ahn(x)ψ(i)(x)dx; (b) use (1/√π)cosnxand (1/√π)sinnx;Mis diago-
nal; eigenvaluesλk=k/πwith eigenfunctionsψ(k)(x)=(1/√
π)coskx.
23.9 d ̃f/dω=−ω ̃f, leading tof ̃(ω)=Ae−ω
(^2) / 2
. Rearrange the integral as a convolution
and deduce that ̃h(ω)=Be−^3 ω
(^2) / 2
;h(t)=Ce−t
(^2) / 6
, where resubstitution and
Gaussian normalisation show thatC=
√
2 /(3π).
23.11 p=k 0 H/(1− 2 πk 0 ),q=k 1 Hc/(1−πk 1 )andr=k 1 Hs/(1−πk 1 ),
whereH=∫ 2 π
0 h(z)dz,Hc=∫ 2 π
0 h(z)coszdz,andHs=∫ 2 π
0 h(z)sinzdz. Positive
values ofk 1 (≈π−^1 ) are most likely to cause a conference breakdown.
23.13 For eigenvalue 0 :f(x)=0for|x|<aorf(x) is such that
∫a
−af(y)dy=0.
For eigenvalue 2a:f(x)=μS(x, a)withμaconstantandS(x, a)≡[H(a+x)−
H(x−a)], whereH(z) is the Heaviside step function.
Takef(x)=g(x)+cGS(x, a), whereG=∫a
−ag(z)dz. Show thatc=λ/(1−^2 aλ).
23.15 y(x)=x^2 −(3I 3 x+I 2 )expx.