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```EIGENFUNCTION METHODS FOR DIFFERENTIAL EQUATIONS
17.14
Express the solution of Poisson’s equation in electrostatics,
∇2 φ(r) = −ρ(r)/0 ,
17.15
where ρ is the non-zero charge density over a ﬁnite part of space, in the form of
an integral and hence identify the Green’s function for the ∇2 operator.
In the quantum-mechanical study of the scattering of a particle by a potential,
a Born-approximation solution can be obtained in terms of a function y(r) that
satisﬁes an equation of the form
(−∇2 − K 2 )y(r) = F(r).
−3/2
Assuming that yk (r) = (2π)
exp(ik·r) is a suitably normalised eigenfunction of
−∇2 corresponding to eigenvalue k 2 , ﬁnd a suitable Green’s function GK (r, r ). By
taking the direction of the vector r − r as the polar axis for a k-space integration,
show that GK (r, r ) can be reduced to
∞
1
w sin w
dw,
4π 2 |r − r | −∞ w 2 − w02
where w0 = K|r − r |.
[ This integral can be evaluated using a contour integration (chapter 24) to give
(4π|r − r |)−1 exp(iK|r − r |). ]
17.1
17.3
17.5
17.7
17.9
17.11
17.13
17.15
Express the condition h|h ≥ 0 as a quadratic equation in λ and then apply the
for no real roots, noting that f|g + g|f is real. To put a limit on
condition
y cos2 kx dx, set f = y 1/2 cos kx and g = y 1/2 in the inequality.
Follow an argument similar to that used for proving the reality of the eigenvalues,
but integrate from x1 to x2 , rather than from a to b. Take x1 and x2 as two
successive zeros of ym (x) and note that, if the sign of ym is α then the sign of ym (x1 )
is α whilst that of ym (x2 ) is −α. Now assume that yn (x) does not change sign in
the interval and has a constant sign β; show that this leads to a contradiction
between
the signs of the two sides of the identity.
(a) y =
an Pn (x) with
1
n + 1/2
f(z)Pn (z) dz;
an =
b − n(n + 1) −1
3
(b) 5x3 =
2P3 (x)+3P1 (x), giving a1 = 1/4
and a3 = 1, leading to y = 5(2x −x)/4.
(a) No, gf ∗ dx = 0; (b) yes; (c) no, i f ∗ gdx = 0; (d) yes.
1/2
The normalised
eigenfunctions are (2/π) sin nx, with n an integer.
y(x) = (4/π) n odd [(−1)(n−1)/2 sin nx]/[n2 (κ − n2 )].
λn = (n + 1/2)2 π 2 , n = 0, 1, 2, . . . .
(a) Since yn (1)ym (1) = 0, the Sturm–Liouville boundary conditions are not satisﬁed and the appropriate weight
has to be justiﬁed by inspection. The
√ function
−x/2
2e
sin[(n
+ 1/2)πx], with ρ(x) = ex .
normalised eigenfunctions
are
−x/2
e
sin[(n
+
1/2)πx]/(n
+ 1/2)3 .
(b) y(x) √
= (−2/π 3 ) ∞
n=0
−1/2
2 2
yn (x) = 2x
sin(nπ ln x) with λn = −n π ;
√
e√
−(nπ)−2 1 2x−1 sin(nπ ln x) dx = − 8(nπ)−3 for n odd,
an =
0
for n even.
Use the form of Green’s function that is the integral over all eigenvalues of the
‘outer product’ of two eigenfunctions corresponding to the same eigenvalue, but
with arguments r and r .
576
```
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