Hints and answers

20.9 HINTS AND ANSWERS
20.25
The Klein–Gordon equation (which is satisfied by the quantum-mechanical wavefunction Φ(r) of a relativistic spinless particle of non-zero mass m) is
∇2 Φ − m2 Φ = 0.
Show that the solution for the scalar field Φ(r) in any volume V bounded by
a surface S is unique if either Dirichlet or Neumann boundary conditions are
specified on S.
20.9 Hints and answers
20.1
20.3
20.5
20.7
20.9
20.11
(a) Yes, p − 4p − 4; (b) no, (p − y)2 ; (c) yes, (p2 + 4)/(2p2 + p).
Each equation is effectively an ordinary differential equation, but with a function
of the non-integrated variable as the constant of integration;
(a) u = xy(2 − ln x); (b) u = x−1 (1 − ey ) + xey .
(a) (y 2 − x2 )1/2 ; (b) 1 + f(y 2 − x2 ), where f(0) = 0.
u = y + f(y − ln(sin x)); (a) u = ln(sin x); (b) u = y + [y − ln(sin x)]2 .
General solution is u(x, y) = f(x + y) + g(x + y/2). Show that 2p = −g (p)/2,
and hence g(p) = k − 2p2 , whilst f(p) = p2 − k, leading to u(x, y) = −x2 + y 2 /2;
u(0, 1) = 1/2.
p = x2 + 2y 2 ; u(x, y) = f(p) + x2 y 2 /2.
2
(a) u(x, y) = (x2 + 2y 2 + x2 y 2 − 2)/2; u(2, 2) = 13. The line y = 1 cuts each
characteristic in zero or two distinct points, but this causes no difficulty with
the given boundary conditions.
(b) As in (a).
(c) The solution is defined over the space between the ellipses p = 2 and p = 11;
(2, 2) lies on p = 12, and so u(2, 2) is undetermined.
√
(d) u(x, y) = (x2 + 2y 2 )1/2 + x2 y 2 /2; u(2, 2) = 8 + 12.
(e) The line y = 0 cuts each characteristic in two distinct points. No differentiable
form of f(p) gives f(±a) = ±a respectively, and so there is no solution.
(f) The solution is only specified on p = 21, and so u(2, 2) is undetermined.
(g) The solution is specified on p = 12, and so u(2, 2) = 5 + 12 (4)(4) = 13.
20.13
20.15
20.17
20.19
20.21
The equation becomes ∂2 f/∂ξ∂η = −14, with solution f(ξ, η) = f(ξ)+g(η)−14ξη,
which can be compared with the answer from the previous question; f1 (z) = 10z 2
and f2 (z) = 5z 2 .
u(x, y) = f(x + iy) + g(x − iy) + (1/12)x4 (y 2 − (1/15)x2 ). In the last term, x and
y may be interchanged. There are (infinitely) many other possibilities for the
specific PI, e.g. [ 15x2 y 2 (x2 + y 2 ) − (x6 + y 6 ) ]/360.
E = p2 /(2m), the relationship between energy and momentum for a nonrelativistic particle; u(r, t) = A exp[i(p · r − Et)/], a plane wave of wave number
k = p/ and angular frequency ω = E/ travelling in the direction p/p.
(a) c = v ± α where α2 = T /ρA;
(b) u(x, t) = a cos[k(x − vt)] cos(kαt) − (va/α)
sin[k(x − vt)] sin(kαt).
√ 1 x(CR/t)1/2
(a) V0 1 − (2/ π) 2
exp(−ν 2 ) dν ;
(b) consider the input as equivalent to V0 applied at t = 0 and continued and
−V0 applied at t = T and continued;
1 x[CR/(t−T )]1/2
2
2V0
V (x, t) = √
exp −ν 2 dν;
π 12 x(CR/t)1/2
(c) For t T , maximum at x = [2t/(CR)]1/2 with value
711
V0 T exp(− 12 )
.
(2π)1/2 t
PDES: GENERAL AND PARTICULAR SOLUTIONS
20.23
20.25
(a) Parabolic, open, Dirichlet u(x, 0) given, Neumann ∂u/∂x = 0 at x = ±L/2
for all t;
(b) elliptic, closed, Dirichlet;
(c) elliptic, closed, Neumann ∂u/∂n = σ/0 ;
(d) hyperbolic, open, Cauchy.
Follow
an argument similar to that in section 20.7 and argue that the additional
term m2 |w|2 dV must be zero, and hence that w = 0 everywhere.
712