Analytical Methods: Solutions 1
1.
1
∂
1
∂
(sin θvθ f ) +
(vφ f ) + sin2 θ cos 2φ = 0 with
a sin θ ∂θ
a sin θ ∂φ
vθ = a sin θ cos θ cos 2φ,
vφ = −a sin θ sin 2φ.
First we need to substitute the v terms in and tidy up the equation:
cos 2φ ∂
∂
(sin2 θ cos θf ) −
(sin 2φf ) + sin2 θ cos 2φ = 0
sin θ ∂θ
∂φ
sin θ cos θ cos 2φ
∂f
∂f
− sin 2φ
− 3f sin2 θ cos 2φ + sin2 θ cos 2φ = 0
∂θ
∂φ
Now we look for characteristics: curves on which
sin θ cos θ cos 2φ
∂
∂
d
− sin 2φ
= g(θ, φ)
∂θ
∂φ
dr
We can decouple the equations by dividing through by cos 2φ to give the
two parametric equations
dθ
= sin θ cos θ
dr
dφ
sin 2φ
=−
dr
cos 2φ
The φ equation integrates easily:
Z
Z
2 cos 2φ
dφ = −2
dr
ln sin 2φ = −2r + C 0
sin 2φ
sin 2φ = Ce−2r .
The θ equation is a little harder:
Z
Z
Z
sin2 θ + cos2 θ
cos θ
sin θ
dr =
dθ =
+
dθ = ln sin θ − ln cos θ
sin θ cos θ
sin θ
cos θ
Our characteristic is given parametrically by
sin 2φ = Ce−2r ,
r = ln tan θ,
sin 2φ tan2 θ = C.
This curve satisfies the two equations
dθ
= sin θ cos θ
dr
dφ
sin 2φ
=−
dr
cos 2φ
and so our original PDE becomes
cos 2φ
dθ ∂f
dφ ∂f
+ cos 2φ
− 3f sin2 θ cos 2φ + sin2 θ cos 2φ = 0
dr ∂θ
dr ∂φ
df
− 3f sin2 θ + sin2 θ = 0
dr
We need to substitute sin2 θ in terms of r before solving:
tan θ = er
tan2 θ = e2r
cos2 θ =
1
(1 + e2r )
sin2 θ =
e2r
(1 + e2r )
df
− 3e2r f + e2r = 0
dr
This ODE has general solution
(1 + e2r )
f = F (C)(1 + e2r )3/2 +
1
3
Finally we need to return to the original variables θ and φ, eliminating
C and r from the solution. We already know r = ln tan θ and C =
sin 2φ tan2 θ so the final solution is
1
f = F (sin 2φ tan2 θ) sec3 θ + .
3
2.
∂u
∂u
+ u2
= 0.
∂t
∂x
This is a nonlinear first-order PDE. We look for characteristics of the form
x = x(r)
t = t(r)
along which
du
= 0.
dr
We look at the equation
dx
= u2
dt
for constant u, and see the curve family
x = u2 r + x0
t = r.
On each of these u is a constant, so u depends only on x0 and not on r:
u = F (x0 ).
We can rearrange the characteristic curve as x0 = x − u2 t and thus the
general implicit solution is
u = F (x − u2 t).
Now we want to apply the initial conditions: u(x, 0) =
p
√
x = F (x)
u = (x − u2 t).
√
x gives
The boundary condition u(0, t) = 0 is now automatically satisfied.
We can rearrange our implicit solution to make it explicit:
p
u = (x − u2 t)
r
u =x−u t
2
2
2
u (1 + t) = x
u(x, t) =
3. y 00 + 2εy 0 + (1 + ε2 )y = 1, with y(0) = 0 and y(π/2) = 0.
Put y = y0 + εy1 + ε2 y2 :
y000
εy100
ε2 y200
+ 2εy00
+ 2ε2 y10
+ y0
+ εy1
+ ε2 y2
+ ε2 y0
= 1
= 0
= 0
x
.
(1 + t)
Leading order: y000 + y = 1 gives y0 = 1 + A0 cos x + B1 sin x.
Boundary conditions: A0 = −1, B1 = −1. The leading-order solution is
y0 = 1 − cos x − sin x.
Order ε: y100 + 2y00 + y1 = 0 becomes y100 + y1 = −2 sin x + 2 cos x.
The general solution is y1 = x sin x + x cos x + A1 cos x + B1 sin x and
applying the boundary conditions gives A1 = 0 and B1 = −π/2:
y1 = (x − π/2) sin x + x cos x.
Order ε2 : y200 + 2y10 + y2 + y0 = 0 becomes
y200 + y2 = − sin x − (1 − π) cos x − 2x cos x + 2x sin x − 1.
After a little more work we obtain the general solution
y2 = (π/2)x sin x − 1 − (1/2)x2 sin x − (1/2)x2 cos x + A2 cos x + B2 sin x.
Applying the boundary conditions fixes A2 = 1 and B2 = 1 − π 2 /8, so
y2 = (π/2)x sin x−1−(1/2)x2 sin x−(1/2)x2 cos x+cos x+(1−π 2 /8) sin x.
The first three terms of the solution are
y = 1 − cos x − sin x + ε[(x − π/2) sin x + x cos x]
− ε2 [1 + (x2 /2 − πx/2 − 1 + π 2 /8) sin x + (x2 /2 − 1) cos x].
Z
ε
dx
.
2 − x2 + cos ε − cos x)1/2
(ε
0
Make a change of variables x = εz to give
Z 1
ε dz
I=
2
2
2
1/2
0 (ε − ε z + cos ε − cos (εz))
4. I =
and now expand the cosine terms, keeping terms up to order ε4 (the “1”
terms cancel):
Z 1
ε dz
I =
1 2 2
1 2
4
4 4
6 1/2
2
2
2
0 (ε − ε z − 2 ε + ε /24 − [− 2 ε z + ε z /24] + O(ε ))
Z 1
dz
=
2 − 1/2 + ε2 /24 + z 2 /2 − ε2 z 4 /24 + O(ε4 ))1/2
(1
−
z
0
√
Z 1
2 dz
=
2
2
4
4 1/2
0 (1 − z + ε (1 − z )/12 + O(ε ))
√
Z 1
2 dz
=
2
1/2
2
(1 + ε (1 + z 2 )/12 + O(ε4 ))1/2
0 (1 − z )
Now we can expand the bracket (1 + ε2 (1 + z 2 )/12 + O(ε4 ))−1/2 :
Z 1 √
2 dz
I =
(1 + ε2 (1 + z 2 )/12 + O(ε4 ))−1/2
2 )1/2
(1
−
z
0
Z 1 √
2 dz
(1 − ε2 (1 + z 2 )/24 + O(ε4 ))
=
(1
−
z 2 )1/2
0
From here to the end is just calculus: substitute z = sin θ and after some
manipulation we obtain:
π
ε2
I=√
1−
+ O(ε4 ) .
16
2
5. ut + uux = 0.
We scale s = εa t, z = εb x and v = εc u. The equation becomes
εa−c vs + εb−2c vvz = 0,
so we have a balance if c = b − a. Then since the quantities
t−b/a x
t−c/a u = t(a−b)/a u
are invariant, we can pose a solution (with b = ma)
u = t(b−a)/a f (ξ) = tm−1 f (ξ)
ξ = t−b/a x = t−m x.
The resultant ODE is
[f (ξ) − mξ]f 0 (ξ) + (m − 1)f (ξ) = 0.
If we are given the inital condition
u(x, 1) =
x + (x2 − 1)1/2
2
then that fixes the function
ξ + (ξ 2 − 1)1/2
2
and we determine m from the ODE: m = 1/2. Then our solution is
−1/2
xt
+ (x2 t−1 − 1)1/2
x/t + [(x/t)2 − t−1 ]1/2
u = t−1/2
=
.
2
2
f (ξ) =
Using the method of characteristics, the characteristic curves are given by
dx
=u
dt
with implicit solution
t=r+1
x = ur + x0 .
u = F (x + u − ut).
The initial condition
x + (x2 − 1)1/2
x + (x2 − 1)1/2
gives F (x) =
2
2
with implicit solution
u(x, 1) =
2u = (x + u − ut) + ((x + u − ut)2 − 1)1/2
which rearranges to
x±
√
x2 − t
.
2t
Taking the positive root in order to match the initial condition, we obtain
the same solution as before:
√
x + x2 − t
x/t + [(x/t)2 − t−1 ]1/2
u=
=
.
2t
2
−4tu + 4xu − 1 = 0
2
u=