Analytical Methods: Solutions 2
1. 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 ′ (ξ) + (m − 1)f (ξ) = 0.
If we are given the inital condition
x + (x2 − 1)1/2
u(x, 1) =
2
then that fixes the function
ξ + (ξ 2 − 1)1/2
f (ξ) =
2
and we determine m from the ODE: m = 1/2. Then our solution is
xt−1/2 + (x2 t−1 − 1)1/2 x/t + [(x/t)2 − t−1 ]1/2
u = t−1/2 = .
2 2
Using the method of characteristics, the characteristic curves are given by
dx
=u t=r+1 x = ur + x0 .
dt
with implicit solution
u = F (x + u − ut).
The initial condition
x + (x2 − 1)1/2 x + (x2 − 1)1/2
u(x, 1) = gives F (x) =
2 2
with implicit solution
2u = (x + u − ut) + ((x + u − ut)2 − 1)1/2
which rearranges to
√
2 x±x2 − t
−4tu + 4xu − 1 = 0 u= .
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
2. εx3 + x2 + (2 − ε)x + 1 = 0.
Scale x ∼ δ and note that the εx term is always smaller than the 2x term.
Further, if x2 ≫ 1 then x2 ≫ 2x; if x2 ≪ 1 then 2x ≪ 1 so the 2x term
only dominates if two other terms balance. We are now comparing the
following terms:
[A] εδ 3 [B] δ 2 [C] 1.
For small δ, [C] dominates. [B] catches up first at δ = 1. Then [A]
catches up with [B] when εδ 3 = δ 2 , δ = ε−1 . The distinguished scalings
are x ∼ 1 and x ∼ ε−1 .
We solve first for the regular root(s): x = x0 + εx1 + ε2 x2 + · · ·
Substiting in gives
x20 + 2x0 + 1 = 0
εx3
0 + 2εx0 x1 + 2εx1 − εx0 = 0
2 2
3ε x0 x1 + 2ε2 x0 x2 + 2ε2 x2
1 + 2ε2 x2 − ε2 x1 = 0
The leading order term gives x0 = −1. At order ε we have
−1 − 2x1 + 2x1 + 1 = 0
which is automatically satisfied. At order ε2 we obtain
3x1 − 2x2 + 2x2 + 2x2 − x1 = 0
1 2x1 (1 + x1 ) = 0
which is satisfied either by x1 = −1 or x1 = 0. In fact x = −1 is an exact
root:
εx3 + x2 + (2 − ε)x + 1 = (x + 1)(εx2 + (1 − ε)x + 1)
so no further terms are available for the other root.
Looking for the singular root, we pose: x = ε−1 x−1 + x0 + · · ·
and substituting gives
ε−2 x3
−1 + ε−2 x2
−1 = 0
3ε−1 x2 x0
−1 + 2ε−1 x−1 x0 + 2ε−1 x−1 = 0
At order ε−2 we obtain x−1 = −1 (recall the leading x term is strictly
order 1 so 0 is not a valid solution); at order ε−1 we get x0 = 2. Thus the
three roots are x = −1 − 2ε + O(ε2 ); x = −1; and x = −ε−1 + 2 + O(ε).
3. εx4 − x2 − x + 2 = 0.
First we look for scalings. One of x2 or 2 is always at least as large as x
so we only consider
[A] εδ 4 [B] δ 2 [C] 1
At small δ [C] is largest, and it is equalled first by [B] when δ = 1. For
larger δ, [A] reaches [B] when εδ 4 = δ 2 i.e. δ = ε−1/2 .
Look at the regular root(s) first (and assume a regular expansion):
x = x0 + εx1 + · · ·
gives
− x2
0 − x0 + 2 = 0
εx4
0 − 2εx0 x1 − εx1 = 0
At order 1 we have
x2 + x0 − 2 = 0
0 (x0 + 2)(x0 − 1) = 0 x0 = 1 or x0 = −2.
If x0 = 1 the next order gives 1 − 3x1 = 0 so x ∼ 1 + ε/3.
If x0 = −2 the next order gives 16 + 3x1 = 0 so x ∼ −2 − 16ε/3.
Now we move on to the singular roots and the scaling suggests an expan-
sion in ε1/2 :
x = ε−1/2 x0 + x1 + ε1/2 x2 + · · ·
Substituting gives
ε−1 x40 − ε−1 x20 = 0
4ε−1/2 x3 x1
0 − 2ε−1/2 x0 x1 − ε−1/2 x0 = 0
At leading order we have
x4 − x2 = 0
0 0 x2 (x0 + 1)(x0 − 1) = 0
0 x0 = 1 or x0 = −1.
If x0 = 1 then the next order gives 2x1 − 1 = 0 so x ∼ ε−1/2 + 1/2.
If x0 = −1 then the next order gives −2x1 + 1 = 0 so x ∼ −ε−1/2 + 1/2.
4. xe−x = ε. Define f (x) = xe−x . This function is positive for x > 0; zero
at both x = 0 and x → ∞; and f (1) = e−1 ≫ ε so we expect two roots,
one in 0 1 depends more strongly on the exponential than on the
x term, so we try a logarithmic scaling. Let us try the values of f (x) when
x = x0 ln (1/ε):
• If x0 = 1 then f (x) = ε ln (1/ε) ≫ ε.
• If x0 = 2 then f (x) = 2ε2 ln (1/ε) ≪ ε.
These two points bracket the root so we know the scaling is correct. We
begin our expansion
x = x0 ln (1/ε) + δ1 x1 + · · ·
and substitute it in, using L1 = ln (1/ε), to obtain:
ε = εx0 (x0 L1 + δ1 x1 + · · · ) exp [δ1 x1 + · · · ] = εx0 x0 L1 exp [δ1 x1 ] + · · ·
To match the powers of ε we need x0 = 0; then to make the logarithm
terms work we need
δ1 x1 = −L2 δ1 = L2 , x1 = −1.
in which we have used L2 = ln L1 .
The beginning of the expansion is
x ∼ ln (1/ε) − ln (ln (1/ε)) + · · ·
t x+c(t−t′ )
1
5. u = F (x′ , t′ ) dx′ dt′ .
2c 0 x−c(t−t′ )
The easy part is the derivatives wrt x: for ease, split the integral in two:
t x+c(t−t′ ) t x−c(t−t′ )
1 ′ ′ ′ 1′
u= F (x , t ) dx dt − F (x′ , t′ ) dx′ dt′
2c 0 0 2c 0 0
Then
t t
∂u 1
= F (x + c(t − t′ ), t′ ) dt′ − F (x − c(t − t′ ), t′ ) dt′
∂x 2c 0 0
2 t t
∂ u 1
= Fx (x + c(t − t′ ), t′ ) dt′ − Fx (x − c(t − t′ ), t′ ) dt′
∂x2 2c 0 0
Partial derivatives wrt t must be taken with more care, as both inner and
outer integrals have limits which depend on t:
x+c(t−t) x−c(t−t)
∂u 1 1
= F (x′ , t) dx′ − F (x′ , t) dx′
∂t 2c 0 2c 0
t t
1 1
+ cF (x + c(t − t′ ), t′ ) dt′ − (−c)F (x − c(t − t′ ), t′ ) dt′
2c 0 2c 0
The first two terms here cancel and we are left with
t
∂u 1
= F (x + c(t − t′ ), t′ ) + F (x − c(t − t′ ), t′ ) dt′
∂t 2 0
2
∂ u 1
= (F (x + c(t − t), t) + F (x − c(t − t), t))
∂t2 2
1 t
+ cFx (x + c(t − t′ ), t′ ) − cFx (x − c(t − t′ ), t′ ) dt′
2 0
Putting these together gives the required result:
∂2u ∂2u
2
− c2 2 = F (x, t).
∂t ∂x
6. ∂ 2 u/∂t2 − e2x ∂ 2 u/∂x2 . Here c(x) = ex so the characteristics are given by
dt
= ±ex t = ±ex + α
dx
Through the point x = 0 , t = 1, the two equations become
t = ex and t = 2 − ex .
∂2f ∂2f ∂f
7. − − ε cos xf = x with f (x, 0) = (x, 0) = 0.
∂t2 ∂x2 ∂t
Set f = f0 + εf1 + · · · .
∂ 2 f0 ∂ 2 f0
− = x
∂t2 ∂x2
2 2
∂ f1 ∂ f1
ε 2 − ε 2 − ε cos x f0 = 0
∂t ∂x
It is useful to note that the solution of the inhomogeneous wave equation
with c = 1:
∂2f ∂2f
− = F (x, t)
∂t2 ∂x2
is ′
1 t x+t−t
f (x, t) = p(x + t) + q(x − t) + F (x′ , t′ ) dx′ dt′
2 0 x−t+t′
and applying the boundary conditions f (x, 0) = ∂f /∂t(x, 0) = 0 to this
form gives
′
1 t x+t−t
f (x, t) = F (x′ , t′ ) dx′ dt′ .
2 0 x−t+t′
Order 1. The inhomogeneous wave equation
∂ 2 f0 ∂ 2 f0 ∂f0
− =x f0 (x, 0) = (x, 0) = 0
∂t2 ∂x2 ∂t
has the solution
t x+t−t′ t x+t−t′
1 ′ ′ 1 ′ 1 ′2
f0 (x, t) = x dx dt = x dt′
2 0 x−t+t′ 2 0 2 x−t+t′
t
t
= 1
x(t − t′ ) dt′ = x(tt′ − 2 t′2 ) 0
= 2 xt2 .
1
0
f0 (x, t) = 1 xt2 .
2
Order ε. We are solving
∂ 2 f1 ∂ 2 f1
− = f0 cos x = 2 x cos xt2
1
∂t2 ∂x2
with the same zero boundary conditions as before. The solution is
t x+t−t′
1
f1 (x, t) = x′ cos x′ t′2 dx′ dt′
4 0 x−t+t′
which becomes, after tedious but straightforward work,
f1 (x, t) = 2 t2 (x cos x − 2 sin x) − x cos x + 4 sin x
1
1
+ 1 (x + t) cos (x + t) + 2 (x − t) cos (x − t) − 2 sin (x + t) − 2 sin (x − t).
2