Download 1 M2AA1 Diffferential Equations: Problem Sheet 4 1. Consider a 2

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M2AA1 Diffferential Equations: Problem Sheet 4
1. Consider a 2 × 2 matrix A. Let λ1 , λ2 be its eigenvalues. Let φt (p) the solution of
x0 = Ax and x(0) = p. (Called the flow). Assume that λ1 < 0 < λ2 . Show that
W s (0) = {p ∈ R2 ; φt (p) → 0 as t → ∞} and W u (0) = {p ∈ R2 ; |φt (p)| → 0 as t →
−∞} are both lines through 0. Draw R2 with the curve t 7→ φt (p) for a few choices
of p (these curves are called orbits). What you see is usually called a saddle-point.
Answer: Since the eigenvalues are distinct, there exists a matrix T so that Λ :=
T −1 AT is diagonal. Indeed, take T so that its first column is the eigenvector v1
associated to Λ1 and the second column the 2nd eigenvector v2 . Then Λ := T −1 AT =
λ1 0
. It follows that for Λ the set {x; eΛt x → 0 as t →→ ∞} corresponds
0 λ2
to the horizontal axis and {x; eΛt x → 0 as t →→ −∞} to the vertical axis. Since
eAt = T eΛt T −1 the set W s (0) =< v1 > and W u (0) =< v2 >.
2. Consider a n×n matrix A and let λ1 , . . . , λn be its eigenvalues. Let φt (p) the solution
of x0 = Ax and x(0) = p. (Called the flow). Assume that Re(λi ) < 0 for i = 1, . . . , k
and Re(λi ) > 0 for i = k + 1, . . . , n. Show that the set W s (0) = {p ∈ Rn ; φt (p) →
0 as t → ∞} is a k-dimensional linear space through 0.
Answer: Take T so that T −1 AT takes the complex Jordan form J, see the end of
Chapter 3. If all eigenvalues are different J is diagonal and then you can easily
compute eJ and the result follows as in question 1. Next consider Jj as in the Jordan
form and assume that it is a k × k matrix. Then

 tλj
e
tetλj (t2 /2)etλj . . .
(tk /k!)etλj
 0
. . . (tk−1 /k − 1!)etλj 
tetλj
etλj








..
tλ
etJj =  0
j

0
e
.
.
.
.








tλ
j
0
...
e
If Re(λj ) < 0 then this goes to zero as t → ∞ and if Re(λj ) > 0 then this goes to
zero as t → −∞. It follows that the same holds for eAt x for each x in the space
spanned by the generalised eigenspaces corresponding to Ji with Re(λj ) < 0.
3. Take x − ey sin(z) = 0. At what points can you write z as a function of x and y?
Compute ∂z/∂x and ∂z/∂y.
Answer: Write F (x, y, z) = x − ey sin(z). Since ∂F/∂z = ey cos(z), the Implicit
Theorem gives that you can do this when cos(z) 6= 0. Differentiating w.r.t. x and
1
y gives 1 − ey cos z ∂z/∂x = 0 and −ey sin z − ey cos z ∂z/∂y = 0. This implies
∂z/∂x = 1/(ey cos(z)) and ∂z/∂y = − tan z.
4. Consider the system of equations x2 + y 2 + z 2 = 5 and xy + yz = 2. Can you write
y and z as a function of x near (x, y, z) = (0, 1, 2)? Determine y 0 , z 0 at this point.
Answer: Write F (x, y, z) = (x2 + y 2 + z 2 − 5, xy + yz − 2) and compute the derivative
w.r.t. y, z. Then we obtain the matrix
2y
2z
x+z y
This matrix is invertible near (x, y, z) = (0, 1, 2).
Differentiating with respect to x gives
0 0
y
2y
2z
2x
=
+
0
z0
x+z y
y
and this gives an explicit expression for y 0 , z 0 in terms of x, y, z.
5. (a) Consider y 0 = 2xy.
P Solven this equation by substituting the power series expansion y(x) = n≥0 an x into this equation and determining the coefficients
an .
P
P
P
n+1 =
n
n−1 =
y0 =
n≥0 an 2x
n≥0 an+1 (n + 1)x and 2xy =
n≥0 an nx
P
n
n≥1 an−1 2x . It follows that a1 = 0, a0 is arbitrary and an+1 (n + 1) = an−1 2
2an−1
for n ≥ 1 (i.e. an+1 =
for n ≥ 1). By induction one gets a2n = a0 /n! and
n+1
a2n+1 = 0 for n ≥ 0. Hence y(x) = a0 exp(x2 ). Since the equation y 0 = 2xy is
linear, and y(0) = a0 it was to be expected that the general solution is of the
form a0 times some solution.
(b)
i. Consider y 0 = sin(x)y.
P Solve this equation by substituting the power series
expansion y(x) = P n≥0 an xn into this equation and determine a0 , a1 , a2 .
n
0
Answer: y(x) =
n≥0 an x . We only need to solve a0 , a1 , a2 take y =
2
3
a1 + 2a2 x + 3aa x + h.o.t. and sin(x)y = (x − x /3! + . . . )(a0 + a1 x + a2 x2 +
h.o.t.) = (a0 x + a1 x2 + (a2 − a0 /3!)x3 ). Comparing terms of the same order
gives a1 = 0, 2a2 = a0 . It follows that a0 is arbitrary and that a2 = a0 /2.
Remark: That a0 is arbitrary is not surprising: the initial problem y 0 =
sin(x)y, y(0) = a0 has a a (unique) solution. Since y 0 = sin(x)y is linear,
it follows that the solution is of the form y(x) = a0 (1 + b1 x + b2 x2 + . . . ).
Since we are not asked to compute a3 , a4 , . . . we can ignore the terms with
x3 , x4 , . . . . (Note however, that an is always some expression in terms of
a0 , a1 , . . . , an−1 ).
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ii. Consider y 0 = sin(x)y. Can you derive from the global Picard theorem that
the solution x 7→ y(x) of y 0 = sin(x)y exists for all x ∈ R?
Answer: if we take f (x, y) = sin(x)y then the differential equation y 0 =
dy
= f (x, y) is the one we are discussing. In the global Picard theorem we
dx
need to have that y 7→ f (x, y) is Lipschitz. In this case |f (x, y1 )−f (x, y2 )| =
sin(x)|y1 −y2 | ≤ |y1 −y2 |. So we can indeed apply the global Picard theorem,
and get that the solution exists for all x ∈ R.
dx
dy
Note that sometimes when we write x0 =
and sometimes y 0 =
.
dt
dx
So what variable is differentiated against should always be clear from the
context.
6. Consider 4x2 y 00 − 8x2 y 0 + (4x2 + 1)y = 0.
(a) 4x2 y 00 − 8x2 y 0 + (4x2 + 1)y = 0 is the same (for x 6= 0) as 4y 00 − 8y 0 + ((4x2 +
1)/x2 )y = 0. Explain why this has a regular singularity.
Answer: The term 8 corresponding to y 0 has no pole and the pole of (4x2 +1)/x2
is of order 2 at x = 0. This implies that by definition 4x2 y 00 −8x2 y 0 +(4x2 +1)y =
0 has a regular singularity at x = 0. At every other point the differential equation
has no singularity.
P
(b) Find solutions as power series by substituting y(x) = xm i≥0 ai xi . Show that
this power series is equal to x1/2 ex .
Answer: Plugging this in the equation gives for the lower order xm−2 term:
x2 [4m(m − 1)a0 xm−2 + (m + 1)ma1 xm−1 + . . . ]
−8x2 [a0 mxm−1 + a1 (m + 1)xm + . . . ]
+(4x2 + 1)[a0 xm + a1 xm+1 + . . . ] = 0
The lowest order term is xm and the coefficient for this is
m(m − 1) + m · 0 + 1/4 = 0.
This quadratic
expression has only one solution, namely m = 1/2. Substitute
P
1/2
i
y=x
i≥0 ai x gives the sum over j ≥ 0 of
(2j + 1)(2j − 1)aj xj+1/2 − (8j + 4)aj xj+3/2 + 4aj xj+5/2 + aj xj+1/2 .
Gathering terms of the same power gives
X
[(2j + 1)(2j − 1)aj − (8j − 4)aj−1 + 4aj−2 + aj ]xj+1/2 +
j≥2
3
[4a1 − 4a0 ]x3/2 .
This gives
a1 = a0 , aj =
(2j − 1)aj−1 − aj−2
.
j2
This means that a0 is arbitrary,
a2 = a0 /2! , a3 = a0 /3! , a4 = a0 /4!
and so on. Now if we choose a0 = 1 we obtain
y(x) = x1/2 (1 + x + x2 /2! + x3 /3! + . . . ) = x1/2 ex .
Of course we could choose a0 some other number, but we would only obtain
a0 x1/2 ex (this follows from linearity of the differential equation).
(c) Show that log(x)y1 (x) = log(x)x1/2 ex is also a solution by plugging this in the
differential equation.
Answer: [log(x)x1/2 ex ]0 = x−1/2 ex + (1/2) log(x)x−1/2 ex + log(x)x1/2 ex ] and
[log(x)x1/2 ex ]00 = [(−1/2)x−3/2 ex +x−1/2 ex ]+[(1/2)x−3/2 ex +(1/2)(−1/2) log(x)x−3/2 ex +
(1/2) log(x)x−1/2 ex ]+[x−1/2 ex +(1/2) log(x)x−1/2 ex +log(x)x1/2 ex ]. Simplifying
this and plugging this in 4x2 y 00 − 8x2 y 0 + (4x2 + 1)y gives zero.
A shorter, more efficient, way to check this is as follows: [log(x)y1 ]0 = (1/x)y1 +
log(x)y10 and [log(x)y1 ]00 = (−1/x2 )y1 + (2/x)y10 + log(x)y100 . This means 4x2 y 00 −
8x2 y 0 + (4x2 + 1)y = [−4y1 + 8xy10 + 4x2 log(x)y100 ] + [−8xy1 − 8x2 log(x)y10 ] +
(4x2 + 1) log(x)y1 . Using the fact that y1 is a solution, this is equal to [−4y1 +
8xy10 ] + [−8xy1 ]. Since y1 (x) = x1/2 ex this is equal to 0 and so indeed log(x)y1
is a solution.
In conclusion, the general solution is of the form ax1/2 ex + bx1/2 ex log(x) where
we can choose a, b arbitrarily.
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