Solutions to HW 1
Mathav Murugan
1.
Solution.(a) Suppose f (y ) = y 37 is UL. Then there exists L R such
37
37
that |y1 y2 | L|y1 y2 |. Substitute y1 = 1 and y2 = 0, to get
1
L 1. Substitute y1 = (1 + L) 36 (this exists because 1 + L > 0) and
1
37
y2 = 0,
Math 3230
Quiz 4 Solutions
By E.W.
1a. Suppose the partial derivative f /y of f (t, y) exists and is continuous
on the rectangle R = cfw_(t, y)| |t t0 | a, |y y0 | b. We know R is
compact, and a continuous function on a compact set is bounded. That
is, we
Math 3230
Fall 2014
Professor Rand
Quiz No.2
Friday 9/19/14
1. Given the matrix A,
A=
1
1
0
0
1
3
0
0
1 1
1 2
2
1
0
2
Find a matrix T and a Jordan form matrix J such that
T 1AT = J
Note: It is not necessary to nd T 1 .
Hint: = 2, m = 4.
2. Given the mat
Math 3230
HW 2 Solutions
by E.W.
1. The ODE
x + w2x = 0
(1)
x(t) = c1 sin wt + c2 cos wt.
(2)
x =0
(3)
x(t) = t.
(4)
has the general solution
The ODE
has the particular solution
Since eq.(1) approaches eq.(3) as w 0, we would expect that the solution
(2)
Math 3230
Quiz 2 Solutions
By E.W.
1. Given the matrix A,
1
1
A=
0
0
1
3
0
0
1
1
2
0
1
2
1
2
we nd a matrix T such that T 1 AT = J, where J is in Jordan form.
Hint: = 2, m = 4.
First, we nd the eigenspace of the eigenvalue = 2. We solve the equation
(
Math 3230
Fall 2014
Professor Rand
Quiz No.1
Friday 9/5/14
1. Find the general solution of the second order ODE
9x 6x + x = 0
(1)
2. Find the general solution of the third order ODE
x x=0
(2)
3. A third order constant coecient homogeneous ODE has the part
Math 3230
Fall 2014
Professor Rand
Homework No.1
Due on Wednesday 9/3/14
1a. Find the general solution of the ODE
x + 5x + 6x = 0
(1)
1b. Find the particular solution of eq.(1) whch satises the initial conditions
x(0) = 0
and
x (0) = 1
(2)
2. Find the gen
Math 3230
Fall 2014
Professor Rand
Homework No.2
Due on Friday 9/12/14
1. The ODE
x + w2 x = 0
(1)
x(t) = c1 sin wt + c2 cos wt
(2)
x =0
(3)
x(t) = t
(4)
has the general solution
The ODE
has the particular solution
Since eq.(1) approaches eq.(3) as w 0, w
Math 3230
Fall 2014
Professor Rand
Homework No.4 (Revised)
Due on Friday 10/10/14
1. You may recall this matrix from Homework No.2 where you solved y = My with y(0) =
M=
Now solve y = My + f(t) for f(t) =
1
1
et cos t
0
1
0
1
1
and y(0) =
1
0
These exerci
Math 3230
Fall 2014
Professor Rand
Homework No.5 REVISED
Due on Friday 10/24/14
1. This problem is from Z&T, p.143: For each of the following equations, describe the regions
in the x y plane, where the equation is hyperbolic, parabolic or elliptic.
1a.
2u
Math 3230
HW 1 Solutions
by E.W.
1a. We have the ODE
x + 5x + 6x = 0.
(1)
t
Assume a solution of the form x(t) = e . Substituting this into (1) and
factoring out et , we get the characteristic equation
2 + 5 + 6 = 0.
The two roots are 1 = 3 and 2 = 2. The
Math 3230
Fall 2014
Professor Rand
Uniqueness Proof
Reference: Brauer and Nohel p.125-126, p.31
Suppose there were two solutions to the scalar initial value problem
y = f(t, y),
y(t0) = y0
Call them y = 1(t) and y = 2 (t). Then
t
1 (t) = y0 +
t0
t
f(s, 1
Math 3230
Quiz 3 Solutions
By E.W.
1. Solve for y(t):
dy
= Ay + f (t),
dt
y(0) = y0
where
A=
2
0
1
2
,
e2t
e2t
f (t) =
The solution is
,
1
0
y0 =
t
Z(t s)f (s)ds
y(t) = z(t) +
0
where Z(t) = eAt is the fundamental solution matrix of the homogeneous
proble
Math 3230 , Professor Rand
JORDAN CANONICAL FORM
In. many applications, it is desirable to be able ta transfcsm a
given n x n matrix A to an n x n matrix J'which has a simple (6r "can»
onical) form, via the transformation
TdAT a J
where T is an a x n matr
Math 3230
Fall 2014
Professor Rand
Quiz No.4
Friday 10/17/14
1a. If the partial derivative f of f(t, y) exists and is continuous on the recty
angle R = cfw_(t, y)| |t t0| a, |y y0 | b, show that f satises a Lipschitz
condition in R.
1b. Compute a Lipschit
Math 3230
Fall 2014
Professor Rand
Quiz No.3
Friday 10/3/14
1. Solve for y(t):
dy
= Ay + f(t),
dt
where
A=
2 1
0 2
,
f(t) =
Useful formula:
y(0) = y0
e2t
e2t
,
y0 =
1
0
t
y(t) = z(t) +
0
Z(t )f( )d
2. The ODE
x + w2 x = sin t
(1)
has the general solution
Solutions to selected problems
HW 2
Mathav Murugan
1
Lemma 1. Let x = (x1 , x2 ) R2 . Then x
Proof. Note that x
lemma.
2
1
2 x
2
2
1
2x
2
= (|x1 | |x2 |)2 0. This proves the
Note that continuous functions are bounded on nite box (since box is
compact and
Solutions to selected problems
HW 3
Mathav Murugan
2
Lemma 1. For any vector v V, |Lv | |L|v |
Proof. Note that if v = 0, the inequality is trivially true. Let v = 0.
Then
|Lv | =
|v |L
= |v | L
v
|v |
v
|v |
|L|v |
The rst equation follows from linearit
Solutions to selected problems
HW 4
Mathav Murugan
24 Verify that A2 = I . Then A3 = A2 A = IA = A and A4 = (A2 )2 =
if n = 4k
I
A
if n = 4k + 1
(I )2 = I . This impies An =
I if n = 4k + 2
A if n = 4k + 3
25
exp(tA) =
i=0
ti i
A
i!
=
k=0
=
k=0
t4k+1
t4k+
Solutions to selected problems
HW 5
Mathav Murugan
2 By the remark at the end of theorem 2.10, we can chose = 0 for
the matrix A and have that | exp(tA)| K .This implies that |y (t)|
| exp(t t0 )A)|y (t0 )| K |y (t0 )|. Hence every solution is bounded.
3
Solutions to selected problems
HW 6
Mathav Murugan
1 Let u1 and u2 two solutions to the boundary value problem. Let w =
u1 u2 . Then we have
(u1 u2 )
u1 u2
u1 u2
w
w =
=
=
=
(1)
in and
w=0
(2)
in . This gives
w2 dv =
wwdv
(3)
w
=
=
w
d
n
( w).( w)dv
(4)
Solutions to selected problems
HW 7
Mathav Murugan
1
(a) By assumption
An rn + Bn rn cos n + Cn rn + Dn rn sin n
u(r, ) = A0 + B0 ln r +
n=1
We will use the boundary conditions to get the coecients. By the
boundary conditions on the annulus we get
f () =
Solutions to selected problems
HW 8
Mathav Murugan
12.3 The heat equation ut = u implies that the steady state solution u
is harmonic. By theorem 12.2, it is unique. We use the fundemental
solution to construct the solution u(r) = RT|0 . It is easy to ver
Solutions to selected problems
HW 9
Mathav Murugan
5.3
(a) Straightforward computation.
(b) Let > 0 and M > 0. Let y = 2 > 0. Let g (n) = exp(ny)1 .
2n2
Note that g (n) as n (by lHpitals rule). Therefore g (n) >
o
M for all n > n0 for some n0 N. Choose n
Solutions to selected problems
HW 10
Mathav Murugan
1 Let x0 R. At time t > T (x0 ) = |x0| + l the backward cone at (x0 , t)
contains [l, l] cfw_0. Therefore
u(x0 , t) =
1
1
(x0 + t) + (x0 t) +
2
2
1
2
=U
x0 + t
( )d
x0 t
l
( )d
=
l
2 Equipartition of e
Solutions to selected problems
HW 11
Mathav Murugan
1 We are given ut uxx = 0, 0 < x < L and t > 0, u(x, 0) = (x),
(0) = a, (L) = b, u(0, t) = a + ct and u(L, t) = b + ct. Use the
x
hint and Dene v (x, t) = u(x, t) a + (b a) L ct + c x(L2 x) . Note
that v