Math 425 / AMCS 525
Assignment 1 solutions
Dr. DeTurck
Due Tuesday, January 20, 2016
1. Find the general solution of x0 (t) + x2 sin t = 0.
This is not a linear equation because of the x2 , but it is separable. Rewrite it as
dx
= −x2 sin t
dx
and then separate:
dx
= − sin t dt
x2
and integrate:
Z
Z
dx
=
sin t dt
x2
1
− = cos t − C
x
1
x=
C − cos t
2. Solve the initial-value problem x0 (t) + x(t) cos t = 0, x(π) = 100.
The differential equation is both separable and linear, so we’ll solve it as a linear equation this
time; it is of the form x0 + p(t)x = q(t) where p(t) = cos t and q(t) = 0. The general solution is:
Z
Z
R
R
R
x = e− p qe p = e− cos t dt 0 dt = Ce− sin t
(recall that in this method you must only add a constant of integration to the “big” integral). Since
x = 100 when t = π, we have
100 = Ce− sin π = Ce0 = C
so C = 100 and the solution of the initial-value problem is
x = 100e− sin t .
3. Find the general solution: 2y 00 + 5y 0 + 2y = 0.
We need the roots of the polynomial 2r2 + 5r + 2 = (2r + 1)(r + 2), which are r = −2 and r = − 21 .
So the general solution is
1
y = c1 e−2t + c2 e− 2 t .
2
laplace equation on the cylinder
4. Find the solution of the initial-value problem: 5y 00 + 8y 0 + 5y = 0, y(0) = 1, y 0 (0) = 0.
This time we need the roots of the polynomial 5r2 + 8r + 5, which doesn’t factor in an obvious
way. So:
√
−8 ± 6
4 3
−8 ± 64 − 100
=
= − ± i.
r=
10
10
5 5
So the general solution is
4
y = e− 5 t c1 cos 35 t + c2 sin 35 t .
And since we’ll need it, we calculate:
4
4
y 0 = − 45 e− 5 t c1 cos 35 t + c2 sin 53 t + e− 5 t − 35 c1 sin 35 t + 35 c2 cos 35 t
4
= e− 5 t 53 c2 − 45 c1 cos 35 t + 35 c2 + 45 c1 sin 35 t .
Since e0 = 1, cos 0 = 1 and sin 0 = 0, it’s easy to see that y(0) = 1 implies that c1 = 1. Next,
y 0 (0) = 0 becomes 53 c2 − 45 = 0, so c2 = 34 , and the solution of the initial-value problem is
4
y = e− 5 t cos 35 t + 43 sin 53 t .
5. Solve the following system of differential equations for x(t) and y(t):
x0 (t) = x(t) − 4y(t),
y 0 (t) = x(t) + y(t),
subject to the initial conditions x(0) = 1 and y(0) = 1.
We’ll solve this one twice, first using elimination (i.e., clever algebra), and then using matrices.
Rearrange the second equation as x = y 0 − y, and then differentiate both sides to get x0 =
y − y 0 . Replace the x and x0 in the first equation with these expressions in terms of y to get
y 00 − y 0 = y 0 − y − 4y, or y 00 − 2y 0 + 5y = 0. So to find y, we have to find the roots of the polynomial
r2 − 2r + 5 = 0, which are
√
2 ± 4 − 20
r=
= 1 ± 2i
2
and so y = c1 et cos 2t + c2 e2 sin 2t. Next, from x = y 0 − y, we get
x = c1 et cos 2t − 2c1 et sin 2t + c2 et sin 2t + 2c2 et cos 2t − c1 et cos 2t + c2 et sin 2t
00
= 2c2 et cos 2t − 2c1 et sin 2t
From x(0) = 1 we conclude 2c2 = 1 or c2 =
solution of the problem is
1
2,
x = et cos 2t − 2et sin 2t,
and from y(0) = 1 we conclude c1 = 1. So the
y = et cos 2t + 21 et sin 2t.
For the matrix version of the solution, we rewrite the system as
0 x
1 −4
x
=
y
1
1
y
3
math 425
We need to find the eigenvalues and eigenvectors of the matrix, so we calculate
1−λ
−4
det
= (1 − λ)2 + 4 = λ2 − 2λ + 5
1 1−λ
which is the same polynomial we had before, so the eigenvalues are λ = 1 ± 2i.
To find the eigenvector corresponding to the eigenvalue λ = 1 − 2i we need the kernel of the
matrix
2i −4
−2i
, which is spanned by
.
1 2i
1
Two linearly independent solutions of the system are then the real and imaginary parts of
−2i
−2i
−2 sin 2t
−2 cos 2t
(1−2i)t
t
t
t
e
= e (cos 2t − i sin 2t)
=e
+ ie
.
1
1
cos 2t
− sin 2t
Thus
x
y
−2 sin 2t
2 cos 2t
= e t c1
+ c2
.
cos 2t
sin 2t
Then x(0) = y(0) = 1 implies that
which tells us that c2 =
1
2
1
1
=
2c2
c1
and c1 = 1, and so the solution of the initial-value problem is
x
y
=e
t
−2 sin 2t + cos 2t
cos 2t + 21 sin 2t
.
6. Prove that the solution of the initial-value problem u00 + cu = 0, u(0) = a, u0 (0) = b for c < 0
exists (easy — just write it down) and is unique (to do this, “factor” the operator and then apply
the theorem on page 4 of the notes twice).
Since c < 0, to simplify notation, let c = −k 2 , so the equation becomes u00 − k 2 u = 0 and (since
the roots of r2 − k 2 are r = ±k), the general solution of this equation is
u = c1 ekt + c2 e−kt .
Now we need
u(0) = c1 + c2 = a
and
u0 (0) = kc1 − kc2 = b.
Some easy algebra yields
c1 =
ka + b
2k
and
c2 =
ka − b
2k
and so the solution exists because
u(t) =
(with k =
√
−c) works.
ka + b kt ka − b −kt
e +
e
2k
2k
4
laplace equation on the cylinder
To prove uniqueness, we start the usual way and suppose there are two solutions to the problem,
call them u1 and u2 , and let v(x) = u1 (x) − u2 (x). Then v will satisfy:
v 00 − k 2 v = 0,
v(0) = 0, v 0 (0) = 0.
Using the hint, “factor” the differential equation as (D + k)(D − k)v = 0, where D stands for taking
the derivative. If we let w = (D − k)v = v 0 − kv, then w satisfies
(D + k)w = w0 + kw = 0,
w(0) = v 0 (0) − kv(0) = 0.
This is an initial-value problem for a first order equation, to which the theorem in the notes applies.
Therefore the unique solution of this problem is w = 0.
Since w = 0, we have that v satisfies:
v 0 − kv = 0,
v(0) = 0.
But this is again an initial-value problem for a first order equation, the unique solution of which is
v = 0. Therefore the difference between two solutions of the original problem must be zero, so we
have proved uniqueness.
7. (a) Torricelli’s law states that fluid will leak out of a small hole at the base of a container at
a rate proportional to the square root of the height of the fluid’s surface from the base. Suppose
that a cylindrical container is initially filled to a depth of one foot. If it takes one minute for three
quarters of the fluid to leak out, how long will it take for all of the fluid to leak out?
(b) It is desired to design a “water clock” by making a container that is in the shape of some
surface of revolution with a small hole in the bottom, so that as the water empties out of the hole,
the water level in the container falls at a constant rate. What should be the shape of the container?
(a) First we have to translate the words into a differential equation: “[F]luid leak[s] out . . . at [a]
dV
, where V (t) is the volume of fluid in the tank at time t. “proportional to the square
rate” means
dt
√
root of the height” means −k y for some constant k (we put in the minus sign because the volume
of fluid in the container is decreasing), and where y(t) is the height of the fluid in the container. So
we can write Torricelli’s law as
dV
√
= −k y.
dt
dV
dy
k
For a cylindrical tank of radius r, we have V (t) = πr2 y(t), so
= πr2 . So let K =
and
dt
dt
πr2
we can write the initial-value problem for y(t):
dy
√
= −K y,
dt
y(0) = 1, and we also know that y(1) =
(and for the record, y has units of feet, and t has units of minutes).
The differential equation is separable, and we calculate:
dy
√ = −K dt
y
√
2 y = C − Kt
1
4
5
math 425
Now y(0) = 1 gives us C = 2, and then y(1) = 14 says that 1 = 2 − K, which gives us that K = 1.
Therefore the solution of the initial-value problem is
√
2 y =2−t
and it is easy to see that the container will be empty (i.e., y = 0) when t = 2 minutes.
(b) For a general surface of revolution, say we’re rotating the curve r = f (y) around the y-axis,
Then the volume up to height y is
Z y
Z y
π(f (α))2 dα
πr2 dy =
V (y) =
0
0
where we have used α as a “dummy variable” in the integral.
Now, the “water level falls at a constant rate” means that
We can combine this with Torricelli’s law,
C
dy
= C for some constant C.
dt
dV
dV
dV dy
√
= k y and the chain rule
=
to write
dt
dt
dy dt
dV
√
= k y.
dy
Moreover, we know how V depends on y, so we can use the fundamental theorem of calculus to
write:
Z y
d
dV
=
π(f (α))2 dα = π(f (y))2 .
dy
dy 0
Combining the last two equations gives us
√
Cπ(f (y))2 = k y
or
r √
k y
√
f (y) =
=A4y
Cπ
for a suitably chosen constant A. So the curve that should be rotated to obtain our water clock is
√
given by r = A 4 y, or y = Br4 .
8. For a function u(x, y) of two variables, its Laplacian is defined to be ∆u = uxx + uyy . Which
radial functions (i.e., functions of the polar coordinate r but independent of θ) are harmonic (i.e.,
satisfy the PDE ∆u = 0, or uxx + uyy = 0) ?
This is first and foremost an exercise in the chain rule for partial derivatives! If u(x, y) = f (r),
where r2 = x2 + y 2 then
∂u
df ∂r
∂u
df ∂r
=
and
=
.
∂x
dr ∂x
∂y
dr ∂y
If we take the partial derivative of both sides of r2 = x2 + y 2 with respect to x. then we get
2r
∂r
= 2x ,
∂x
so
∂r
x
= .
∂x
r
6
laplace equation on the cylinder
Similarly,
∂r
y
= ,
∂y
r
and so we can write:
x
y
and uy = f 0 (r) .
r
r
Now we can take the derivative again, using the product, quotient and chain rules together with the
fact that we know ∂r/∂x and ∂r/∂y:
ux = f 0 (r)
∂ 0 x
f (r)
∂x
r
∂r x
1
x ∂r
00
= f (r)
+ f 0 (r) − f 0 (r) 2
∂x r
r
r ∂x
x2
1
x2
= f 00 (r) 2 + f 0 (r) − f 0 (r) 3
r
r
r
0
2
2
f
(r)
x
x
1− 2 .
= f 00 (r) 2 +
r
r
r
uxx =
Similarly,
uyy = f 00 (r)
y2
f 0 (r)
+
2
r
r
1−
y2
r2
.
Therefore
00
∆u = uxx + uyy = f (r)
x2 + y 2
r2
f 0 (r)
+
r
x2 + y 2
f 0 (r)
00
2−
=
f
(r)
+
r2
r
So we have to solve
1
f 00 + f 0 = 0.
r
1
We do this in two jumps: let g = f 0 , so we need g 0 + g = 0, in other words
r
dg
dr
=−
g
r
which implies g =
c1
.
r
Integrate g to get
f (r) = c1 ln r + c2 .
These are the functions we seek (the “radial harmonic functions” on
R2).