MATH 215/255, Homework 7
d
1. Consider the statement: “ dt
u (t − a) = δ(t − a) ”. Give an argument why this is the case.
Rt
Answer.
Reason 1: Let f (t) = −∞ δ(t − a)dt. Then by the definition of the delta function,
f (t) = 0 if t < a and f (t) = 1 if t > a. So f (t) = u (t − a) .
Reason 2: Take any ”test function” g(t) which is differentiable and has compact support (i.e. g(t)
is zero for sufficiently large t). Then integrate by parts:
Z ∞
Z ∞
Z ∞
d
g 0 (t) = g(a).
u(t − a)g 0 (t)dt = −
u (t − a) g(t) = −
dt
a
−∞
−∞
R∞
d
But the delta function is precisely a function such that −∞ δ(t − a)g(t)dt = g(a). So dt
u(t − a) =
δ(t − a).
2. Solve
x00 = δ(t − 1),
x(0) = 1, x0 (0) = 0.
Sketch the graph of the resulting function.
Answer. Solution 1 (using Laplace transform): Letting X(s) = L(x(t)) we get
s2 X − sx(0) − x0 (0) = e−s
X=
1 e−s
+ 2
s
s
Taking the inverse Laplace transform then yields
x(t) = 1 + u (t − 1) (t − 1)
1, 0 < t < 1
=
t, t > 1
The graph is:
2
1.5
1
0.5
0
0
1
2
Solution 2 (optional, using jump conditions): For t < 0, we solve x00 = 0 with x = 1, x0 = 0 at
t = 0 to get x(t) = 1, 0 < t < 1. At t = 1, we have the jump conditions x (1+ ) = x (1− ) and
x0 (1+ ) − x0 (1− ) = 1, which yields x (1+ ) = 1, x0 (1+ ) = 1; moreover x00 = 0 for t > 0. Thus
x (t) = t for t > 1. So we obtain
1, 0 < t < 1
x(t) =
t, t > 1
with the graph as shown.
1
3. Solve
x00 + x = δ(t − π) − δ(t − 2π),
x(0) = 0, x0 (0) = 1.
Sketch the graph of the resulting function.
Answer. Taking Laplace’s transform we get
X=
e−πs − e−2πs
1
+
2
1+s
1 + s2
so that
x(t) = sin(t) + sin(t − π)u(t − π) − sin(t − 2π)u(t − 2π)

sin(t), 0 < t < π

sin(t) + sin(t − π), π < t < 2π
=

sin(t) + sin(t − π) − sin(t − 2π), t > 2π

 sin(t), 0 < t < π
0, π < t < 2π
=

− sin(t), t > 2π
The graph is shown below.
1
0.5
0
-0.5
-1
0
5
10
15
4. Find the general solution to the system x0 = Ax, where A is as specified below. Make sure to write
the solution in purely real form.
1 0
(a) A =
.
−4 3
1
Answer. The two eigenvalues are λ = 1, 3. The corresponding eigenvectors are
and
2
0
. So the general solution is
1
x=a
1
2
et + b
0
1
e3t
where a, b are arbitrary constants.
1 −5
(b) A =
.
1 −3
2
Answer. trace = −2, det = 2, so characteristic equation λ2 + 2λ + 2 = 0 or (λ + 1) + 1 = 0
so that λ = −1 ± i.
2
x
2−i
−5
x
satisfies
=
y
1
−2 − i
y
0. The second row reads x = (2 + i)y so that one of the eigenvalue/eigenvector pair is
2+i
λ = −1 + i, v =
;
1
Now take λ = −1+i, the corresponding eigenvector v =
the other is complex conjugate. Now expand:
(2 + i) eit e−t
2 cos t − sin t + i (cos t + 2 sin t)
λt
−t
ve =
=e
eit e−t
cos t + i sin t
so that the general solution is given by x = a Re veλt + b Im veλt ,
2 cos t − sin t
cos t + 2 sin t
x= a
+b
e−t
cos t
sin t
where a, b are arbitrary constants.


1
0 0
(c) A =  −2 1 2 
−2 −2 1
Answer. Note that the matrix has a lower-triagonal block-structure. This means that one of
the eigenvalues
isλ = 1 and the other two eigenvalues are the eigenvalues of the lower-block
1 2
2
matrix
. The latter has characteristic equation (λ − 1) + 4 = 0 so that λ = 1 ± 2i
−2 1
[equivalently, compute the characteristic polynomial by row-expanding for det (A − λI) along
the first row]. The corresponding eigenvectors are:



x
0
0 0
• λ = 1 :  −2 0 2   y  = 0, so that x = z, x = −y so the eigenvector is
z
−2 −2 0

1
v =  −1  .
1



−2i
0
0
x
2   y  = 0, so that the first row reads x = 0 and the
• λ = 1 + 2i :  −2 −2i
−2 −2 −2i
z
 
0
second row reads 2iy = 2z so that v =  1  .
i


0
• λ = 1 − 2i : v =  1  (i.e. complex conjugate).
−i
When λ = 1 + 2i, we compute

0
veλt = et  e2it
ie2it



0
 = et  cos (2t) + i sin(2t)  .
− sin(2t) + i cos(2t)
The general solution is therefore






1
0
0
x = aet  −1  + bet  cos 2t  + cet  sin 2t  .
1
− sin 2t
cos 2t
3
(d) A =
−5
3
−3
1
2
Answer. The characteristic equation is λ2 + 4λ + 4 = 0 ⇐⇒ (λ + 2) = 0 so there is a
double eigenvalue λ = −2. The eigenvector satisfies
−3 −3
x
=0
(1)
3
3
y
1
whose only solution, up to a constant multiple, is v =
. So it’s a defective eigenvalue!
−1
So we need to find a vector w such that (A − λI)w = v. That is, we need to solve
−3 −3
x
1
=
.
3
3
y
−1
A solution is given by x = 31 , y =
−2
3 .
v=
To summarize,
1
1/3
,
, w=
−1
−2/3
and therefore the general solution is
x = ave−2t + b (w + tv) e−2t
for arbitrary constants a, b.
5. A 3x3 real
has three eigenvalues. One of them is λ1 = −1 and its corresponding
 matrix

 eigenvector
0
1
is v1 =  1  . Another is λ2 = 1 + i and its corresponding eigenvector is v2 =  2  . (a) What
0
i
is the third eigenvalue and its corresponding eigenvector? (b) Find the
general
solution
to x0 = Ax.
 
1
(c) Find the solution to x0 = Ax subject to initial condition x (0) =  0  .
0
Answer. (a) The third
pair is the complex conjugate of the second, that is
 eigenvalue/eigenvector

1
λ3 = 1 − i and v2 =  2  .
−i
(b) In complex form, the general solution is



 

1
1
0
x = a  1  e−t + b  2  et+it + c  2  et−it
i
−i
0
(c) Plugging in t = 0, we obtain a linear system to solve for a, b, c :


  
0 1 1
a
1
 1 2 2  b  =  0 
0 i −i
c
0
By inspection (or otherwise), the solution is b = c = 1/2,



0

x (t) = −2  1  e−t + et 
0
 

0
= −2  1  e−t + et 
0
4
a = −2 so that

it
−it
e +e
2
2eit +2e−it
2
ieit −ie−it
2

cos t
2 cos t  .
− sin t


6. Two tanks are connected by two pumps: one pump pushes the liquid from tank A to tank B at
the rate of 6 liters/hour while the other pushes from tank B to tank A at the same rate. Initially,
both tanks contain 2 liters of liquid and tank A contains pure water while tank B has a mixture of
80% water and 20% pollutant. (a) Find the concentration of pollutant in tanks A and B after ten
minutes. (b) Find the concentration of pollutant in tanks A and B after a very long time.
Answer. Let x(t) be the concentration of pollutant in tank A and let y(t) be the concentration of
pollutant in tank B. Then we have:
6
6
dy
6
6
dx
= y − x,
= x− y
dt
2
2
dt
2
2
x(0) = 0, y(0) = 0.2
−3 3
; A=
.
3 −3
1
1
The eigenvalues of A are λ = 0, −6 with the corresponding eigenvectors
and
; the
1
−1
general solution is then
1
1
~x(t) = a
+b
e−6t ;
1
−1
or in matrix form, ~x0 = A~x, ~x (0) =
0
0.2
, where ~x =
x
y
the constants a, b satisfy
a
1
1
+b
1
−1
=
0
0.2
so that a = 0.1, b = −0.1. Therefore
x (t) = 0.1 1 − e−6t ;
y (t) = 0.1 1 + e−6t .
After ten minutes, t = 1/6, the concentrations are
x(0.5) = 0.1 1 − e−1 = 0.0632;
y(0.5) = 0.1 1 + e−1 = 0.1378;
Note that e−6t → 0, as t → ∞ so that x, y → 0.1 as t → ∞.
5
(2)