Math 5120
Homework 3 Solutions
1. (a) Letting a = 0 in the solution gives u(0, t) = ceλt . Setting this equal to the boundary
condition and substituting the derived expression for u(a, t), we get
Z ∞
Ra
λt
u(0, t) = ce =
β(a)ceλt e− 0 (λ+µ(b)) db da.
0
Dividing by
ceλt
yields
Z
∞
β(a)e−
Ra
0
(λ+µ(b)) db
da = 1.
0
(b) Let f (λ) equal the left-hand side of the above equation. To show that f (λ) is a strictly decreasing function of λ, we need to show that for every x and y Rsatisfying x < y, fR(x) > f (y).
a
a
So, given x < y, it follows that x+µ(b) < y +µ(b), and so e− 0 (x+µ(b)) db > eR− 0 (y+µ(b)) db .
− 0a (x+µ(b)) db
>
Therefore,
R a since we know that β(a) > 0 for all a, it’s also true that β(a)e
− 0 (y+µ(b)) db
for all a. Then evaluating the integrals of both sides of the inequality
β(a)e
with respect to a will also satisfy the inequality. Thus, f (x) > f (y).
lim f (λ) = 0
λ→+∞
lim f (λ) = +∞
λ→−∞
Now, since
lim f (λ) > lim f (λ) and f (λ) is strictly decreasing, there is no point at
λ→−∞
λ→+∞
which f can equal the same value more than once. So the solution to f (λ) = 1 must be
unique for β(a) ≥ β0 > 0.
Ra
− 0 (λ+µ(b)) db
and u(a, 0) = u0 (a), then ce−
(c) Since u(a, 0)
R a = ce
c = u0 (a)e 0 (λ+µ(b)) db .
Ra
0
(λ+µ(b)) db
= u0 (a), and so
(d) Assuming the given the parameter values, we can solve for λ using the equation in part
(a). Here λ ≈ −0.008. Also, c = 106 e0.002a , and substituting this into the solution gives
u(a, t) = 106 e−0.008t with the boundary condition u(0, t) = 1.05 · 106 e−0.008t . The following
picture shows the change in the population according to the solution for various times t.
Notice that since λ < 0, the population density for each age group decreases as time
increases.
5
10
x 10
9
8
7
u(a,t)
6
5
4
3
2
1
0
0
10
20
30
40
1
50
age a
60
70
80
90
100
2. First note that the solution to the one-dimensional diffusion equation with initial condition
c(x, 0) = αδ(x − y) and D = 1 is
(x−y)2
α
c(x, t) = √ e− 4t .
2 πt
(a) To find x(t), we need to solve the inequality c(x, t) ≥ 0.1α for
of t. For y = 0
q x in terms
√
and x ≥ 0, the boundary of the region of influence is x(t) = −4t ln(0.2 πt).
(b)
Time evolution of the boundary x(t)
2.5
2
x(t)
1.5
1
0.5
0
0
1
2
3
4
t
5
6
7
8
(c) Note that for all t > t∗ = 25
π , x(t) is not defined, and so the inequality in (a) is not satisfied,
leaving the region of influence empty.
2