Homework from Section 2.3
1. Exercise 2.3.2
(a) We consider the BVP: φ00 + λφ = 0, with φ(0) = 0 and φ(π) = 0.
Side Remark: In this exercise, we should practice making the three cases for λ.
√
SOLUTION: The characteristic equation is r2 + λ = 0, or r = ± −λ.
• If λ = 0, we have one real root (r = 0) and the solution is C1 x + C2 . Applying
the boundary conditions, we get C1 = C2 = 0, and we get only the trivial
solution.
• If λ < 0, then we have two real solutions and we can write the general solution
as the following (getting practice with sinh and cosh):
√
√
C1 cosh( −λx) + C2 sinh( −λx)
Using the boundary conditions,
φ(0) = 0
⇒
φ(π) = 0
⇒
C1 + 0 = 0
√
C2 sinh( −λπ) = 0
Since λ 6= 0, this is true only if C2 = 0. In this case, we also get only the
trivial solution.
• Last case is λ > 0, in which case we get two complex solutions to the characteristic equation, and the general solution is:
√
√
φ(x) = C1 cos( λx) + C2 sin( λx)
Applying the boundary conditions, φ(0) = 0 implies C1 = 0, so the remaining
condition gives us:
√
C2 sin( λπ) = 0
Either C2 = 0 (which leads us back to the trivial solution), or the sine evaluates to zero. The sine evaluates to zero if:
√
λπ = nπ for n = 1, 2, 3, · · · ⇒ λ = n2 for n = 1, 2, 3, · · ·
Our solutions are therefore of the form:
φn = Cn sin(nx) for n = 1, 2, 3, · · ·
2. Exercise 2.3.3(b,c)
In Section 2.3.3, we are given the heat equation with zero boundary conditions and
varying initial functions. It is fine if you are already very familiar with separation of
variables to start with the general form of the solution:
u(x, t) =
∞
X
2
Bn e−k(nπ/L) t sin
n=1
1
nπ
x
L
where
nπ
2ZL
f (x) sin
x
Bn =
L 0
L
HOWEVER, if you’re not exactly sure where any of those terms come from, you should
be able to get the general solution using separation of variables and the integral formulas for the sine and cosine.
(b) In this case, the coefficients Bn are going to be zero except for n = 1 and n = 3
(by the orthogonality of the sines):
2L
2ZL
3 sin2 (πx/L) dx = 3
=3
L 0
L2
And similarly, B3 = −1. Therefore, the full solution is:
B1 =
2
u(x, t) = 3e−k(π/L) t sin
πx
3πx
2
− e−k(9π/L) t sin
L
L
(c) We already know the formula for the coefficients:
u(x, t) =
∞
X
Bn e
−k(n]pi/L)2 t
n=1
nπ
x
sin
L
where
3π
2ZL
nπ
2 cos
x sin
x
L 0
L
L
(We’ll be computing these in Chapter 3, that’s why the little footnote was there).
Bn =
3. Exercise 2.3.4 Here we have our friend the heat equation,
ut = kuxx
with zero boundary conditions and initial temperature profile f (x).
(a) Let’s compute the total heat energy in the rod as a function of time.
In this case, we are meant to use the general formula for the solution:
E(t) =
Z L
cρu(x, t)A dx = cρA
0
∞
X
n=1
−k(nπ/L)2 t
Bn e
"Z
L
0
nπ
sin
x dx
L
#
The quantity in the square brackets is:
−L
nπ L −L
cos
x =
(cos(nπ) − 1)
nπ
L
nπ
0
Now, we should simplify further: cos(nπ) = (−1)n , so that
nπ
L(1 + (−1)n+1 )
sin
x dx =
L
nπ
0
Substitute that expression back into the square brackets for the energy, and we
note that this is a function of time.
Z L
2
(b) What is the flow of heat energy out of the rod at x = 0? At x = L?
SOLUTION: With the general solution, we can compute ux (0, t) and ux (L, t),
since the flux is φ(0, t) = −K0 ux (0, t) and φ(L, t) = −K0 ux (L, t).
∞
X
−k(nπ/L)2 t
Bn e
n=1
so that:
ux (x, t) =
∞
X
Bn e
n=1
so that
ux (0, t) =
ux (L, t) =
nπ
x
sin
L
nπ
cos
x
L
L
−k(nπ/L)2 t nπ
∞
π X
2
nBn e−k(nπ/L) t
L n=1
∞
π X
2
(−1)n nBn e−k(nπ/L) t
L n=1
(c) What relationship should exist between parts (a) and (b)? (See back of text). We
don’t have to actually show this relationship here.
3