MATH 267
Due: Feb 1, 2012, in the class
ASSIGNMENT # 4
You have SEVEN problems to hand-in. Hand in written solutions for grading at the BEGINNING of the
lecture on the due date. Illegible, disorganized or partial solutions will receive no credit. Staple your HW.
P∞
1. Express the following functions as the complex Fourier series of the form k=−∞ ck eikx for 2π-periodic
P∞
kπ
functions or more generally of the form k=−∞ ck ei L x for 2L-periodic functions. (Hint: First determine
the periods of the functions. Also you may want to see the online notes ”Fourier series” page 3 and 4,
Example 3.)
(a) cos(2x) + sin(3x)
(b) 1 + sin(2πx) + 2 cos(πx)
(c) cos(x + π/3).
Solution
(a): Note that eix = cos x + i sin x. Thus, cos x = 12 (eix + e−ix ), and sin x =
1
ix
2i (e
− e−ix ). Therefore,
1 i2x
1
(e + e−i2x ) + (ei3x − e−i3x )
2
2i
1
1
1
1
= e−i3x + e−i2x + ei2x + ei3x
2i
2
2
2i
∞
X
=
ck eikx
cos(2x) + sin(3x) =
k=−∞
where
c−3 =
1
,
2i
1
1
1
, , c2 = , c3 =
2
2
2i
ck = 0, for all k 6= −3, −2, 2, 3.
c−2 =
(b): Now, the function has period 2 = 2 for L = 1. Similarly as in (a) we get
1
1 i2πx
(e
− e−i2πx ) + 2 · (eiπx + e−iπx )
2i
2
1 −i2πx
1
−iπx
iπx
=− e
+e
+1+e
+ ei2πx
2i
2i
∞
X
=
ck eikπx
1 + cos(2x) + sin(3x) = 1 +
k=−∞
where
c−2 = −
1
,
2i
1
,
2i
ck = 0, for all k =
6 −2, −1, 0, 1, 2.
c−1 = 1,
c0 = 1,
c1 = 1,
c2 =
(c):
1 i(x+π/3)
e
+ e−i(x+π/3)
2
1 iπ/3 ix 1 −iπ/3 −ix
= e
e + e
e
2
√ 2
√
1 1
3 ix 1 1
3 −ix
= ( +i
)e + ( − i
)e
2 2
2
2 2
2
∞
X
=
ck eikπx
cos(x + π/3) =
k=−∞
1
where
c−1
√
√
3
3
1
1
= −i
, c1 = + i
4
4
4
4
ck = 0, for all k 6= −1, 1.
2. In each case below, sketch
the graph of the 2π-periodic function f (x) and find the complex Fourier series
P∞
of f (x) (i.e. of the form k=−∞ ck eikx ). Also sketch the graph of the Fourier series. You must indicate
which value the functions take at the points of jump continuity. Also, the graphs should be drawn at
least for interval from −4π to 4π. (Hint: It will help to see Theorem 1 in page 3 of the online notes
‘Fourier Series”. Also, for computing the Fourier series, you may want to use the integration by parts
Z −int
Z
e
te−int
+
dt.
te−int dt = −
ni
ni
Be careful with n = 0 case. )
(a) f (x) is periodic with period 2π, and
x + π,
π,
f (x) =
−π ≤ x ≤ 0
0 < x < π.
(b) f (x) is periodic with period 2π, and
1,
0,
f (x) =
0≤x<π
π ≤ x < 2π.
(c) f (x) is periodic with period 2π and f (x) = |x| for −π < x < π.
Solution
We omit the sketch of graphs. For the sketch of the graph of Fourier series, be careful at the points of
(x−)
discontinuity: there the Fourie series takes the value f (x+)+f
.
2
(a) :
We use the formula
ck =
1
2π
Z
π
f (x)e−ikx dx
−π
Then,
1
ck =
2π
=
1
2π
=
1
2π
Z π
1
(x + π)e
dx +
πe−ikx dx
2π 0
−π
Z 0
Z 0
Z π
1
1
xe−ikx dx +
πe−ikx dx +
πe−ikx dx
2π −π
2π 0
−π
Z 0
Z π
1
xe−ikx dx +
πe−ikx dx
2π
−π
−π
Z
0
−ikx
Now two cases:
Case k = 0: Then,
1
c0 =
2π
Z
0
1
xdx+
2π
−π
Z
π
πdx =
−π
1 2 0
3π
x +π =
.
4π
4
−π
Case k 6= 0:
First,
1
2π
Z
π
−π
πe−ikx dx =
i
i
1 h e−ikx iπ
1 h −ikπ
1 h
=
e
− eikπ =
(−1)k − (−1)k = 0
2 −ik −π
−2ik
−2ik
2
Second,
1
2π
Z
0
xe−ikx dx
−π
Z 0 −ikx i
e
1 h xe−ikx 0
−
dx .
(used integration by parts)
+
2π
ki −π
ki
−π
e−ikx 0 i
1 h xe−ikx 0
.
−
dx
=
−
2π
ki −π
(ki)2
−π
1 h πeikπ 1 − eikπ i
−
+
=
2π
ki
k2
h
k
1
π(−1)
1 − (−1)k i
=
−
(used eikπ = (−1)k )
+
2π
ki
k2
( 1
− 2ki
k even = π2, ±4, ±6, · · · ,
= 1 hπ 2 i
k odd = ±1, ±3, ±5, · · · .
2π ki + k2
=
Therefore,
(
ck =
1
− 2ki
h
1
2π
Thus, the complex Fourier series is
X
π
− +
4
k even = π2, ±4, ±6, · · · ,
2
k2
i
π
ki
+
−
1 ikx
e
+
2ki
k=±2,±4,±6,···
k odd = ±1, ±3, ±5, · · · .
X
k=±1,±3,±5,···
1 hπ
2i
+ 2 eikx
2π ki k
(b):
First, by periodicity f (x) = 0 on −π ≤ x < 0.
We use the formula
1
ck =
2π
(Note that we can also use ck =
1
2π
R 2π
0
Z
π
f (x)e−ikx dx
−π
f (x)e−ikx dx, instead. )
Then,
Z 0
Z π
1
1
−ikx
0·e
dx +
1 · e−ikx dx
ck =
2π −π
2π 0
Z π
1
=
1 · e−ikx dx.
2π 0
Now two cases:
Case k = 0: Then,
c0 =
1
2π
Z
π
1dx =
0
1 π
1
x = .
2π 0
2
Case k 6= 0:
1 h e−ikx iπ
−
dx
2π
ki
0
1 h (−1)k
1i
=
+
−
(used e−ikπ = (−1)k )
2π
ki
ki
(
0
k even = ±2, ±4, ±6, · · · ,
=
1
k odd = ±, 1, ±3, ±5, · · · .
πki
ck =
3
Thus, the Fourier series:
1
2
X
+
k=±1,±3,±5,···
1 ikx
e
πki
(c):
We use the formula
1
2π
ck =
Z
π
f (x)e−ikx dx
−π
Then,
1
2π
Z
1
=
2π
Z
ck =
0
1
2π
Z
1
−xe−ikx dx +
2π
−π
Z
−xe−ikx dx +
−π
0
π
xe−ikx dx
0
π
xe−ikx dx
0
Now two cases:
Case k = 0: Then,
Z π
1
xdx
2π 0
−π
1 2 π
π
1 0
x = .
= − x2 +
4π
4π
2
−π
0
c0 =
1
2π
Z
0
−xdx +
Case k 6= 0:
Z 0 −ikx i
Z π −ikx i
1 h −xe−ikx π
1 h xe−ikx 0
e
e
dx +
dx .
(used integration by parts)
ck =
−
+
2π
ki −π
ki
2π
ki
ki
0
−π
0
e−ikx 0 i
1 h xe−ikx π e−ikx π i
1 h xe−ikx 0
=
+
−
+
−
.
2π
ki −π (ki)2 −π
2π
ki 0
(ki)2 0
1 h πeikπ 1 − eikπ i
1 h πe−ikπ e−ikπ − 1 i
=
−
+
+
−
2π ki
k2
2π
ki
k2
h
h
k
ki
k
1 − (−1)
π(−1)
(−1)k − 1 i
1 π(−1)
1
−
−
+
=
+
(used eikπ = (−1)k )
2π
ki
k2
2π
ki
k2
(
0
k even = π2, ±4, ±6, · · · ,
=
− πk2 2 k odd = ±1, ±3, ±5, · · · .
Thus, the complex Fourier series is
−
π
+
4
X
−
k=±1,±3,±5,···
2 ikx
e
πk 2
3. In the following problems t is a real variable.
(a) Let z be a complex number. Show that z is a real number if and only if z = z̄ where z̄ is the complex
conjugate of z. (Hint: write z = x + iy where x, y are real. Now recall the definition of complex
conjugate.)
(b) Show that for complex numbers z, w, the complex conjugates satisfy z + w = z + w.
(c) Show that if f (t) is given by a Fourier series
f (t) =
∞
X
n=−∞
4
cn eint .
then
f (t) =
∞
X
c−n eint .
n=−∞
(Hint: Use (b) and see what is the complex conjugate of eint . Also, use the fact: zw = z w. )
P∞
(d) Suppose the Fourier series n=−∞ cn eint is real-valued for all real numbers −∞ < t < ∞. Show
that ck = c−k . (Hint: Use (a) and (c) ) .
Solution
(a): z = x + iy and z = x − iy: here of course x, y are real. Note that z = x + iy is real is equivalent to
1
(z − z), thus y = 0 is equivalent to z = z. Thus z is real if and only if z = z.
that y = 0. But, see y = 2i
(b): Let z = x1 + iy1 and w = x2 + iy2 (here of course, x1 , x2 , y1 , y2 are all real. Then, z + w =
(x1 + x2 ) + i(y1 + y2 ). Thus,
z + w = x1 + x2 − i(y1 + y2 ) = x1 − iy1 + x2 − iy2 = z + w.
(c): From (b), we see that
f (t) =
∞
X
∞
X
cn eint =
n=−∞
cn eint .
n=−∞
(Here we used zw = z w)
Now, eint = e−int , thus,
∞
X
f (t) =
cn e−int
n=−∞
Now, we can change the summation index n to −n, then
∞
X
f (t) =
c−n eint
n=−∞
getting the desired identity.
(d): From (a), f (t) is real if and only if f (t) = f (t). Thus, if f (t) is real-valued, then
∞
X
cn eint = f (t) = f (t) =
n=−∞
∞
X
c−n eint .
n=−∞
Now, compare the Fourier coefficient for each n, we get
cn = c−n .
P∞
int
4. Suppose a real-valued function f (t) is given by a Fourier series
for all real numbers
n=−∞ cn e
P∞
int
−∞ < t < ∞. Recall e = cos(nt) + i sin(nt). Use this to show that the function f (t) = n=−∞ cn eint
can be expressed as the trigonometric Fourier series
∞
a0 X f (t) =
+
an cos(nt) + bn sin(nt) .
2
n=1
You have to demonstrate how you perform each of your steps. Also, express an and bn using cn ’s for
n = 0, 1, 2, 3, · · · . (Hint: You may look at page 5 and 6 in the online notes ”Fourier series”.)
5
Solution We use the identity
eint = cos(nt) + i sin(nt)
Thus,
∞
X
int
cn e
∞
X
=
n=−∞
=
cn cos(nt) + i sin(nt)
n=−∞
∞
X
cn cos(nt) +
n=−∞
∞
X
cn i sin(nt)
n=−∞
Rewrite
∞
X
cn cos(nt) = c0 cos 0 +
n=−∞
∞
X
cn cos(nt) +
n=1
= c0 +
∞
X
∞
X
c−n cos(−nt)
n=1
(cn + c−n ) cos(nt)
(since cos(−nt) = cos(nt) and cos 0 = 1).
n=1
and also
∞
X
cn i sin(nt) = c0 i sin 0 +
n=−∞
∞
X
cn i sin(nt) +
=
c−n i sin(−nt)
n=1
n=1
∞
X
∞
X
(cn − c−n )i sin(nt)
(since sin(−nt) = − sin(nt) and sin 0 = 0).
n=1
These two expressions together, we get
f (t) =
∞
X
cn eint = = c0 +
n=−∞
= c0 +
∞
X
(cn + c−n ) cos(nt) +
n=1
∞ h
X
∞
X
(cn − c−n )i sin(nt)
n=1
i
(cn + c−n ) cos(nt) + (cn − c−n )i sin(nt)
n=1
Comparing with
f (t) =
∞
a0 X +
an cos(nt) + bn sin(nt) .
2
n=1
we see
a0 = 2c0 .
For n = 1, 2, 3, cdots,
bn = (cn − c−n )i.
an = (cn + c−n ),
Optional:
Now use the fact (from problem 3 (d))
cn = c−n
thus
cn = c−n
we see that a0 = 2c0 is real because c0 = c0 , and an = cn + cn = 2Re(cn ) and bn = (cn − cn )i = −2Im(cn ).
P∞
int
5. Suppose a real-valued function f (t) is given by a Fourier series
for all real numbers
n=−∞ cn e
−∞ < t < ∞. Suppose in addition that f (t) is an even function. Here, f (t) is called an even function if
f (t) = f (−t) for all −∞ < t < ∞.
6
(a) Show that cn = c−n for all integer n. (Hint: what happens if t is replaced with −t in the Fourier
series?)
P∞
(b) Now recall eint = cos(nt) + i sin(nt). Use this to show that the even function f (t) = n=−∞ cn eint
can be expressed as the Fourier cosine series
∞
a0 X
+
an cos(nt).
2
n=1
f (t) =
You have to demonstrate how you perform each of your steps. Also, express an using cn ’s for
n = 0, 1, 2, 3, · · · . (Hint: Use (a) as well as Problem 3 (d) and the result of Problem 4.)
Solution (a) See
∞
X
f (−t) =
cn ein(−t)
n=−∞
Here, the right hand side
∞
X
∞
X
cn ein(−t) =
n=−∞
−∞
X
=
=
cn e−int
n=−∞
c−n eint
n=∞
∞
X
c−n eint
n=−∞
Here, in the second last line, we replaced the summation index n with −n.
P∞
Now compare this with f (−t) = f (t) = n=−∞ cn eint then, for each n,
cn = c−n .
(b): We use the identity
eint = cos(nt) + i sin(nt)
Thus,
∞
X
cn eint =
n=−∞
=
∞
X
n=−∞
∞
X
cn cos(nt) + i sin(nt)
cn cos(nt) +
n=−∞
∞
X
cn i sin(nt)
n=−∞
Rewrite
∞
X
cn cos(nt) = c0 cos 0 +
n=−∞
∞
X
cn cos(nt) +
n=1
= c0 +
= c0 +
∞
X
∞
X
(cn + c−n ) cos(nt)
n=1
∞
X
c−n cos(−nt)
n=1
2cn cos(nt)
(since cos(−nt) = cos(nt) and cos 0 = 1)
(since f (t) is even thus by (a) cn = c−n .)
n=1
and also
∞
X
cn i sin(nt) = c0 i sin 0 +
n=−∞
∞
X
cn i sin(nt) +
n=1
=
∞
X
∞
X
c−n i sin(−nt)
n=1
(cn − c−n )i sin(nt)
(since sin(−nt) = − sin(nt) and sin 0 = 0)
n=1
=0
(since f (t) is even thus by (a) cn = c−n .)
7
These two expressions together, we get
f (t) =
∞
X
cn eint = = c0 +
n=−∞
∞
X
2cn cos(nt)
n=1
Comparing with
∞
a0 X
+
an cos(nt).
2
n=1
f (t) =
we see
a0 = 2c0 .
For n = 1, 2, 3, · · · ,
an = 2cn
P∞
int
for all real numbers
6. Suppose a real-valued function f (t) is given by a Fourier series
n=−∞ cn e
−∞ < t < ∞. Suppose in addition that f (t) is an odd function. Here, f (t) is called an odd function if
f (t) = −f (−t) for all −∞ < t < ∞.
(a) Show that cn = −c−n for all integer n. (Hint: what happens if t is replaced with −t in the Fourier
series?)
P∞
(b) Now recall eint = cos(nt) + i sin(nt). Use this to show that the odd function f (t) = n=−∞ cn eint
can be expressed as the Fourier sine series
∞
X
f (t) =
bn sin(nt).
n=1
You have to demonstrate how you perform each of your steps. Also, express bn using cn ’s for
n = 0, 1, 2, 3, · · · . (Hint: Use (a) as well as Problem 3 (d) and the result of Problem 4 .)
Solution
(a) See
∞
X
f (−t) =
cn ein(−t)
n=−∞
Here, the right hand side
∞
X
cn e
in(−t)
=
n=−∞
=
=
−∞
X
∞
X
cn e−int
n=−∞
c−n eint
n=∞
∞
X
c−n eint
n=−∞
Here, in the second last line, we replaced the summation index n with −n.
P∞
P∞
Now compare this with f (−t) = −f (t) = − n=−∞ cn eint = n=−∞ −cn eint then, for each n,
−cn = c−n
thus
cn = −c−n .
(b): We use the identity
eint = cos(nt) + i sin(nt)
8
Thus,
∞
X
int
cn e
∞
X
=
n=−∞
=
cn cos(nt) + i sin(nt)
n=−∞
∞
X
cn cos(nt) +
n=−∞
∞
X
cn i sin(nt)
n=−∞
Rewrite
∞
X
cn cos(nt) = c0 cos 0 +
n=−∞
∞
X
cn cos(nt) +
n=1
= c0 +
∞
X
∞
X
c−n cos(−nt)
n=1
(cn + c−n ) cos(nt)
(since cos(−nt) = cos(nt) and cos 0 = 1)
n=1
(since f (t) is even thus by (a) cn = −c−n .)
= c0 + 0
=0
(since f (t) is even thus by (a), c0 = − − c−0 which implies c0 = 0).
and also
∞
X
cn i sin(nt) = c0 i sin 0 +
n=−∞
∞
X
cn i sin(nt) +
n=1
=
=
∞
X
c−n i sin(−nt)
n=1
(cn − c−n )i sin(nt)
n=1
∞
X
∞
X
2icn sin(nt)
(since sin(−nt) = − sin(nt) and sin 0 = 0)
(since f (t) is even thus by (a) cn = −c−n .)
n=1
These two expressions together, we get
f (t) =
∞
X
∞
X
cn eint = =
n=−∞
2cn i sin(nt)
n=1
Comparing with
f (t) =
∞
X
bn cos(nt).
n=1
we see for n = 1, 2, 3, · · · ,
bn = 2cn i
7. Suppose f (t) is a real-valued 2π-periodic function on the whole real line R. Suppose f (t) is an even
function, i.e. f (t) = f (−t) for all −∞ < t < ∞, and that
(
2t
for 0 ≤ t < π2 ,
f (t) =
2(π − t) for π2 < t < π.
(a) Draw the graph of f (t) on the interval −5π < t < 5π. Also, compute the (complex) Fourier series
of f (t).
(Hint: It will help to see Theorem 1 in page 3 of the online notes ‘Fourier Series”. Also, for
computing the Fourier series, you may want to use the integration by parts
Z
Z −int
te−int
e
−int
+
dt.
te
dt = −
ni
ni
Be careful with n = 0 case. )
9
P∞ (b) Find the trigonometric Fourier series of f (t) of the form: f (t) = a20 + n=1 an cos(nt) + bn sin(nt) .
Of course, you have to determine the coefficients an and bn . (Hint: You may want to use the
results in Problem 4 or 5.)
Solution
(a) : We use that the function is even f (−t) = f (t), so that we also have
(
−2t
for − π2 < t ≤ 0,
f (t) =
2(π + t) for −π < t < − π2 .
We omit the sketch of the graph. (Basically, draw the graph for the interval −π < t < π and repeat the
same pattern.
Then (after sketching the graph), we find that the function is also π-periodic, because the pattern for
−π/2 < t < π/2, i.e.
(
−2t for − π2 < t ≤ 0,
f (t) =
2t
for 0 ≤ t < π2 .
repeats.
π
P∞
P∞
Thus, f (t) has the complex Fourier series k=−∞ ck eik π/2 t = k=−∞ ck ei2kt . (Here, L = π/2.) Note
that this expression will really help for the following computations. If we used the original period 2π
instead, then the computations would be twice longer: still the result does not change: but, may be in
different expression....
P∞
For the complex Fourier series k=−∞ ck e2ikt (← this is for π periodic functions), use the formula
ck =
1
π
π/2
Z
f (t)e−i2kt dt.
−π/2
Let us compute
Z
π/2
f (t)e−i2kt dt
−π/2
Now two cases:
Case k = 0: Then,
Z
π/2
f (t)e
−ikt
Z
0
−2tdt +
dt =
−π/2
Z
−π/2
2
π/2
2tdt
0
= π /2
Case k 6= 0: We use
Z
−i2kt
te
te−i2kt
dt = −
+
2ki
Z
e−i2kt
dt
2ki
(Notice k is in the denominator, but this is fine in this case since we are in the case k 6= 0.)
A long computation goes as follows:
Z
π/2
f (t)e−i2kt dt =
−π/2
Z
0
=
−2te−i2kt dt +
−π/2
Z
0
10
π/2
2te−i2kt dt
Now,
0
Z
−2te−i2kt dt
−π/2
Z 0
n te−i2kt 0
e−i2kt o
+
dt
= −2 −
2ki −π/2
−π/2 2ki
o
n
π ikπ h e−i2kt i0
= −2 −
e +
i4k
4k 2 −π/2
io
n
π ikπ
1 h
= −2 −
e + 2 1 − eikπ
i4k
4k
π/2
Z
2te−i2kt dt
0
n te−i2kt π/2 Z π/2 e−i2kt o
+
=2 −
dx
2ki 0
2ki
0
n
π −ikπ h e−i2kt iπ/2 o
=2 −
e
+
i4k
4k 2 0
n
io
π −ikπ
1 h −ikπ
=2 −
−1 .
e
+ 2 e
i4k
4k
Thus,
Z
π/2
f (t)e−ikt dt =
−π/2
Z
0
−2te−ikt dt +
=
π/2
Z
−π/2
2te−ikt dt
0
i
i
−2 h
2 h
= 2 1 − eikπ + 2 e−ikπ − 1
4k
4k
i
−1 h
ikπ
(since eikπ = e−ikπ = (−1)k . )
= 2 1−e
k
i
−1 h
= 2 1 − (−1)k
k
(
−2
k = odd,
2
= k
0
k = even.
Thus,
1
ck =
π
(
Z
−2
k2 π
0
π
f (t)e−ikt dt
−π
k = odd ±1, ±3, · · · ,
k = even = ±2, ±4, · · · .
Finally, the complex Fourier series is
X
f (t) = π 2 /2 +
k=±1,±3,±5,···
−2 i2kt
e
k2 π
P∞ (b): For the trigonometric Fourier series, f (t) = a20 + n=1 an cos(nt) + bn sin(nt) , let us use the results
in Problem 5. Note that the function is even, so we have Fourier cosine series (if we see f (t) as π-periodic
funciton)
f (t) = a0 /2 +
∞
X
k=1
11
ak cos(2kt)
(Note that if we use the period 2π still the result does not change: but, may be in different expression....)
Now use 5 (a), to see
ak = 2ck =

2

π
for k = 0,
for k = ±1, ±3, ±4, · · · ,
otherwise.
−4
 k2 π

0
Therefore, we have
X
f (t) = π 2 /2 +
k=±1,±3,±5,···
8. (Do NOT hand-in this problem.)

−1,





0,
f (x) =





1,
−4
cos(2kt)
k2 π
Let f (x) be the 2π-periodic function determined by the following:
for −π ≤ x <
−2π
3
and
− π3 ≤ x < 0
−2π
3
−π
3
and
π
3
and
2π
3
for
≤x<
0≤x<
for
π
3
≤x<
2π
3
≤x<π
Draw the graph of f (x) and compute its (complex) Fourier series.
Solution See: Problem 4 in
http://www.iam.ubc.ca/˜ sospedra/a3MATH267-sol.pdf
9. (Do NOT hand-in this problem.) Let f (x) be the 2π-periodic function defined on −π ≤ x < π by
π + x, for −π ≤ x < 0
f (x) =
π − x, for 0 ≤ x < π
(a) Draw the graph of f (x) for −3π ≤ x ≤ 3π.
(b) Find the Fourier series of f (x) in the complex form:
f (x) =
∞
X
cn einx
n=−∞
(c) Find the Fourier series of f (x) in the trigonometric form (equivalently its real Fourier series):
∞
f (x) =
a0 X
+
an cos(nx) + bn sin(nx)
2
n=1
Solution
See: Problem 2 in
http://www.iam.ubc.ca/˜ sospedra/a3MATH267-sol.pdf
10. (Do NOT hand-in this problem.) The temperature, u(x, t), in a metal rod of length 2L satisfies
ut = uxx ,
−L < x < L,
t > 0.
The ends of the rod at x = −L and x = L satisfy the boundary conditions
u(−L, t) = u(L, t)
and ux (−L, t) = ux (L, t),
t > 0.
(a) Use the method of separation of variables to obtain the solution u(x, t) satisfying the initial condition
u(x, 0) = f (x), −L < x < L.
(b) Let L = π. Find the solution u(x, t) in part (a) when the rod is initially at a constant temperature
u(x, 0) = π.
(c) What is the limit of u(x, t) in part (b) when t → ∞?
Solution
See Problem 1 in
http://www.iam.ubc.ca/˜ sospedra/a3MATH267-sol.pdf
12