MATH 338 - Assignment 6
Solutions
Section 3.6: *19 We consider the heat equation with c = 1, so ut = uxx , for 0 < x < 1
(that is, L = 1). Our boundary conditions are ux (0, t) = u(0, t) and ux (1, t) = u(1, t),
with generic initial condition u(x, 0) = f
MATH 338 - Assignment 3
Solutions
* The trigonometric system with arbitrary period p is orthogonal. In particular,
p
sin
p
n
x sin
p
m
x
p
dx = 0
if m = n. Prove this.
Recall that
1
sin a sin b = [cos(a b) cos(a + b)]
2
So
p
sin
p
n
x sin
p
1 p
(n m)
(n +
MATH 338 - Assignment 1
Solutions
*2. p. 6, #13. We are asked to solve ux + (sin x)uy = 0 by the method of
characteristic curves. Here p(x, y) = sin x, so we have to solve dy/dx = sin x.
This gives y = cos x + C, or y + cos x = C. Thus u = f (y + cos x) s
MATH 338 - Assignment 4
Solutions
Section 2.4: *17
(a) This is straightforward.
(b) We are asked for the sine series, so we need bn .
bn =
=
=
p
2
p
f (x) sin
0
a
2
p
0
2h
ap
n
x dx
p
h
n
2
x sin
x dx +
a
p
p
a
x sin
0
p
a
h
n
(x p) sin
x dx
ap
p
n
2h
x d
MATH 338 - Assignment 9
Solutions
Sec 4.3: 11* We are asked to solve the boundary value problem
u
= c2
t
2 u 1 u
1 2u
+ 2 2 , 0 < r < a, 0 < < 2, t > 0 ,
+
r2
r r r
u(a, , t) = 0 , 0 < < 2, t > 0 ,
u(r, , 0) = f (r, ) ,
0 < r < a, 0 < < 2 .
(1)
(2)
(3)
MATH 338 - Assignment 2
Solutions
*1. [3] . . . for the simple case with c = 1 and L = 1, the functions
un (x, t) = sin(nx) cos(nt)
solve the boundary value problem
utt = uxx ,
u(0, t) = 0 , u(1, t) = 0 for all t > 0
(1)
(a) Does
1
1
1
u(x, t) =
sin(x) co
MATH 338 - Midterm Solutions
1. We consider the following boundary value problem for the heat equation
with c = 1 and L = 1:
ut = uxx ,
ux (0, t) = u(0, t) ,
0 < x < 1,
t>0
ux (1, t) = u(1, t)
for all t > 0
u(x, 0) = f (x)
a) Using separation of variables
MATH 338 - Assignment 7 Solutions
Throughout this course weve found a lot of solutions, to various problems,
that are innite series: they consist of an innite number of terms, added
up. Well, that immediately raises the issue of convergence: when you add
MATH 338 - Assignment 5
Solutions
Section 3.4: *4: We are asked to apply dAlemberts formula when f (x) = 0 and
g(x) = 1 for c = 1. The odd extension of g(x) is
1
0<x<1
1 1 < x < 0
g (x) =
Now we nd an expression for the antiderivative G(x) of g (x); remem
MATH 338 - Assignment 10
Solutions
Sec 6.1: 6* We can prove the orthogonality of 1, 1 x, (2 4x + x2 )/2 with respect
to ex on [0, ) by integrating directly, using techniques such as integration by parts.
But we can observe something more general, concerni
MATH 338 - Assignment 8
Solutions
Sec 4.7: 20* We have the DE x2 y +xy +(4x4 1 )y = 0, which as it stands is not quite a
4
Bessel equation. Let z = x2 ; then x = z, and y(x) = y(x(z) = Y (z). Using the chain
rule gives y (x) = Y (z)2x, and then y (x) = Y
Name:
MATH 338 - Quiz 7
1
1. Let f (x) =
e x
0
0<x<
.
< x 0
(a) Convince yourself that for any x > 0, the kth derivative f (k) (x) is a linear combi1
e x
nation of terms of the form m , m an integer.
x
This is straightforward.
1
1
1 1
(b) Show that m+1 (