Summer School, 2012 Page 4 of 26
1. [14 marks total]
{a} [2 marks] Find the general solution ydr) of the diffmonrial equation
11; + 9y 2 U. Summer School, 2012 Page 5 01 26
{h} [3 marks] Find a particular solution yplfli) of
I! + 9.1} I sin 1?.
Jla
MATH2065: INTRO TO PDEs
Summer School 2014
Tutorial Questions 3
Questions marked with the dagger symbol are intentionally more challenging.
1. Find the Laplace transforms of each of the following functions.
(a) t3 e2t
(b) t sin 4t
(c) H (t 3)
(d) e3t sin
MATH2065: INTRO TO PDEs
Summer School 2014
Tutorial Questions 2
Questions marked with the dagger symbol are intentionally more challenging.
1. Use the method of undetermined coecients to nd a particular solution to each of the
following inhomogeneous dier
MATH2065: INTRO TO PDEs
Summer School 2014
Tutorial Questions 1
1. Show that the ordinary dierential equation
dx
= x,
dt
= constant
has the general solution
x = A et ,
where A is a constant.
Hence solve
(a) dx/dt = 3x,
x = 2 when t = 0.
(b) dy/dz = 5y,
y
MATH2065: INTRO TO PDEs
Summer School 2014
Tutorial Questions 6
1. Determine the eigenvalues and corresponding eigenfunctions of the dierential equation
d2
+ = 0
dx2
d
(L) = 0. Analyze the three cases with real :
with boundary conditions (0) = 0 and
dx
MATH2065: INTRO TO PDEs
Summer School 2014
Tutorial Questions 4
1. An electrical circuit in which a resistor (R) and an inductor (L) are connected in series
with a voltage source (V (t) has a resulting current ow i(t). Using Kirchos Laws, it is
possible t
MATH2065: INTRO TO PDEs
Summer School 2014
Tutorial Questions 5
1. To practice your partial dierentiation, verify that
u(x, t) = sin (2x) e4
2 kt
is a solution to the heat equation
u
2u
=k 2.
t
x
2. Verify that u(x, t) = sin (x ct) is a solution to the wa
Summer School 2014
Tutorial Questions 9
Questions marked with the dagger symbol are intentionally more challenging.
1. For the following functions, sketch the Fourier series of f (x) on the interval L x L
and determine the Fourier coecients:
(a) f (x) = x
MATH2065: INTRO TO PDEs
Summer School 2014
Tutorial Questions 7
1. Solve Laplaces equation
2 2
+ 2 =0
x2
y
for (x, y ) in the square 0 < x < L, 0 < y < L, subject to the boundary conditions
(x, 0)
(x, L)
(0, y )
(L, y )
=
=
=
=
0 for 0 x L,
0 for 0 x L,
0
Summer School 2014
Tutorial Questions 13
This Tutorial Set contains a selection of questions covering material over the semester.
1. Consider the wave equation
2u
2u
= c2 2
;
t2
x
u(x, 0) = f (x) ,
u
(x, 0) = 0 .
t
< x < ,
(a) Let U (, t) = F cfw_u(x, t)
Summer School 2014
Tutorial 8
There is no new material in this tutorial; Instead, you are being given a chance to revise
previous work.
1. Consider the initial value problem
d
5 = sin t + (t + 1) e5t
dt
;
(0) = 0 ,
where the unknown is (t). In First Year
MATH2065: INTRO TO PDEs
Summer School 2014
Tutorial Solutions 1
1. The general solution has been covered in lectures.
(a) x = 2e3t .
(b) y = 10e5(1z) .
(c) The transformed equation is dX/dt = 3X , with solution X = Ae3t or 43x = Ae3t .
4
Hence A = 4 and t
Summer School 2014
Tutorial Questions 10
Questions marked with the dagger symbol are intentionally more challenging.
1. (a) Calculate the Fourier series for the 2 -periodic extension of the function
f (x) = x2 ,
Answer:
2
3
+4
(1)n
n=1 n2
< x .
cos nx.
Summer School 2014
Tutorial Questions 12
Questions marked with the dagger symbol are intentionally more challenging.
1. Determine the solution to
2u
u
=2
<x< ,
t
x
2
u(x, 0) = 70 ex /2 .
t > 0,
by directly applying the Fourier transform method to the PDE.
Summer School 2014
Tutorial Questions 11
Questions marked with the dagger symbol are intentionally more challenging.
1. If is a positive constant, show that the Fourier transform of e|x| is given by
F e|x| =
1
.
2 + 2
2. If F ( ) = e| , > 0, determine the
MATH2065: INTRO TO PDEs
Summer School 2014
Tutorial Solutions 2
1. (a) Try yp = Aex , where A is a constant to be determined. Dierentiating this form gives
yp = Aex
and yp = Aex .
1
Substituting into the ODE we nd A = 3 . A particular solution is therefor
MATH2065: INTRO TO PDEs
Summer School 2014
Tutorial Solutions 4
1. (a) Notice that the voltage input is zero for 0 < t < a, is V0 between a and b, and is
zero thereafter so this is a square pulse. Denoting I (s) = L cfw_i(t), and applying the
Laplace tran
MATH2065: INTRO TO PDEs
Summer School 2014
Tutorial Solutions 5
1. Taking partial derivatives of u:
u
2
= 4 2 k sin 2xe4 kt ,
t
u
2
= 2 cos 2xe4 kt ,
x
2u
2
= 4 2 sin 2xe4 kt .
2
x
The last of these when multiplied by k gives
u
t
as required.
2. Let u(x,
MATH2065: INTRO TO PDEs
Summer School 2014
Tutorial Solutions 3
6
(each time we multiply by t, it is equivalent to dierentiating the Laplace
(s + 2)4
transform with respect to s)
8s
(b) 2
(same reason as above)
(s + 16)2
1. (a)
e3s
(directly from the tabl
MATH2065: INTRO TO PDEs
Summer School 2014
Tutorial Solutions 6
1. We have the equation
(x) + (x) = 0 ,
which is a second-order constant-coecient, homogeneous ODE. While this is exactly the
equation we have looked at in class, the boundary conditions are
MATH2065: INTRO TO PDEs
Summer School 2014
Tutorial Solutions 7
1. A product solution (x, y ) = X (x) Y (y ) is assumed, and substitution into the PDE yields
X (x)Y (y ) + X (x)Y (y ) = 0
X (x)
Y (y )
=
= ,
X (x)
Y (y )
where is the separation constant. (
Summer School 2014
Tutorial Solutions 9
1. Here, we will just focus on determining the Fourier coecients.
(a) Within the interval (L, L), we have the Fourier series
nx
nx
x a0 +
an cos
+
bn sin
.
L
L
n=1
n=1
a0 = 0 since f (x) is odd. Similarly, an = 0 fo
Summer School 2014
Tutorial Solutions 8
1. The homogeneous solution satises
dh
5h = 0.
dt
The auxiliary equation is 5 = 0, so = 5. Therefore
h (t) = Ce5t
where C is an arbitrary constant.
Let p1 (t) satisfy
dp1
5p1 = sin t.
dt
For a solution, try
p1 (t)
Summer School 2014
Tutorial Solutions 10
1. (a) The bn Fourier coecients are all zero since f (x) is even. The period of the function
here is 2 , and hence we use our standard Fourier series formulas with 2L = 2 , i.e.,
with L = . Thus,
a0 =
=
=
=
1
x2 dx
Summer School 2014
Tutorial Solutions 11
1. Let f (x) = e|x| . Using the Fourier transform denition
|x|
Fe
1
=
2
1
=
2
e|x| eix dx
0
ex eix dx +
(+i )x 0
1
2
ex eix dx
0
(+i )x
1e
1e
+
2 + i 2 + i 0
1
1
1
1
=
2 + i 2 + i
1 + i i
=
2 (i + )(i )
1
1
2
=
.
Summer School 2014
Tutorial Solutions 12
1. Let U (, t) = F cfw_u(x, t), that is, the Fourier transform of the solution with respect to x,
while treating t as a parameter. Fourier transforming this heat equation, we get
U
= (i )2 U (, t) = 2 U (, t) .
t
T
Summer School 2014
Tutorial Solutions 13
1. (a) Using the table of Fourier transforms and (i )2 = 2 immediately gives the stated
equation.
(b) As and ODE in t the transformed equation has general solution A cos ct + B sin ct.
In the PDE the function U (,