CHAP.
BOUNDARY
1]
Thus
VALUE
B
y(x, t)
is a solution.
-2- cos 2
.
m11'X
Sin
Since this solution is bounded, the condition
In order to satisfy the last condition,
of superposition to obtain the solution
y(x,
17
PROBLEMS
Iy(x, t) I
5 sin
0)
m11't
<
3 sin
11'X -
M
is automatically satisfied.
411'x,
we first use the principle
(5)
y(x, t)
Then putting
t = 0 we arrive at
y(x, 0)
6 sin
This is possible if and only if BI = 6,
(5) is
y(x, t) 6 sin
ml
11'X -
= 2,
Bz
3 sin
-3,
mz
3 sin
411'x
=
11'X cos 411't -
411'X
= 8. Thus the required solution
cos 1611't
(6)
This boundary value problem can be interpreted physically in terms of the vibrations of a string.
The string has its ends fixed at x = 0 and x = 2 and is given an initial shape f(x) = 6 sin 11'X 3 sin 411'x. It is then released so that its initial velocity is zero. Then (6) gives the displacement
of any point x of the string at any later time t.
1.25.
au
azu
Solve
at = 2 axz,'
lu(x, t)j
< M.
0
< x < 3, t> 0, given that
u(O, t)
= u(3,
t)
= 0,
u(x,O)
= f(x),
This problem differs from Problem 1.23 only in the condition u(x,O) = f(x).
In seeking to
satisfy this last condition we see that taking a finite number of terms, as in (1) of Problem 1.23,
Thus we are led to assume that infinitely many terms are
will be insufficient for arbitrary f(x).
'"
taken, i.e.
u(x, t)
~
Bme-Zm211'2t/9
sin m11'X
m=l
The condition
u(x,O)
=
f(x)
3
then leads to
co
•
f(x)
m~l
Bm
m7TX
sIn-3-
or the problem of expansion of a function into a sine series.
Fourier series, will be considered in detail in the next chapter.
Supplementary
MATHEMATICAL
1.26.
OF PHYSICAL
expansions, or
Problems
PROBLEMS
If a taut, horizontal string with fixed ends vibrates in a vertical plane under the influence of gravity, show that its equation is
where
1.27.
FORMULATION
Such trigonometric
g
is the acceleration due to gravity.
A thin bar located on the x-axis has its ends at
bar is f(x), 0 < x < L, and the ends x = 0, x =
respectively. Assuming the surrounding medium
cooling applies, show that the partial differential
point at any time is given by
au
at
and write the corresponding boundary conditions.
= 0 and x = L. The initial temperature of the
L are maintained at constant temperatures Tl' Tz
is at temperature Uo and that Newton's law of
equation for the temperature of the bar at any
x
x2
BOUNDARY VALUE PROBLEMS
18
[CHAP. 1
1.28.
Write the boundary conditions in Problem 1.27 if (a) the ends x = 0 and x = L are insulated,
(b) the ends x = 0 and x = L radiate
into the surrounding medium according to Newton's law
of cooling.
1.29.
The gravitational potential v at any point (x, y, z) outside of a mass m located at the point (X, Y, Z)
is defined as the mass m divided by the distance of the point (x, y, z) from (X, Y, Z). Show that
v satisfies Laplace's equation V2v = O.
1.30.
Extend the result of Problem 1.29 to a solid body.
1.31.
A string has its ends fixed at x = 0 and x = L. It is displaced a distance h at its midpoint and
then released. Formulate a boundary value problem for the displacement y(x, t) of any point x
of the string at time t.
CLASSIFICATION OF PARTIAL DIFFERENTIAL EQUATIONS
1.32. Determine whether each of the following partial differential equations is linear or nonlinear, state
the order of each equation, and name the dependent and independent variables.
(a)
(b)
1.33.
+
a2u
ax2
(x2
a3</>
x2
+
2 a2u
ax ay
a2u
ay2
- at2
0
a2T
a2T
ax2
ay2
ay2
x2-(a)
-
Show that
z(x, y)
-(x2au -1)+ au
ax2 +
a2u
(I)
ay2
ar
+
1
az
as
+ay (y2
2xy-ax
ay2
-1)-
EQUATIONS
= 4e-3x cos 3y is a solution to the boundary value problem
a2z
ax2
+
v(x, y)
=
(a)
Show that
(b)
Find a particular
ay2
-
xF(2x
0,
z(x,
+ y)
solution satisfying
= 0,
"/2)
is a general solution of
v(l, y)
=
u
1.37.
Find a partial differential equation having general solution
z
=
exf(2y - 3x),
(b)
GENERAL SOLUTIONS OF PARTIAL DIFFERENTIAL
v.
xav _ 2xav
ax
ay
=
f(2x
F(x - 3y)
+ y) +
+ G(2x + y).
g(x - 2y)
EQUATIONS
az
a2z
+ ay = o.
(a)
Solve
(b)
Find the particular
x ax ay
z
4e-3x
y2.
Find a partial differential equation having general solution
(a)
=
z(x,O)
1.36.
1.38.
az
(e)
(d)
a2u
a2u a2u
a2z
a2z a2u =
a2z
a2u
a2</>
x0x-Mau
+ >0
3y
+_ y-1)a2</>
a2u
a2u
a2u = (e)
0 4+_axy2_
+(M2
2xy-ax2
ay 0,
a2z
1.35.
-
ay3
a2y _ 4 a2y
SOLUTIONS OF PARTIAL DIFFERENTIAL
1.34.
-
Classify each of the following equations as elliptic, hyperbolic or parabolic.
_
axay
ax
ay
(c)
</>
-+-
+ y2) a4T
az4
ax2
ax
a</>
solution for which
z(x,O)
x5
+x
_ 68
x '
z(2,y)
3y4.
Z2
BOUNDARY VALUE PROBLEMS
CHAP. 1]
1.39.
Find general solutions of each of the following.
(a)
=
iJ2u
iJx2
iJ2u
iJy2
(b)
iJ2z
(d)
1.40.
19
°
2 iJxiJy
+
iJu
iJx
=
iJu
2 iJy
3u
2--+-
(e)
°
(c)
iJ2z
iJ2z
oX iJy
iJy2
°
Find general solutions of each of the following.
(a)
iJu
iJx
+
=
2iJu
iJy
X
(c)
iJ4u
iJx4
+
iJx3 iJy
3~
(d)
(b)
1.41.
Solve
1.42.
Show that a general solution of
=
2~
iJx iJy
+
4
2iJ2z
iJy2
sin y
X
16.
iJ2v
iJr2
+ gr
is
iJv
iJr
F(r - ct)
v
+ G(r + ct)
r
SEPARATION OF VARIABLES
1.43.
2u, ,
y- iJx'
Solve each of the fo1l0wing boundary value problems by the method of separation
iJu
iJu
iJuiJ2u
317X
-_ 3
iJx2'
iJ2u
3iJu
u0)
iJu
iJ2u
u(x,
U(17,
t)t)0, =
=- 0,
2=
3x 2iJuu(4,
u(2,
=- u(x,O)
0,sin
iJu= 2iJU
(a)iJx2' ux(O,
=10e-X
3e-5x
4e-x
u(x,O)
t)u(x,O)
=
u(O,
t) t)u(x,O)
-u(O,
u(x,O)
0,
0, =
= 8e-2x
60, sin
17; -
+
+
917X
=
u(x,O)
4 2e-3x
sin
5x
36e-4",
sin 17X
++
8 cosT
of variables.
- 6 cosT
(b)(I)
(e)
(c)
(g)
(d)
1.44.
Solve and give a physical interpretation
if
iJ2y
iJt2
=
(a)
f(x)
iJ2y
4 iJx2'
y(O,
= 5 sin
iJu
iJ2u
iJt
iJx2 -
=
t)
17X,
(b)
y(5,
t) = 0,
Solve
1.46.
Suppose that in Problem 1.24 we have
can be solved if we know how to expand
2u
if
u(O,
y(x,O)
=
0,
= 3 sin 217X - 2 sin
f(x)
1.45.
1.47.
to the boundary value problem
t)
0,
u(3, t)
y(x,O)
f(x)
0,
=
Yt(x,O)
(0
f(x)
<x
< 5, t>
0)
517X.
2 sin
u(x, 0)
= f(x), where ° <
in a series of sines.
x
< 2.
17X-
sin
417X.
Show how the problem
Suppose that in Problem 1.25 the boundary conditions are u",(O, t) = 0, u(3, t) = 0, u(x,O) = f(x).
Show how the problem can be solved if we know how to expand f(x) in a series of cosines. Give
a physical interpretation of this problem.