Powers D.L. Boundary Value Problems and Partial Differential Equations
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378 Chapter 6 Laplace Transform
from which we find the solution,
u(x, t) =
cos(πt) −
1
π
sin(πt)
sin(πx).
Example 3.
Now we consider a problem that we know to have a more complicated solu-
tion:
∂
2
u
∂x
2
=
∂u
∂t
, 0 < x < 1, 0 < t,
u(0, t) = 1, u(1, t) = 1, 0 < t,
u(x, 0) = 0, 0 < x < 1.
The transformed problem is
d
2
U
dx
2
=sU, 0 < x < 1,
U(0, s) =
1
s
, U(1, s) =
1
s
.
The general solution of the differentialequationiswellknowntobeacombi-
nation of sinh(
√
sx) and cosh(
√
sx). Application of the boundary conditions
yields
U(x, s) =
1
s
cosh
√
sx
+
(1 −cosh(
√
s)) sinh(
√
sx)
s sinh(
√
s)
=
sinh(
√
sx) +sinh(
√
s(1 −x))
s sinh(
√
s)
.
This function rarely appears in a table of transforms. However, by extending
the Heaviside formula, we can compute an inverse transform.
When U is the ratio of two transcendental functions (not polynomials) of s,
we wish to write
U(x, s) =
A
n
(x)
1
s −r
n
.
In this formula, the numbers r
n
are values of s for which the “denominator” of
U is zero, or, rather, for which |U(x, s)| becomes infinite; the A
n
are functions
of x but not s .Fromthisformweexpecttodetermine
u(x, t) =
A
n
(x) exp(r
n
t).
This solution should be checked for convergence.
The hyperbolic sine (also the cosh, cos, sin, and exponential functions) is
not infinite for any finite value of its argument. Thus U(x, s) becomes infinite
6.3 Partial Differential Equations 379
only where s or sinh(
√
s) is zero. Because sinh(
√
s) =0 has no real root besides
zero, we seek complex roots by setting
√
s =ξ +iη (ξ and η real).
The addition rules for hyperbolic and trigonometric functions remain valid
for complex arguments. Furthermore, we know that
cosh(iA) = cos(A), sinh(iA) =i sin(A).
By combining the addition rule and these identities we find
sinh(ξ +iη) = sinh(ξ ) cos(η) +i cosh(ξ) sin(η).
This function is zero only if both the real and imaginary parts are zero. Thus ξ
and η must be chosen to satisfy simultaneously
sinh(ξ ) cos(η) = 0, cosh(ξ ) sin(η) = 0.
Of the four possible combinations, only sinh(ξ) = 0 and sin(η) = 0produce
solutions. Therefore ξ =0andη =±nπ (n = 0, 1, 2,...); whence
√
s =±inπ, s =−n
2
π
2
.
Recall that only the value of s,notthevalueof
√
s, is significant.
Finally then, we have located r
0
= 0, and r
n
=−n
2
π
2
(n = 1, 2,...). We
proceed to find the A
n
(x) by the same method used in Section 6.2. The com-
putations are done piecemeal and then the solution is assembled.
Part a. (r
0
= 0.) In order to find A
0
we multiply both sides of our proposed
partial fractions development
U(x, s) =
∞
n=0
A
n
(x)
1
s −r
n
by s − r
0
= s and take the limit as s approaches r
0
= 0. The right-hand side
goes to A
0
. On the left-hand side we have
lim
s→0
s
sinh(
√
sx) + sinh(
√
s(1 −x))
s sinh(
√
s)
=x +1 −x =1 =A
0
(x).
Thus the part of u(x, t) corresponding to s = 0is1· e
0t
= 1, which is easily
recognized as the steady-state solution.
Part b. (r
n
=−n
2
π
2
, n = 1, 2,....) For these cases, we find
A
n
=
q(r
n
)
p
(r
n
)
,
380 Chapter 6 Laplace Transform
where q and p are the obvious choices. We take
√
r
n
=+inπ in all calculations:
p
(s) = sinh
√
s
+s
1
2
√
s
cosh
√
s
,
p
(r
n
) =
1
2
inπ cosh(inπ) =
1
2
inπ cos(nπ),
q(r
n
) = sinh(inπx) +sinh
inπ(1 −x)
= i
sin(nπx) +sin
nπ(1 −x)
.
Hence the portion of u(x, t) that arises from each r
n
is
A
n
(x) exp(r
n
t) = 2
sin(nπx) +sin(nπ(1 − x))
nπ cos(nπ)
exp
−n
2
π
2
t
.
Part c. On assembling the various pieces of the solution, we get
u(x, t) = 1 +
2
π
∞
1
sin(nπx) +sin(nπ(1 − x))
n cos(nπ)
exp
−n
2
π
2
t
.
The same solution would be found by separation of variables but in a slightly
different form.
Example 4.
Now consider the wave problem
∂
2
u
∂x
2
=
∂
2
u
∂t
2
, 0 < x < 1, 0 < t,
u(0, t) = 0,
∂u(1, t)
∂x
=0, 0 < t,
u(x, 0) = 0,
∂u(x, 0)
∂t
=x, 0 < x < 1.
The transformed problem is
d
2
U
dx
2
=s
2
U −x, 0 < x < 1,
U(0, s) = 0, U
(1, s) = 0,
and its solution (by undetermined coefficients or otherwise) gives
U(x, s) =
sx cosh(s) −sinh(sx)
s
3
cosh(s)
.
The numerator of this function is never infinite. The denominator is zero at
s = 0ands =±i(2n − 1)π/2(n = 1, 2,...). We shall again use the Heaviside
formula to determine the inverse transform of U.
6.3 Partial Differential Equations 381
Part a.
(r
0
= 0.) The limit as s approaches zero of sU(x, s) may be found by
L’Hôpital’s rule or by using the Taylor series for sinh and cosh. From the latter,
sU(x, s) =
sx
1 +
s
2
2
+···
−
sx +
s
3
x
3
6
+···
s
2
1 +
s
2
2
+···
=
s
3
x
2
−
x
3
6
−···
s
2
1 +
s
2
2
+···
→0.
Thus, in spite of the formidable appearance of s
3
in the denominator, s = 0is
not really a significant value and contributes nothing to u(x, t).
Part b. It is convenient to take the remaining roots in pairs. We label
±i(2n −1)π
2
=±iρ
n
.
The derivative of the denominator is
p
(s) = 3s
2
cosh(s) +s
3
sinh(s),
p
(±iρ
n
) =±i
3
ρ
3
n
sinh(±iρ
n
)
= ρ
3
n
sin(ρ
n
)
since sinh(iρ) = i sin(ρ) and (±i)
4
= 1. The contribution of these two roots
together may be calculated using the exponential definition of sine:
q(iρ
n
)
p
(iρ
n
)
exp(iρ
n
t) +
q(−iρ
n
)
r
(−iρ
n
)
exp(−iρ
n
t)
=
−sinh(iρ
n
x) exp(iρ
n
t) +sinh(iρ
n
x) exp(−iρ
n
t)
ρ
3
n
sin(ρ
n
)
=
sin(ρ
n
x)
ρ
3
n
sin(ρ
n
)
i
−exp(iρ
n
t) +exp(−iρ
n
t)
=
2sin(ρ
n
x) sin(ρ
n
t)
ρ
3
n
sin(ρ
n
)
.
Part c. The final form of u(x, t), found by adding up all the contributions
from Part b, is the same as would be found by separation of variables
u(x, t) = 2
∞
1
sin(ρ
n
x) sin(ρ
n
t)
ρ
3
n
sin(ρ
n
)
.
382 Chapter 6 Laplace Transform
EXERCISES
1. Find all values of s, real and complex, for which the following functions are
zero.
a. cosh(
√
s);
c. sinh(s);
e. cosh(s) +s sinh(s).
b. cosh(s);
d. cosh(s) −s sinh(s);
2. Find the inverse transforms of the following functions in terms of an infi-
nite series.
a.
1
s
tanh(s);
b.
sinh(sx)
s cosh(s)
.
3. Find the transform U(x, s) of the solution of each of the following prob-
lems.
a.
∂
2
u
∂x
2
=
∂u
∂t
, 0 < x < 1, 0 < t,
u(0, t) = 0, u(1, t) =t, 0 < t,
u(x, 0) = 0, 0 < x < 1;
b.
∂
2
u
∂x
2
=
∂u
∂t
, 0 < x < 1, 0 < t,
u(0, t) = 0, u(1, t) =e
−t
, 0 < t,
u(x, 0) = 1, 0 < x < 1.
4. Solve each of the problems in Exercise 3, inverting the transform by means
of the extended Heaviside formula.
5. Solve each of the following problems by Laplace transform methods.
a.
∂
2
u
∂x
2
=
∂u
∂t
, 0 < x < 1, 0 < t,
u(0, t) = 0, u(1, t) = 1, 0 < t,
u(x, 0) = 0, 0 < x < 1;
b.
∂
2
u
∂x
2
=
∂u
∂t
, 0 < x < 1, 0 < t,
u(0, t) = 0, u(1, t) = 0, 0 < t,
u(x, 0) = 1, 0 < x < 1.
6.4 More Difficult Examples 383
6.4 More Difficult Examples
The technique of separation of variables, once mastered, seems more straight-
forward than the Laplace transform. However, when time-dependent bound-
ary conditions or inhomogeneities are present, the Laplace transform offers
a distinct advantage. Following are some examples that display the power of
transform methods.
Example 1.
A uniform insulated rod is attached at one end to an insulated container of
fluid. The fluid is circulated so well that its temperature is uniform and equal
to that at the end of the rod. The other end of the rod is maintained at a con-
stant temperature. A dimensionless initial value–boundary value problem that
describes the temperature in the rod is
∂
2
u
∂x
2
=
∂u
∂t
, 0 < x < 1, 0 < t,
∂u(0, t)
∂x
=γ
∂u(0, t)
∂t
, u(1, t) = 1, 0 < t,
u(x, 0) = 0, 0 < x < 1.
The transformed problem and its solution are
d
2
U
dx
2
=sU, 0 < x < 1,
dU
dx
(0, s) = sγ U(0, s), U(1, s) =
1
s
,
U(x, s) =
cosh(
√
sx) +
√
sγ sinh(
√
sx)
s(cosh(
√
s) +
√
sγ sinh(
√
s))
=
q(s)
p(s)
.
Aside from s = 0, the denominator has no real zeros. Thus we again search
for complex zeros by employing
√
s =ξ +iη. The real and imaginary parts of
the denominator are to be computed by using the addition formulas for cosh
and sinh. The requirement that both real and imaginary parts be zero leads to
the equations
cosh(ξ ) +ξγ sinh(ξ )
cos(η) − ηγ cosh(ξ ) sin(η) = 0, (1)
ηγ sinh(ξ) cos(η) +
sinh(ξ ) +ξγ cosh(ξ)
sin(η) = 0. (2)
We may think of these as simultaneous equations in sin(η) and cos(η).Because
sin
2
(η) + cos
2
(η) = 1, the system has a solution only when its determinant is
zero. Thus after some algebra, we arrive at the condition
1 +ξ
2
γ
2
+η
2
γ
2
sinh(ξ ) cosh(ξ ) +ξγ
sinh
2
(ξ) +cosh
2
(ξ)
=0.
384 Chapter 6 Laplace Transform
The only solution occurs when ξ = 0, for otherwise both terms of this equa-
tion have the same sign.
After setting ξ =0, we find that Eq. (1) reduces to
tan(η) =
1
ηγ
,
for which there is an infinite number of solutions. We shall number the posi-
tive solutions η
1
,η
2
,.... Now, we have found that the roots of p(s) are r
0
= 0
and r
k
=(iη
k
)
2
=−η
2
k
. The computation of the inverse transform follows.
Part a. (r
0
= 0.) The limit of sU(x, s) as s tendstozeroiseasilyfoundtobe
1. Thus this root contributes 1 ·e
0t
=1tou(x, t).
Part b. (r
k
=−η
2
k
.) First, we compute
p
(s) =cosh
√
s
+
√
sγ sinh
√
s
+
1
2
√
s(1 +γ)sinh
√
s
+
1
2
γ s cosh
√
s
.
Using the fact that cosh(
√
r
k
) +
√
r
k
γ sinh(
√
r
k
) =0, we may reduce the fore-
going to
p
(r
k
) =
−1
2γ
1 +γ +η
2
k
γ
2
cos(η
k
).
Hence the contribution to u(x, t) of r
k
is
q(r
k
)
p
(r
k
)
exp(r
k
t) =−2γ
cos(η
k
x) −η
k
γ sin(η
k
x)
(1 +γ +η
2
k
γ
2
) cos(η
k
)
exp
−η
2
k
t
.
Part c. The construction of the final solution is left to the reader. We note
that an attempt to solve this problem by separation of variables would find
difficulties, for the eigenfunctions are not orthogonal.
Example 2.
Sometimes one is interested not in the complete solution of a problem, but
only in part of it. For example, in the problem of heat conduction in a semi-
infinite solid with time-varying boundary conditions, we may seek that part of
the solution that persists after a long time. (This may or may not be a steady-
state solution.) Any initial condition that is bounded in x gives rise only to
transient temperatures; these being of no interest, we assume a zero initial con-
dition. Thus, the problem to be studied is
∂
2
u
∂x
2
=
∂u
∂t
, 0 < x < ∞, 0 < t,
u(0, t) = f (t), 0 < t,
u(x, 0) = 0, 0 < x < ∞.
6.4 More Difficult Examples 385
The transformed equation and its general solution are
d
2
U
dx
2
=sU, 0 < x, U(0, s) = F(s),
U(x, s) = A exp
−
√
sx
+B exp
√
sx
.
We make two further assumptions about the solution: first, that u(x, t) is
bounded as x tends to infinity and, second, that
√
s means the square root
of s that has a nonnegative real part. Under these two assumptions, we must
choose B = 0andA = F(s),making
U(x, s) = F(s) exp
−
√
sx
.
In order to find the persistent part of u(x, t), we apply the Heaviside inver-
sion formula to those values of s having nonnegative real parts, because a value
of s with negative real part corresponds to a function containing a decaying
exponential — a transient. To understand this fact, consider the pair
f (t) = 1 −e
−βt
+α sin(ωt),
F(s) =
1
s
−
1
s +β
+
αω
s
2
+ω
2
.
Now we return to the original problem with this choice for f (t).Thevalues
of s for which U(x, s) = F(s) exp(−
√
sx) becomes infinite are 0, ±iω, −β.The
last value is discarded, because it is negative. Thus, the persistent part of the
solution is given by
A
0
e
0t
+A
1
e
iωt
+A
2
e
−iωt
,
and the coefficients are found from
A
0
= lim
s→0
sF(s) exp
−
√
sx
=1,
A
1
= lim
s→iω
(s −iω)F(s) exp
−
√
sx
=
α
2i
exp
−
√
iωx
,
A
2
= lim
s→−iω
(s +iω)F(s) exp
−
√
sx
=−
α
2i
exp
−
√
−iωx
.
We also need to know that the roots of ±i with positive real part are
√
i =
1
√
2
(1 +i),
√
−i =
1
√
2
(1 −i).
Thus the function we seek is
386 Chapter 6 Laplace Transform
1 +
1
2i
exp
iωt −
ω
2
(1 +i)x
−
α
2i
exp
−iωt −
ω
2
(1 −i)x
=1 +α exp
−
ω
2
x
sin
ωt −
ω
2
x
.
Example 3.
If a steel wire is exposed to a sinusoidal magnetic field, the boundary value–
initial value problem that describes its displacement is
∂
2
u
∂x
2
=
∂
2
u
∂t
2
−sin(ωt), 0 < x < 1, 0 < t,
u(0, t) = 0, u(1, t) = 0, 0 < t,
u(x, 0) = 0,
∂u
∂t
(x, 0) = 0, 0 < x < 1.
The nonhomogeneity in the partial differential equation represents the effect
of the force due to the field. The transformed equation and its solution are
d
2
U
dx
2
=s
2
U −
ω
s
2
+ω
2
, 0 < x < 1,
U(0, s) = 0, U(1, s) = 0,
U(x, s) =
ω
s
2
(s
2
+ω
2
)
cosh(
1
2
s) −cosh
s(
1
2
−x )
cosh(
1
2
s)
.
Several methods are available for the inverse transformation of U.Anobvi-
ous one would be to compute
v(x, t) =
L
−1
cosh(
1
2
s) −cosh
s(
1
2
−x )
s
2
cosh(
1
2
s)
and write u(x, t) as a convolution
u(x, t) =
t
0
sin
ω
t −t
v
x, t
dt
.
The details of this development are left as an exercise.
We could also use the Heaviside formula. The application is now rou-
tine, except in the interesting case where cosh(iω/2) = 0, that is, where ω =
(2n −1)π , one of the natural frequencies of the wire.
Let us suppose ω = π ,so
U(x, s) =
π
s
2
(s
2
+π
2
)
cosh(
1
2
s) −cosh
s(
1
2
−x)
cosh(
1
2
s)
.
6.4 More Difficult Examples 387
At the points s = 0, s =±iπ, s =±(2n −1)iπ , n = 2, 3,..., U(x, s) becomes
undefined. The computation of the parts of the inverse transform correspond-
ing to the points other than ±iπ is easily carried out. However, at these two
troublesome points, our usual procedure will not work. Instead of expecting a
partial-fraction decomposition containing
A
−1
s +iπ
+
A
1
s −iπ
and other terms of the same sort, we must seek terms like
A
−1
(s +iπ) +B
−1
(s +iπ)
2
+
A
1
(s −iπ) +B
1
(s −iπ)
2
,
whose contribution to the inverse transform of U would be
A
−1
e
−iπt
+B
−1
te
−iπt
+A
1
e
iπt
+B
1
te
iπt
.
One can compute A
1
and B
1
, for example, by noting that
B
1
= lim
s→iπ
(s −iπ)
2
U(x, s)
,
A
1
= lim
s→iπ
(s −iπ)
U(x, s) −
B
1
(s −iπ)
2
and similarly for A
−1
and B
−1
. The limit for B
1
is not too difficult. For example,
B
1
= lim
s→iπ
π
s
2
(s +iπ)
cosh(
1
2
s) −cosh
s(
1
2
−x )
cosh(
1
2
s)
/(s −iπ)
=
π
−π
2
(2iπ)
cosh(
1
2
iπ)−cosh
iπ(
1
2
−x )
1
2
sinh(
1
2
iπ)
=
−1
π
2
cos
π
1
2
−x
=
−1
π
2
sin(πx).
The limit for A
1
is rather more complicated but may be computed by
L’Hôpital’s rule. Nevertheless, because B
−1
= B
1
,wealreadyseethatu(x, t)
contains the term
B
1
te
iπt
+B
−1
te
−iπt
=−
2t
π
2
sin(πx) cos(πt),
whose amplitude increases with time. This, of course, is the expected reso-
nance phenomenon.