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```Solved Sample Problems
1
The following differential equations arise in dealing with the problems noted. Find the
general solution of each equation assuming k 6= 0
a) Vibration of a beam.
d4 y
− k4y = 0
dx4
b) Beam on an elastic foundation.
d4 y
+ 4k 4 y = 0
dx4
c) Bending of a plate.
2
d4 y
2d y
−
2k
+ k4y = 0
dx4
dx2
a) The characteristic equation is
r4 − k4 = 0
with roots k, −k, ik and −ik so the general solution is
y = c1 exp(kx) + c2 exp(−kx) + c3 cos(kx) + c4 sin(kx)
b) The characteristic equation is
r4 + 4k 4 = 0
with roots (1 + i)k, (1 − i)k, (−1 + i)k and (−1 − i)k so the general solution is
y = exp(kx)(c1 cos(kx) + c2 sin(kx)) + exp(−kx)(c3 cos(kx) + c4 sin(kx))
1
c) The characteristic equation is
r4 − 2k 2 r2 + k 4 = 0
with roots k and −k (repeated), so the general solution is
y = exp(kx)(c1 + c2 x) + exp(−kx)(c3 + c4 x)
2
Express the function z π in the form
z π = exp(π log z)
√
and find the principal value when z = (1 + i)/ 2 = √12 + i √12 in the form a + ib.
rπ (cos(πθ) + i sin(πθ)) =
Let z = r exp(iθ) = r(cos(θ) + i sin(θ)) thus z π = rπ exp(iπθ) =
√
rπ (cos(π(θP + 2kπ)) + sin(π(θP + 2kπ))). Now using z = (1 + i)/ 2 (which implies θP = π4 )
with k = 0 yields the principal value cos(π 2 /4) + i sin(π 2 /4).
3
Obtain the solution of the following differential equation in terms of Maclaurin series
dy
d2 y
+x −y =0
2
dx
dx
P
n
Let f (x) = ∞
n=0 An x to obtain
0
(2A2 − A0 )x +
∞
X
[(n + 2)(n + 1)An+2 + nAn − An ]xn = 0
n=1
From the first coefficient A0 is an arbitrary constant; then for n = 1 with A1 arbitrary
constant, A3 = 0. For n ≥ 1
An+2 = −
(n − 1)
An
(n + 2)(n + 1)
2
Must now consider two cases: n even and n odd. The general solution is
f (x) = A1 x + A0 [1 + x2 /2 − x4 /24 + x6 /240 + ...]
4
Obtain the solution of the following differential equation in terms of Maclaurin series
Ly =
d2 y
−y =0
dx2
Assume the solution y(x) can be represented by a power series around x = 0, i.e.
y(x) =
∞
X
Ak xk
k=0
Differentiating
∞
X
d2 y
=
k(k − 1)Ak xk−2
dx2 k=0
and substituting in the original ODE
Ly =
∞
X
k(k − 1)Ak xk−2 −
k=0
∞
X
Ak xk = 0
k=0
Term by term comparison produces the recurrence relation
k(k − 1)Ak = Ak−2
for k = 2, 3, .... Therefore only two of the constant coefficients (A0 and A1 ) are independent
and the required solution of the problem is
1
1
1
1 5
x + ...) =
y(x) = A0 (1 + x2 + x4 + ...) + A1 (x + x3 +
2
24
6
120
= A0 cosh x + A1 sinh x = A0 y1 (x) + A1 y2 (x)
where y1 (x) = cosh x and y2 (x) = sinh x are two linearly independent solutions.
3
5
Use the method of Frobenius to obtain a general solution of the following differential equation
valid near x = 0
x
dy
d2 y
+ 2 + xy = 0
2
dx
dx
P
n+s
Let y = f (x) = ∞
. Thus
n=0 An x
xf (x) = x
∞
X
An x
n+s
∞
X
=
n=0
An x
n+s+1
n=0
=
∞
X
An−1 xn+s
n=1
where instead of the dummy index n, the dummy n − 1 was used.
Similarly
∞
∞
X
X
dy
0
n+s−1
= 2f (x) = 2
(n + s)An x
=2
(n + 1 + s)An+1 xn+s
2
dx
n=0
n=−1
Where the dummy index has been changed from n to n + 1 in order to have x raised to the
power n + s in the sum.
Finally,
∞
∞
X
X
d2 y
00
n+s−1
(n + s − 1)(n + s)An x
=
(n + s)(n + 1 + s)An+1 xn+s
x 2 = xf (x) =
dx
n=0
n=−1
Where the dummy index has also been changed from n to n + 1.
Now, substituting the above into the original ODE yields
∞
X
(n + s)(n + 1 + s)An+1 xn+s +
n=−1
∞
X
2(n + 1 + s)An+1 xn+s +
n=−1
∞
X
An−1 xn+s = 0
n=1
And division by xn+s yields the general recurrence relationship
∞
X
(n + 1 + s)(n + s)An+1 +
n=−1
∞
X
2(n + 1 + s)An+1 +
n=−1
∞
X
An−1 = 0
n=1
The indicial equation is now obtained from the above with n = −1 and it is
s(s − 1)A0 + 2sA0 = 0
Assuming A0 is arbitrary but 6= 0 one obtains the roots s1 = 0 and s2 = −1. Since s1 −s2 = 1
we must expect either no solution or a complete one for s2 = −1 and of course, one solution
for s1 = 0.
4
Now, when n = 0 the recurrence relation becomes
(1 + s)sA1 + 2(1 + s)A1 = A1 (1 + s)(s + 2) = 0
which is always true when s = −1, regardless of the value of A1 , therefore A1 is also arbitrary
and we necessarily have a complete solution.
With n ≥ 1
An+1 = (−1)
1
An−1
(n + 1 + s)(n + 2 + s)
which yields, for s = −1,
An+1 = (−1)
1
An−1
n(n + 1)
Specifically, for n = 1,
1
A2 = (−1) A0
2
for n = 2
1
A3 = (−1) A1
6
for n = 3
A4 = (−1)
1
1
1
A2 = (−1) (−1) A0
12
12
2
A5 = (−1)
1
1
1
A3 = (−1) (−1) A1
20
20
6
for n = 4
The recurrence relationship can finally be represented by the following two equations in
more compact form
A2l = (−1)l
1
A0 ,
(2l)!
l≥0
and
A2l+1 = (−1)l
1
A1 ,
(2l + 1)!
5
l≥0
6
Apply Frobenius method to find the general solution of the equation:
Ly = 2x2
dy
d2 y
− x + (1 + x)y = 2x2 y 00 − xy 0 + (1 + x)y = 0
2
dx
dx
Since x = 0 is a regular singular point, we expect solutions of the form
y(x) = x
s
∞
X
k
Ak x =
k=0
∞
X
Ak xk+s
k=0
Here
0
y(x) =
∞
X
(k + s)Ak xk+s−1
k=0
and
00
y(x) =
∞
X
(k + s − 1)(k + s)Ak xk+s−2
k=0
Substituting into the given ODE yields
∞
X
2(k + s − 1)(k + s)Ak xk+s −
k=0
∞
X
∞
X
(k + s)Ak xk+s +
k=0
Ak xk+s +
k=0
∞
X
Ak xk+s+1 = 0
k=0
and a change in the dummy index of the fourth sum readily makes the x in all sums have
the same power
∞
X
2(k + s − 1)(k + s)Ak xk+s −
k=0
∞
X
∞
X
(k + s)Ak xk+s +
k=0
Ak xk+s +
k=0
∞
X
Ak−1 xk+s = 0
k=1
The above can also be written as
[2(s − 1)s − s + 1]A0 xs +
−
∞
X
k=1
∞
X
2(k + s − 1)(k + s)Ak xk+s
k=1
(k + s)Ak x
k+s
+
∞
X
Ak x
k=1
k+s
+
∞
X
k=1
6
Ak−1 xk+s = 0
I.e.
∞
X
[2(s − 1)s − s + 1]A0 xs +
[{2(k + s − 1)(k + s) − (k + s) + 1}Ak + Ak−1 ]xk+s = 0
k=1
Collecting all terms who have as common factor the lowest power of x (i.e. xs ) yields the
indicial equation
2s2 − 3s + 1 = 0
with roots s1 = 1 and s2 = 1/2.
And the recurrence relation is
Ak = −
2(s +
k)2
1
Ak−1
− 3(s + k) + 1
Finally the corresponding solutions are
y1 (x) = x[1 +
∞
X
(−1)k xk
]
[(2k + 1)(2k − 1)...5 × 3]k!
k=1
and
y2 (x) = x1/2 [1 +
∞
X
(−1)k xk
]
k=1 [(2k − 1)(2k − 3)...3 × 1]k!
7
Apply Frobenius method to find the general solution of the equation:
Ly =
2
2d y
x
dx2
+ (x2 + x)
dy
−y =0
dx
Here the indicial equation is s2 − 1 = 0 with solutions s1 = +1, s2 = −1 and a series
solution is assured for s1 . The recurrence formula is
(s + k − 1)[(s + k + 1)Ak + Ak−1 ] = 0
and for s = s1 = 1 this is
Ak = −
Ak−1
k+2
7
so that the associated solution is
1
1
1
x2 −
x3 + ...] =
y1 = xA0 [1 − x +
3
3×4
3×4×5
e−x − 1 + x
= 2A0
x
Now for s = s2 = −1 the recurrence formula becomes
(k − 2)(kAk + Ak−1 ) = 0
which yields A1 = −A0 when k = 1, 0 = 0 when k = 2 (i.e. A2 can be anything we want,
including zero) and Ak = −Ak−1 /k for k ≥ 3. With this, the solution associated with s = −1
becomes
1−x
y2 = A0
x
8
Evaluate the following quantities, from the series definitions, to three place accuracy a)
(1)
J1 (0.3) b) Y0 (0.2) c) J0.75 (0.2) d) I2 (1) e) H0 (0.2) f) J10 (0.5)
At most three terms in each series are required for the desired accuracy.
1 3
1
x + 384
x5 + ... = 0.1483.
a) For x = 0.3, J1 (0.3) = 12 x − 16
b) Y0 (0.2) = −0.07380 + 0.6366 ln x + (0.1776 − 0.1591 ln x)x2 +
(−0.016073 + 0.009947 ln x)x4 + ... = −1.081.
c) J0.75 (0.2) = 0.6469x3/4 − 0.0924x11/4 + 0.0042x19/4 + ... = 0.1924
d) I2 (1) = 0.1357.
(1)
e) H0 (0.2) = 0.99 − 1.081i.
f) J10 (0.5) = 0.4539
9
Obtain the general solution of the following equation in terms of Bessel functions or, if
possible, in terms of elementary functions
x
dy
d2 y
− 3 + xy = 0
2
dx
dx
8
Answers: As described in the class notes, the most general ODE satisfied by Bessel
functions has the form
2
2d y
x
dx2
+ x(a + 2bxr )
dy
+ [c + dx2s − b(1 − a − r)xr + b2 x2r ]y = 0
dx
The given equation can be put in this form simply multiplying by x yielding the following
x2
dy
d2 y
− 3x + x2 y = 0
2
dx
dx
where a = −3, b = 0, c = 0, d = 1 and s = 1.
The solution now can be written down directly as
y(x) = x2 Zp (x) = x2 (c1 Jp + c2 Yp )
with
s
1 1−a 2
) −c=2
p=
(
s
2
Alternatively, the compatible form above can be obtained directly from Bessel’s equation
X2
dY
d2 Y
+ (X 2 − P 2 )Y = 0
+X
2
dX
dX
by using the transformed variables X = d1/2 xs /s = x, and Y = y/x2 and P = 2.
The solution to the above Bessel’s equation is
Y (X) = c1 JP (X) + c2 YP (X)
and using the original variables the desired solution is
y(x) = c1 x2 J2 (x) + c2 x2 Y2 (x)
10
A uniform string is unrestrained from transverse motion at x = 0 while at x = L is attached
to a yielding support of modulus k, the ends of the string being constrained against appreciable movement parallel to the axis of rotation. Show that the nth critical speed ωn is given
by
µn
ωn =
L
s
9
T
ρ
where cot(µn ) = αµn and α = T /kL.
ρ
d2 y
+ ω2 y = 0
2
dx
T
subject to the free end condition at x = 0
y 0 (0) = 0
and, since a yielding support is nothing but an elastic spring which exerts a restoring force
proportional to its stretch, (i.e. T y 0 = −ky),
T
Ly 0 (L) = αLy 0 (L) = −y(L)
kL
at x = L.
Inspection shows that the functions
r
y1 (x) = cos(
ρ
ωx)
T
and
r
y2 (x) = sin(
ρ
ωx)
T
both satisfy the ODE, therefore, the general solution is
r
r
ρ
ρ
y(x) = c1 y1 (x) + c2 y2 (x) = c1 cos(
ωx) + c2 sin(
ωx)
T
T
The stated boundary conditions require that, c2 = 0 and that
r
µ=
ρ
ωL
T
be the roots of
cot(µ) = αµ
10
11
Reduce the following equation
x
dy
d2 y
+ 2 + (x + λ)y = 0
2
dx
dx
to the standard form
d dy
(p ) + (q + λr)y = 0
dx dx
Answer: The given equation has the general form
a0 (x)
d2 y
dy
+ a1 (x) + [a2 (x) + λa3 (x)]y = 0
2
dx
dx
with a0 (x) = x, a1 (x) = 2, a2 (x) = x, a3 (x) = 1. To transform it into standard form set
Z
p(x) = exp( (a1 /a0 )dx) = x2
q(x) = (a2 /a0 )p(x) = x2
and
r(x) = (a3 /a0 )p(x) = x
to obtain
d 2 dy
[x
] + [x2 + λx]y = 0
dx dx
12
Determine the Fourier coefficients of the following expansions for 0 < x < π,
a)
f (x) = 1 =
∞
X
0
An sin(nx)
f (x) = 1 =
∞
X
0
An sin(nx)
11
Now multiply left and right hand side by sin(mx) and then integrate from 0 to π, i.e.
Z
Z
∞
πX
0
0
π
0
Z
f (x) sin(mx)dx =
An sin(nx) sin(mx)dx =
∞ Z
X
π
0
0
π
0
1 sin(mx)dx =
An sin(nx) sin(mx)dx
However the sin(nx) functions constitute an orthogonal set and thus on sum on the right
hand side above only the term for which m = n survives (all others are equal to zero).
Therefore
Z
π
0
Z
sin(nx)dx =
π
0
An sin(nx)2 dx = An
Z
π
0
sin(nx)2 dx
so that
Rπ
An =
R π0
0
(
sin(nx)dx
=
sin2 (nx)dx
0
4
nπ
n = even
n = odd
b)
f (x) = x =
∞
X
0
An sin(nx)
2
An =
π
Z
0
π
(
x sin(nx)dx =
− n2
2
n
n = even
n = odd
13
Expand the following function in a Fourier series of period 2L over the interval (−L, L) and
in each case sketch the function represented by the series in the interval (−3L, 3L).


 0
(x < 0)
(0 < x < L/2)
f (x) =  x

L − x (x > L/2)
Answer: The complete Fourier sine-cosine series representation is
f (x) = Ao +
∞
X
(An cos(
n=1
nπx
nπx
+ Bn sin(
)
L
L
12
where
A0 =
An =
and
Bn =
1
L
Z
Z L
1 ZL
1 Z L/2
L
(
f (x)dx =
xdx +
(L − x)dx) =
2L −L
2L 0
4
L/2
Z
L
−L
f (x) cos(
1
nπx
)dx = [
L
L
L/2
0
x cos(
nπx
)dx +
L
Z
L
nπx
)dx] =
L
L/2
(
0
(neven)
n−1
4L 1
(−1) 2 (nodd )
π 2 n2
(L − x) cos(
Z L
1 Z L/2
1ZL
nπx
nπx
nπx
)dx = [
)dx +
)dx] =
f (x) sin(
x sin(
(L − x) sin(
L −L
L
L 0
L
L
L/2


(nodd )
 0
0
(n = 2m)


8L 1
− π2 n2 (n = 4m − 2)
Finally, for −L ≤ x ≤ L,
f (x) ≈
∞
∞
X
n−1 1
2L X
nπ
nπ
L 4L
1
− 2
x)
+
...
cos(
(−1) 2 2 sin( x)
2
2
8
π n=2;n=4m−2 n
L
π n=1;n=odd
n
L
14
The temperature at a point in space is T = xy + yz + zx.
a) Find the direction in which the temperature changes most rapidly with distance from
(1, 1, 1). What is the maximum rate of change?
b) Find the derivative of T in the direction of the vector 3i − 4k at (1, 1, 1).
Answer: a) Here ∇T =√(y + z)i + (x + z)j + (y + x)k. The maximum rate of change at
(1, 1, 1) is |∇T (1, 1, 1)| = 2 3 and direction cosines are
1
1
1
∇T
= √ i + √ j + √ k = cos αi + cos βj + cos γk
|∇T |
3
3
3
b) The required derivative is
∇T (1, 1, 1) ·
2
3i − 4k
=−
|3i − 4k|
5
13
15
For each of the following vector functions F, determine whether ∇φ = F has a solution and
determine it if it exists.
a) F = 2xyz 3 i − (x2 z 3 + 2y)j + 3x2 yz 2 k
b) F = 2xyi + (x2 + 2yz)j + (y 2 + 1)k
a) Here ∇φ = F requires ∇ × F = 0 which is not the case here, so no solution.
b) Here ∇ × F = 0 so that
φ(x, y, z) = x2 y + y 2 z + z + c
16
a) Evaluate the surface integral
Z Z
S
F · dσ
where F = xi − yj + zk and S is the lateral surface of the cylinder x2 + y 2 = 1 between the
planes z = 0 and z = 1, using cylindrical coordinates.
b) Check your result using the divergence theorem.
Answer: a) With x = r cos θ, y = r sin θ and z = z, S is given by r = 1, 0 ≤ z ≤ 1 and
0 ≤ θ ≤ 2π. Also, F = r cos θi − r sin θj + zk.
Further,
dσ = ur dθdz = (cos θi + sin θj)dθdz = ur dθdz
Therefore
Z Z
S
F · dσ =
Z Z
(cos2 θ − sin2 θ)dθdz =
0≤θ≤2π;0≤z≤1
Z z=1
=
z=0
Z
[
θ=2π
cos 2θdθ]dz = 0
θ=0
b) Let S1 = bottom of cylinder, S2 = top of cylinder, C = total surface of cylinder, then
Z Z
S
F · dσ +
Z Z
S1
F · dσ +
Z Z
Z Z Z
=
Z Z
S
F · dσ +
Z Z
x2 +y 2 ≤1
−zdxdy +
Z Z
S2
F · dσ =
∇ · FdV ==
Z Z
x2 +y 2 ≤1
14
zdxdy =
S
I I
C
F · dσ =
Z Z Z
dV π =
F · dσ + 0 + π
Thus,
RR
S
F · dσ = 0.
17
For the system of coordinates u, φ, θ defined by
x = au sin φ cos θ
y = bu sin φ sin θ
z = cu cos φ
show that the element of volume is of the form
dτ = abcu2 sin φdudφdθ
Answer: The element of volume in the u1 , u2 , u3 coordinate system is given by
dτ = (U1 · U2 × U3 )du1 du2 du3
Introducing u1 = u, u2 = φ, and u3 = θ and recalling that for i = 1, 2, 3
Ui =
∂r
∂ui
one has
∂x
∂y
∂z
+j
+k
=
∂u1
∂u1
∂u1
= ia sin φ cos θ + jb sin φ sin θ + kc cos φ
U1 = i
∂x
∂y
∂z
+j
+k
=
∂u2
∂u2
∂u2
= iau cos φ cos θ + jbu cos φ sin θ − kcu sin φ
U2 = i
∂x
∂y
∂z
+j
+k
=
∂u3
∂u3
∂u3
= −iau sin φ sin θ + jbu sin φ cos θ + k0
U3 = i
15
Therefore,
∂x ∂y ∂z ∂x ∂x ∂x ∂u1 ∂u1 ∂u1 ∂u1 ∂u2 ∂u3 ∂x ∂y ∂z ∂y ∂y ∂y ∂(x, y, z)
=|
|
=
U1 · U2 × U3 = ∂u
∂u1 ∂u2 ∂u3 =
∂u
∂u
∂(u, φ, θ)
∂x2 ∂y2 ∂z2 ∂z ∂z ∂z ∂u ∂u ∂u ∂u ∂u ∂u 1
2
3
3
3
3
a sin φ cos θ au cos φ cos θ − au sin φ sin θ = abcu2 sin φ
= b sin φ sin θ bu cos φ sin θ bu sin φ cos θ
c cos φ − cu sin φ 0
so that
dτ = abcu2 sin φdudφdθ
18
Locate and identify all relative maxima, minima and saddle points of the following function
f (x, y) = x3 y 3 − 3x − 3y
Here fx = 3x2 y 3 − 3, fy = 3x3 y 2 − 3, fxx = 6xy 3 , fyy = 6x3 y and fxy = 9x2 y 2 .
2
= 36 − 81 < 0 so,
Note that fx = fy = 0 when x = y = 1 and that there fxx fyy − fxy
19
Determine when, if ever, f (x, y) = x2 + y 2 takes on a minimum or a maximum value under
the constraint xy = 1.
The constraint can be written down as g(x, y) = xy − 1 = 0. Therefore, the auxiliary
function in this case is
φ(x, y) = f (x, y) + λg(x, y) = x2 + y 2 + λ(xy − 1)
16
with this
φx = 2x + λy
φy = 2y + λx
and
φλ = xy − 1 = g(x, y)
For an extremum it is necessary that φx = φy = φλ = 0. So φx = 0 when x = −(λ/2)y
and φy = 0 when y = −(λ/2)x so that y(λ2 − 4) = 0 and either y = 0 or λ = ±2. However
y = 0 is not possible because xy = 1. For the same reason, λ = 2 is impossible too. Last,
if λ = −2 then x = y and either (x, y) = (1, 1) or (x, y) = (−1, −1). These two points are
minima (check!).
20
Determine the function u(x) that minimizes the following functional
Z
I=
π/2
0
F (x, u, u0 )dx =
Z
π/2
0
[(
du 2
) − u2 + 2xu]dx
dx
subject to the conditions
u(0) = 0
and
u(π/2) = 0
Answer: Since F = (u0 )2 − u2 + 2xu, then
∂F
= 2u0
0
∂u
and
∂F
= −2u + 2x
∂u
17
Therefore, the Euler equation is
d
∂F
d ∂F
( 0) −
=
(2u0 ) − (−2u + 2x) = 0
dx ∂u
∂u
dx
i.e.
d2 u
+u−x=0
dx2
Integrating twice yields the general solution
u(x) = C1 cos(x) + C2 sin x + x
Finally, introducing the boundary conditions gives the desired solution as
π
u(x) = x − ( ) sin(x)
2
21
Show that the following three arbitrary functions of the specific function u = x2 + 2y
f1 (u) = u2 + 4 = (x2 + 2y)2 + 4 = z1 (x, y)
f2 (u) = sin(u) = sin(x2 ) cos(2y) + cos(x2 sin(2y) = z2 (x, y)
(x2 + 2y) + 2
u+2
=
= z3 (x, y)
3u + 5
3(x2 + 2y) + 5
are all general solutions of the partial differential equation
∂z
∂z
=x
∂x
∂y
f3 (u) =
Here, for i = 1, 2, 3 one has
∂fi ∂u
∂zi
=
= 2xfi0
∂x
∂u ∂x
and
∂fi ∂u
∂zi
=
= 2fi0
∂y
∂u ∂y
Eliminating fi0 yields the stated PDE.
18
22
The temperature T is maintained at 0 at three edges of a square plate and at 100 at the
fourth edge. The plate is 100 cm in side. For steady state conditions
a) Find T (x, y) in the plate.
b) Calculate T (50, 50).
a) Assume Tp = X(x)Y (y). Substituting into Laplace’s equation yields
−
1 d2 Y
1 d2 X
=
= k2
X dx2
Y dy 2
with general solutions
X(x) = a sin kx + b cos kx
Y (y) = αeky + βe−ky
Using the three homogeneous boundary conditions reduces the above to
Xn (x) = an sin(
nπ
x)
100
and
Yn (y) = αn (eky − e−ky ) = αn sinh(
nπ
y)
100
Now, substituting Tp = X(x)Y (y) and superimposing for all n yields
T (x, y) =
∞
X
An sin(
n=1
nπ
nπ
x) sinh(
y)
100
100
Finally, introducing the initial condition T (x, 100) = 100 gives
100 =
∞
X
n=1
An sin(
∞
∞
X
X
nπ
nπ
nπ
x) sinh(nπ) =
x) =
x)
[An sinh(nπ)] sin(
cn sin(
100
100
100
n=1
n=1
which is the Fourier sine series representation of 100 with the Fourier coefficients
2
cn = An sinh(nπ) =
100
Z
I.e.
100
0
(
An =
100 sin(
0
400
nπ sinh nπ
19
100
nπ
x)dx = 2
[1 − cos(nπ)]
100
nπ
n even
n odd
With the above the desired solution can now be rearranged to
T (x, y) =
∞
X
cn sin(
n=1
nπ
y)
nπ sinh( 100
x)
100 sinh(nπ)
with
(
cn =
0
400
nπ
n even
n odd
b) To find T (50, 50) use the first four terms in the above series to get
T (x, y) =
π
3π
sinh( 100
sinh( 100
50)
50)
π
400
3π
400
sin(
50)
+0+
sin(
50)
+0=
π
100
sinh(π)
3π
100
sinh(3π)
= 25.37 − 0.3812 = 24.98
23
A string of length L = 1 m is clamped at its two ends and initially displaced from its
equilibrium position by the function
u(x, 0) = kx(L − x)
and then released. Find an expression for the displacement of the string as a function of
distance and time u(x, t).
The desired displacement is the solution to the 1D wave equation
1 ∂2u
∂ 2u
=
∂x2
c2 ∂t2
subject to the boundary conditions
u(0, t) = u(L, t) = 0
and the initial conditions
u(x, 0) = kx(L − x)
20
∂u(x, t)
|t=0 = 0
∂t
Assuming u(x, t) = X(x)T (t) and substituting into the PDE yields
T 00 + c2 λ2 T = 0
and
X 00 + λ2 X = 0
where λ is a constant. Admissible solutions to the above are
Tn (t) = An cos(
ncπ
ncπ
t) + Bn sin(
t)
L
L
and
Xn (x) = D sin(
nπ
x)
L
for n = 1, 2, .... After linear combination the solution becomes
u(x, t) =
∞
X
sin(
n=1
nπ
ncπ
ncπ
x)[an cos(
t) + bn sin(
t)]
L
L
L
Substituting back and introducing the second initial condition yields
∞
X
nπ ncπ
∂u
=0=
bn
sin( x)
∂t
L
L
n=1
which requires necessarily that bn = 0 for n = 1, 2, ....
Introducing the first initial condition gives
u(x, 0) = kx(L − x) =
∞
X
n=1
an sin(
nπ
x)
L
Now, multiplying by sin(mπx/L) and integrating from 0 to L gives
Z
L
0
Z L
mπ
mπ 2
x)dx = an
x) dx
kx(L − x) sin(
sin(
L
L
0
which yields
(
am =
0
4kL2
[1
m3 π 3
m even
− (−1) ] m odd
m
21
Finally, changing the summation index from m back to n with the relationship m = 2n+1
such that n = 0, 1, 2... to avoid the even m’s yields after rearrangement
u(x, t) =
∞
(2n + 1)cπ
(2n + 1)π
8kL2 X
1
x) cos(
t)
sin(
3
3
π n=0 (2n + 1)
L
L
or since L = 1
u(x, t) =
∞
8k X
1
sin((2n + 1)πx) cos((2n + 1)πct)
3
π n=0 (2n + 1)3
Note the above solution really consists of a superposition of traveling waves since
1
sin(mπx) cos(mπct) = [sin(mπ(x − ct)) + sin(mπ(x + ct))]
2
24
For a freely vibrating square membrane of side l , supported along the boundary x = 0,
x = l, y = 0, y = l, obtain the deflection w(x, y, t) in the form
w=
∞ X
∞
X
m=1 n=1
sin(
nπy
mπx
) sin(
)[amn cos(ωmn t) + bmn sin(ωmn t)]
l
l
where
ωmn = π
√
s
m2
+
n2
T
ρl2
Answer: The function w(x, y, t) must satisfy the wave equation
1 ∂ 2w
∂ 2w ∂ 2w
+
=
∂x2
∂y 2
α2 ∂t2
where α2 = T /ρ.
Assume that wp = X(x)Y (y)Θ(t) and substitute to get
1 00 1 00
1 1
X + Y = 2 Θ00 = −c2
X
Y
α Θ
22
and
1 00
X = −k 2
X
introducing the boundary conditions gives
Xm (x) = bm sin(
mπ
x)
l
and
Yn (x) = en sin(
nπ
y)
l
and
c=
π√ 2
m + n2
l
so that
T π2
1 00
2
Θ = − 2 (m2 + n2 ) = ωmn
Θ
ρl
with solution
Θ(t) = f cos ωmn t + g sin ωmn t
combining the above and relabeling the constants leads to the required expression.
25
The temperatures at the ends x = 0 and x = L = 100 of a 100 cm long rod with insulated
sides are held at temperatures of 0 and 100 , respectively until reaching steady state. Then,
the temperatures at the ends are interchanged. Find T (x, t).
Answer: The solution to the problem of T = T1 at x = 0 and T = T2 at x = 100, is
∞
n2 π 2 α
xX
nπ
an sin( x)e− l2 t
T (x, t) = T1 + (T2 − T1 )
l n=1
l
where
2
2ZL
nπ
(T2 cos nπ − T1 )
f (x) sin( x)dx +
an =
L 0
L
nπ
23
This can be applied directly noticing that here f (x) = 100x/L is the initial condition (obtained from the previous steady state) and that T1 = 100 and T2 = 0. The Fourier coefficients
are then
200
1 1002
200
(−1)n ] −
=
[(−1)n+1 − 1] =
an = [−
50
nπ
nπ
nπ
(
−400/nπ
0 odd
The result is
X
n2 π 2 α
nπ
400
sin(
x)e− 1002 t =
100
n≥1,even nπ
T (x, t) = 100 − x −
X 400
(2m)2 π 2 α
2mπ
x)e− 1002 t =
100
m=1 2mπ
∞
m2 π 2 α
mπ
200 X 1
sin(
x)e− 2500 t
= 100 − x −
π m=1 m
50
= 100 − x −
sin(
26
Find the function T (x, t) in 0 ≤ x < ∞ satisfying
1 ∂T
∂ 2T
=
2
∂x
α ∂t
and subject to
T (0, t) = f (t)
T (x → ∞, t) = 0
and
T (x, 0) = 0
using the Laplace Transform method.
Taking Laplace transforms one gets
s
d2 T̄ (x, s)
− T̄ (x, s) = 0
2
dx
α
24
even
T̄ (0, s) = f¯(s)
T̄ (x → ∞, s) = 0
The solution of this problem is
√
T̄ (x, s) = f¯(s)e−x s/α = f¯(s)ḡ(x, s) = L[f (t) ∗ g(x, t)]
Inversion then produces
T (x, t) = f (t) ∗ g(x, t) =
Z
t
0
f (τ )g(x, t − τ )dτ
Inversion of ḡ(x, s) to get g(x, t) finally gives
T (x, t) = √
x
4πα
Z
t
τ =0
x2
f (τ )
]dτ
exp[−
(t − τ )3/2
4α(t − τ )
If f (t) = T0 = constant, the solution is
T̄ (x, s) =
q
T0
exp(−x s/α)
s
and inverting the transform yields
x
T (x, t) = T0 erf c( √ )
2 αt
where the complementary error function erf c(z) is given by
2 Z∞
exp(−z 02 )dz 0
erf c(z) = √
π z
27
Find a bounded solution u(r, t) of the following problem using the Laplace transform method.
∂u
∂ 2 u 1 ∂u
=
+
2
∂r
r ∂r
∂t
subject to
u(r, 0) = 0
25
and
u(a, t) = u0
where t > 0, 0 < r < a with a = constant.
Taking Laplace transforms of the terms in the given PDE yields
d2 ū(r, s) 1 dū(r, s)
= sū(r, s)
+
dr2
r dr
This is a Bessel-type equation with general solution given by
√
√
ū(r, s) = AJ0 (ir s) + BY0 (ir s)
Since we look for a bounded solution, necessarily B = 0. Now transformation of the boundary
condition at r = a yields
ū(a, s) =
u0
s
and combining with the above gives
A=
1
u0
√
s J0 (ia s)
so that the solution for ū(r, s) becomes
√
J0 (ir s)
√
ū(r, s) = u0
sJ0 (ia s)
Finally, the inverse transformation gives
2
u(r, t) = u0 [1 − 2
2
∞
X
e−λn t/a J0 (λn r/a)
n=1
where λn are the roots of
J0 (λn ) = 0
26
λn J1 (λn )
]
```
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