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Midterm Exam (Open Book(s), Open Notes, Calculator, Computer):

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Midterm Exam (Open Book(s), Open Notes, Calculator, Computer):
Rensselaer at Hartford
MANE-7010 Mathematics of Engineering and Science
Prof. Gutierrez-Miravete
Fall 2003
Mid-Term Exam Answers
1.- Use the method of Frobenius to find Frobenius power series solutions of the following differential equation.
d2 y
+y =0
dx2
Show all the steps in your analysis including the derivation and final forms of the indicial
equation and the recurrence formula. Express your final result in terms of elementary functions if
possible.
Hints. Modify dummy summation index such that xk+s−2 appears in all sums. Note also the
remarks made in the second paragraph in p. 134 in your text. Remember power series formulae
for sin x, cos x, sinh x, cosh x.
Answer
Note there are no singular points anywhere for this equation so two linearly independent soP
k
exist. However, the problem asks to use the
lutions of the form y(x) = ∞
0 Ak x are certain toP
k+s . Substituting and rearranging as
method of Frobenius so assume then that y(x) = ∞
0 Ak x
suggested in the hint yields
(s − 1)sA0 xs−2 + s(s + 1)A1 xs−1 +
∞
X
[(k + s − 1)(k + s)Ak + Ak−2 ]xk+s−2 = 0
k=2
Since A0 is arbitrary, the first term yields the indicial equation (s − 1)s = 0 with roots s1 = 1,
s2 = 0. The smaller root confirms our initial expectation and using it on the second term (see
second paragraph on p. 134, text) indicates that A1 is also arbitrary. With the same root, the
recurrence relation becomes
Ak−2
Ak = −
(k − 1)k
Substituting into the Frobenius sum and rearranging yields finally
y(x) = A0 [1 −
x 2 x4 x6
x3 x5 x7
+
−
+ ...] + A1 [x −
+
−
+ ...] = A0 cos(x) + A1 sin(x)
2!
4!
6!
3!
5!
7!
1
To confirm, one can use the Maple command dsolve(dif f (y(x), x, x) + y(x) = 0, y(x));
2.- a) Show that the function
√
√
y(x) = Ay1 (x) + By2 (x) = A sin( λx) + B cos( λx)
is the general solution of the following differential equation
d2 y
+ λy = 0
dx2
Hint. Check that the individual functions y1 (x) and y2 (x) are indeed solutions by direct differentiation.
b) Obtain the eigenvalues and eigenfunctions associated with the above problem and the boundary conditions
y(0) = 0; y(π) = 0
Answer.
a) Since
√
y100 = −λ sin( λx)
and
√
y200 = −λ cos( λx)
the given expressions are indeed solutions.
√
b) Since y(0) = 0, necessarily B = 0 and since y(π) = 0 then sin( λπ) = 0 which yields the
eigenvalues
λ = n2
and eigenfunctions
φ(x) = sin(nx)
See p. 189 in your text.
3.- Consider the function f (x) defined for x ∈ [−π, π] by
(
f (x) =
0
1
−π ≤x<0
0≤x<π
2
a) Obtain the Fourier sine series representation of this function.
b) Set x = 1, evaluate the first four non-zero terms in the series and compare the result against
the given function.
c) Compare your results above with the Fourier-Bessel series representation given on pp. 229-230
in your text.
Answer.
a) The given function is nor even or odd, so a complete Fourier series is really required. However,
the problem asks for the sine series so taking L = π
f (x) =
∞
X
An sin(
n=1
with coefficients
∞
X
nπ
An sin(nx)
x) =
L
n=1
Rπ
(−1)n − 1
sin(nx)dx
=
2
An = R π0
2
sin(nx) dx
0
nπ
As expected, the series is a good representation of the given function only for x > 0.
b) Evaluating the first four non-zero terms for x = 1 yields 1.0066.
c) The first four non-zero terms of the Fourier-Bessel representation yield 0.9871
As indicated above, the appropriate representation in −L < x < L (with L = π) is instead the
complete Fourier series (see text p. 220) , i.e.
f (x) = A0 +
∞
X
[An cos(nx) + Bn sin(nx)]
n=1
with coefficients
A0 =
An =
1
π
Z
1
2π
Z
π
f (x)dx =
1
2π
f (x) cos(nx)dx =
1
π
−π
π
−π
Z
π
0
Z
0
dx =
1
2
π
cos(nx)dx = 0
and
Bn =
1
π
Z
π
−π
f (x)sin(nx)dx =
1
π
Z
0
π
sin(nx)dx =
(−1)n − 1
nπ
4.- The following potential function describes potential flow near a corner
φ(x, y) = x2 − y 2
3
Use a rectangular Cartesian system of coordinates and;
a) Obtain the velocity field associated with the given potential;
b) Show that the given potential satisfies Laplace’s equation;
c) Use the expression
dψ = −
h3 h1 ∂φ
h2 h3 ∂φ
du1 +
du2
h2 ∂u2
h1 ∂u1
to determine the stream function ψ associated with this flow. Recall that lines of constant value of
ψ are everywhere perpendicular to the equipotential lines φ(x, y) = x2 − y 2 = const.;
d) Sketch the equipotential and the streamlines associated with this flow. Recall the velocity
vector is always tangential to the streamlines.
Hint. For part (c) identify the appropriate expressions for the scale factors h1 , h2 , h3 and the
coordinates u1 , u2 applicable in this case.
Answer.
a) With the given potential and in Cartesian coordinates with u1 = x, u2 = y, h1 = h2 = h3 = 1,
the velocity vector is
v = ∇φ = 2xi − 2yj
b) The given potential satisfies Laplaces’s equation since
∇2 φ = 0
c) From the given expression the differential of the stream function is
dψ = 2ydx + 2xdy = 2d(xy)
so that ψ = 2xy.
d) Focusing on the quadrant x > 0, y > 0 this represents a flow which is parallel to the x− axis
for large x and small y and parallel to the y− axis for small x and large y. (See text p. 316).
4
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