Math 257/316 Section 202
2 questions;
Midterm 1
50 minutes;
February 6
max = 30 points
1. Consider this second order, linear, homogeneous ODE:
2(x
1)y 00 + y 0 + y = 0.
(a) Find the general solution in the form of a power series based at x0 = 0 (just
find the first four non-zero terms). [8 points]
(b) Find (only!) the first two terms of a series
based at x0 = 1 (and valid
p solution
0
for x > 1) satisfying y(1) = 1, limx!1+ x 1y (x) = 1. [8 points]
1
(Blank page)
2
2. Consider the initial-boundary-value problem
8
>
< utt + 2ut = uxx , 0 < x < 1, t > 0,
u(0, t) = 0, u(1, t) = 0
>
:
u(x, 0) = f (x), ut (x, 0) = g(x)
(a) Write clearly the “X problem” and the “T problem” that arise when finding separated variables solutions u(x, t) = X(x)T (t) of the PDE and BC. [5 points]
(b) The solutions of the “X problem” are Xk (x) = sin(k⇡x), k = 1, 2, 3, . . .. Solve
the “T problem” and write the general solution of the PDE and BC [5 points]
(c) Find the solution when f (x) = 0, g(x) = sin(4⇡x).
3
[4 points]
(Blank page)
4
Formula sheet - final exam
constant coefficients
ay 00 + by 0 + cy = 0
ar2 + br + c = 0
y = Aer1 x + Ber2 x
y = Aerx + Bxerx
e x [A cos(µx) + B sin(µx)]
Euler eq
ax2 y 00 + bxy 0 + cy = 0
ar(r 1) + br + c = 0
y = Axr1 + Bxr2
y = Axr + Bxr ln |x|
x [A cos(µ ln |x|) + B sin(µ ln |x|)]
sin2 t + cos2 t = 1
sin t = 12 (1 cos(2t))
cosh2 t sinh2 t = 1
2
sinh t = 12 (cosh(2t) 1)
2
y1 (x) =
n=0
P1
an (x
x!x0
x0 )
n+r1
where a0 = 1.
r2 = 0:
bn (x
n=1
1
X
x0 ) +
n=0
1
X
bn (x
bn (x
x0 )n+r2 where b0 = 1.
x0 )n+r2 for some b1 , b2...
x0 )n+r2 where b0 = 1.
r2 is a positive integer:
x0 ) +
n=0
1
X
y2 (x) = ay1 (x) ln(x
Case 3: If r1
y2 (x) = y1 (x) ln(x
Case 2: If r1
y2 (x) =
The second linerly independent solution y2 is of the form:
Case 1: If r1 r2 is neither 0 nor a positive integer:
x!x0
Regular singular point x0 : Rearrange (?) as:
(x x0 )2 y 00 + [(x x0 )p(x)](x x0 )y 0 + [(x x0 )2 q(x)]y = 0
If r1 > r2 are roots of the indicial equation:
r(r 1) + br + c = 0 where
b = lim (x x0 )p(x) and c = lim (x x0 )2 q(x) then a solution of (?) is
Ordinary point x0 : Two linearly independent solutions of the form:
P1
y(x) = n=0 an (x x0 )n
Series solutions for y 00 + p(x)y 0 + q(x)y = 0 (?) around x = x0 .
ODE
indicial eq.
r1 6= r2 real
r1 = r 2 = r
r = ± iµ
Basic linear ODE’s with real coefficients
sin(↵ ± ) = sin ↵ cos ± sin cos ↵
cos(↵ ± ) = cos ↵ cos ⌥ sin sin ↵.
sinh(↵ ± ) = sinh ↵ cosh ± sinh cosh ↵
cosh(↵ ± ) = cosh ↵ cosh ± sinh sinh ↵.
Trigonometric and Hyperbolic Function identities
Math 257-316 PDE
1
n=1
1
X
bn sin(
n⇡x
),
L
bn =
2
L
0
Z
L
f (x) sin(
ODE: [p(x)y 0 ]0 q(x)y + r(x)y = 0, a < x < b.
0
BC:
↵1 y(a) + ↵2 y 0 (a) = 0,
1 y(b) + 2 y (b) = 0.
0
Hypothesis: p, p , q, r continuous on [a, b]. p(x) > 0 and r(x) > 0 for
x 2 [a, b]. ↵12 + ↵22 > 0. 12 + 22 > 0.
Properties (1) The di↵erential operator Ly = [p(x)y 0 ]0 q(x)y is symmetric
in the sense that (f, Lg) = (Lf, g) for all f, g satisfying the BC, where (f, g) =
Rb
f (x)g(x) dx. (2) All eigenvalues are real and can be ordered as 1 < 2 <
a
· · · < n < · · · with n ! 1 as n ! 1, and each eigenvalue admits a unique
(up to a scalar factor) eigenfunction n .
Rb
(3) Orthogonality: ( m , r n ) = a m (x) n (x)r(x) dx = 0 if m 6= n .
(4) Expansion: If f (x) : [a, b] ! R is square integrable, then
Rb
1
X
f (x) n (x)r(x) dx
f (x) =
cn n (x), a < x < b , cn = a R b
, n = 1, 2, . . .
2 (x)r(x) dx
n=1
a n
Sturm-Liouville Eigenvalue Problems
n⇡x
) dx.
L
PDE: utt = c2 uxx , 1 < x < 1, t > 0 IC: u(x, 0) = f (x), ut (x, 0) = g(x).
R x+ct
1
SOLUTION: u(x, t) = 12 [f (x + ct) + f (x ct)] + 2c
g(s)ds
x ct
D’Alembert’s solution to the wave equation
Sf (x) =
For f (x) defined in [0, L], its cosine and sine series are
Z
1
a0 X
n⇡x
2 L
n⇡x
Cf (x) =
+
an cos(
), an =
f (x) cos(
) dx,
2
L
L
L
0
n=1
Theorem (Pointwise convergence) If f (x) and f 0 (x) are piecewise continuous, then F f (x) converges for every x to 12 [f (x ) + f (x+)].
Parseval’s indentity
Z
1
1 L
|a0 |2 X
|f (x)|2 dx =
+
|an |2 + |bn |2 .
L L
2
n=1
Let f (x) be defined in [ L, L]then
a 2L-periodic
P1its Fourier series F f (x) isn⇡x
function on R: F f (x) = a20 + n=1 an cos( n⇡x
)
+
b
sin(
n
L )
R
RL
1 L
n⇡x
1 L
n⇡x
where an = L L f (x) cos( L ) dx and bn = L L f (x) sin( L ) dx
Fourier, sine and cosine series