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If A is a 4 × 5 matrix, then what must the integers a and b be in order to define a linear transformationT: Ra →Rb byT(x)=Axforx∈Ra?
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dependent
K
V
p
n then
vectors are^
dependent
a
p
5 n^3 p
Inner product spaces f · g =
R (^) b a f^ (t)g(t)dt. G-S applies with this inner product as well.
Cauchy-Schwarz: |u · v| kuk kvk
Triangle inequality: ku + vk kuk + kvk
T
Has n real eigenvalues, always diagonalizable,
orthogonally diagonalizable (A = P DP T^ , P is an
orthogonal matrix, equivalent to symmetry!).
Theorem: If A is symmetric, then any two
eigenvectors from different eigenspaces are orthogonal.
How to orthogonally diagonalize: First diagonalize,
then apply G-S on each eigenspace and normalize.
Then P = matrix of (orthonormal) eigenvectors, D =
matrix of eigenvalues.
Quadratic forms: To find the matrix, put the
x^2 i -coefficients on the diagonal, and evenly distribute
the other terms. For example, if the x 1 x 2 term is 6 ,
then the (1, 2)th and (2, 1)th entry of A is 3.
Then orthogonally diagonalize A = P DP T^.
Then let y = P
T x, then the quadratic form becomes
1 y^21 + · · · + ny^2 n, where i are the eigenvalues.
Spectral decomposition:
1 u 1 u 1 T^ + 2 u 2 u 2 T^ + · · · + nununT
Homogeneous solutions: Auxiliary equation: Replace
equation by polynomial, so y^000 becomes r^3 etc. Then
find the zeros (use the rational roots theorem and long
division, see the ‘Diagonalization-section). ’Simple
zeros’ give you ert, Repeated zeros (multiplicity m)
give you Aert^ + Btert^ + · · · Ztm ^1 ert, Complex
zeros r = a + bi give you
Aeat^ cos(bt) + Beat^ sin(bt).
Undetermined coefficients: y(t) = y 0 (t) + yp(t),
where y 0 solves the hom. eqn. (equation = 0), and yp
is a particular solution. To find yp:
If the inhom. term is Ctmert, then:
yp = t
s (Amt
m · · · + A 1 t + 1)e
rt , where if r is a
root of aux with multiplicity m, then s = m, and if r is
not a root, then s = 0.
If the inhom term is Ct
m e
at sin( t), then:
yp = ts(Amtm^ · · · + A 1 t + 1)eat^ cos( t) +
t
s (Bmt
m · · · + B 1 t + 1)e
rt sin( t), where s = m, if a + bi is also a root of aux with multiplicity m (s = 0 if not). cos always goes with sin and vice-versa, also, you have to look at a + bi as one
entity. Variation of parameters: First, make sure the leading
coefficient (usually the coeff. of y^00 ) is = 1.. Then
y = y 0 + yp as above. Now suppose yp(t) = v 1 (t)y 1 (t) + v 2 (t)y 2 (t), where y 1 and y 2 are your hom. solutions. Then y 1 y 2 y^01 y^02
v
0 1 v 20
=
0 f (t)
. Invert the matrix and
solve for v
0 1 and^ v
0 2 , and integrate to get^ v^1 and^ v^2 , and finally use: yp(t) = v 1 (t)y 1 (t) + v 2 (t)y 2 (t).
Useful formulas:
a b c d
1
= 1 ad bc
d b c a
R R sec(t) = ln^ |sec(t) + tan(t)|, tan(t) = ln |sec(t)|,
R tan^2 (t) = tan(x) x, R ln(t) = t ln(t) t Linear independence: f, g, h are linearly independent if
af (t) + bg(t) + ch(t) = 0 ) a = b = c = 0. To show linear dependence, do it directly. To show linear
independence, form the Wronskian:
fW (t) =
f (t) g(t) f 0 (t) g^0 (t)
(for 2 functions),
fW (t) =
2
4
f (t) g(t) h(t) f 0 (t) g^0 (t) h^0 (t) f 00 (t) g^00 (t) h^00 (t)
3
(^5) (for 3 functions).
Then pick a point t 0 where det(fW (t 0 )) is easy to
evaluate. If det 6 = 0, then f, g, h are linearly independent! Try to look for simplifications before you differentiate.
Fundamental solution set: If f, g, h are solutions and linearly independent. Largest interval of existence: First make sure the
leading coefficient equals to 1. Then look at the domain of each term. For each domain, consider the part of the interval which contains the initial condition. Finally,
intersect the intervals and change any brackets to parentheses. Harmonic oscillator: my^00 + by^0 + ky = 0 (m = inertia, b = damping,
k = stiffness)
To solve x^0 = Ax:
x(t) = Ae ^1 tv 1 + Be ^2 tv 2 + e ^3 tv 3 ( i are your eigenvalues, vi are your eigenvectors) Fundamental matrix: Matrix whose columns are the
solutions, without the constants (the columns are solutions and linearly independent) Complex eigenvalues If = ↵ + i , and v = a + ib.
Then: x(t) = A
e↵t^ cos( t)a e↵t^ sin( t)b
B
e↵t^ sin( t)a + e↵t^ cos( t)b
Notes: You only need to consider one complex
eigenvalue. For real eigenvalues, use the formula above. Also, 1 a+bi
= a bi a^2 +b^2 Generalized eigenvectors If you only find one
eigenvector v (even though there are supposed to be 2 ), then solve the following equation for u: (A I)(u) = v (one solution is enough).
Then: x(t) = Ae
t v + B
te
t v + e
t u
Undetermined coefficients First find hom. solution. Then for xp, just like regular undetermined
coefficients, except that instead of guessing xp(t) = aet^ + b cos(t), you guess aet^ + b cos(t),
where a =
a 1 a 2
is a vector. Then plug into
x
0 = Ax + f and solve for a etc. Variation of parameters First hom. solution
xh(t) = Ax 1 (t) + Bx 2 (t). Then sps xp(t) = v 1 (t)x 1 (t) + v 2 (t)x 2 (t), then solve
W^ f (t)
v^01 v^02
= f , where fW (t) =
⇥ x 1 (t) | x 2 (t)
⇤ .
Multiply both sides by
⇣ Wf (t)
⌘ 1 , integrate and solve
for v 1 (t), v 2 (t), and plug back into xp. Finally, x = xh + xp
Matrix exponential eAt^ =
P 1 n=
Antn n!. To calculate e
At , either diagonalize: A = P DP ^1 ) eAt^ = P eDtP ^1 , where eDt^ is a diagonal matrix with diag. entries e it. Or if A only
has one eigenvalue with multiplicity m, use
eAt^ = e t^
Pm 1 n=
(A I)ntn n!
. Solution of x^0 = Ax
is then x(t) = eAtc, where c is a constant vector.
Case N = 2
Equation: x^00 = Ax, A =
2 1
1 2
Proper frequencies: Eigenvalues of A are:
= 1 , 3 , then proper frequencies ±i, ±
p 3 i (±
square roots of eigenvalues)
Proper modes: v 1 =
sin
(^) ⇡ 3
sin
2 ⇡ 3
=
" p 3 p 2 3 2
,
v 2 =
sin
2
⇡ 3
sin
4 ⇡ 3
=
" p 3 2
p 3 2
Case N = 3
Equation: x^00 = Ax, A =
2
4
2 1 0
1 2 1 0 1 2
3
5
Proper frequencies: Eigenvalues of A:
= 2 , 2
p 2 , 2 +
p 2 , then proper frequencies
±
p 2 i, ±
⇣p 2 +
p 2
⌘ i, ±
⇣p 2
p 2
⌘ i
Proper modes:
v 1 =
2
4
sin
(^) ⇡ 4
sin
2
⇡ 4
sin
3 ⇡ 4
3
(^5) =
2
6 4
p 2 2 p^1 2 2
3
7 5 , v 2 =
2
4
sin
2
⇡ 4
sin
4
⇡ 4
sin
6 ⇡ 4
3
(^5) =
2
4
1 0
1
3
(^5) , v 3 =
2
4
sin
3
⇡ 4
sin
6 ⇡ 4
sin
9
⇡ 4
3
(^5) =
2
6 4
p 2 2 1 p 2 2
3
7 5
General case (just in case!)
Equation: x^00 = Ax,
A =
2
6 6 6 6 6 6 6 4
2 1 0 · · · 0 1 2 1 0 · · · 0 0 1 2 1 0 · · ·
. . .
. . .
. . .
. . .
. . .
. . . 0 0 · · · 1 2 1
0 0 0 0 1 2
3
7 7 7 7 7 7 7 5
Proper frequencies: ± 2 i sin
⇣ k⇡ 2(N +1)
⌘ ,
k = 1, 2 , · · · N
Proper modes: vk =
2
6 6 6 6 6 6 4
sin
⇣ k⇡ N +
⌘
sin
⇣ 2 k⇡ N +
⌘
. . .
sin
⇣ N k⇡ N +
⌘
3
7 7 7 7 7 7 5
Full Fourier series: f defined on ( T, T ): f (x) e
P 1 m=
am cos
⇡mx T
⇡mx T
, where:
a 0 =
1 2 T
R (^) T T f^ (x)dx
am =
1 T
R (^) T T f^ (x) cos^
(^) ⇡mx T
b 0 = 0 bm = 1 T
R (^) T
T
f (x) sin
⇡mx T
Cosine series: f defined on (0, T ): f (x) e
P 1 m=0 am^ cos^
(^) ⇡mx T
, where:
a 0 = 2 2 T
R (^) T
0
f (x)dx (not a typo)
am =
2 T
R (^) T 0 f^ (x) cos^
(^) ⇡mx T
Sine series: f defined on (0, T ): f (x) e
P 1 m=0 bm^ sin^
⇡mx T
, where:
b 0 = 0 bm =
2 T
R (^) T 0 f^ (x) sin^
(^) ⇡mx T
Tabular integration: (IBP:
R f 0 g = f g
R f g^0 ) To
integrate
R f (t)g(t)dt where f is a polynomial, make a table whose first row is f (t) and g(t). Then differentiate f as many times until you get 0 , and antidifferentiate as many times until it aligns with the 0
for f. Then multiply the diagonal terms and do + first term second term etc. Orthogonality formulas: R (^) T T cos^
⇡mx T
sin
⇡nx T
dx = 0 R (^) T T cos^
(^) ⇡mx T
cos
(^) ⇡nx T
dx = 0 if m 6 = n R (^) T T sin^
(^) ⇡mx T
sin
(^) ⇡nx T
dx = 0 if m 6 = n
Convergence: Fourier series F goes to f (x) is f is
continuous at x, and if f has a jump at x, F goes to the average of the jumps. Finally, at the endpoints, F goes to average of the left/right endpoints. Heat/Wave equations:
Step 1: Suppose u(x, t) = X(x)T (t), plug this into PDE, and group X-terms and T -terms. Then
X^00 (x) X(x) =^ , so^ X
00 = X. Then find a differential
equation for T. Note: If you have an ↵-term, put it with T.
Step 2: Deal with X^00 = X. Use boundary conditions to find X(0) etc. (if you have @ @ux , you might have X^0 (0) instead of X(0)).
Step 3: Case 1: =!
2 , then X(x) = Ae!x^ + Be !x, then find! = 0, contradiction. Case 2: = 0, then X(x) = Ax + B,
then eihter find X(x) = 0 (contradiction), or find X(x) = A. Case 3: = !^2 , then X(x) = A cos(!x) + B sin(!x). Then solve for !,
usually! =
⇡m T. Also, if case^2 works, should find cos, if case 2 doesn’t work, should find sin. Finally, = !^2 , and X(x) = whatever you found in
Step 4: Solve for T (t) with the you found. Remember that for the heat equation:
T 0 = T ) T (t) = gAme t. And for the wave
equation:
T 00 = T ) T (t) = Agm cos(!t) + Bgm sin(!t). Step 5: Then u(x, t) =
P 1 m=0 T^ (t)X(x)^ (if case 2 works), u(x, t) =
P 1 m=1 T^ (t)X(x)^ (if case 2 doesn’t work!) Step 6: Use u(x, 0), and plug in t = 0. Then use
Fourier cosine or sine series or just ‘compare’, i.e. if
u(x, 0) = 4 sin(2⇡x) + 3 sin(3⇡x), then Af 2 = 4,
A^ f 3 = 3, and gAm = 0 if m 6 = 2, 3.
Step 7: (only for wave equation): Use
@u @t u(x,^ 0): Differentiate Step 5 with respect to t and set t = 0. Then use Fourier cosine or series or ‘compare’
Nonhomogeneous heat equation: 8 <
:
@u @t =^ ^
@^2 u @x^2 +^ P^ (x) u(0, t) = U 1 , u(L, t) = U 2
u(x, 0) = f (x) Then u(x, t) = v(x) + w(x, t), where:
v(x) =
h U 2 U 1 +
R (^) L 0
R (^) z 0
1 P^ (s)dsdz
i x L +^ U^1 R (^) x 0
R (^) z 0
1 P^ (s)dsdz^ and^ w(x, t)^ solves the hom. eqn: 8 <
:
@w @t =^ ^
@^2 w @x^2 w(0, t) = 0, w(L, t) = 0 u(x, 0) = f (x) v(x) D’Alembert’s formula: ONLY works for wave
