text stringlengths 30 4k | source stringlengths 60 201 |
|---|---|
1) which are removed in the construction and zero
elsewhere. Show that g is Lebesgue integrable and compute its integral.
Solution.
(1) The total length of the intervals is being reduced by a factor of
1/3 each time. Thus l(Ck) = 3k . Thus the integral of f, which is non
negative, is actually
2k
(6.59)
�
fk =
�... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
which is not an
empty set, f (x) = k.
(5) The set F, which is the union of the intervals removed is [0, 1) \ E. Taking
step functions equal to 1 on each of the intervals removed gives an absolutely
summable series, since they are non-negative and the kth one has integral
1/3×(2/3)k−1 for k = 1, . . . . This series... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
has union containing a given
rectange then the sum of the areas of the rectangles is at least as large of
that of the rectangle contained in the union.
(5) Prove the extension of the preceeding result to a countable collection of
rectangles with union containing a given rectangle.
Solution.
(1) For subdivision of... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
. Similarly let C2 be
the corresponding set of end-points of the second intervals of the rectangles.
Now divide each of the rectangles repeatedly using the finite number of
points in C1 and the finite number of points in C2. The total area remains
the same and now the rectangles covering [a1, b1) × [A2, b2) are preci... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
D.
Each of the resulting rectangles is either contained in D or is disjoint from
it. Replace D1 by the (one in fact) subrectangle contained in D. Proceeding
by induction we can suppose that the first N −k of the rectangles are disjoint
and all contained in D and together all the rectangles cover D. Now look at
the ... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
[a1, b1 − δ] × [a2, b1 − δ].
By compactness, a finite number of these open rectangles cover, and hence
there semi-closed version, with the same endpoints, covers [a1, b1 − δ) ×
LECTURE NOTES FOR 18.102, SPRING 2009
39
[a2, b1 − δ). Applying the preceeding finite result we see that
(6.66)
Sum of areas + 2Cδ ≥ Area... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
fj (x)| < ∞ for all x ∈ [0, 1).
(3) Conclude that any continuous function on [0, 1], extended to be 0 outside
this interval, is a Lebesgue integrable function on R.
Solution.
(1) Since the real and imaginary parts of a continuous function are
continuous, it suffices to consider a real continous function f and then
... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
(x)| = |Fn+1(x) − Fn(x)| ≤ 2−n ∀ n > 1.
|fn| ≤ 2−n and so the series is absolutely summable. Moreover, it
Thus
actually converges everywhere on [0, 1) and uniformly to f by (6.68).
�
(3) Hence f is Lebesgue integrable.
40
LECTURE NOTES FOR 18.102, SPRING 2009
(4) For some reason I did not ask you to check that... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
OpencourseWare
2.11 The p diacritic (and the %r tone label)
9 August 2006
Note for this 2006 ToBI tutorial: Prosodic labelling of disfluencies has turned out to be one of
the biggest challenges to the ToBI system, and because these events happen often in spontaneous
speech, extending the system to label them appro... | https://ocw.mit.edu/courses/6-911-transcribing-prosodic-structure-of-spoken-utterances-with-tobi-january-iap-2006/5f9c5ac91b6352207d27efe39712e480_chap2_11.pdf |
from fluent cases of 2 by the use of the p diacritic.
EXAMPLE <display>: Display all the flights from Baltimore to Dallas
1 Timothy Arbisi-Kelm and Sun-Ah Jun (2005) “A comparison of disfluency patterns in normal and stuttered
speech” in Proceedings of Disfluency in Spontaneous Speech Workshop, Aix-en Provence, Fran... | https://ocw.mit.edu/courses/6-911-transcribing-prosodic-structure-of-spoken-utterances-with-tobi-january-iap-2006/5f9c5ac91b6352207d27efe39712e480_chap2_11.pdf |
also had an occurrence of 3p. Note the presence of the phrase accent distinguishing this
interword juncture from the surrounding cases of 2p.
EXAMPLE <amazing>:
um But I had I mean the stuff he knows is kind of
0 0 1p 3 1 4 1
1 1
4 1 1 1
amazing 'coz he does a lot of um environmental
3 1p 1 1 X 0 1 4
1
impa... | https://ocw.mit.edu/courses/6-911-transcribing-prosodic-structure-of-spoken-utterances-with-tobi-january-iap-2006/5f9c5ac91b6352207d27efe39712e480_chap2_11.pdf |
keep in mind is that the ‘cutoff’ described for 1p is not
equivalent to the loss of phonological material from the end of a word. For example, speakers
often apparently delete the final stop from an consonant cluster, as in ‘los’ an’ found’ for ‘lost
and found’. In these cases there is clearly no disfluency and 1p s... | https://ocw.mit.edu/courses/6-911-transcribing-prosodic-structure-of-spoken-utterances-with-tobi-january-iap-2006/5f9c5ac91b6352207d27efe39712e480_chap2_11.pdf |
-L% H*
1 1
4
1
!H* L-L%
is kind of amazing 'coz he does a lot of um
1 1 1
3 1p 1 1 X 0 1 4
L+H* L- %r H* !H*
L-L%
environmental impact stuff
1
2p 4
H*
H*
H-L%
EXAMPLE <connections>: What are the plane sizes for these flights and
4 1
4 1
1
1 1 1
H*
1
L* H-H% H* H* H-L%
do they ha(ve)- do are th... | https://ocw.mit.edu/courses/6-911-transcribing-prosodic-structure-of-spoken-utterances-with-tobi-january-iap-2006/5f9c5ac91b6352207d27efe39712e480_chap2_11.pdf |
as if there had been no interruption.
EXAMPLE <abbreviation>: What is the b- abbreviation n
under
0 1 1 1p
3-
3
3p
H* H- L+H* L- H* !H-
the category d c mean
1 1 1 4
1
H* H* H* L-L%
Especially, %r should not be used after a 3p, where the (re)start of a new intonation contour is
already implicit in the break ... | https://ocw.mit.edu/courses/6-911-transcribing-prosodic-structure-of-spoken-utterances-with-tobi-january-iap-2006/5f9c5ac91b6352207d27efe39712e480_chap2_11.pdf |
ation where there is a phrase accent in the tone tier.
Other disfluency markers:
%r: marks the well formed beginning (restart) of an intonational phrase after a disfluent ending
of the previous intonational phrase
Optional labels:
<: late High Tonal peak | https://ocw.mit.edu/courses/6-911-transcribing-prosodic-structure-of-spoken-utterances-with-tobi-january-iap-2006/5f9c5ac91b6352207d27efe39712e480_chap2_11.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
ESD.70J / 1.145J Engineering Economy Module
Fall 2009
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
ESD.70J Engineering Economy
Fall 2009
Session One
Michel-Alexandre Cardin
Prof. Richard de Neufville
ESD.70J Engineering Eco... | https://ocw.mit.edu/courses/esd-70j-engineering-economy-module-fall-2009/5fa0b84793362aca263cb900be59ba34_MITESD_70Jf09_lec01.pdf |
exam – one final assignment!
ESD.70J Engineering Economy Module - Session 1
5
Course Outline
Four (4) recitation-style sessions:
1. NPV, Data Tables, and Sensitivity Analysis
– today
2. Monte Carlo simulations
3. Modeling uncertainty using different
stochastic models and probability
distributions
4. Analyzing the ... | https://ocw.mit.edu/courses/esd-70j-engineering-economy-module-fall-2009/5fa0b84793362aca263cb900be59ba34_MITESD_70Jf09_lec01.pdf |
Lord Kelvin – British Mathematician, Physicist, and
President of the British Royal Society, c. 1895
Everything that can be invented has been invented.
Charles H. Duell – Commissioner of the U.S. Patent Office,
1899
Reagan doesn’t have the presidential look.
United Artists Executive – dismissing Ronald Reagan for the ... | https://ocw.mit.edu/courses/esd-70j-engineering-economy-module-fall-2009/5fa0b84793362aca263cb900be59ba34_MITESD_70Jf09_lec01.pdf |
!
ESD.70J Engineering Economy Module - Session 1
14
Session one – Big vs. Small
• Objectives:
– Good spreadsheet setup habits to facilitate
Sensitivity Analysis
– Charts
– One-way/Two-way Data Tables
– Goal Seek
ESD.70J Engineering Economy Module - Session 1
15
Proper spreadsheet setup
• Programming/modeling (cid:2... | https://ocw.mit.edu/courses/esd-70j-engineering-economy-module-fall-2009/5fa0b84793362aca263cb900be59ba34_MITESD_70Jf09_lec01.pdf |
with capital cost of $300 million
• No salvage value for Plan A; $300 million salvage value for Plan B
• Discount rate for Plan A is 9%, and 8% for Plan B
• The company will sell each computer for $2,000
• Variable cost for Plan A is $1,280 due to economies of scale;
Variable cost for Plan B is $1,500
• See “Entries”... | https://ocw.mit.edu/courses/esd-70j-engineering-economy-module-fall-2009/5fa0b84793362aca263cb900be59ba34_MITESD_70Jf09_lec01.pdf |
Variable Cost (Thousand dollar)
Variable Cost (Million dollar)
Investment (Million dollar)
Salvage (Million dollar)
Net value (Million dollar)
Discount Factor @ 9.0%
Present Value (Million dollar)
NPV (Million dollar)
0
900
-900
1
-900.0
162.1
1
1
900
300
300
2
600
1.28
384
0
216
0.917431
198.2
2
1
900
600
600
2
1200
1... | https://ocw.mit.edu/courses/esd-70j-engineering-economy-module-fall-2009/5fa0b84793362aca263cb900be59ba34_MITESD_70Jf09_lec01.pdf |
300
2
2
600
600
600
2
1200
1.5
900
300
-150
0.925926
-138.9
0
0.857339
0.0
3
3
900
900
900
2
1800
1.5
1350
300
750
0.793832
595.4
ESD.70J Engineering Economy Module - Session 1
23
Conclusion?
• Which plan is better?
– Plan A: $162.1M OR;
– Plan B: $156.5M
• Express this finding as the difference
between NPVA and NPVB... | https://ocw.mit.edu/courses/esd-70j-engineering-economy-module-fall-2009/5fa0b84793362aca263cb900be59ba34_MITESD_70Jf09_lec01.pdf |
)
– What Excel was presumably invented
for…
• Charts
• Goal seek
ESD.70J Engineering Economy Module - Session 1
27
Data Tables
• Use Data Tables to see how different input values
affect the output
• Data Tables provide a shortcut for calculating,
viewing and comparing multiple versions in one
calculation (what-if ... | https://ocw.mit.edu/courses/esd-70j-engineering-economy-module-fall-2009/5fa0b84793362aca263cb900be59ba34_MITESD_70Jf09_lec01.pdf |
cell below the row
of values
ESD.70J Engineering Economy Module - Session 1
31
One-way Data Table (Cont)
•
Step 3: create new output values
Select the range of cells containing the
formulas and values (no labels!)
–
– Go to “Data” menu, select “Table”
–
–
Data menu ⇒ ‘What-If Analysis’ in Excel 2007
•
Reference “Col... | https://ocw.mit.edu/courses/esd-70j-engineering-economy-module-fall-2009/5fa0b84793362aca263cb900be59ba34_MITESD_70Jf09_lec01.pdf |
questions…
ESD.70J Engineering Economy Module - Session 1
35
Two-way Data Tables
• Same idea as One-way, only now we
explore output dependency on 2 inputs
• Expect a 2-D matrix
ESD.70J Engineering Economy Module - Session 1
36
Two-way Data Tables
• Step 1: Create one column and one row
varying input values for each... | https://ocw.mit.edu/courses/esd-70j-engineering-economy-module-fall-2009/5fa0b84793362aca263cb900be59ba34_MITESD_70Jf09_lec01.pdf |
8
-71.24238649 -56.17608562 -41.1098
-85.99441135 -70.92811048 -55.8618
ESD.70J Engineering Economy Module - Session 1
39
Conditional formatting
• You can vary text appearance with values
• Step 1: Enter value range for varying number
appearance (i.e.: min/max)
• Step 2: Select the target formatting range
• Step 3: ... | https://ocw.mit.edu/courses/esd-70j-engineering-economy-module-fall-2009/5fa0b84793362aca263cb900be59ba34_MITESD_70Jf09_lec01.pdf |
curve of NPVA – NPVB as a
function of discount rate for Plan B…
ESD.70J Engineering Economy Module - Session 1
45
Chart example
• Step 1: Chart Type
– XY(Scatter) with data
points connected
• Step 2: Source Data
• Step 3: Chart Options
– Titles, Legend, etc…
• Step 4: Chart Location
– As object in “Entries”
Sensiti... | https://ocw.mit.edu/courses/esd-70j-engineering-economy-module-fall-2009/5fa0b84793362aca263cb900be59ba34_MITESD_70Jf09_lec01.pdf |
0J Engineering Economy Module - Session 1
49
Additional note: NPV function
• You may use NPV(rate,value1,value2, ...)
– “rate” is the discount rate for one time period
– “value1”, “value2”, ... are values you wish to
discount
– Excel’s NPV function assumes that all cash
flows occur at the END of their time period. I... | https://ocw.mit.edu/courses/esd-70j-engineering-economy-module-fall-2009/5fa0b84793362aca263cb900be59ba34_MITESD_70Jf09_lec01.pdf |
.pdf
• Excel 2007 for PC:
http://office.microsoft.com/en-
us/training/HA102255331033.aspx
ESD.70J Engineering Economy Module - Session 1
55
Next class…
With the deterministic base case NPV
sheet finished, we proceed to Monte
Carlo simulations
CRITICAL PART OF THE COURSE
See you tomorrow!
ESD.70J Engineering Economy ... | https://ocw.mit.edu/courses/esd-70j-engineering-economy-module-fall-2009/5fa0b84793362aca263cb900be59ba34_MITESD_70Jf09_lec01.pdf |
The heat and wave equations in 2D and 3D
18.303 Linear Partial Differential Equations
Matthew J. Hancock
Fall 2006
1 2D and 3D Heat Equation
Ref: Myint-U & Debnath
2.3 – 2.5
§
[Nov 2, 2006]
Consider an arbitrary 3D subregion V of R3 (V
defined at all points x = (x, y, z)
find an equation governing u. The heat ener... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
negative. To compute the total heat energy flowing across the boundaries,
we sum φ nˆ over the entire closed surface S, denoted by a double integral
dS.
Therefore, the conservation of energy principle becomes
S
·
|
·
·
|
� �
d
dt
V
� � �
cρu dV =
nˆ dS +
φ
·
−
S
� �
V
� � �
Q dV
(1)
1
1.1 Di... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
∂2F
∂z2 .
Applying the Divergence theorem to (1) gives
d
dt
V
� � �
cρu dV =
−
V ∇ ·
� � �
φ dV +
Q dV
V
� � �
Since V is independent of time, the integrals can be combined as
cρ
V
� � �
�
∂u
∂t
+
φ
−
∇ ·
Q
dV = 0
�
Since V is an arbitrary subregion of R3 and the integrand is assumed continuous, th... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
l2Q/K0.
2 2D and 3D Wave equation
The 1D wave equation can be generalized to a 2D or 3D wave equation, in scaled
coordinates,
This models vibrations on a 2D membrane, reflection and refraction of electromagnetic
(light) and acoustic (sound) waves in air, fluid, or other medium.
utt =
2
u
∇
(6)
3 Separation of ... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
t > 0,
(7)
(8)
(9)
The 3D Heat Equation implies
′T
T
2X
= ∇ =
X
λ = const
−
(10)
where λ = const since the l.h.s. depends solely on t and the middle X ′′/X depends
solely on x. The 3D wave equation becomes
On the boundaries,
′′ T
T
= ∇
2X
X
=
λ = const
−
X (x) = 0,
∂D
x
∈
The Sturm-Liouville Probl... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
with solution
Tn (t) = αn cos
λn
�
and the corresponding normal mode is un (x, t) = Xn (x) Tn (t).
��
��
t + βn sin
�
λn
t
4
(13)
(14)
(15)
(16)
5 Uniqueness of the 3D Heat Problem
Ref: Guenther & Lee
10.3
§
We now prove that the solution of the 3D Heat Problem
u (x, 0) = f (x) ,
x
D
∈
is unique. Let ... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
D
dV
dt
(t) = 2
v
v
∇
·
nˆdS
2
−
dV
v
2
|
D |∇
� � �
∂D
� �
But on ∂D, v = 0, so that the first integral on the r.h.s. vanishes. Thus
Also, at t = 0,
dV
dt
(t) =
2
−
dV
v
2
|
0
≤
D |∇
� � �
V (0) =
v 2 (x, 0) dV = 0
(17)
(18)
D
� � �
≤
≥
0 and dV /dt
Thus V (0) = 0, V (t)
0, i.e. V (t) is a non... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
·
5
6 Sturm-Liouville problem
§
§
Ref: Guenther & Lee
10.2, Myint-U & Debnath
7.1 – 7.3
Both the 3D Heat Equation and the 3D Wave Equation lead to the Sturm-Liouville
problem
∇
2X + λX = 0,
X (x) = 0,
D,
∂D.
x
x
∈
∈
(19)
6.1 Green’s Formula and the Solvability Condition
Ref: Guenther & L... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
applied to G = v1 and F =
v2 gives
∇
(v1∇
v2)
·
nˆdS =
S
� �
V
� � �
v1∇
�
v1 gives
2 v2 +
v2 · ∇
v1
∇
dV
�
Result (22) applied to G = v2 and F =
∇
(v2∇
v1)
·
nˆdS =
S
� �
V
� � �
2
v1 +
v2∇
v1 · ∇
v2
∇
dV
�
�
Subtracting (23) and (24) gives Green’s Formula (also known as Green’s second iden
tity):
(... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
∇
·
nˆdS =
S
� �
V
� � �
=
v
v
�
2
v +
∇
v
∇
· ∇
v
dV
2 v +
v
2
|
|∇
∇
�
dV.
(26)
V
� � �
Result (26) holds for any smooth function v defined on a volume V with closed smooth
surface S.
�
�
We now apply result (26) to a solution X (x) of the Sturm-Liouville problem (19).
Letting v = X (x), S = ∂D and V... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
> 0. Thus (30)
can be rearranged,
� � �
λ =
� � �
2
X
dV
X 2dV ≥
|
D |∇
D
0
(31)
Since X is real, the the eigenvalue λ is also real.
� � �
∇
∈
X = 0 for all points in D, then integrating and imposing the BC X (x) = 0
X is nonzero
∂D gives X = 0 for all x
∇
dV > 0. Thus,
If
for x
at some points in D, and ... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
1 = 0 = v2 on ∂D and
λ2v2, then Eq. (25) becomes
λ1v1 and
2v2 =
2v1 =
∇
−
∇
−
0 = (λ1 −
λ2)
� � �
D
v1v2dV
Thus if λ1 = λ2,
v1v2dV = 0,
(32)
and the eigenfunctions v1, v2 are orthogonal.
� � �
D
7 Heat and Wave problems on a 2D rectangle, ho
mogeneous BCs
Ref: Guenther & Lee
[Nov 7, 2006]
§
10.2, Myint-... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
positive a constant (we showed above that λ had to
be both constant and positive). We employ separation of variables again, this time
in x and y: substituting v (x, y) = X (x) Y (y) into the PDE (33) and dividing by
X (x) Y (y) gives
Y ′′
Y
+ λ =
X ′′
− X
Since the l.h.s. depends only on y and the r.h.s. only ... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
) = X (x0) = 0
0
x0
x
≤
(37)
(38)
We solved this problem in the chapter on the 1D Heat Equation. We found that for
non-trivial solutions, µ had to be positive and the solution is
Xm (x) = am sin
�
The problem for Y (y) is
mπx
x0
,
�
µm =
mπ
x0
�
2
,
�
m = 1, 2, 3, ...
(39)
′′ Y + νY = 0,
≤
Y (0) = Y ... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
�
9
7.2 Solution to heat equation on 2D rectangle
The heat problem on the 2D rectangle is the special case of (7),
ut =
∂2u
∂x2
+
u (x, y, t) = 0,
∂2u
∂y2 ,
(x, y)
u (x, y, 0) = f (x, y) ,
(x, y)
D,
∈
t > 0,
∂D,
∈
(x, y)
D,
∈
where D is the rectangle D =
. We reverse the
separation of variables (9) ... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
mπx
x0
sin
�
�
nπy
y0
�
�
−λmnt
e
(43)
Setting t = 0 and imposing the initial condition u (x, y, 0) = f (x, y) gives
f (x, y) = u (x, y, 0) =
Amnvmn (x, y) =
Amn sin
∞∞
∞∞
m=1 n=1
�
�
m=1 n=1
�
�
mπx
x0
sin
�
�
nπy
y0
�
�
nπy
where vmn (x, y) = sin
y0
Liouville problem on a rectangle, (33) – (35). M... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
πy
ˆ
y0
dx
dy
�
nπy
y0
0
�
�
�
�
x0y0/4, m = ˆm and n = ˆn
�
=
�
0,
otherwise
10
Thus (44) becomes
f (x, y) vmˆ nˆ (x, y) dA =
A ˆmnˆ
4
x0y0
D
� �
Since ˆm, ˆn are dummy variables, we replace ˆm by m and ˆn by n, and rearrange to
obtain
Amn
=
=
4
x0y0
4
x0y0... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
where D is the rectangle D =
. We reverse the
separation of variables (9) and substitute solutions (16) and (42) to the T (t) problem
(15) and the Sturm Liouville problem (33) – (35), respectively, to obtain
(x, y) : 0
{
x0, 0
y0}
≤
≤
≤
≤
x
y
umn (x, y, t) =
sin
=
sin
mπx
x0
mπx
x0
�
�
�
�
sin
sin
nπy... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
, y)
∞
∞
∞
∞
=1 =1
m n
� �
=
αmn sin
m=1 n=1
� �
�
∞
mπx
x0
∞
sin
�
�
nπy
y0
�
g (x, y) = ut (x, y, 0)
=
λmn
βmnvmn (x, y)
=1 =1
m n
� �
λmnβmn sin
�
�
mπx
x0
sin
�
�
nπy
y0
�
∞
∞
=
=1 =1
m n
� �
sin
�
nπy
y0
mπx
x0
where vmn (x, y) = sin
are the eigenfunctions of the 2D Sturm Li
ouville prob... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
the special case where the subregion D is the unit circle (we may
assume the circle has radius 1 by choosing the length scale l for the spatial coordinates
as the original radius):
The Sturm-Liouville Problem (19) becomes
�
D =
(x, y) : x + y
2
2
1
≤
�
∂2v
∂x2
+
∂2v
∂y2 + λv = 0,
v (x, y) = 0,
(x, y)
D,
∈
x ... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
use separation of variables by substituting
−
w (1, θ) = 0,
π
θ < π.
w (r, θ) = R (r) H (θ)
(49)
(50)
(51)
into the PDE (49) and multiplying by r2/ (R (r) H (θ)) and then rearranging to obtain
r
d
dr
r
�
1
dR
dr R (r)
�
+ λr2 =
d2H
1
− dθ2 H (θ)
Again, since the l.h.s. depends only on r and the r.h.s.... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
.
−
Substituting (51) gives
H (
π) = H (π) ,
−
dH
dθ
π)
(
−
=
dH
dθ
(π) .
(54)
(55)
The solution v (x, y) is also bounded on the circle, which implies R (r) must be
bounded for 0
1.
r
≤
≤
13
�
The problem for H (θ) is
d2H
dθ2 + µH (θ) = 0;
H (
π) = H (π) ,
−
dH
dθ
π)
(
−
=
dH
dθ... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
λr2
−
2
m
Rm =
0;
Rm (1) = 0,
Rm (0)
|
|
<
∞
(57)
We know already that λ > 0, so we can let
�
�
so that (57) becomes
2 d2R¯m
ds2 + s
dR¯m
ds
s
+
2
s
−
s = √λr,
¯
Rm (s) = Rm (r)
m 2
R¯m = 0;
R¯m
√λ = 0,
R¯m (0)
<
(58)
∞
�
�
�
�
The ODE is called Bessel’s Equation which, for each m = 0, 1, 2, ... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
�
�
�
Rm (r) = cmJm
√λr
.
[Nov 9, 2006]
The Bessel Function Jm (s) of the first kind of order m has power series
�
�
(
−
1)k s2k+m
k! (k + m)!22k+m
.
Jm (s) =
k=0
�
(59)
Jm (s) can be expressed in many ways, see Handbook of Mathematical Functions by
Abramowitz and Stegun, for tables, plots, and equations. The... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
is
L2(k+1)+m
(k+1)!(k+1+m)!22(k+1)+m
L2k+m
k!(k+m)!22k+m
=
L2
L2
(k + 1) (k + 1 + m) 4 ≤ (k + 1) 4
2
L
2 (k + 1)
2
�
=
�
�
�
�
�
�
�
Thus for k > N =
,
L/2
⌉
⌈
L2(k+1)+m
(k+1)!(k+1+m)!22(k+1)+m
L2k+m
k!(k+m)!22k+m
<
L
2 (N + 1)
2
< 1
�
�
�
�
�
�
�
�
�
�
Since the upper bound is less than one and is... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
,2 = 5.5001
j1,2 = 7.016
3
j0,3 = 8.6537
j1,3 = 10.173
J0 (s)
J1 (s)
The second BC requires
Rm (1) = Rm
¯
√λ
= Jm
√λ
= 0
�
This has an infinite number of solutions, namely √λ = jm,n for n = 1, 2, 3, .... Thus
the eigenvalues are
�
�
�
λmn = m,n, m, n = 1, 2, 3, ...
j2
with corresponding eigenfunctions Jm (... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
y)
D,
∈
where D is the circle D =
. We reverse the separation of vari
1
}
ables (9) and substitute solutions (14) and (42) to the T (t) problem (13) and the
Sturm Liouville problem (47) – (48), respectively, to obtain
(x, y) : x2 + y2
{
≤
umn (x, y, t) = vmn (x, y) e −λmnt = vmn (x, y) e −J 2
m,n
t
where vmn is... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
βmn.
9 The Heat Problem on a square with inhomoge
neous BC
We now consider the case of the heat problem on the 2D unit square
D =
(x, y) : 0
{
≤
x, y
,
1
}
≤
(62)
16
where a hot spot exists on the left side,
ut =
u (x, y, t) =
2 u,
u0/ε,
0,
∇
�
u (x, y, 0) = f (x, y)
D
(x, y)
∈
x = 0,
{
y
y0|
< ε/2
... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
called Laplace’s equation. Laplace’s equation is an example of
an elliptic PDE. The wave equation is an example of a hyperbolic PDE. The heat
equation is a parabolic PDE. These are the three types of second order (i.e. involving
double derivatives) PDEs: elliptic, hyperbolic and parabolic.
We proceed via separation... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
An equivalent and more convenient way to write this is
X (x) = c3 sinh nπ (1
x) + c4 cosh nπ (1
x)
−
−
Imposing the BC at x = 1 gives
X (1) = c4 = 0
and hence
Thus the equilibrium solution to this point is
X (x) = c3 sinh nπ (1
x)
−
∞
uE (x, y) =
An sinh (nπ (1
n=1
�
x)) sin (nπy)
−
You can check that this sa... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
)
y0+ε/2
y0−ε/2
�
sin (mπy) dy
y0+ε/2
y0−ε/2
�
2u0
cos (mπy)
=
=
=
ε sinh (mπ) − mπ
�
2u0
εmπ sinh (mπ)
4u0 sin (mπy0) sin
εmπ sinh (mπ)
�
mπε
2
�
18
(cos (mπ (y0 −
ε/2))
−
cos (mπ (y0 + ε/2)))
Thus
∞
sin (nπy0) sin
nπε
2
uE (x, y) =
4u0
επ
n sinh (nπ)
�
To solve the transient problem... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
h nπ (1
−
sinh nπ
x)
≈
enπ(1−x)
enπ = e
−nπx
Thus the terms decrease in magnitude (x > 0) and hence uE (x, y) can be approxi
mated the first term in the series,
uE (x, y)
4u0 sin (πy0) sin
sinh (π)
�
≈ επ
πε
2
�
sinh (π (1
−
x)) sin (πy)
A plot of sinh (π (1
the square is approximately
−
x)) sin (πy) is gi... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
−
It turns out there is a much easier way to derive this last result. Consider a plate
with BCs u = u0 on one side, and u = 0 on the other 3 sides. Let α = uE
.
Rotating the plate by 90o will not alter uE
, since this is the center of the plate.
�
Let uEsum be the sums of the solutions corresponding to the BC u = ... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
(πy0)
sin (πy0)
sin (πy0)
u0
4
u0
4
u0
2π
20
for small ε. Thus the steady-state center temperature is hottest when the hot spot is
placed in the center of the side, i.e. y0 = 1/2.
9.5 Solution to inhomogeneous heat problem on square
We now use the standard trick and solve the inhomogeneous Heat P... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
x0 = 1 = y0 in (43); don’t mix up
the side length y0 in (43) with the location y0 of the hot spot in the homogeneous
problem):
∞ ∞
v (x, y, t) =
Amn sin (mπx) sin (nπy) e
−π2(m +n2)t
2
where
m=1 n=1
� �
1
1
Amn = 4
(f (x, y)
0
�
Thus we have found the full solution u (x, y, t).
�
0
−
uE (x, y)) sin (mπx) s... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
,
∇
uE (x, y) = g (x, y) ,
(x, y)
D,
∈
(x, y)
∂D.
∈
As before, switch to polar coordinates via
x = r cos θ,
y = r sin θ,
wE (r, θ) = uE (x, y)
for
The problem for uE becomes
0
r
≤
≤
1,
π
−
≤
θ < π.
1 ∂
r ∂r
r
∂wE
∂r
+
1 ∂2wE
r2 ∂θ2 = 0,
�
w (r,
�
π) = w (r, π) ,
−
0
r
≤
≤
1,
π
−
≤
θ < π
wθ (r,
... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
d
R (r) dr
dR
dr
r
�
�
=
The problem for H (θ) is, as before,
d2H
1
− dθ2 H (θ)
= µ
d2H
dθ2 + µH (θ) = 0;
H (
π) = H (π) ,
−
dH
dθ
π)
(
−
=
dH
dθ
(π) ,
22
with eigen-solutions
Hm (θ) = am cos (mθ) + bm sin (mθ) ,
m = 0, 1, 2, ...
The problem for R (r) is
0 = r
d
dr
dR
dr −
�
r
�
2R
... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
θ)) ,
m = 0, 1, 2, ...
The full solution is the infinite sum of these over m,
uE (x, y) = wE (r, θ) =
We still need to find the Am, Bm.
Imposing the BC (69) gives
∞
∞
m=0
�
r m (Am cos (mθ) + Bm sin (mθ))
(70)
(Am cos (mθ) + Bm sin (mθ)) = gˆ (θ)
(71)
The orthogonality relations are
m=0
�
π
cos (mθ) sin (nθ)... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
spot on boundary
Suppose
gˆ (θ) =
�
u0
π
θ
θ0+π −
≤
otherwise
≤
0
θ0
which models a hot spot on the boundary. The Fourier coefficients in (72) are thus
A0 =
Bm =
u0
2π
,
Am =
u0
mπ (θ0 + π)
sin (mθ0)
u
−mπ (θ
0
+ π)
0
(cos (mθ0)
1)m)
(
−
−
Thus the steady-state solution is
uE =
u0 +
2π
∞
r m
m=1
�
�... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
for some interesting cases. We draw the level
curves (isotherms) uE = const as solid lines. Recall from vector calculus that the
uE, is perpendicular to the level curves. Recall also
gradient of uE, denoted by
uE. Thus heat flows along
from the physics that the flux of heat is proportional to
uE. Note that the heat... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
hence
u nl = 0. Thus there is no flux across lines of symmetry. Equivalently,
u is parallel
∇
to the lines of symmetry. Thus the lines of symmetry are flow lines. Identifying the
lines of symmetry help draw the level curves, which are perpendicular to the flow
lines. Also, lines of symmetry can be thought of as an in... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
curves and heat flow lines. Note that the temperature at the hot point
is infinite. See plot. The heat flux lines must go from a point of hot temperature to
a point of low temperature. Since the temperature along boundary is zero except at
25
u=0
u=0
u=0
u=u0
/S
u
u
0
ST0
u of
T0
T0 S
T0 o S
Fig... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
∇
v (x, y, t) = 0,
2 v,
v (x, y, 0) = f (x, y)
−
(x, y)
(x, y)
D
∈
∂D
∈
uE (x, y)
This is the Heat Problem with homogeneous PDE and BCs. We found the solution
to this problem above in (61),
v (x, y, t) =
∞ ∞
m=1 n=1
� �
Jm (rjm,n) (αmn cos (mθ) + βmn sin (mθ)) e −λmnt
where αmn and βmn are found from orthogo... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
−π
�
Note that the curve
radius R centered at (x0, y0).
{
(x0 + R cos θ, y0 + R sin θ) :
π
−
≤
θ < π
}
traces the circle of
Proof: To prove the Mean Value Property, we first consider Laplace’s equation
(74) on the unit circle centered at the origin (x, y) = (0, 0). We already solved this
problem, above, when we ... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
−
which is a disc of radius R centered at (x, y) = (x0, y0). Since B(x0,y0) (R)
Laplace’s equation (74) holds on this disc. Thus
�
�
2
v = 0,
∇
(x, y)
∈
B(x0,y0) (R) .
D,
⊆
(78)
We make the change of variable
xˆ =
x
x0 ,
−
R
yˆ =
y
y0 ,
−
R
uE (ˆx, yˆ) = v (x, y)
(79)
to map the circle B(x0,y0) (R) into t... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
at any point (x0, y0) in D equals the mean value of the temperature around any
disc centered at (x0, y0) and contained in D.
12 Maximum Principle for Laplace’s Eq.
Ref: Guenther & Lee
8.4, Myint-U & Debnath
8.2
§
§
Theorem [Maximum Principle] Suppose v (x, y) satisfies Laplace’s equation
in a 2D domain D,
2
v =... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
circle
must have value v (x0, y0). Since R > 0 is arbitrary, this holds for all values in the disc
B(x0,y0)(R). Keep increasing R until the disc hits the boundary of D. Then v (x0, y0)
is a value of v along the boundary of D.
A similar argument holds for the minimum of v. �
Corollary If v = 0 everywhere on the bou... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
problem
∇
2
v + λv = 0,
v = 0,
D
∂D
x
x
∈
∈
and consider the effect of the shape of the domain on the eigenvalues.
Definition Rayleigh Quotient
R (v) =
vdV
v
· ∇
v2dV
D ∇
D
(80)
Theorem Given a domain D
� � �
R3 and any function v that is piecewise smooth
� � �
on D, non-zero at some points on the inter... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
an orthonormal basis of eigenfunctions on D, i.e. all the functions φn
∇
2φn + λnφn = 0,
φn = 0,
D
∂D
x
x
∈
∈
29
and
φnφmdV =
1, m = n
0, m = n
�
� � �
D
We can expand v in the eigenfunctions,
∞
v (x) =
Anφn (x)
=1
n
�
where the An are constants. Assuming we can differentiate the... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
−
� � �
D
2vdV =
v
∇
2
λnAn
∞
n=1
�
(85)
Substituting (83) and (85) into (80) gives
R (v) =
� � �
D ∇
D
vdV
v
· ∇
v2dV
=
∞
n=1 λnA2
∞
A2
n=1 n
n
�
We assume the eigenfunctions are arranged in increasing order. In particular, λn ≥
Thus
∞
λ1A2
n=1
∞
A2
n=1 n
∞
A2
n=1 n = λ1.
∞
A2
n=1 n
n = λ1 ... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
�
In other words, the domain D
smaller eigenvalue.
that contains the sub-domain D is associated with a
�
�
Proof: Note that the Sturm-Liouville problems are
∇
∇
2 v + λv = 0,
v = 0,
2
v + λ
v = 0,
v = 0,
��
�
(x, y)
(x, y)
D
∂D
∈
∈
(x, y)
(x, y)
D
∂D
�
∈
∈
Let v1 be the eigenfunction corresponding to λ1 on ... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
without proof]: the eigenfunction(s) corresponding to the
are nonzero in the interior of D
From the useful fact, ˆv1 cannot be an eigenfunction corresponding to λˆ1 on D
,
�
(outside D). Thus, as the previous theorem states,
�
(87)
λˆ1 < R (ˆv1) .
since it is zero in the interior of D
�
�
.
31
�
Since ˆv1 = 0... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
for a circle of radius R, we’d rescale by the change of
rˆ where ˆr goes from 0 to 1. Thus
variable ˆr = r/R, so that Jm
on the circle of radius R, the smallest eigenvalue is
�
= Jm
√λr
��
λ
2
R
�
�
λ01 =
J0,1
R
2
,
J0,1 = 2.4048.
�
Suppose the rectangle is actually a square of side length 2R. Then
�
λ11 =
... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
since v (r) is smooth on D2, v (R) = 0 (zero on the
�
boundary of D2), and v is nonzero in the interior.
�
32
13.1 Faber-Kahn inequality
Thinking about the heat problem on a 2D plate, what shape of plate will cool the
slowest? It is a geometrical fact that of all shapes of equal area, the circle (... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
x
∈
∈
(89)
Nodal lines are the curves where the eigenfunctions of the Sturm-Liouville problem
are zero. For the solution to the vibrating membrane problem, the normal modes
unm (x, y, t) are zero on the nodal lines, for all time. These are like nodes on the 1D
string. Here we consider the nodal lines for the squar... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
eigenfunction with eigenvalue λmn, for any constants A, B. The nodal lines for fnm
can be quite interesting.
Examples: we draw the nodal lines on the interior and also the lines around the
boundary, where vnm = 0.
(i) m = 1, n = 1.
πx
a
This is positive on the interior and zero on the boundary. Thus the nodal lin... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
/a2, this is a solution to SL
problem (89) on D with λ = λ13. To find the nodal lines, we use the identity
sin 3θ = (sin θ)
4 sin2 θ
to write
−
3
−
�
�
v13 = sin
v31 = sin
v13 −
v31 = 4 sin
sin
sin
sin
�
�
�
�
πy
a
πy
a
πy
a
πx
a
πx
a
πx
a
�
�
�
3
4 sin2
−
πy
a
� �
3
� �
−
sin2
�
�
��
πx
a
��
sin2 ... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
with the nodal lines that intersect the interior
of the square plate:
y =
x + ka
y = x,
y =
x + a
−
v31 are the sides and diagonals of the square.
Thus the nodal lines of v13 −
Let DT be the isosceles right triangle whose hypotenuse lies on the bottom hor
v31 is zero on the boundary ∂DT ,
izontal side of the ... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
boundary ∂Dc, positive on the interior of Dc, and thus
the eigenfunction corresponding to the first (smallest) eigenvalue λ13 of the Sturm-
Liouville problem (89) on Dc.
πy
a
sin2
3
2
−
=
�
�
�
�
.
(vii) Find the first eigenvalue on the right triangle
DT 2 =
0
y
≤
≤
√3x,
x
0
≤
≤
�
1
.
�
Note that separat... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
the eigenvalues are given
by
λmn = π2
2
m
1
+
2
n
3
=
π2
3
�
�
3m 2 + n 2
We create a table of 3m2 + n2 and look for repeated values:
�
�
n, m
1
2
3
4
5
1
4
7
12
19
28
2
13
16
21
28
37
4
3
28
31
36
43
The smallest repeated value of 3m2 + n2 is 28:
λ31 = λ24 = λ15 = 28π2/3
The linear com... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
we seek the next largest repeated value
of 3m2 + n2 and continue as above. If we can find (A, B, C) such that v = 0 on
y = √3x, but v is zero on the interior of DT 2, then λ31 is not the smallest eigenvalue
and we must pursue another method of finding the smallest eigenvalue. The Rayleigh
quotient can help us identif... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
1.
(iii) m = 1 , n = 1 and sine.
v11S = J0 (rj1,1) sin θ
The nodal lines are the boundary (circle of radius 1) and the line θ =
(horizontal diameter).
π, 0, π
−
15 Steady-state temperature in a 3D cylinder
Suppose a 3D cylinder of radius a and height L has temperature u (r, θ, z, t). We
assume the axis of the c... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
2
v =
∇
1 ∂
r ∂r
r
∂v
∂r
�
�
+
1 ∂2v
r ∂θ2 +
2
∂2v
∂z2
We separate variables as
v (r, θ, z) = R (r) H (θ) Z (z)
37
so that (90) becomes
1
d
rR (r) dr
dR (r)
dr
r
�
�
+
1
r2H(θ)
d2H(θ)
dθ2
+
1
Z (z)
d2Z (z)
dz2
= 0.
Rearranging gives
1
d
rR (r) dr
dR (r)
dr
r
�
�
+
1
r2H(θ)
d2H(θ)... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
, up to a multiplicative
constant,
Zn (z) = sin
,
λn =
,
n = 1, 2, 3, ...
nπz
L
2
nπ
L
�
�
Multiplying Eq. (92) by r2 gives
�
�
r
d
R (r) dr
dR (r)
dr
r
�
−
�
2
λnr =
1
−H(θ)
d2H(θ)
dθ2 = µ
(93)
where µ is constant since the l.h.s. depends only on r and the middle only on θ.
We assume g (θ, z) is smo... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
of the first and second
�
�
�
�
Rmn (r) = c1mIm
r
λn
+ c2mKm
r
λn
Since the Im’s are regular (bounded) at r = 0, while the Km’s are singular (blow up),
and since Rmn (r) must be bounded, we must have c2m = 0, or
�
�
�
�
�
�
Rmn (r) = c1mIm
r
λn
= c1mIm
�
Thus the general solution is, by combining constants,
�... | https://ocw.mit.edu/courses/18-303-linear-partial-differential-equations-fall-2006/5faa7f7c21719c38fb701bc79ef4a29f_pde3d.pdf |
(R|Ψ) = cos Ψ + sin Ψ = 2 1/2 cos(Ψ - π/4)
2. Identify Sample Space
0
π/2 ψ
3. Probability Law over Sample Space:
Invoke isotropy implying uniformity of
angle
fΨ(ψ)
2/ π
0
π/2 ψ
4. Find CDF
(cid:139) FR(r) = P{R < r} = P{21/2 cos(Ψ-π/4) < r}
(cid:139) FR(r) = P{R < r} = P{cos(Ψ-π/4) < r/ 21/2 }
g(Ψ)
1
r/21/2
1/21/2... | https://ocw.mit.edu/courses/1-203j-logistical-and-transportation-planning-methods-fall-2006/5fb3828ccf1c35471a4ae82804586d50_lec3.pdf |
(cid:139) Median R = 1.306
(cid:139) E[R] = 4/ π = 1.273
(cid:139) σR/E[R] = 0.098, implies very robust
A Quantization Problem
NYC Marine Transfer Station
Tug
Delivers
LIGHTS
Tug
Picks Up
HEAVIES
LIGHT and HEAVY
Barges Stored
Loading
Loading
Barge
Barge
Loading
Loading
Barge
Barge
Barges
Shifted
By Hand
Or Tug
Refuse... | https://ocw.mit.edu/courses/1-203j-logistical-and-transportation-planning-methods-fall-2006/5fb3828ccf1c35471a4ae82804586d50_lec3.pdf |
) = fD(d)(1) = fD(d), d > 0, 0 <θ <1
θ
1
0
K = 0
K = 1
K = 2
1
2
K = 3
3
d
3. Joint Probability Distribution
a) D and Θare independent.
b) Θis uniformly distributed over [0, 1]
fD,Θ(d, θ) = fD(d) fΘ(θ) = fD(d)(1) = fD(d), d > 0, 0 <θ <1
x
θ
1
K = 0
K = 1
K = 2
1-x
0
1
x
1-x
2
K = 3
3
d
4. Working in the Joint Sample ... | https://ocw.mit.edu/courses/1-203j-logistical-and-transportation-planning-methods-fall-2006/5fb3828ccf1c35471a4ae82804586d50_lec3.pdf |
18.445 Introduction to Stochastic Processes
Lecture 17: Martingle: a.s convergence and Lp-convergence
Hao Wu
MIT
15 April 2015
Hao Wu (MIT)
18.445
15 April 2015
1 / 10
Recall
Martingale : E[Xn | Fm] = Xm for n ≥ m.
Optional Stopping Theorem : E[XT ] = E[X0] ?
Today’s goal
a.s.martingale convergence
Doob’s maximal ineq... | https://ocw.mit.edu/courses/18-445-introduction-to-stochastic-processes-spring-2015/5fe9145062f4c67f4f675b6e18bc2065_MIT18_445S15_lecture17.pdf |
the website.
Corollary
Let X = (Xn)n≥0 be a non-negative supermartingale. Then Xn
converges a.s. to some a.s. finite limit.
Hao Wu (MIT)
18.445
15 April 2015
4 / 10
Examples
Example 1 Let (ξj )j≥1 be independent random variables with mean
zero such that ∞ E[|ξj |]
. Set
< ∞
(cid:80)
j=
1
X0 = 0, Xn =
ξj .
n
(cid:88)
j=... | https://ocw.mit.edu/courses/18-445-introduction-to-stochastic-processes-spring-2015/5fe9145062f4c67f4f675b6e18bc2065_MIT18_445S15_lecture17.pdf |
n Xk . Then
λP[X ∗ ≥n
λ] ≤ E[Xn1[X ∗ ≥n
λ]] ≤ E[Xn].
Theorem
Let X = (Xn)n≥0 be a non-negative submartingale. Define
X ∗
n = max0≤k ≤n Xk . Then, for all p > 1, we have
||X ∗||n p ≤
p
p − 1
||X
n||p.
Recall Hölder inequality : p > 1, q > 1 and 1/p + 1/q = 1, then
E[|XY |] ≤ E[|X p 1| ] /p × E[|Y q 1| ] /q.
Hao Wu (MIT)
... | https://ocw.mit.edu/courses/18-445-introduction-to-stochastic-processes-spring-2015/5fe9145062f4c67f4f675b6e18bc2065_MIT18_445S15_lecture17.pdf |
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