Lesson: Music in the Golden Age and in “El Retrato Vivo”

Description

Lesson: Music in the Golden Age and in “El Retrato Vivo” Ronda Music By Fran Morón, BYU MA Student Date: January 23, 2008 Objectives: • Students will be able to recognize the folkloric Ronda music from Spain. • Students will be able to recognize the musical instruments used in the Golden Age. • Students will be able to find similarities with this medieval type of music and the music they hear today.

• quedo romances de amor luego
• tuna compostelana
• ronda music
• little sample
• dancing to the sound of ronda music
• people during intermissions
• music

Sujets

• Compostela
• Golden
• Santiago de Compostela
• Little
• People
• Romances

Informations

• Publié par: nuthang
• Nombre de lectures: 29
• Langue: English

Extrait

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M342 PDE: THE DIVERGENCE THEOREM
MICHAEL SINGER
1. STATEMENT OF THE DIVERGENCE THEOREM
nLet R be a bounded open subset ofR with smooth (or piecewise smooth) boundary
nR.Let X=(X ;:::;X ) be a smooth vector field defined inR ,oratleastin R[ R.Let1 n
n be the unit outward pointing normal of R. Then the divergence theorem states:
(1) divXdV = X ndA
R R
where
X X X1 2 n
divX = X = + + + ;
x x x1 2 n
ndV is the element of volume inR and dA is the element of surface area on R.
1.1. Suitable domains. Examples of suitable bounded domains R include: if n = 1, inter-
vals (a;b);if n = 2, rectanglesfa < x< b ;a < y< bg, discs, and pieces of discs such1 1 2 2
as half discs, quarter discs etc.; ifn = 3, boxesfa < x< b ;a < y< b ;a < z< bg,1 1 2 2 3 3
balls, half balls, etc. We shall seldom go beyond 3 dimensions in this course.
1.2. Construction of n and ndA. If n =1and R =(a;b), then vectors are just real
numbers and n =−1at x = a and =+1at x = b.
If n = 2, the normal is got by rotating the tangent vector through 90 (in the correct
direction so that it points out!). The quantity tds can be written (dx;dy) along the surface,
so that
(2) ndA := nds=(dy;−dx):
Here t is the tangent vector along the boundary curve and ds is the element of arc-length.
If n = 3, then we have to decide how the boundary of R is to be described. You may
recall that if R is described as a level set of a function of 3 variables (i.e. R =fx :
F(x)= 0g), then a vector pointing in the direction of n is gradF. We shall use the case
where F = z− f (x;y) and R corresponds to the inequality z< f (x;y). Then
(− f ;− f ;1)x y 2 2 1=2(3) n = ; dA=(1 + f + f ) dxdy:x y2 2 1=2(1 + f + f )x y
Hence the quantity ndA is simpler than either n or dA separately:
(4) ndA=(− f ;− f ;1)dxdy:x y
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2 MICHAEL SINGER
2. THE DIVERGENCE THEOREM IN 1 DIMENSION
In this case, vectors are just numbers and so a vector field is just a function f (x).
Moreover, div = d=dx and the divergence theorem (if R=[a;b]) is just the fundamental
theorem of calculus:
b
(df=dx)dx = f (b)− f (a)
a
3. THE DIVERGENCE THEOREM IN 2 DIMENSIONS
Let R be a 2 dimensional bounded domain with smooth boundary and letC = R be its
boundary curve. Recall Green’s theorem states:
( Q− P)dxdy = Pdx+Qdy:x y
R C
This is the same as the two dimensional divergence theorem if we take the vector field
(X ;X ) with X = Q and X =−P. For then it reads1 2 1 2
divXdxdy = ( X + X )dxdy = −X dx+ X dyx 1 y 2 2 1
R R C
= (X ;X ) (dy;−dx)= X nds1 2
C R
where we have used (2).
4. THE DIVERGENCE THEOREM IN 3 DIMENSIONS
We shall give a ‘proof’ of this theorem in stages.
4.1. The divergence theorem for a box. Consider the box R =fa < x< b ;a < y<1 1 2
b ;a < z< bg.Let u be a function of x=(x;y;z). For each fixed (y;z) the fundamental2 3 3
theorem of calculus gives
b1
u (x)dx = u(b ;y;z)− u(a ;y;z)x 1 1
a1
Now integrating with respect to y and z,
(5) f (x)dV = [ f (b ;y;z)− f (a ;y;z)]dxdy;x 1 1
R S1
where S =fa < y< b ; a < z< bg. This is just the divergence theorem for the vector1 2 2 3 3
field X=(u;0;0)! To see this, note divX = f for this vector field, so the LHS of (5) is1
certainly divXdV.Now R is a union of six rectangles in parallel pairsR
S =fx = a ;a < y< b ; a < z< bg; S =fx = b ;a < y< b ; a < z< bg;11 1 2 2 3 3 12 1 2 2 3 3
parallel to the (y;z) plane,
S =fa < x< b ;y = a ; a < z< bg; S =fa < x< b ;y = b ; a < z< bg;21 1 1 2 3 3 22 1 1 2 3 3
parallel to the (x;z) plane, and
S =fa < x< b ;a < y< b ; z = ag; S =fa < x< b ;a < y< b ; z = bg;31 1 1 2 2 3 32 1 1 2 2 3
parallel to the (x;y) plane. It looks complicated, and a diagram would tell the story much
better. Draw one for yourself. Moreover we have that
n =−i on S ;n = i on S ;11 12Z
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M342 PDE: THE DIVERGENCE THEOREM 3
n =−j on S ;n = j on S ;21 22
n =−k on S ;n = k on S :31 32
So for X=(u;0;0), X n = 0 on the four faces S ;S ;S ;S , whereas21 22 31 32
X n =−u(a ;y;z) on S ; X n = u(b ;y;z) on S :1 11 1 12
This is precisely the combination of signs on the RHS of (5), so that this really is the
divergence theorem for X=(u;0;0) and this R.
In precisely analogous fashion, the divergence theorem for X=(0;v;0) and for X =
(0;0;w) is verified. Adding these results, we obtain the divergence theorem for the box,
with any vector field X=(u;v;w).
4.2. Cutting lemma. Consider now a bounded domain R decomposed as a union of 2
subdomains R and R , with a common interface S . Typical example: an apple cut in1 2 0
half. Let R = S and write S = S [ S ,sothat1 2
R = S [ S ; R = S [ S :1 1 0 2 2 0
Let the normal of S be denoted n , the normal of S be denoted n and the normal of1 1 2 2
S , pointing into R be n . (Draw a picture.) In particular, the outward drawn0 2 0
normal of R is equal to n along S and the outward drawn normal of R is equal to−n1 0 0 2 0
along S .0
We claim that if the divergence theorem holds for the pieces R and R , then it holds1 2
for R. To see this, let X be a smooth vector field, and apply the divergence theorem for R1
and R , taking careful note of the sign of n as in the previous paragraph. We get2 0
divXdV = X ndA + X n dA; divXdV = X ndA− X n dA:0 0
R S S R S S1 1 0 2 2 0
Adding, the contributions from S cancel out and so0
divXdV = divXdV + divXdV = X ndA + X ndA = X ndA;
R R R S S S1 2 1 2
just as required.
4.3. Dissection argument. With the aid of the divergence theorem for boxes and the
cutting lemma, one can imagine proving the divergence theorem by slicing a given domain
R into small boxes. We know the divergence theorem for boxes, so by the cutting lemma,
we know it for any domain that can be cut up into boxes. But most domains have a curved
boundary, so the whole of R is unlikely to be a union of boxes. It is not uncommon to argue
that by taking the boxes to be smaller and smaller you can approximate any reasonable
domain R better and better, and hence taking some sort of limit, the divergence theorem
follows for any such domain.
If you are not satisfied with this argument, read on.
24.4. Divergence theorem for regions with a curved boundary. Let DR be a bounded
domain with piecewise smooth boundary D, and consider the region
3(6) R =f(x;y;z)2R : (x;y)2 D;0< z< f (x;y)g
where f is a smooth function in D that is continuous up to D. We shall prove the diver-
gence theorem for this region R.¶
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4 MICHAEL SINGER
The motivation for considering this kind of R is that it is intuitively plausible that any
3reasonable domain inR can be split up as a union of subdomains each of which is either
a box or one like (6). By ‘like’ here, I mean that you may have to permute the roles of
x, y and z in the definition. For example, if D were itself a rectangle, then R would be a
box with 5 flat sides and one curved side. The flat sides are given by the vertical planes
through the sides of D, plus the bottom face z = 0. The curved side corresponds to the
surface z = f (x;y).
In general the boundary of R consists of 3 pieces, S , S and S , say, where the bottom0 1 2
face
(7) S =f(x;y;0) : (x;y)2 Dg; ndA =−kdxdy0
the ‘vertical’ side
(8) S =f(x;y;z) : (x;y)2 D;0 z f (x;y)g; ndA=(dydz;−dxdz;0)1
and the top face
(9) S =f(x;y; f (x;y)) : (x;y)2 Dg; ndA=(− f ;− f ;1)dxdy:2 x y
In (8) and (9) we have used (2) and (4). S may naturally consist of several pieces, but for1
the purposes of the proof it is enough to think of R as consisting of S , S and S .0 1 2
We shall now prove the divergence theorem for R. We shall do it for vector fields
X=(0;0;u) and X=(v;0;0). The argument for a vector field with x andz coordinates
zero is very similar to that for (v;0;0) and will be omitted. The general result follows by
addition, just as for the box.
The easiest case is X=(0;0;u).ThendivX = u ,andz

f (x;y)
divXdV = u dz dxdy = u(x;y; f (x;y))dxdy− u(x;y;0)dxdy:z
R D z=0 D D
Now X n = 0 in this case over S . So, taking into account (7) and (9), this equation can1
be rewritten as
divXdV = X ndA + X ndA = X ndA:
R S S R0 2
Now we consider the case X=(v;0;0).Pick w so that
(10) w(x;y;z)= v(x;y;z):z
We have divX = v = w = w. Hencex x z z x

f (x;y)
(11) divXdV = wdz dxdyz x
R D z=0
= [w (x;y; f (x;y))− w (x;y;0)]dxdy:x x
D
We will use Green’s theorem to turn this into a boundary integral, but note first that
w (x;y; f (x;y)) the partial derivative of w with respect to x, evaluated at the point (x;y; f (x;y)),x
is notthesameas [w(x;y; f (x;y))] , the partial derivative of w(x;y; f (x;y)) with respect tox
x! In