text stringlengths 16 3.88k | source stringlengths 60 201 |
|---|---|
onecker delta and λ and µ are positive elastic constants of the material,
called Lame coefficients. The corresponding (free) energy density E of the body associated
with deformation, obtained from the relation
is therefore
σij =
∂E
∂eij
,
1
E = λe2
2
ii + µe2
ij.
(270)
(271)
As stated above, the sum eii = Tre is related ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
(cid:18) 1
µ
1
3
+
1 (cid:19)
3K
p.
(275)
(276)
The component ezz gives the lengthening of the rod, and the coefficient of p is the coefficient
of extension. Its reciprocal is Young’s modulus
Y =
9Kµ
3K + µ
.
(277)
The components exx and eyy give the relative compression of the rod in the transverse
direction. The ratio of... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
our result from these. Rather, we shall use our
knowledge of variational calculus to calculate afresh the energy of a bent plate, and set
about varying that energy.
When a plate is bent, it is stretched at some points and compressed at others: on the
convex side there is evidently an extension and on the concave side t... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
∂w
∂x
∂w
∂y
where the constants of integration were chosen so as to make ux = uy = 0 for z = 0.
Knowing ux and uy we can now determine all the components of the strain tensor:
exx = −z
∂2w
∂x2 ,
eyy = −z
∂2w
∂y2
exy = −z
∂2w
∂x∂y
,
(290)
55
exz = eyz = 0,
ezz =
zν
1 − ν
∂2w
∂x2
+
∂2w
y2
∂
.
We now calculate the ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
2D problem. Using the calculus of variations we have that the energy of the distorted beam
is
δE
δw
This expression must equal the force f (x) applied to deform the plate:
Y h3
12(1 − ν2)
d4w
.
dx4
=
Y h3
d4w
12(1 − ν2) dx4
= f (x).
The simplest boundary conditions are if the edges are clamped, in which ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
)
conserved. This simple argument shows that what we really need to do is figure out how to
generalise our random walkers to something that conserves momentum and energy.
To rigorously justify the macroscopic equations of motion for a fluid (i.e., a collection
of particles interac... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
their number.
The continuum hypothesis implies that it is possible to attach a definite meaning to
the notion of value ‘at a point’ of the various fluid properties such as density, velocity
and temperature, and that in general values of these quantities are continuous functions of
position and time. There is ample observ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
the continuity equation, consider Newton’s laws
for a particular moving element of fluid:
d (cid:90)
dt V (t)
ρudV = −
(cid:90)
(cid:90)
pndS +
S(t)
V (t)
f dV,
(302)
where V (t) is the volume of the element enclosed by the surface S(t), f is the density of
body forces, such as gravity ρg, and p is a pressure force. The... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
scalar potential ∇ψ. It is therefore usual to redefine
pressure as p + ψ → p. This implies that gravity simply modifies the pressure distribution
in the fluid and does nothing to change the velocity. However, we cannot do this if ρ is not
constant or if we have a free surface (as we shall see later with water waves).
Assu... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
δ(v − vi)∇xiδ(x − xi) · x˙ i + δ(x − xi)∇viδ(v − vi) · v˙ i}
= −∇x
N
(cid:88)
i=1
δ(v − vi)δ(x − xi) · vi − ∇v
N
(cid:88)
i=1
δ(x − xi)δ(v − vi) ·
F i
m
(311)
59
where, in the last step, we inserted Newton’s equations and used that
∂
∂xi
δ(x − xi) = −
∂
∂x
δ(x − xi)
(312)
Furthermore, making use of the defining propert... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
(x − xj) · ∇vf
(314)
In the second line, we have again exploited the properties of the delta function which allow
us to replace xi by x. Also note the appearance of the convective derivative on the lhs.;
the above derivation shows that it results from Newton’s first equation.
To obtain the hydrodynamic equations from (... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
+ v · ∇ (cid:104)f (cid:105) = −∇v · [G(x)(cid:104)f (cid:105) + C]
(316)
where the collision-term
C(t, x, v) :=
(cid:88)
xj (cid:54)=x
(cid:104)H(x − xj)f (t, x, v)(cid:105)
(317)
represents the average effect of the pair interactions on a fluid particle at position x.
We now define the mass density ρ, the velocity field ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
velocity space) that vanishes since
for physically reasonable interactions [G(x)(cid:104)f (cid:105) + C] → 0 as |v| → ∞. We thus recover the
mass conservation equation
∂
∂t
ρ + ∇ · (ρu) = 0.
(321)
To obtain the momentum conservation law, lets multiply (316) by v and subsequently
integrate over v,
(cid:90)
dv3
m
(cid:1... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
g(x) := G(x)/m is the force per unit mass (acceleration) and the last term
(cid:90) (cid:88)
(cid:90)
c(t, x) = dv3C = dv3
(cid:104)H(x − xj)f (t, x, v)(cid:105)
(324)
(325)
encodes the mean pair interactions. Combining (323) and (324), we find
xj =(cid:54) x
(cid:18) ∂
∂t
ρ
The symmetric tensor
(cid:19)
+ u · ∇ u = −∇ ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
∇yϕ(x − y)]f (t, x, v)(cid:105)
d3y [∇ρ(t, y)] (cid:104)ϕ(x − y)f (t, x, v)(cid:105)
(330)
62
ϕ 3
0a
m
ϕ0a3
m
ϕ0a3
m2
ϕ0a3
2m2
∂
∂t
∂
∂t
(cid:18)
In general, it is impossible to simplify this further without some explicit assumptions about
initial distribution P that determines the average (cid:104) · (cid:105). There... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
manifestation of the well-known hierarchy problem,
encountered in all14 attempts to derive hydrodynamic equations from microscopic models.
More precisely, the hierarchy problem means that the time evolution of the nth-moment
depends on that of the higher moments. The standard approach to overcoming this obstacle
is to ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
337a)
(337b)
In this case, one has to solve for (p, u).
13 The Navier-Stokes Equations
In the previous section, we have seen how one can deduce the general structure of hydro
dynamic equations from purely macroscopic considerations and and we also showed how
one can derive macroscopic continuum equations from an un... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
matrix σ with components σij , which has the property that the stress acting on a surface S
with unit normal n is just σijnj . For an arbitrary fluid element, the net force arising from
surface stresses is
σ · ndS =
( · σ)dV.
S
V
(338)
We must therefore determine the form of σ in order derive our equations of moti... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
the form of the viscous tangential stresses, we reason that these must
arise from relative motion between fluid elements. Thus the stress should somehow depend
on \u, which will only be nonzero if there are velocity gradients. Note that \u is also a
tensor, and can be written explicitly as
\u =
⎛
∂xux ∂yux ∂zux
⎝ ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
vorticity tensor. To see this, let’s consider a flow that
is rotating, but not deforming, and also a flow that is deforming, but not rotating. In two
dimensions a rotating flow is u ∝ (−y, x) and a deforming flow is u ∝ (x, y). For the
rotating flow it can be shown that the antisymmetric part \ua is non-zero, and for the... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
how hard
you are shearing it. Fortunately, it happens that most simple fluids are Newtonian under
ordinary conditions. So for water, oil, air etc. it is often possible to approximate fluids as
being Newtonian. Non-Newtonian also happens frequently in nature (e. g. liquid crystals)
and gives rise to fascinating flow ph... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
both important and useful to make sure that all
of the assumptions have been clearly stated. Is it true in general that only one number
is sufficient to completely characterise the viscosity? Stated another way, the viscous ef
fects in a fluid capture the macroscopic consequences of dissipative collisions between fluid ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
problems, but to directly compare solutions to experiments we first
need to think a little about boundary conditions.
There are several types of boundaries that occur in practice, though the most common is
simply the solid wall. The equations of motion for the fluid specify what happens everywhere
except for the fluid... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
breaks down). The no-slip
67
condition says that dissipative processes are so strong that the tangential component of the
velocity actually vanishes there! This is known to be true from experimental studies of the
motion of fluid near walls. It is extremely important, however, to understand that this is a
phenomen... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
= 0 in the upper half plane. The boundary conditions are
that u(0, t) = U for t > 0 (no-slip), and we expect that u will vanish as y → ∞ (since it
vanishes initially).
We look for a similarity solution of the form u(y, t) = f (y/
√
νt) = f (η). The logic is the
same as we employed in examining the diffusive spreadi... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
i.e., the velocity is always the same function of (y/
profile becomes stretched out and the effects
distance νt from the boundary.
√
√
νt). As time progresses, the velocity
of motion are largely confined to within a
The solution above would not have been so simple if, for instance, an upper boundary
were present. I... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
time taken for viscous diffusion to
act over the gap between the plates, and gives an order of magnitude estimate for the time
taken to set up a steady profile. And we have obtained all this information, without having
to do any difficult mathematics!
13.4 The Reynolds number
For an incompressible flow, we have establi... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
��usion equation.
An important parameter that indicates the relative importance of viscous and inertial
forces in a given situation is the Reynolds number. Suppose we are looking at a problem
where the characteristic velocity scale is U0, and the characteristic length scale for variation
of the velocity is L. Then ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
ascal, it moves a distance equal to the thickness of
the layer between the plates in one second. The dynamic viscosity of water (T = 20 ◦C) is
µ
= 1.0020 × 10−3 Pa · s.
Kinematic viscosity When dealing with incompressible fluids of constant density, then
it’s often convenient to consider the kinematic viscosity ν, ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
Sperm cells move by inducing a wave-like
deformation in a thin flagellum or cilium, whereas algae and other organisms move by
beating two or more cilia in a synchronized manner (see slides).
Because of their tiny size, these microswimmers operate at very low Reynolds number,
i.e., inertial and turbulent effects are negli... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
, by replacing Eq. (354) with Eq. (359), it is assumed that for small Reynolds numbers
˜Re(t, x) := |(cid:37)(u · ∇)u|/(µ∇2u) (cid:39) U L((cid:37)/µ) (cid:28) 1 one can approximate
(cid:37) [∂tu + (u · ∇)u] − µ∇2u (cid:39) −µ∇2u
The consistency of this approximation can be checked a posteriori by inserting the solutio... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
boundary conditions (360) reads18
ui(t, x) = Uj
(cid:20)
p(t, x) =
3
2
µa
(cid:18)
δji +
3
a
4
|x|
Ujxj + p .
|x|3
∞
(cid:19)
+
xjxi
|x|2
1
4
a3
|x|3
(cid:18)
δji − 3
(cid:19)(cid:21)
,
xjxi
|x|2
(365a)
(365b)
If the particle is located at X(t), one has to replace xi by xi −Xi(t) on the rhs. of Eqs. (365).
Parameterizi... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
above that, in 3D, the fundamental solution to the Stokes equations for a point
force at the origin is given by the Oseen solution
where
ui(x) = Gij(x) Fj ,
p(x) =
Fjxj
4π|x|3
+ p ,
∞
Gij(x) =
1
8πµ |x|
(cid:18)
δij +
(cid:19)
,
xixj
x|2
|
(370a)
(370b)
It is interesting to compare this result with corresponding 2D sol... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
�erence between 3D and 2D hydrodynamics has been confirmed experimentally
for Chlamydomonas algae.
74
14.5 Force dipoles
In the absence of external forces, microswimmers must satisfy the force-free constraint. This
simplest realization is a force-dipole flow, which provides a very good approximation for the
mean flow fiel... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
cid:96)
2πµ
= −
(cid:18)
(cid:20)
−δij
(cid:18)
−
δik
xk
|x|2 +
x
knk
i |x|2 + ni
n
xj
|x|2 + δjk
xjnj
|x|2
+ nknk
F +
j
xi
|x|2 − 2
xi
|x|2 − 2
xixjxk
|x|4
nkxixjxknj
|x|4
(cid:19)
(cid:19)(cid:21)
and, hence,
where xˆ = x/|x|.
u(x) =
F (cid:96)
πµ|x|
2
(cid:2)2(n · xˆ)2 − 1(cid:3) xˆ
(380)
3D case To compute the dipo... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
�i)
|
=
(383a)
(383b)
(383c)
(cid:1).
(384)
Inserting this expression into (379), we obtain the far-field dipole flow in 3D
u(x) =
F (cid:96)
4πµ x 2
|
|
(cid:2)
3(n · xˆ)2 − 1 xˆ.
(cid:3)
(385)
Experiments show that Eq. (385) agrees well with the mean flow-field of a bacterium.
Upon comparing Eqs. (380) and (385), it beco... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
�x + ey∂y + ez∂z,
(386b)
76
and, using the orthonormality ej · ek = δjk, the Laplacian is obtained as
∆ = ∇ · ∇ = ∂2 + ∂2
y + ∂2
z .
x
(386c)
One therefore finds for the vector-field divergence
∇ · u = ∂iui = ∂xux + ∂yuy + ∂zuz
(386d)
and the vector-Laplacian
∂2
2
x + ∂y ux + ∂2
xu
z ux
2
∆u = ∂i∂iu = ∂2
2
xuy + ∂y uy ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
∂φuφ + ∂zuz.
1
r
The Laplacian of a scalar function f (r, φ, z) is given by
∇2
f =
1
r
∂r(r∂rf ) + ∂2 +
1
2 φf
r
∂2
z f
and the Laplacian of a vector field u(r, φ, z) by
∇2u = Lrer + Lφeφ + Lzez
77
(387e)
(387f)
(387g)
(387h)
where
Lr =
Lφ =
Lz =
1
r
1
r
1
r
∂
1 2
∂r(r∂rur) + ∂φur + ∂
2
r
1
2
r
1
2 φ z + ∂2
r
∂r(r∂ruz)... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
the fluid motion, let us consider pressure
driven flow along a cylindrical pipe of radius R pointing along the z-axis. Assume that the
flow is rotationally
symmetric about the z-axis and constant in z-direction, u = uz(r)ez,
where r = x2 + y2 is the distance from the center. For such a flow, the incompressibil-
ity conditi... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
channel radius, signaling that the presence of boundaries can
substantially suppress hydrodynamic flows. To illustrate this further, we next consider a
useful approximation that can help to speed up numerical simulations through an effective
reduction from 3D to 2D flow.
14.6.3 Hele-Shaw flow
Consider two quasi-infinit... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
subsequently averaging
along the z-direction20, yielding
0 = \ · U ,
0 = −\P + µ\2U − κU
(397)
12µ/H 2 and \ is now the 2D gradient operator. Note that compared with
where κ =
unconfined 2D flow in a free film, the appearance of the κ-term leads to an exponential
damping of hydrodynamic excitations. This is analog... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
we expect the solution to be
cylindrically symmetric
p(x, t) = p(r, t) ,
u(x, t) = uφ(r, t)eφ,
(398)
with only a component of velocity in the angular direction eφ. This component will only
depend on r, not the angular coordinate or the distance along the axis of the cylinder
(because the cylinder is assumed to b... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
(400a)
and the boundary conditions that
uφ(0, t) = 0 ,
uφ(R, t) = 0
∀t > 0.
(400b)
This is done using separation of variables. Since the lhs. of Eq. (399b) features a first-
order time derivative, which usually suggests exponential growth or damping, let’s guess a
solution of the form
Putting this into the govern... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
The equation becomes
ξ2F "" + ξF " + ξ2 − 1 F = 0.
Even with the factors of ξ included, this problem is not more conceptually difficult, though
it does require knowing solutions to the equation. It turns out that the solutions are
called Bessel Functions. You should think of them as more complicated versions of sines... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
and so has a countably infinite
number of zeros.) Our solution is therefore
√
√
uφ(r, t) =
∞
−νλ2 t/R2
n
Ane
J1(λnr/R).
(406)
n=1
To determine the An’s we require that the initial conditions are satisfied. The initial con
dition is that
uφ(r) = ωr.
(407)
Again, we now think about what we would do if the abov... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
Each of the terms in the sum is decreasing exponentially in time. The smallest value
of λn decreases the slowest. It turns out that this value is λ1 = 3.83. Thus the spin down
time should be when the argument of the exponential is of order unity, or
t ∼
R2
.
νλ2
1
82
(413)
This is our main result, and we shoul... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
atoms, each of which has a molecular spin. The energetics of
the interaction between the spins is that each spin produces a magnetic field which tries to
align the neighboring spins. A popular microscopic model for a magnet imagines the spins
confined to a regular lattice, and then ascribes an energy
U =
Jij si · sj ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
Using the calculus of variations the function M (x) that minimises the energy satisfies
We already know that
d2M
dx2
ν
− bM + 2cM 3 = 0.
M = 0, ±
b
2c
(416)
(417)
are three constant solutions. Now if ν = 0 there is no penalty for orientation change
throughout the system, and for b > 0 the entire system has ma... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
The different behaviour arises because if ν is nonzero the entire system is forced to match
the imposed boundary conditions at the edges. Setting ν = 0 is therefore called a singular
perturbation.
16.2 An elementary algebraic equation
As another example of a singular perturbation, consider the solution of the algebr... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
the ‘outer’ condition u(1) = 2. The
approximate solution satisfying the outer condition is therefore
u = x + 1.
(428)
In the outer region the approximate solution and the true solution are very close. However,
in a region close to x = 0 they differ greatly. We call this the boundary layer. It arises
because the sm... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
this is not
possible and we must resort to approximations like those used here. The inner solution
is valid within a boundary layer of thickness E and matches to the outer solution. Once
again we see that ignoring the term multiplied by E in the original problem is a singular
perturbation; no matter how small E is,... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
ν
=
U L Re
(434)
where Re is the Reynolds number. For an airplane typical values are U =400mph, L=5m
and ν=0.1cm2/s, giving Re = 108 . Thus, the inertial forces are eight orders of magnitude
larger than viscous forces, so it is seems very reasonable that we can neglect them. Doing
so, we are left with the Euler ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
u
2
(439)
The constancy of H in irrotational flow is a famous result, and has many simple qualitative
consequences. It requires that the pressure in a fluid is smaller when the velocity is larger,
and is a statement of conservation of energy when viscous dissipation can safely be ignored.
What are the consequences of... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
.
(443b)
(444)
The first term concerns changes in the velocity field and is identically zero since
(cid:90)
Du
Dt
· dl =
(cid:90)
1
ρ
−\p · dl =
(cid:90)
1
ρ
(\ × \p) · ndA = 0.
(445)
The second term relates to stretching of the loop and is zero for essentially geometric reasons.
One has that
= du
(446)
sin... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
,
4πr
(449)
24More generally, DΓ/Dt = 0 still holds for barotropic ideal fluids with conservative body forces.
25Read Acheson (p. 168) for an interesting discussion of this.
88
where c would be the charge in an electrostatic problem. What does this solution correspond
to for us? The velocity... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
g(r) = Ark + Br−k .
f = C + Dθ.
g(r) = A + B ln r.
89
(455b)
(455c)
(456)
So, the most general solution is
φ(r, θ) = (A0 + B0lnr)(C0 + D0θ) +
(Ck sin kθ + Dk cos kθ)(Akr k + Bkr
−k)
∞
0
The corresponding velocity field is
k=1
u = \φ =
∂φ
∂r
ˆr +
1 ∂φ ˆ
θ.
r ∂θ
Putting in our general sol... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
φ → u0x as r → ∞,
∂φ
∂r
= 0 at r = R.
(462a)
(462b)
Using the general solution found above, the first boundary condition requires that
(cid:18)
φ = u0r +
(cid:19)
D1B1
r
cos θ + A0(C0 + D0θ).
(463)
The second boundary condition gives us
φ = D0θ + u0 cos θ r +
(cid:18)
(cid:19)
R2
r
,
(464)
where we have... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
wing;
FL = Γρ u0 £.
(468)
Thus, the lift on the cylindrical wing is zero (unless it is spinning, so that Γ = 0!). The lift
force due to rotation is also known as Magnus force.
Our earlier discussion of singular perturbations suggests that D’Alembert’s paradox for
inviscid flows arises as a consequence of the fact ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
��ow. Besides it’s physical convenience, another
great thing about the stream function is the following. By definition
u =
∂ψ
∂y
=
∂φ
,
∂x
v = − =
∂ψ
∂x
∂φ
,
∂y
(471a)
(471b)
where φ is the velocity potential for an irrotational flow. Thus, both φ and ψ obey the well
known Cauchy-Riemann equations of compl... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
=
∂v/∂x, the Cauchy-Riemann equations. If this is true then f (z) is said to be analytic and
we can simply differentiate with respect to z in the usual manner. For our simple example
f (z) = z2 we have that df /dz = 2z (confirm for yourself that z2 is analytic as there are
many functions that are not, e.g., |z| is no... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
example consider uniform flow at an angle α to the x-axis. The corresponding
complex potential is w = u0ze
. Using the above relation,
this tells us that u = u0 cos α and v = u0 sin α.
. In this case dw/dz = u0e
−iα
−iα
We can also determine the complex potential for flow past a cylinder since we know that
(cid:18)
... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
), then W (Z) = w(F −1(Z)) is also analytic,
and may be considered a complex potential in the new co-ordinate system. Because W (Z)
and w(z) take the same value at corresponding points of the two planes it follows that Ψ
and ψ are the same at corresponding points. Thus streamlines are mapped into streamlines.
In pa... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
constant however and doesn’t change the velocity.
Finally there is the non-trivial Joukowski transformation,
Z = F (z) = z +
2c
z
.
What does this do to the circle? Well, z = aeiθ becomes
Z = ae +
iθ
2
c
a
e −iθ = (a +
2
c
) cos θ + i(a −
a
2
c
a
) sin θ.
Defining X = Re(Z), Y = Im(Z), it is easily shown th... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
If we choose c = R, then the ellipse collapses to a flat plate. The velocity components in
the Z plane are
U − iV =
which can also be written as
dW
dZ
= u0 cos α −
i(Γ + 2πu0Z sin α)
2π Z2 − 4R2
√
,
(489)
(490)
U − iV =
dw/dz
dZ/dz
=
(cid:16)
u0 e
−iα − e
(cid:17)
iαR2
iΓ
z2 − 2πz
1 − R2/z2
On the surface... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
we have to rotate in
the opposite direction to cancel out the singularity at θ = π.
20.2 Flow past an aerofoil
What if we could now construct a mapping with a singularity just on one side ? This we
can do by considering a shifted circle, that passes through z = R but encloses z = −R. In
this case we obtain an aero... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
� and
FD = 0 are unaffected by the wing shape. To this end, we first derive Blasius’ lemma and
then the Kutta-Joukowski theorem.
20.3 Blasius’ lemma
To derive Blasius’ lemma, we consider the force acting per unit length on the wing, f =
F /(cid:96) = (fx, fy), which is obtained by integrating the pressure over the (now a... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
idy)
2
= (v2
ydx − 2ivxvydx + vxidy − v2
= v2dx − v2
x
xdy + v2
ydx − 2iv2
xdx − v2
= v2
xidy − v2
ydx − v2
xdx + v2
= v2
x + v2
= (v2
x + v2
y)dx − (v2
y)idy
= (v2
2
x + vy)(dx − idy)
= |v|2 dz¯
xidy − v2
yidy
yidy + 2vxvydy
yidy + 2v2
ydx
¯f = i
(cid:73)
ρ
2
v2 dz = i
(cid:73) (cid:18) dw (cid:19)2
dz
ρ
2
dz
which is... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
0. The residue is the coefficient
k=−∞ akz . The residue theorem states that for a positively oriented
k
(cid:73)
γ
f (z)dz = 2πi a−1.
97
Computing the integral on the rhs. gives
dw
dz
dz =
(vx − ivy)(dx + idy)
=
(vxdx + vydy) + i
(vxdy − vydx)
(511)
The last integral vanishes as the boundary is stream line, see E... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
of airfoils, we have neglected viscosity which led to d’Alembert’s
paradox. To illustrate further the importance of boundary layers, we will consider one
more example of substantial geophysical importance, where the dynamics of laminar flows
is completely controlled by the boundary layer. In the process of deriving t... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
ective term is of order U 2/L whereas the coriolis force is of order ΩU . We will assume
that ΩU (cid:29) U 2/L so that the equation of motion is effectively
∂u
∂t
= −
1
ρ
∇ · u = 0.
∇p + ν∇2u − 2Ω × u,
(518a)
(518b)
21.2 Steady, inviscid flow
Now let’s consider flow at high Reynolds number. The Reynolds number is now ΩL2... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
of a rotating liquid must
be two-dimensional, what happens if one attempts to make a three dimensional motion
by, for example, pushing a three dimensional object through the flow with a small uniform
velocity? At the beginning of his paper he points out three possibilities:
1. The motion in the liquid is never steady.
2... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
small so that we can use the limit of a stationary, inviscid fluid.
More precisely, we consider the ratio
Ek =
ν∇2u
Ω × u
=
ν
ΩL2
(cid:28) 1.
100
(522)
This dimensionless number is called the Ekman number. The flow is then strictly two-
dimensional, and the Navier-Stokes equation simplifies to
−
\p = 2Ω × u.
1
ρ
Tak... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
deduce about atmospheric flows. For an
atmospheric flow, the analogue of Ω is not the earth’s rotation speed ωˆ, but instead Ω =
ωˆ sin φ, where φ is the longitude. Now, this shows that the effective Ω changes sign in
the northern and southern hemisphere. What does this imply for the dynamics? When
Ω > 0 the velocity ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
.
22.1 A small deviation
Suppose ΩT =Ω and ΩB = Ω + E. Now there is no way to satisfy the no slip condition on
both the top and bottom while having the whole flow spin at angular velocity Ω. Let’s move
first to the rotating frame, and try to compute the secondary flow that is induced. Clearly,
101
without viscosity... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
527b)
(527c)
(527d)
Here we have made the boundary layer approximation that ∂/∂z (cid:29) ∂/∂x, ∂/∂y.
From the continuity equation we deduce that w is much smaller than the velocity com-
ponents parallel to the boundary so that ∂pI /∂z = 0, and the equations become
−2Ω(v − vI ) = ν
2Ω(u − uI ) = ν
,
∂2u
∂z2
∂2v
∂z2
.
(... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
�z
=
−
∂u
∂x
−
∂v
∂y
=
(cid:18) ∂vI
∂x
∂u
− I
∂y
(cid:19)
e−z∗
sin z∗.
(533)
Integrating from z∗ = 0 to ∞ gives
w =
1 (cid:18) Ω (cid:19)−1/2 (cid:18) ∂vI
∂x
2
ν
∂uI
∂y
(cid:19) ωˆI=
2
(cid:114) ν
Ω
,
−
(534)
where ωˆI is the vorticity in the inviscid flow. Thus, if ωˆI > 0 (i.e., the bottom boundary is
moving slower th... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
container both are happening. Since uI , vI and wI are all independent of z
then so is ωˆI . Thus, the only way the experiment could work is if the induced value of ωˆI
from both cases matches. This implies that
ωˆI = ΩT + ΩB.
(536)
With ΩB = 0 and ΩT = (cid:15) we have that ωˆI = (cid:15). Thus, the flow in the inner r... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
= 2Ω[w(L)
− w(0)].
(539)
The velocity is equal and opposite at the two boundaries (flow is leaving both boundary
layers), and has magnitude (ν/Ω)1/2ωˆI /2. Thus
√
∂ωˆI
∂t
2
= −
Ων
L
ωˆI ,
(540)
implying that vorticity is decreasing in the interior with a characteristic decay time L/2
Ων.
For the coffee cup this gives us ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
� is the gravitational potential such that g = −∇χ. This can be integrated to give
the unsteady Bernoulli relation
∂φ
∂t
+
1
2
(∇φ)2 +
p
ρ
+ χ = C(t).
(542)
Here, C(t) is a time dependent constant that does not affect the flow, which is related to φ
only through spatial gradients. The surface is h(x, t) and we have the k... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
small, so that we can linearise the equations. The linearised system of equations consists
of Laplace’s equation and the boundary conditions at y = 0:
=
∇2φ = 0
∂h
∂t
∂φ
∂t
∂φ
∂y
= −gh +
(x, 0, t),
γ
ρ
∂2h
∂x2
,
These conditions arise because we have Taylor expanded terms such as
v(x, h, t) = v(x, 0, t) + hvy(x, 0, t),... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
a raindrop fall into water.
23.2 Properties of the dispersion relation
When a group of waves travels across the surface of water each particular wave crest moves
faster than the group as a whole, i.e., if you look closely then wave crests within the
disturbance appear to move through it. Why is this? The answer is that... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
waves in deep
water, we have
Alternatively for capillary waves
cg =
d
dk
(gk)
1
2 =
1
2
cp.
cg =
d
dk
(γk3/ρ)
1
2 =
3
2
cp.
For comparison, the dispersion relation for sound waves, ω = ck, tells us that
h(x, t) =
(cid:90) +∞
−∞
b eik(x±ct)
k
dk.
So we start with an arbitrary disturbance, and this perturbation just move... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
and regain Laplace’s
If M < 1 we can make a simple change of variables X = x/
equation. Thus everything can be solved using our conformal mapping techniques. However,
if M > 1 then the original equation now looks like a wave equation, with y replacing t,
yielding solutions of the form
√
˜φ(x, y) = Φ(x − y M 2 − 1)
(cid... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
(cid:48)+ω(k)t(cid:48)
(cid:48)
]eΓkt Θ(t − t(cid:48)) δ(x(cid:48) − U t(cid:48))
ˆ
dk h e
k
i[kx−ω(k)t]−Γkt
(cid:90) U t
−∞
dx h e{−i[(k−ω(k)/U ]+(Γ/U )}x(cid:48)
(cid:48) ˆ
k
To simplify things a bit, let’s focus on spatial perturbations at time t = 0
h(x, 0) ∝
(cid:90) ∞
ˆdk h eikx
k
(cid:90) 0
−∞
dx(cid:48)
e{−i[(k... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
,
(568)
where U0 = U − ω(cid:48)(k0) is the difference between the boat velocity and the group velocity of
the Fourier component, and the contour C includes the real k-axis with other contributions
vanishing. If we are considering gravity waves, then U0 is a positive quantity. The integral
has to be evaluated around a c... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
1D treatment needs to be extended to account for the V-shaped wake
behind a boat, also known as the Kelvin wedge. In 2D, the disturbance generated by the
boat is
h(x, t|
x , t ) = dk h(k)eik(x−x')e iω(k)(t−t') Γk(t−t'
'
e
ˆ
'
).
(570)
tribute significantly are those whose phase velocity in
As before, the only wav... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
of a large solitary elevation, a rounded, smooth and well-defined heap of water, which
continued its course along the channel apparently without change of form or diminution of
speed. I followed it on horseback, and overtook it still rolling on at a rate of some eight or
nine miles an hour [14km/h], preserving its or... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
eweg and de Vries
(KdV) in 1895. According to their theory, the spatio-temporal evolution of weakly nonlinear
shallow water waves is described by
∂th =
(cid:114) g
l
3
2
(cid:18) h2
2
+
2
3
∂x
h +
(cid:19)
σ
∂2
3 xh
where h(t, x) is the surface profile of the wave and
σ =
(cid:96)3
3
−
γ(cid:96)
ρg
(573)
(574)
with (cid... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
ky and Kruskal showed how one can derive this equation in the continuum
limit from the FPU model. They argued that two KdV solitons could collide and penetrate
each other without exchanging energy. This explains intuitively why the FPU chain model
does not lead to thermalization. Miura et al.28 provided a general rigor... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
6.801/6.866: Machine Vision, Lecture 10
Professor Berthold Horn, Ryan Sander, Tadayuki Yoshitake
MIT Department of Electrical Engineering and Computer Science
Fall 2020
These lecture summaries are designed to be a review of the lecture. Though I do my best to include all main topics from the
lecture, the lectures ... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/617445f0e31836831b40d42cb2f11a10_MIT6_801F20_lec10.pdf |
could then determine object surface orientation along. We can integrate along this direction (the profile direction) to find
height z, but recall that we gain no information from the other direction.
For the Hapke example, we have a rotated coordinate system governed by the following ODEs (note that xi will be the vari... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/617445f0e31836831b40d42cb2f11a10_MIT6_801F20_lec10.pdf |
given step size. For small steps, this approach will be
accurate enough, and accuracy here is not too important anyway since we will have noise in our brightness measurements.
Recall brightness of Hapke surfaces is given by:
r
E =
p
f rac1 + psp + qsq 1 + p2 + q2rs · √
nˆ · ˆs
nˆ · vˆ
r
=
=
cos θi
cos θe
1
... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/617445f0e31836831b40d42cb2f11a10_MIT6_801F20_lec10.pdf |
, q)
Let us now consider taking a small step δx, δy in the image, and our goal is to determine how z changes from this small step.
We can do so using differentials and surface orientation:
δz =
∂z
∂x
δx +
∂z
∂y
δy = pδx + qδy
Therefore, if we know p and q, we can compute δz for a given δx and δy. But in additi... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/617445f0e31836831b40d42cb2f11a10_MIT6_801F20_lec10.pdf |
2
⎤
⎥ δx
⎦
δy
⎡
px py
⎤
=
⎣
⎦
qy
qx
δx
δy
= H
δx
δy
Where H is the Hessian of second spatial derivatives (x and y) of z. Note that from Fubuni’s theorem and from our Taylor
expansion from last lecture, py = qx.
Intuition:
2
• First spatial derivatives
∂z
∂x
and ∂z describe the surface o... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/617445f0e31836831b40d42cb2f11a10_MIT6_801F20_lec10.pdf |
=
r +
Ey =
=
=
s +
+
∂R ∂q
∂q ∂x
+
∂R ∂q
∂q ∂y
s
∂R ∂p
∂p ∂x
∂R
∂q
∂R ∂p
∂p ∂y
∂R
t
∂q
In matrix-vector form, we have:
Ex
Ey
r
s
=
⎡ ⎤
⎡ ⎤
∂R
∂R
∂p
∂p
s ⎢ ⎥
⎢ ⎥
⎣ ⎦ = H ⎣ ⎦
t
∂R
∂q
∂R
∂q
Notice that we have the same Hessian matrix that we had derived for our surface orient... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/617445f0e31836831b40d42cb2f11a10_MIT6_801F20_lec10.pdf |
ighthand side is our gradient of the reflectance
map in gradient space (p, q), and ξ is the step size. Intuitively, this is the direction where we can “make progress”. Substituting
this equality into our update equation for p and q, we have:
δp
δq
= H
ξ
Rp
Rq
Ex
Ey
δξ
=
Therefore, we can formal... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/617445f0e31836831b40d42cb2f11a10_MIT6_801F20_lec10.pdf |
step in the image (x, y) depends on the gradient of the reflectance map in (p, q).
• Analogously, the step in the reflectance map (p, q) depends on the gradient of the image in (x, y).
• This system of equations necessitates that we cannot simply optimize with block gradient ascent or descent, but rather a
process in ... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/617445f0e31836831b40d42cb2f11a10_MIT6_801F20_lec10.pdf |
factor when we update these variables after taking a step. With this, our updates
become:
1. δx ← ps
2. δy ← qs
3. δz ← (psp + qsq) = rsE2 − 1
Which are consistent with our prior results using our Hapke surfaces. Next, we will apply this generalized approach to Scanning
Electron Microscopes (SEMs):
4
1.2.2 App... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/617445f0e31836831b40d42cb2f11a10_MIT6_801F20_lec10.pdf |
Our solution generates characteristic strips that contain
information about the surface orientation. To continue our analysis of this SfS problem, in addition to defining characteristic
strips, it will also be necessary to define the base characteristic.
1.3 Base Characteristics
Recall that the characteristic strip t... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/617445f0e31836831b40d42cb2f11a10_MIT6_801F20_lec10.pdf |
sets
of base characteristics can be thought of like a wavefront propagating outward from the initial curve. We expect the
solutions corresponding to these base characteristics to move at similar “speeds”.
1.4 Analyzing “Speed”
When considering how the speeds of these different solution curves vary, let us consider w... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/617445f0e31836831b40d42cb2f11a10_MIT6_801F20_lec10.pdf |
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