text stringlengths 30 4k | source stringlengths 60 201 |
|---|---|
for representing the solution. In both cases, the central idea is that since the equation is
linear, it is possible to “break down” any initial state into a linear combination of simpler
problems. By solving the simpler problems explicitly it is then possible to reconstruct the
general solution.
6.1 Fourier method
This... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
∝ 1/λ, with λ the wavelength, is in this superposition.
25
(cid:90) ∞
1
2π −∞
(cid:90) ∞
The solution of the original problem is therefore
n(x, t) =
(cid:90) ∞
1
2π −∞
nˆ0(k)eikx−Dk2tdk.
(109)
We note that the high-wavenumber components (which correspond to sharp gradients) are
rapidly damped, emphasising the smoothin... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
90) ∞
−∞
e− (x+ikσ2)2
22σ
dx.
(112)
(113)
To calculate the above integral, which involves a complex integrand, we use the Cauchy
integral formula. It states that for a complex function f (z) ∈ C, z ∈ C, integration along a
closed path in the complex plane is zero, provided that f (z) has no poles inside the path:
(cid:... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
2σ dz .
Inserting this into Eq. (115), we obtain
nˆ0(k) =
2 2
2
e− k σ
√
2πσ2
(cid:90) ∞
−∞
e− x2
22σ dx.
(117)
(118)
This means that we can basically drop the imaginary part in the original integral, Eq. (113).
From 18.01 we know that
(cid:90) ∞
e−x2 √
=
√
π,
−∞
so by making a change of variable y = x/ 2σ2 we have tha... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
e
2π
−∞
(cid:104)
−(Dt+σ2/2)
e
k−
ix
(cid:105)2
2
2(Dt+σ /2) dk.
(123)
This is essentially the same integral that we had before, so we drop the imaginary part and
change the integration variable, giving the result
n(x, t) =
−
2
x
4(Dt+σ2/2)
e
4π(Dt + σ2/2)
(cid:112)
.
(124)
This is the solution of the diffusion equation... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
area that
is centered exactly at the position x0. The definition of δ is that given any function4 f (x),
(cid:90) ∞
−∞
f (x(cid:48)) δ(x − x(cid:48))dx(cid:48) = f (x).
(126)
We can represent the initial distribution of particles n(x, 0) = n0(x) as a superposition
of δ-functions
n0(x) =
(cid:90) ∞
−∞
n
(cid:48)
0(x ) δ(... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
t) so
that G(x − x(cid:48), 0) = δ(x − x(cid:48)), and
n(x, t) =
(cid:90) ∞
−∞
G(x − x(cid:48), t) n0(x(cid:48)) dx(cid:48).
Plugging this into the diffusion equation we see that
n0 x(cid:48) ∂G(x − x(cid:48), t
)
dx(cid:48) = D
(cid:90) ∞
(cid:90) ∞
(
)
−∞
∂t
n0(x(cid:48))
∂2G(x − x(cid:48), t)
∂x2
dx(cid:48).
−∞
(128)... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
as when you take two spatial derivatives. This means
that the characteristic length scale over which n varies is of order
t. Now, since the
initial distribution δ is perfectly localised, we expect that at time t, G(x − x(cid:48)) will have a
characteristic width
t. Thus, we guess a (so-called) similarity solution
√
√
G... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
2
Cancelling out the time factors and integrating this equation once gives
− F y = DF (cid:48).
1
2
This equation can be immediately integrated to give F (y) = F
0e−y
2
/
4D, and thus
G(x − x(cid:48)
F0
, t) = √
t
e− (x−x(cid:48))2
4Dt
,
where the constant F0 = 1/
√
4πD is determined by requiring that
(cid:82)
dx G = 1... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
the mean square displacement of Brow-
nian particles grows linearly in time.
6.3 Zero-flux solution: Sedimentation
Consider spherical particles diffusing under the effect of a constant drift velocity u in one
dimension, described by the conservation law
with current
∂n
∂t
= −
∂
∂x
Jx
Jx = un − D n.
∂
∂x
30
(140)
(141)
In... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
∗
7 Linear stability analysis and pattern formation
7.1 Linear stability analysis of fixed points for ODEs
Consider a particle (e.g., bacterium) moving in one-dimension with velocity v(t), governed
by the nonlinear ODE
d
dt
v(t) = −(α + βv2)v =: f (v).
(148)
We assume that the parameter β is strictly positive, but... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
) δv = f (cid:48)(v∗) δv,
we find that the growth of the perturbation δv(t) is governed by the linear ODE
which has the solution
d
dt
δv(t) = f (cid:48)(v∗) δv(t),
δv(t) = δv(0) ef (v∗)t.
(cid:48)
(149)
(150)
(151)
(152)
If f (cid:48)(v∗) > 0, then the perturbation will grow and the fixed point is said to be linearly
uns... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
,
∂n
∂x
(0, t) =
∂n
∂t
(L, t) = 0.
This dynamics defined by Eqs. (154) conserves the total ‘mass’
N (t) =
(cid:90) L
0
dx n(x, t) (cid:17) N0,
and a spatially homogeneous stationary solution is given by
n0 = N0/L.
32
(154a)
(154b)
(155)
(156)
To evaluate its stability, we can consider wave-like perturbations
n(x, t) = ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
to ensure stability. The appearance of higher-order spatial derivatives means
that this model accounts for longer-range effects than the diffusion equation. This becomes
immediately clear when one writes a (159) in a discretized form as necessary, for example,
when trying to solve this equation numerically on a space-tim... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
0 to a < 0 or γ0 > 0 to γ0 < 0, which both lead to
non-zero flow patterns, must be connected to the microscopic self-swimming speed v0 of the
33
bacteria. Assuming a linear relation, this suggests that, to leading order, a0 = δ (cid:0)αv0 where
δ > 0 is a passive damping contribution and αv0 > 0 the active part, and si... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
) (cid:0)(2a + bψ ) + γ jkj2
|
±
±
0
|
(cid:2)
−
4
+ γ2jkj
|
|
(cid:3)
.
(162)
(163)
(164)
(165)
In both cases, k-modes with σ > 0 are unstable. From Eqs. (163) and (165), we see
immediately that γ2 > 0 is required to ensure small-wavelength stability of the theory and,
furthermore, that non-trivial dynamics can be exp... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
8) −(cid:28) (2π)2γ
2/L , increasingly more
complex quasi-stationary structures arise; qualitatively similar patterns have been observed
in excited granular media and chemical reaction systems.
2
(159) must be invariant under ψ ! −→ ψ, implying that U (ψ) = U (−ψ) and, therefore, b = 0
in (160). Intuitively, the transf... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
(Fig. 2a) – such self-assembled hexagonal vortex lattices have
indeed been observed experimentally for highly concentrated spermatozoa of sea urchins
(Strongylocentrotus droebachiensis) near a glass surface (Fig. 2b).
35
Figure 2: Effect of symmetry-breaking in the Swift-Hohenberg model. (a) Stationary hexag-
onal latt... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
, let us consider q(t, x) = (u(t, x), v(t, x)), D = diag(Du, Dv) and
R = (F (u, v), G(u, v)), then
ut = Duuxx + F (u, v)
vt = Dvvxx + G(u, v)
(167a)
(167b)
In general, (F, G) can be derived from the reaction/reproduction kinetics, and conservation
laws may impose restrictions on permissible functions (F, G). The fixed p... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
∂uG(u∗, v∗) ,
G∗
v = ∂vG(u∗, v∗).
Solving this eigenvalue equation for σ, we obtain
σ± =
1
2
(cid:26)
(cid:0)(Du + Dv)k2 + (F ∗
u + G∗
v) (cid:6)
(cid:113)
4F ∗
v G∗
u + [F ∗
u (cid:0) G∗
2
v + (Dv (cid:0) Du)k2]
(cid:27)
. (171)
In order to have an instability for some finite value k, at least one of the two eigenvalue... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
A
−
−C + AE
B
.
(173)
(174a)
(174b)
It is straightforward to verify that, for suitable choices of A, B, C, D, the model exhibits a
range of unstable k-modes.
8 Variational Calculus
In this part of the course, we consider the energetics governing the shape of water droplets,
soap films, bending beams etc. For syst... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
t. To develop techniques for continuous systems we consider an
energy functional U [f (x)], which is a function of a function. Several sorts of difficulties arise
(mathematical and otherwise) when one tries to think about physical problems described
by this energy using the same notions as those for finite dimensional ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
[h + δh] − L[h] (cid:39) √
(cid:20)
h(cid:48)
1 + h(cid:48)2
(cid:21)x2
δh
−
(cid:90) x2
(cid:18)
dx
x1
x1
√
h(cid:48)
1 + h(cid:48)2
(cid:19)(cid:48)
δh.
(178)
(179)
The first term on the right hand side is identically zero, as we require δh to be zero at
either end (since we know our starting and ending point). Otherw... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
A functional I is called linear, if it satisfies
I[af + bg] = aI[f ] + bI[g]
39
(181b)
(182)
for arbitrary numbers a, b and functions f, g. Obviously, both examples in Eq. (181) are
linear. Typical examples of nonlinear functionals are action functionals, such as
S[x, x˙ ] =
(cid:90) t
0
ds
(cid:104) m
2
(cid:105)
x˙ (... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
δ
δf
(F [g(f )]) =
δ
δf
g(F [f ]) =
δ(F [g(f )])
δg
dg(F [f ])
dF
dg(x)
df (x)
δ(F [f ])
δf
(187b)
(187c)
(187d)
As a nice little exercise, you can use the above properties to prove that the exponential
functional
satisfies the functional differential equation
(cid:82)
F [f ] = e dx f (x)c(x)
δF [f ]
δf (y)
= c(y) F [f ]... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
(cid:48) δ(cid:48)(x − y)
dx
−
d
dx
∂f (cid:21)
∂Y (cid:48)
δ(x − y)dx.
Equating this to zero, yields the Euler-Lagrange equations
0 =
∂f
∂Y
−
d
dx
∂f
∂Y (cid:48)
(191)
(192)
In fact, the relation might even indicate a maximum.
It should be noted that the condition δI/δY = 0 alone is not a sufficient condition for
It is ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
at position x. Thus
(cid:112)v = 2g(h0 − h(x)).
(194)
41
By definition v = ds/dt so that the time taken to go from A to B is
(cid:90) B
A
dt =
(cid:90) B
A
ds
2g[h0 − h(x)]
.
(cid:112)
We know that ds = 1 + h(cid:48)2, so that the time taken is
√
The integrand
T [h] =
(cid:115)
dx
(cid:90) B
A
1 + h(cid:48)2
2g(h0 − h)... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
8)
(199)
(200)
(201)
where C = (2a)− 1
2 . To evaluate this integral we substitute h0 − h = 2a sin2 θ and obtain
2
(cid:90)
x = 2a
sin2 dθ + x0 = a(θ − sin θ) + x0.
θ
2
(202)
We have thus found a parametric representation for the desired curve of most rapid descent:
x = x0 + a(θ − sin θ),
h = h0 − a(1 − cos θ).
(203)
T... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
90)
(cid:90)
E[y] = γ
ds = γ
(cid:112)
1 + y(cid:48)2.
dx
(205)
∂B
∂B
We have already solved the problem to determine the curve that connects two points while
satisfying δE/δy = 0, the solution being a straight line. Now we need to introduce the
constraint of a fixed volume, and this is done by adding a Lagrange multipl... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
is simply
this to a one dimensional problem. The problem to be solved is now
1 + r(cid:48)2dx, and we see that through symmetry we have reduced
√
E[r] = 2πγ
(cid:90) a
−a
(cid:112)
dx r
!
1 + r(cid:48)2 = min.
Putting the integrand into the Euler-Lagrange equation gives
(cid:112)
1 + r(cid:48)2 −
d
dx
(cid:18) rr(cid:4... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
(x) = r0 for the perfect cylinder. The cylinder has surface energy and we look for an
extremum by considering the functional
E[r] = γ
(cid:90)
dx 2πr
(cid:112)
1 + r(cid:48)2 − λ
(cid:90)
dx πr2,
(214)
where the second part is just the volume constraint.
We shall perturb the shape and show that if the wavelength of dis... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
, yielding
(cid:90) (cid:20)
E[r] = γ
2π(r0 + (cid:15) cos kx) 1 +
(cid:18)
(cid:15)2k2 sin2 kx
1
2
(cid:19)
−
π
r0
(r0 + (cid:15) cos kx)2 dx.
(cid:21)
(217)
(218)
Expanding everything out, we find that terms linear in (cid:15) cancel out, and we are left with
(cid:90)
E[r] = γ
πr0dx + (cid:15) γπ
2
(cid:90) (cid:20)
2... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
and the system is unstable. Since k = 2π/λ this can be rearranged
to say that the system is unstable to wavelengths λ greater 2πr0, the circumference of
the cylinder. This was Plateau’s results, which Rayleigh found to be wrong by a factor
of
2, due to hydrodynamic effects within the cylinder (i.e., λ = 2 2πr0 is the co... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
riefly recall a few basics of differential
geometry.
10.1 Differential geometry of curves
Consider a continuous curve r(t) ∈ R3, where t ∈ [0, T ]. The length of the curve is given by
L =
� T
0
dt r˙ (t)
||
||
(222)
where r˙ (t) = dr/dt and || · || denotes the Euclidean norm. The local unit tangent vector is
defi... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
), b(0)}, the Frenet frames along
the curve can be obtained by solving the Frenet-Serret system
1
||r˙ ||
t˙
0
n˙ = −κ
κ
0
0 −τ
˙b
0
t
τ n .
b
0
(228a)
The above formulas simplify if t is the arc length, for in this case ||r˙ || = 1.
As a simple example (which is equivalent to our shortest path prob... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
∈ U ⊆ R2. Alternatively, if one chooses Cartesian coordinates (s1, s2) =
(x, y), then it suffices to specify
or, equivalently, the implicit representation
z = f (x, y)
Φ(x, y, z) = z − f (x, y).
(233a)
(233b)
The vector representation (232) can be related to the ‘height’ representation (233a) by
9γ carries units of energ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
36)
(237a)
(237b)
where fx = ∂xf and fy = ∂yf . For later use, we still note that the inverse of the metric
tensor is given by
g−1 = (g−1
ij ) =
(cid:18)
1
x + f 2 −y
1 + f 2
1 + f 2
y −fxfy
1 + f 2
fyf
x
x
(cid:19)
.
(237c)
Assuming the surface is regular at (s1, s2), which just means that the tangent vectors F 1
and ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
= F yx = 0 ,
F yy = 0
(242a)
fxx
fxy
fyy
yielding the curvature tensor
(Rij) =
(cid:18)
N · F xx N · F xy
N · F yx N · F yy
(cid:19)
=
(cid:113)
1
1 + f 2
x + f 2
y
(cid:19)
(cid:18)
fxx fxy
fyy
fyx
(242b)
Denoting the eigenvalues of the matrix g−1 · R by κ1 and κ2, we obtain for the mean
curvature
H =
1
2
(κ1 ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
hence χ(S2) = 2,
whereas a two-dimensional torus M = T2 has g = 1 handle and therefore χ(T2) = 0.
Equation (245) implies that, for any closed surface, the integral over K is always a
constant. That is, for closed membranes, the first integral in Eq. (245) represents just a
trivial (constant) energetic contribution.
10.3... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
fyy
= −2H.
(250)
Thus, minimal surfaces satisfy
H = 0
⇔
κ1 = −κ2,
(251)
implying that each point of a minimal surface is a saddle point.
10.4 Helfrich’s model
Assuming that lipid bilayer membranes can be viewed as two-dimensional surfaces, Hel-
frich proposed in 1973 the following geometric curvature energy per unit ar... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
= 1,
(255)
(256)
(257)
corresponding to the first term on the rhs. of Eq. (254).
For open membranes with boundary ∂M , there is no volume constraint and a plausible
energy functional reads
Eo =
dA E + σ
dA + γ
ds,
∂M
(258)
where γ is the line tension of the boundary. In this case, variation yields not only the
c... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
denoted by
u = u(x1, x2, x3), where
2, x;
;
ui = xi
− xi.
(259)
The vector u is called the displacement vector.
When a body is deformed the distance between its points change. Let’s consider two
points very close together. If the vector joining them before is dxi, the vector joining them in
11Here, we made use... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
is not really linear). It is often very useful to separate pure shear from pure compression
effects, which can be achieved by rewriting12
(cid:18)
eij =
eij −
δij
3
(cid:19)
ell
+
δij
3
(cid:18)
ell =
eij −
(cid:19)
∇ · u
+
δij
3
δij ∇ · u.
3
(263)
The first part in parentheses has a vanishing trace and therefore represe... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
7). For instance, if there is an external
force per unit area, fˆ, acting over the surface, then we require
ˆ
σiknk = fi,
(268)
where n is the outward unit normal on the surface.
11.3 Hooke’s law
In general, we would like to use Eqs. (267) to predict the deformation of a solid body under
a given force distribution. Tha... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
)
1
2
1
3
where K and µ positive constants, respectively called the modulus of compression and the
modulus of rigidity. In 3D, K is related to the Lame coefficients by13
K = λ +
2
3
µ
(274)
13In 2D, this relation becomes K = λ + µ.
53
11.4 A simple problem
Consider the simple case of a beam. Let the beam be along the z-... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
K and µ are always positive, Poisson’s ratio can vary between -1 and 1 . Note that a
negative value corresponds to pulling on the beam and it getting thicker! Now we see why
we use Y and ν; they are easier to measure. Inverting these formulae, we get
2
µ =
Y
2(1 + ν)
,
K =
Y
3(1 − 2ν)
.
The free energy then becomes
E =... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
the
neutral surface is given by uz = w(x, y). For further calculations we note that since the
plate is thin, comparatively small forces on the surface are needed to bend it. These forces
are always considerably less than the internal stresses caused in the deformed beam by the
extension and compression of its parts. Th... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
,
ezz =
zν
1 − ν
∂2w
∂x2
+
∂2w
y2
∂
.
We now calculate the free energy of the plate, using our general formula,
2
E = z
Y
+ ν
1
1
2(1
− ν)
∂2w
∂x2
+
2
∂ w
∂y2
2
+
∂2w
∂x∂y
2
−
∂2w ∂2w
∂x2 ∂y2
.
(291)
(292)
2
Integrating from − h to h , where h is the thickness of
2
an area element gives the free energy pe... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
0,
dw
dx
= 0
at the edges. The first of these expresses the fact that the edge of the plate undergoes no
deformation, and the second that it remains horizontal. For more details, see chapter 2 in
Theory of Elasticity, Landau & Lifschitz.
12 Towards hydrodynamic equations
The previous classes focussed on the cont... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
’s laws) it is necessary to find a way of
passing in detail from the microscopic (quantum) mechanical description, to the macroscopic
description. The ideas behind this are highly related to what we have already done for the
random walker, albeit with another level of complexity.
12.1 Euler equations
Instead of obtainin... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
of
position and time. There is ample observational evidence that common real fluids move as if
they were continuous, under normal conditions and indeed for considerable departures from
normal conditions. However, some of the properties of the equivalent continuous media need
to be determined empirically, and cannot be d... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
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 pressure force is a normal
force per unit area (usually compressive) exerted across the sur... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
if we realise that the gravitational force, being conservative,
can be written as the gradient of a 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 t... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
∂
∂t
f =
=
N
(cid:88)
i=1
N
(cid:88)
i
d
dt
[δ(x − xi)δ(v − vi)]
{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) = −
∂... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
) · G(x) +
j(cid:54)=i
(cid:88)
xj (cid:54)=x
H(x − xj)
= − G(x) +
(cid:88)
xj (cid:54)=x
H(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 derivat... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
density f with
respect to P(Γ0), formally expressed as
(cid:104)f (t, x, v)(cid:105) = dP(Γ0) f (t, x, v).
(cid:90)
(315)
60
Averaging Eq. (314) and using the fact that integration over initial conditions commutes
with the partial differentiations, we have
m
(cid:18) ∂
∂t
(cid:19)
+ v · ∇ (cid:104)f (cid:105) = −∇v · [... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
= m d3v (cid:104)f (t, x, v)(cid:105) vivj,
and the trace of Σ defines the local kinetic energy density
(cid:15)(t, x) :=
1
2
Tr(ρΣ) =
m (cid:90)
2
d3v (cid:104)f (t, x, v)(cid:105) |v|2.
Integrating Eq. (316) over v, we get
ρ + ∇ · (ρu) = − dv3 ∇v · [G(x)(cid:104)f (cid:105) + C] ,
(cid:90)
∂
∂t
(319)
(320)
but the rhs... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
∂
∂t
∂
∂t
∂
∂t
∂
∂t
(cid:18) ∂
∂t
(ρu) + ∇ · (ρΣ)
=
=
(ρu) + ∇ · (ρuu) + ∇ · [ρ(Σ − uu)]
= ρ
u + u
(321)
= ρ
∂
∂t
ρ + u∇ · (ρu) + ρu · ∇u + ∇ · [ρ(Σ − uu)]
(cid:19)
+ u · ∇ u + ∇ · [ρ(Σ − uu)]
(323)
The rhs. of (322) can be computed by partial integration, yielding
(cid:90)
− dv3 v∇
v · [G(x)(cid:104)f (cid:105) + C] =... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
let us assume that the pair interaction force
H can be derived from a pair potential ϕ, which means that H(r) = −∇rϕ(r). Assuming
further that H(0) = 0, we may write
c(t, x) = − dv3
(cid:90) (cid:88)
xj (t)
(cid:104)[∇xϕ(x − xj)]f (t, x, v)(cid:105)
(328)
Replacing for some function ζ(x) the sum over all particles by t... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
the case when interactions are very short-range so that we can approximate the
potential by a delta-function,
ϕ(r) = φ0a3δ(r),
(331)
where ϕ0 is the interaction energy and a3 the effective particle volume. In this case,
(cid:90)
(cid:90)
c(t, x) = −
dv3
d3y [∇ρ(t, y)] (cid:104)δ(x − y)f (t, x, v)(cid:105)
= −
= −
= −
(c... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
m
I,
(334)
where T is the temperature and k the Boltzmann constant. For this closure condition,
Eqs. (333a) and (333b) become to a closed system for ρ and u.
Traditionally, and in most practical applications, one does not bother with microscopic
derivations of Ξ; instead one merely postulates that
Ξ = −pI + µ( (cid:62)... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
purely macroscopic considerations and and we also showed how
one can derive macroscopic continuum equations from an underlying microscopic model.
For the remainder of this course, we will return to the macroscopic viewpoint developed in
Sec. 12.
13.1 Viscosity
A main insight from the discussion in the previous sec... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
ndS =
( · σ)dV.
S
V
(338)
We must therefore determine the form of σ in order derive our equations of motion.
64
Firstly, suppose there were no tangential stresses on a fluid element. The normal stresses
are just pressures. In this case the stress tensor would just be
σ =
⎛
−p
⎝ 0
0
0
−p
0
0
... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
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
⎝ ∂xuy ∂yuy ∂zuy
∂xuz ∂yuz ∂zuz
⎞
⎠ .
(342)
e immediately have a problem bec... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
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
deforming flow th... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
uid is non-Newtonian or not depends on 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... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
of viscosity, and since our derivation of the
viscous force is phenomenological, it is 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
fect... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
basis of this mathematicians can
go looking for solutions. To become familiar with the equations we shall find some simple
solutions to well known 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 p... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
velocity at the boundary. If there
is viscosity then it is possible to demand another condition. The condition which is mostly
accepted to be true is called the no slip boundary condition (there is important ongoing
research which aims at determining if and when such a condition breaks down). The no-slip
67
condi... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
the nonlinear term has vanished because of the form of the velocity field.
The equation is a diffusion equation for the velocity in which ν is the diffusion coefficient.
The initial condition is that u = 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... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
and boundary conditions was essential to our finding this
solution. We see that at different times the velocity profiles are all geometrically similar
2
e −s /4ds
u(y, t) = U
(352)
.
√
(cid:90)
y/
νt
68
i.e., the velocity is always the same function of (y/
profile becomes stretched out and the effects ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
ask, how long does it take for this profile to be realised.
Well, we know that viscous diffusion is responsible for setting up the profile, and the di
mensions of ν are L2/T . Since the separation of the plates is h, then we obtain a timescale
by forming the combination h2/ν. This is roughly the time taken for viscous ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
in the equation are large, and which are small. In
different limits the Navier-Stokes equations contain all of the important classes of partial
differential equations (i.e., diffusion equation, Laplace’s equation, wave equations) which
are usually considered. In the next lecture we shall find an example which has within... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
�ow equations for very viscous flow. At high Reynolds
number one ends up with the Euler equations.
The Reynolds number can be varied by changing the viscosity of fluid. In practice, one
distinguishes two types of viscosities.
Dynamic viscosity The SI physical unit of dynamic viscosity µ is the Pascal×second
[µ] = 1 ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
U0 10 µm/s
⇒
Re 10−5.
This is a huge difference and allows for considerable mathematical simplifications.
14 Low-Reynolds number limit
In this section, we look at the limit of Re → 0 which is relevant to the construction of
microfluidic devices and also governs the world of swimming microbes.
70
Bacteria an... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
.
(359a)
(359b)
The four equations (359) determine the four unknown functions (u, p). However, to uniquely
identify such solutions, these equations must still be endowed with appropriate initial and
boundary conditions, such as for example
(cid:40)
u(t, x) = 0,
p(t, x) = p ,∞
as
|x| → ∞.
(360)
Note that, by neglecting ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
which has the inverse
(cid:18)
G (x) = 8πµ x δ
|
|
−1
jk
x
− jxk
|x|2
2
jk
(cid:19)
,
(362a)
(362b)
(363)
as can be seen from
GijG−1
jk
(cid:19) (cid:18)
(cid:18)
xixj
ij +
=
δ
|x|2
= δik − ixk
x
2|x|2 +
xixk
2|x|2 +
= δik −
δjk −
−
xixk
|x|2
xixk
2|x|2
xjxk
2|x|2
xixj
|x|2
(cid:19)
xjxk
2|x|2
= δik.
(364)
14.3 Stokes’... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
= aiei
where θ ∈ [0, π], φ ∈ [0, 2π), one finds that on this boundary
u(t, a(θ, φ)) = U ,
µ
a2 Uj aj(θ, φ) + p∞,
p(t, a(θ, φ)) =
3
2
(366a)
(366b)
17Proof by insertion.
18Proof by insertion.
72
corresponding to a no-slip boundary condition on the sphere’s surface. The O(a/|x|)-
contribution in (365a) coincides with the... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
j
2π|x|2
+ p
∞
,
x = (x, y)
(371a)
where
Jij(x) =
1
4πµ
(cid:20)
−δij ln
(cid:18) |x|
a
(cid:19)
+
xixj
x|2
|
(cid:21)
(371b)
with a being an arbitrary constant fixed by some intermediate flow normalization condi-
tion. Note that (371) decays much more slowly than (370), implying that hydrodynamic
interactions in 2D free... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
ij
1
|x|
x
ij |x|2 +
δ
k
(cid:20)
−
(cid:21)
xixj
|x|2
(cid:18) xixj
|x|2
(cid:19)(cid:21)
(cid:18)
δik
xj
|x|2 + δjk
xi
|x|2 − 2
xixjxk
x|4
|
(cid:19)(cid:21)
.
(374)
To check the incompressibility condition, note that
∂iJij =
=
1
4πµ
1
4πµ
= 0,
(cid:20)
−δij
xi
|x|2 +
x
− j
|x|2 + 2
(cid:18)
(cid:18)
δii
xj
|x|2 + δj... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
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 field generated by an indiv... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
n with
+
x+
ui(x) = − k
2πµ
F (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
(3... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
xˆkxˆixˆj
|x|2
ikxˆj + Πjkxˆi)
|
=
(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 Eq... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
represented in the form
u(x) = ux(x, y, z) ex + uy(x, y, z) ey + uz(x, y, z) ez.
(386a)
The gradient vector is given by
∇ = ex∂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
∇... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
φ, z) er + uφ(r, φ, z) eφ + uz(r, φ, z) ez.
(387d)
The gradient vector takes the form
∇ = er∂r + eφ ∂φ + ez∂z,
1
r
yielding the divergence
∇ · u =
1
r
∂r(rur) + ∂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,... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
u
1
2
r − u
r φ.
(388)
Physically, the term u2
urer + uφeφ + uzez and some of the unit vectors change with φ (e.g., ∂φeφ = −er).
φ/r corresponds to the centrifugal force, and it arises because u =
14.6.2 Hagen-Poiseuille flow
To illustrate the effects of no-slip boundaries on the fluid motion, let us consider pressure
dri... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
maximal at center of the pipe
u+
z =
P (0) − P (L)
4µL
R2
78
(392)
(393)
and the average transport velocity is
uz =
1 R
πR2
0
uz(r) 2πrdr = 0.5u+
z .
(394)
Note that, for fixed pressure difference and channel length, the transport velocity uz de
creases quadratically with the channel radius, signaling that the pres... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
(395)
corresponding to a parabolic flow profile in the vertical direction that accounts for no-slip
boundaries at the walls; in particular, in the mid-plane
u(x, y, H/2) = U (x, y).
3
2
(396)
We would like to obtain an effective equation for the effective 2D flow U (x, y). This can
be achieved by inserting ansatz (395... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
1: how long does it take a cup of coffee (or glass of water) to spin down if you start
by stirring it vigorously? To proceed, we need a model of a coffee cup. For mathematical
simplicity, let’s just take it to be an infinite cylinder occupying r ≤ R. Suppose that at
t = 0 the fluid and cylinder are spinning at an angula... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
equation to be satisfied by the eφ-component is
and the vertical equation is
∂uφ
∂t
= ν
∂2uφ
∂r2
+
1 ∂uφ
r ∂r
−
uφ
r2
,
0 =
1 ∂p
.
ρ ∂z
(399b)
(399c)
The last equation of these three is directly satisfied by our solution ansatz, and the first
equation can be used to compute p by simple integration over r ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
).
This equation looks complicated. However, note that if the factors of r weren’t in this
equation we would declare victory. The equation would just be F "" + k2/νF = 0, which
has solutions that are sines and cosines. The general solution would be Asin(k/ νr) +
Bcos(k/ νr). We would then proceed by requiring that (... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
For more information, see for example the book Elementary Applied
Partial Differential Equations, by Haberman (pp. 218-224). Now let’s satisfy the boundary
(404)
22Bessel functions Jα(x) of order α are solutions of
(
2 ff
x J + xJ + x − α J = 0
)
2
f
81
condition uφ(R, t) = 0. Since we have that
uφ ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
J1(λmr/R) dr =
R
0
ωr2J1(λmr/R)dr.
(408)
Using the identities
and
we get
R
0
rJ1(λnr/R)J1(λmr/R) dr =
R2
2
J2(λn)2δnm,
R
0
ωr2J1(λmr/R)dr =
ωR3
λm
J2(λm),
An = −
2ωR
λnJ0(λn)
,
where we have used the identity J0(λn) = −J2(λn). Our final solution is therefore
(409)
(410)
(411)
uφ(r, t) = −
∞
2ωR... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
by a factor of 100 (roughly the difference
between water and motor oil) then it will take roughly a factor of 100 shorter to spin down.
Note that for these predictions to be accurate, one must start with the same angular velocity
for each case.
In your problem set you are asked to look at the spin down of a coffee cu... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
to take the interaction
constants Jij to be a constant.
If one assumes that the local spins vary on a length scale much longer than a lattice
spacing, then it is possible to derive a macroscopic analogue of the above energy. A complete
derivation of this includes the effect of random thermal fluctuations and is beyon... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
ization ±
with any
orientation. If the system has boundaries then the magnetization must match the boundary
conditions, but is otherwise free to be orientated however it wants.
t
b
2c
What happens, however, if ν
0?=
If we multiply both sides of (3) by M ', then the
equilibrium condition can be integrated to gi... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
is simply x = −c/b. Now we make a small change, and consider the equation
Using the quadratic formula,
Ex2 + bx + c = 0,
√
−b ± b2 − 4Ec
.
2E
x =
84
(421)
(422)
In the limit E → 0
x ≈ − , −
c
b
2b − 2Ec
.
2E
(423)
and the latter solution can be further approximated as −b/E if E is very small. I... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
close to x = 0 they differ greatly. We call this the boundary layer. It arises
because the small parameter E multiplies the highest derivative in the equation, and by
ignoring this term we lower the order of the system and are unable to satisfy both boundary
conditions.
We need to find an approximate ‘inner’ solution... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
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, there exists a region in which it has a significant
affect on the solution. This idea was d... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
4)
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 equations for an ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
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