# Math 5120 Homework 2.1 Solutions ```Math 5120
Homework 2.1 Solutions
1. If D = f (c), and we also suppose that J = −D∇c (i.e. Fick’s law holds) and σ = 0 (i.e. no
sources/sinks), then the 1d diffusion equation is
∂
∂c
∂c
=−
−D
∂t
∂x
∂x
∂c
∂
−f (c)
=−
∂x
∂x
∂c ∂c
∂2c
= f ′ (c) &middot;
&middot;
+ f (c) 2
∂x ∂x
∂x
2
2
∂c
∂ c
.
= D 2 + f ′ (c)
∂x
∂x
2. To derive the two-dimensional conservation equation, first let the flux be represented by J(x, y) =
(Jx , Jy ). The number of particles in time ∆t that cross through a given area is defined by N =
(J &middot; m)A∆t, where m is normal to the given area. Therefore, we can think of the concentration as
the number of particles per volume. If we consider a small rectangular region, where particles can
only move in the x- and y-directions, we can define the volume of the region to be ∆V = ∆x&middot;∆y&middot;h,
where h is the height of the region. Then the change in the concentration, ∆c, in the small
rectangular region beginning at (x0 , y0 ) is given by
∆N
[J(x0 , y0 ) &middot; (1, 0)]∆y &middot; h &middot; ∆t [J(x0 + ∆x, y0 ) &middot; (1, 0)]∆y &middot; h &middot; ∆t
=
−
∆V
∆x &middot; ∆y &middot; h
∆x &middot; ∆y &middot; h
[J(x0 , y0 ) &middot; (0, 1)]∆x &middot; h &middot; ∆t [J(x0 , y0 + ∆y) &middot; (0, 1)]∆x &middot; h &middot; ∆t
−
+
∆x &middot; ∆y &middot; h
∆x &middot; ∆y &middot; h
Jx (x0 , y0 ) − Jx (x0 + ∆x, y0 ) Jy (x0 , y0 ) − Jy (x0 , y0 + ∆y)
∆c
=
+
&plusmn; σ(x, y)
⇒
∆t
∆x
∆y
∂c
∂Jx ∂Jy
⇒
=−
+
&plusmn; σ(x, y) = −∇ &middot; J &plusmn; σ
∂t
∂x
∂y
∆c =
3. (a) The initial flux is given by J = c0 α∇Ψ, which, if α = 1 and c0 = 4, is J0 = (−4x, −4y).
(b) If a bar of unit length is placed at (1, 1) parallel to the x-axis, then the normal vector for
particles being attracted to the origin is (0, −1). Therefore, the number of particles crossing
this bar during a time interval of length ∆t is N = [J0 &middot; (0, −1)]1∆t = 4∆t, where it’s
assumed that ∆t is a small interval just after the field is turned on.
1
Homework 2.2 Solutions
L2
2D
ln La be associated with random motion and T = Ls be associated with direct
q
motion, where L = A
π . Recall that the bacterial doubling time is about 5 h.
1. (a) Let τ =
i. If the macrophages move twice as fast, then s = 6, and τ ≈ 20 h and T ≈ 47 min.
ii. If the target (surface area) is twice as big, then a = 40, and τ ≈ 58 h and T ≈ 1.5 h.
iii. If the area of the alveolus is half as big, then A = 0.5(2.5 &times; 105 ), and τ ≈ 34 h and
T ≈ 1 h.
In all cases, direct motion toward the target occurs faster than the bacteria double, while
random motion is a considerably slower process.
(b)
L2
L s
τ
=
ln &middot;
T
2D a√ L
√
L L
A
=
ln = − √ ln( Aπ)
ǫs a
ǫs π
R0 =
√
√
√
If R0 = 1, then s = − ǫ√Aπ ln( Aπ). If s &gt; 0, then it must be the case that ln( Aπ) &lt; 0,
which can only happen provided A &lt; π1 . So, in order for random motion to be as fast as
direct motion, the area in question must be sufficiently small.
2. The steady-state solution of the 1d diffusion equation satisfies
∂2c
d2 c
∂c
=D 2 =0 ⇒
= 0.
∂t
∂x
dx2
This differential equation has solution c(x) (no t dependence) subject to the given boundary
conditions. To solve for c(x), we integrate:
dc
d2 c
=0 ⇒
= k1 ⇒ c(x) = k1 x + k2 .
2
dx
dx
To solve for constants k1 and k2 , use the boundary conditions: c(0) = k2 = 0 and c(L) = k1 L =
C0 ⇒ k1 = CL0 . Therefore,
x
c(x) = k1 x + k2 = C0 ,
L
C0
which is a line with slope L that passes through the origin.
c(x)
C0
0
L
0
x
The total number of particles in the interval [0, L] is
Z L
Z L
x
C0 1 2 L C0 1 2 1
C0 dx =
c(x) dx =
x
&middot; L = C0 L.
=
L
L 2
L 2
2
0
0
0
2
```