Problem 5.3.3
Multiplying the ODE through by H HxL we get
But,
2 f
f
H HxL + H HxL a HxL + h HxL 8l b HxL + g HxL< f = 0
x2
x
2 f
f
H f
H HxL = JH N -
x2
x
x
x x
Substituting into the previous expression and collecting terms gives
f
f
H
JH N + JH a -

Problem 5.3.9 (c)
If the equation is equidimensional then it will have solutions of the form fH xL ~ x p . This implies then that (by substitution) pH p - 1L + p + l = 0, p2 = - l ! If l > 0, then p = i l . thus the solution is of the form Note that x = f

Problem 5.3.9(a)
We are given
2 f
f
x2 + x + l f = 0
x2
x
If we divide through by 1 x we obtain
2 f
f
l
x + + f = 0
x2
x
x
This equation can be arranged as
f
l
Jx N + f = 0
x
x
x
which is in the conventional Sturm- Liouville form for the operator.
(1)

Problem 5.3.4(b)
We assume the form for the solution u Hx, tL = f HxL h HtL
(1)
Substituting into the PDE gives
1 h 1 i 2 f f y = jk - V0 z = - l2 j z h t f k x2 x cfw_ The appropriate eigenvalue problem is
(2)
2 f f k - V0 + l2 f = 0, f H0L = 0, f HLL =

Analysis of Surface Reaction in a
Cylindrical Pore
BGHiggins UCDavis May_ 08 HDraftL
Pore Diffusion with Adsorption and Reaction
Consider a cylindrical pore of radius R with catalytic sites on the wall of the pore. The species continuity equation is given

Problem 3.4.9
Since u H0L = 0 and u HLL = 0, we can differentiate the series for u Hx, tL with respect to t :
u
np
= I M bn Cos Hn p x LL
x
L
(1)
n=1
Since this is a cosine series we can differentiate it again term-by-term:
2 u
np 2
= - I M bn Sin Hn p

Problem 3.5.4(a)
The function Cosh HxL is represented by a Fourier Sine series. Thus in general we cannot differentiate a
Fourier Sine series term-by-term. Let
f HxL = Cosh HxL
(1)
Then from Eqn 3.4.11(RH) we have
Now
Thus
f' HxL
(2)
1
np
2
~ @f HLL - f H

Problem 3.2.2(f)
Let us first sketch the function , by taking L=0.5 1 0.8 0.6 0.4 0.2 -0.4-0.2 0.2 0.4 0.6 0.8 Figure 1 The periodic extension of this function is then 1 0.8 0.6 0.4 0.2 1
-2
-1 Figure 2
1
2
Note that the periodic extension is neither even

Problem 2.5.1(e)
The problem statement is : 2 u 2 u + = 0 x2 y2
BC1 : u H0, yL = 0 BC2 : u HL, yL = 0
(1)
We will use separation of variables and look for a solution of the form u Hx, yL = h HxL f HyL. Substituting this form for the solution into the PDE

Problem 3.3.10
We make use of the definition of even and odd functions. An even function can be represented as
1
fe HxL = @f HxL + f H-xLD
2
1
fo HxL = @f HxL - f H-xLD
2
For convenience let's define x = -x so that our definitions for odd and even functio

Problem 2.3.8
(a) For an equilibrium solution, we must solve
2
d us
k = aus
dx 2
a > 0
(1)
The general solution to this equation is
a
a
us HxL = c1 SinhH, xL + c2 CoshH, xL
k
k
The BCs yield
us H0L = 0, us HLL = 0 c1 = c2 = 0
(2)
(3)
Hence there is a uni

Problem 2.5.3
The problem statement that we must consider is:
1
u
1 2 u
Jr N + = 0
r r
r
r2 q2
The appropriate periodic BCs that must imposed are
(1)
BC1 : u Hr, -pL = u Hr, pL
u
u
BC2 : Hr, -pL = Hr, pL
q
q
We assume that the solution can be found be s

Problem 2.4.7
The mathematical statement is given by 1 u 1 2 u Jr N + = 0, r r r r2 q2 0 < r < a, 0 < q < 2 p
BC1 : u Ha, qL = f HqL
u u BC4 : Hr, pL = Hr, - pL q q Note that BC2 arises because at r = 0 is a singular point of the PDE and we require that t

Problem 2.3.1(d)
The PDE is given by
u
k
u
= Hr2 L
t
r r
r
We let u Hr, tL = G HtL f HrL and on subsitution we get
dG
k
d
df
f = G Hr2 L = -l
dt
r dr
dr
Thus the ODEs we need to solve are:
dG
= -l k G ,
dt
d
df
Hr2 L = -lrf
dr
dr
(1)
(2)
(3)

Problem 2.4.3
If we let x = x - p , then the problem is defined over a symmetric domain - p < x < p. The ODE in terms of x becomes 2 f = - lf x2 with BCs f H- pL = f Hp L (1)
The general solution to this eigenvalue problem is
Applying the BCs give
! ! f H

Problem 2.3.2 (d) The eigenvalue problem we need to address is f H0L = 0
df dx d f + l f = 0 d x2
2
BC1: BC2:
The general solution will depend on the sign of l : (i) l > 0 ! ! f HxL = C1 Cos I l xM + C2 Sin I l xM f H0L = C1 = 0
df HLL = 0 = C2 dx
HLL =

Problem 1.5.13 The spherically symmetric heat equation is given by
k u = Hr2 u L t r 2 r r d d Hr2 u L = 0, dr dr
(1)
If the system is at steady state, we can set the time derivative to zero and then solve the following BVP u H1L = 0, u H4L = 80
du r2 =

Problem 1.5.3
(a) For this part we need to make use of the following relations: y = [email protected] x Now differentiating the first equation in (1) with respect to x and then again with respect to y gives r2 = x2 + y2 , r r x 2 r = 2 x = = [email protected] x x r r r y 2 r =

Problem 1.4.1(f) The differential equation and BCs at steady state are given by where Q K0 = x2 . The general solution (after integrating twice) is Applying the BCs gives
x u HxL = - + c1 x + c2 12
4
d Q du u + , u H0L = T, HLL = 0 Ko dx dx2
2
(1)
(2)
c2

Problem 1.2.4
(a) We consider an arbitrary slice of thickness Dx located between x = a and x = b. If u Hx, tL is the density or concentaration per unit volume in that slice and is the mass flux per unit area, then the material balance is j
d u Hx, tL A H

Primer on Method of Characteristics:
First Order Partial Differential Equations
BGH/UCDavis/ECH256/March7_06
Introduction
First order partial differential equations in chemical engineering arise in models of processes that are dominated
by convection rath

Problem 10
The Fitzhugh-Nagumo model decribes two chemical species, an activator u and an
inhibitor v. The PDEs that describe the reaction diffusion system in the region
0 < x < L are
u
u
= ! - f Hu; r L - r Hv - uL
2
2
t
x
v
v
= - Hv - uL
2
2
t
BC1:

NLDProblem 11
A dimensionless model that is used to describe the logistic behavior of a fishery with a constant harvest rate is
given by
U
2mU
= F HU, mL U H1 - UL -
t
1+ 2U
where the parameter m 0 and U 0.
(1)
(i) Find the steady states for this mode

NLDProblem 2
Consider the following dynamical system
x
= m x - x2 + 2 x y
t
y
= Hm - 1L y + y2 + x y
t
(1)
(i) From inspection deduce the maximum number of equilibrium points (you may assume that the
equilibrium points are real).
(ii) Determine all the

Problem 6
In this problem you will analyze the effect of diffusion on the stability of the steady
state u 1 = u 2 = 0 found in Problem 2. The governing equations are
1
= D 21 + f Hu 1 , u 2 L
u
t
2 u
x
2
= D 22 + g Hu 1 , u 2 L
u
t
BC1:
BC2:
with
2 u
x

NLDProblem 3
Consider the following evolution equation
u1
= u1 - 4 m u2 - u2 2
t
(1)
du2
= 2 u1 H1 - u2 L
dt
(i) Determine the steady state solutions.
(ii) Determine whether there are any bifurcation points, (static or Hopf)
(iii) Determine the stabili

Problem 7
Suppose fishing is regulated within a zone H km from a country's shore (taken to
be a straight line) but outside of this zone over-fishing is so excessive that the
population is effectively zero. If the fish reproduce logistically, disperse by d

Department of Chemical Engineering & Materials Science
University of California, Davis
ECH 259 Final Exam
December 10, 2003
(Closed book, 1 sheet of notes)
Problem 1 (20 points)
Consider the following PDE
T
1
T
= Hr L + A,
t
r r
r
IC:
BC1
0<r<1
T Hr, 0L

Department of Chemical Engineering & Materials Science
University of California, Davis
ECH 259 MidTerm
(Closed Book, One Sheet of Notes)
Midterm Exam Solution
Total Marks: 55 points
Problem 1 (15 points)
Consider the following transient heat conduction pr

NLD Problem 1
Consider the following dynamical system
x
= f1 Hx, y, mL = x - 2 y + m x
t
(1)
y
= f2 Hx, y, mL = 3 x - y - x2
t
(i) Determine the steady state solutions and plot these solutions in the (x - m) and (y - m) planes
(ii) Determine whether the