7 An Example of Reaction–Diffusion in Dimension n = 3
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72
2 Diﬀusion
Then, writing the Laplace operator in cylindrical coordinates39 , N solves the equation
1
Nt = κ Nrr + Nr + Nzz
(2.121)
r
with the same initial and boundary conditions of N . By maximum principle, we
know that there exists only one solution, continuous up to the boundary of the
cylinder. To ﬁnd an explicit formula for the solution, we use the method of separation of variables, ﬁrst searching for bounded solutions of the form
N (r, z, t) = u (r) v (z) w (t) ,
(2.122)
satisfying the homogeneous Dirichlet conditions u (R) = 0 and v (0) = v (h) = 0.
Substituting (2.122) into (2.121), we ﬁnd
1
u (r) v (z) w (t) = κ[u (r) v (z) w (t) + u (r) v (z) w (t) + u (r) v (z) w (t)].
r
Dividing by N and rearranging the terms, we get,
v (z)
w (t)
u (r) 1 u (r)
=
−
+
.
Dw (t)
u (r)
r u (r)
v (z)
(2.123)
The two sides of (2.123) depend on diﬀerent variables so that they must be equal
to a common constant b. Then for v we have the eigenvalue problem
v (z) − bv (z) = 0
v (0) = v (h) = 0.
2
2
2
= − mh2π , m ≥ 1 integer, with corresponding
The eigenvalues are bm ≡ −νm
eigenfunctions
vm (z) = sin νm z.
The equation for w and u can be written in the form:
u (r) 1 u (r)
w (t)
2
+ νm
+
=
Dw (t)
u (r)
r u (r)
(2.124)
where the variables r and t are again separated. This forces the two sides of (2.124)
to be equal to a common constant μ. Then the equation for u is
1
u (r) + u (r) − μu (r) = 0
r
(2.125)
with
u (R) = 0 and
39
See Appendix C.
u bounded in [0, R] .
(2.126)
2.7 An Example of Reaction–Diﬀusion in Dimension n = 3
73
The (2.125) is a Bessel equation of order zero with parameter −μ (see Sect. 6.4.2);
conditions (2.126) force40 μ = −λ2 < 0. The only bounded solution of (2.125),
(2.126) is J0 (λr), where
∞
J0 (x) =
k
(−1)
x
2
2
k=0
(k!)
2k
is the Bessel function of ﬁrst kind and order zero. To match the boundary condition u (R) = 0 we require J0 (λR) = 0. Now, J0 has an inﬁnite number of positive
simple zeros41 λn , n ≥ 1:
0 < λ1 < λ2 < . . . < λn < . . . .
Thus, if λR = λn , we ﬁnd inﬁnitely many solutions of (2.125), given by
λn r
R
un (r) = J0
and μ = μn = −λ2n /R2 . Finally, for w we have the equation
2
w (t) = D(μn − νm
)w (t)
that gives
2
wmn (t) = exp D μn − νm
t ,
c ∈ R.
To summarize, we have determined so far a countable number of solutions of the
diﬀerential equation, namely
Nmn (r, z, t) = un (r) vm (z) wmn (t) =
λn r
λ2
2
sin νm z exp −κ νm
= J0
+ n2
R
R
t ,
all satisfying the homogeneous Dirichlet conditions. It remains to satisfy the initial
condition. Due to the linearity of the problem, we look for a solution obtained by
In fact, write Bessel’s equation (2.125) in the form (ru ) − μru = 0. Multiplying by u
and integrating over (0, R) , we have
40
R
R
ru
udr = μ
0
u2 dr.
(2.127)
0
Integrating by parts and using (2.126), we get
R
ru
0
udr =
ru u
R
0
−
R
0
(u )2 dr = −
R
(u )2 dr < 0
0
and from (2.127) we get μ < 0.
The zeros of the Bessel functions are known with a considerable degree of accuracy.
The ﬁrst ﬁve zeros of J0 are: 2.4048. . . , 5.5201. . . , 8.6537. . . , 11.7915. . . , 14.9309. . . ..
41
74
2 Diﬀusion
superposition of the Nmn , that is
∞
cmn Nmn (r, z, t) .
N (r, z, t) =
n,m=1
Then, we choose the coeﬃcients cmn in order to have
∞
∞
cmn Nmn (r, z, 0) =
n,m=1
cmn J0
n,m=1
λn r
R
sin
mπ
z = N0 (r, z) .
h
(2.128)
Since N0 (r, 0) = N0 (r, h) = 0, formula (2.128) suggests an expansion of N0 in sine
Fourier series with respect to z. Let
cm (r) =
h
2
h
0
N0 (r, z) sin
mπ
z,
h
m ≥ 1,
and
∞
N0 (r, z) =
cm (r) sin
m=1
mπ
z.
h
Then (2.128) shows that, for ﬁxed m ≥ 1, the cmn are the coeﬃcients of the
expansion of cm (r) in the Fourier-Bessel series
∞
cmn J0
n=1
λn r
R
= cm (r) .
We are not really interested in the exact formula for the cmn , however we will
come back to this point in Remark 2.10 below.
In conclusion, recalling (2.120), the analytic expression of the solution of our
original problem is the following:
∞
N (r, z, t) =
cmn J0
n,m=1
λn r
R
exp
2
γ − κνm
−κ
λ2n
R2
t sin νm z.
(2.129)
Of course, (2.129) is only a formal solution, since we should check in which sense
the boundary and initial condition are attained and that term by term diﬀerentiation can be performed. This can be done under reasonable smoothness properties
of N0 and we do not pursue the calculations here.
Rather, we notice that from (2.129) we can draw an interesting conclusion on
the long range behavior of N . Consider for instance the value of N at the center
of the cylinder, that is at the point r = 0 and z = h/2; we have, since J0 (0) = 1
2.7 An Example of Reaction–Diﬀusion in Dimension n = 3
2
and νm
=
2
m π
h2
2
,
∞
N
75
h
0, , t =
cmn exp
2
n,m=1
γ−κ
m2 π 2
λ2
− κ n2
2
h
R
t sin
mπ
.
2
The exponential factor is maximized for m = n = 1, so, assuming c11 = 0, the
leading term in the sum is
γ−κ
c11 exp
2
π2
λ2
− κ 12
2
h
R
t .
λ2
If now γ − κ πh2 + R12 < 0, each term in the series goes to zero as t → +∞ and
the reaction dies out. On the opposite, if
γ−κ
that is
π2
λ21
+
h2
R2
> 0,
γ
λ2
π2
> 2 + 12 ,
κ
h
R
(2.130)
the leading term increases exponentially with time. To be true, (2.130) requires
that the following relations be both satisﬁed:
h2 >
κπ 2
γ
and R2 >
κλ21
.
γ
(2.131)
Equations (2.131) give a lower bound for the height and the radius of the cylinder. Thus we deduce that there exists a critical mass of material, below which the
reaction cannot be sustained.
y
1
0.75
0.5
0.25
0
0
12.5
25
37.5
50
x
-0.25
Fig. 2.12 The Bessel function J0
76
2 Diﬀusion
Remark 2.10. A suﬃciently smooth function f, for instance of class C 1 ([0, R]), can
be expanded in a Fourier-Bessel series, where the Bessel functions J0 λRn r , n ≥ 1,
play the same role of the trigonometric functions. More precisely42 , setting R = 1
for simplicity, the functions J0 (λn r) satisfy the following orthogonality relations:
1
0
xJ0 (λm x)J0 (λn x)dx =
where
∞
k
cn =
k=0
(−1)
k! (k + 1)!
Then
0
m=n
m=n
1 2
2 cn
λn
2
2k+1
.
∞
fn J0 (λn x)
f (x) =
(2.132)
n=0
with the coeﬃcients fn assigned by the formula
fn =
2
c2n
1
0
xf (x) J0 (λn x) dx.
The series (2.132) converges in the following least square sense: if
N
SN (x) =
fn J0 (λn x)
n=0
then
1
lim
N→+∞
0
[f (x) − SN (x)]2 xdx = 0.
(2.133)
In Chap. 6, we will interpret (2.133) from the point of view of Hilbert space theory.
2.8 The Global Cauchy Problem (n = 1)
2.8.1 The homogeneous case
In this section we consider the global Cauchy problem
ut − Duxx = 0
u (x, 0) = g (x)
in R× (0, ∞)
in R,
(2.134)
where g, the initial datum, is given. We conﬁne ourselves to the one dimensional
case; techniques, ideas and formulas can be extended without too much eﬀort to
the n−dimensional case.
42
See e.g. [23], R. Courant, D. Hilbert, vol. 1, 1953.
2.8 The Global Cauchy Problem (n = 1)
77
The problem (2.134) models the evolution of the temperature or of the concentration of a substance along a very long (inﬁnite) bar or channel, respectively,
given the initial (t = 0) distribution.
By heuristic considerations, we can guess what could be a candidate solution.
Interpret u as a linear mass concentration. Then, u (x, t) dx gives the mass inside
the interval (x, x + dx) at time t.
We want to determine the evolution of u (x, t), due to the diﬀusion of a mass
whose initial concentration is given by g.
Thus, the quantity g (y) dy represents the mass concentrated in the interval
(y, y + dy) at time t = 0. As we have seen, Γ (x − y, t) is a unit source solution,
representing the concentration at x at time t, due to the diﬀusion of a unit mass,
initially concentrated in the same interval. Accordingly,
ΓD (x − y, t) g (y) dy
gives the concentration at x at time t, due to the diﬀusion of the mass g (y) dy.
Thanks to the linearity of the diﬀusion equation, we can use the superposition
principle and compute the solution as the sum of all contributions. In this way,
we get the formula
u (x, t) =
R
ΓD (x − y, t) g (y) dy = √
1
4πDt
e−
(x−y) 2
4Dt
g (y) dy.
(2.135)
R
Clearly, one has to check rigorously that, under reasonable hypotheses on the initial
datum g, formula (2.135) really gives the unique solution of the Cauchy problem.
This is not a negligible question. First of all, if g grows too much at inﬁnity, more
2
than an exponential of the type eax , a > 0, in spite of the rapid convergence to
zero of the Gaussian, the integral in (2.135) could be divergent and formula (2.135)
loses any meaning.
Even more delicate is the question of the uniqueness of the solution, as we will
see later.
Remark 2.11. Formula (2.135) has a probabilistic interpretation. Let D = 12 and
let B x (t) be the position of a Brownian particle, started at x. Let g (y) be the
gain obtained when the particle crosses y. Then, we can write:
u (x, t) = E x [g (B x (t))]
where E x denotes the expected value with respect to the probability P x, with
density Γ (x − y, t)43 .
In other words: to compute u at the point (x, t) , consider a Brownian particle starting at x, compute its position B x (t) at time t, and ﬁnally compute the
expected value of g (B x (t)).
43
See Appendix B.
78
2 Diﬀusion
2.8.2 Existence of a solution
The following theorem states that (2.135) is indeed a solution of the global Cauchy
problem under rather general hypotheses on g, satisﬁed in most of the interesting
applications. Although it is a little bit technical, the proof can be found at the end
of this section.
Theorem 2.12. Assume that there exist positive numbers a and c such that
|g (x)| ≤ ceax
2
for all x ∈ R.
Let u be given by formula (2.135) and T <
hold.
1
4aD .
(2.136)
Then, the following properties
i) There are positive numbers c1 and A such that
|u (x, t)| ≤ CeAx
2
for all (x, t) ∈ R × (0, T ] .
ii) u ∈ C ∞ (R × (0, T ]) and in the strip R × (0, T ]
ut − Duxx = 0.
iii) Let (x, t) → (x0 , 0+ ) . If g is continuous at x0 then u (x, t) → g (x0 ).
Remark 2.13. The theorem says that, if we allow an initial data with a controlled
exponential growth at inﬁnity expressed by (2.136), then (2.135) is a solution in the
strip R× (0, T ). We will see that, under the stated conditions, (2.135) is actually
the unique solution.
In some applications (e.g. to Finance), the initial datum grows at inﬁnity no
more than c1 ea1 |x| . In this case (2.136) is satisﬁed by choosing any positive number
a and a suitable c. This means that there is really no limitation on T, since
T <
1
4Da
and a can be chosen as small as we like.
Remark 2.14. The property ii) shows a typical and important phenomenon connected with the diﬀusion equation: even if the initial data is discontinuous at
some point, immediately after the solution is smooth. The diﬀusion is therefore a
smoothing process.
In Fig. 2.13, this phenomenon is shown for the initial data g (x) = χ(−2,0) (x) +
χ(1,4) (x), where χ(a,b) denotes the characteristic function of the interval (a, b). By
iii ), if the initial data g is continuous in all of R, then the solution is continuous
up to t = 0, that is in R×[0, T ).
2.8 The Global Cauchy Problem (n = 1)
79
1
0.5
0
1
0.8
0.6
0.4
0.2
0
−2
−1
t
1
0
2
3
4
x
Fig. 2.13 Smoothing eﬀect of the diﬀusion equation
2.8.3 The nonhomogeneous case. Duhamel’s method
The diﬀerence equation (or the total probability formula)
p (x, t + τ ) =
1
1
p (x − h, t) + p (x + h, t) ,
2
2
that we found in Sect. 4.2 during the analysis of the symmetric random walk,
could be considered a probabilistic version of the mass conservation principle: the
mass located at x at time t + τ is the sum of the masses diﬀused from x + h and
x − h at time t; no mass has been lost or added over the time interval [t, t + τ ].
Accordingly, the expression
1
1
p (x, t + τ ) − [ p (x − h, t) + p (x + h, t)]
2
2
(2.137)
could be considered as a measure of the lost/added mass over the time interval
from t to t + τ. Expanding with Taylor’s formula as we did in Sect. 4.2, keeping
h2 /τ = 2D, dividing by τ and letting h, τ → 0 in (2.137), we ﬁnd
pt − Dpxx .
Thus the diﬀerential operator ∂t − D∂xx measures the instantaneous density
production rate.
Suppose now that, from time t = 0 until a certain time t = s > 0, no mass is
present and that at time s a unit mass at the point y (inﬁnite density) appears.
We know that we can model this kind of source by means of a Dirac measure at
y, that has to be time dependent since the mass appears only at time s. We can
write it in the form
δ2 (x − y, t − s) .
80
2 Diﬀusion
Thus, we are led to the nonhomogeneous equation
pt − Dpxx = δ2 (x − y, t − s)
with p (x, 0) = 0 as initial condition. What could be the solution? Until t = s nothing happens and after s we have δ2 (x − y, t − s) = 0. Therefore it is like starting
from time t = s and solving the problem
pt − Dpxx = 0,
x ∈ R, t > s
with initial condition
p (x, s) = δ2 (x − y, t − s) .
We have solved this problem when s = 0; the solution is ΓD (x − y, t). By the time
translation invariance of the diﬀusion equation, we deduce that the solution for
any s > 0 is given by
p (x, t) = ΓD (x − y, t − s) .
(2.138)
Consider now a distributed source on the half-plane t > 0, capable to produce
mass density at the time rate f (x, t). Precisely, f (x, t) dxdt is the mass produced44
between x and x+dx, over the time interval (t, t+dt). If initially no mass is present,
we are lead to the nonhomogeneous Cauchy problem
vt − Dvxx = f (x, t)
v (x, 0) = 0
in R× (0, T )
in R.
(2.139)
As in Sect. 2.8.1, we motivate the form of the solution at the point (x, t) using
heuristic considerations.
Let us compute the contribution dv to v (x, t) due to a mass f (y, s) dyds. It is
like having a source term of the form
f ∗ (x, t) = f (x, t) δ2 (x − y, t − s)
and therefore, recalling (2.138), we have
dv (x, t) = ΓD (x − y, t − s) f (y, s) dyds.
(2.140)
We obtain the solution v (x, t) by superposition, summing all the contributions
(2.140). We split it into the following two steps:
•
We sum over y the contributions for ﬁxed s, to get the total density at (x, t) ,
due to the diﬀusion of mass produced in the time interval (s, s+ds). The result
is w (x, t; s) ds, where
w (x, t; s) =
R
44
ΓD (x − y, t − s) f (y, s) dy.
Negative production (f < 0) means removal.
(2.141)
2.8 The Global Cauchy Problem (n = 1)
•
81
We sum the above contributions for s ranging from 0 to t:
t
v (x, t) =
0
R
ΓD (x − y, t − s) f (y, s) dyds.
The above construction is an example of application of the Duhamel’s method,
that we state below:
Duhamel’s method. The procedure to compute the solution u of problem (2.139)
at the ponit (x, t) consists in the following two steps:
1. Construct a family of solutions of homogeneous Cauchy problems, with variable
initial time s, 0 ≤ s ≤ t, and initial data f (x, s).
2. Integrate the above family with respect to s, over (0, t).
Indeed, let us examine the two steps.
1. Consider the family of homogeneous Cauchy problems
wt − Dwxx = 0
w (x, s; s) = f (x, s)
x ∈ R, t > s
x ∈ R,
(2.142)
where the initial time s plays the role of a parameter.
The function Γ y,s (x, t) = ΓD (x − y, t − s) is the fundamental solution of the
diﬀusion equation that satisﬁes for t = s the initial condition
Γ y,s (x, s) = δ (x − y) .
Hence, the solution of (2.142) is given by the function (2.141):
ΓD (x − y, t − s) f (y, s) dy.
w (x, t; s) =
R
Thus, w (x, t; s) is the required family.
2. Integrating w over (0, t) with respect to s, we ﬁnd
t
t
w (x, t; s) ds =
v (x, t) =
0
R
0
ΓD (x − y, t − s) f (y, s) dyds.
(2.143)
Using (2.142) we have
t
vt − Dvxx = w (x, t; t) +
0
[wt (x, t; s) − Dwxx (x, t; s)] = f (x, t) .
Moreover, v (x, 0) = 0 and therefore v is a solution to (2.139).
Everything works under rather mild hypotheses on f. More precisely45 :
45
For a proof, see [12], McOwen, 1996.
82
2 Diﬀusion
Theorem 2.15. If f and its derivatives ft , fx , fxx are continuous and bounded
in R × [0, T ), then (2.143) gives a solution v of problem (2.139) in R × [0, T ),
continuous up to t = 0, with derivatives vt , vx , vxx continuous in R × (0, T ).
The formula for the general Cauchy problem
ut − Duxx = f (x, t)
u (x, 0) = g (x)
in R × (0, T )
in R
(2.144)
is obtained by superposition of (2.135) and (2.139). Under the hypotheses on f
and g stated in Theorems 2.12 and 2.15, the function
t
u (x, t) =
R
ΓD (x − y, t) g (y) dy +
R
0
Γ (x − y, t − s) f (y, s) dyds
is a solution of (2.144) in R × (0, T ),
1
,
4Da
T <
continuous with its derivatives ut , ux, uxx.
The initial condition means that
u (x, t) → g (x0 )
as
(x, t) → (x0 , 0)
at any point x0 of continuity of g. In particular, if g is continuous in R then u is
continuous in R × [0,T ).
2.8.4 Global maximum principles and uniqueness.
The uniqueness of the solution to the global Cauchy problem is still to be discussed.
This is not a trivial question since the following counterexample of Tychonov shows
that there could be several solutions of the homogeneous problem. Let
e−t
0
h (t) =
−2
for t > 0
for t ≤ 0.
It can be checked46 that the function
∞
T (x, t) =
k=0
x2k dk
h (t)
(2k)! dtk
is a solution of
ut − uxx = 0
in R× (0, +∞)
with
u (x, 0) = 0
46
Not an easy task! See [7], F. John, 1982.
in R.