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7 An Example of Reaction–Diffusion in Dimension n = 3

# 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

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

z,

h

m ≥ 1,

and

N0 (r, z) =

cm (r) sin

m=1

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

.

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

(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.

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