Solution profiles for some simple combustion models

SOLUTION

PROFILES

FOR SOME SIMPLE COMBUSTION

MODELS*

J. BEBERNES Department

of Mathematics,

University of Colorado, Boulder, CO 80309, U.S.A.

D. EBERLY University of Texas, San Antonio, Texas, U.S.A. and W. FULKS University of Colorado Boulder, Colorado, U.S.A. (Received

3 November

1983; receioed for publication 9 April 1985)

Key wmis and phrases: Solution profiles, Gelfond problems, simple combustion

models.

1. INTRODUCTION Iru’THIS

paper, we study the shape (solution profile) of the solutions of the Gelfand problem

and the perturbed

-Au=SeU,

xEB,

u(x) = 0,

xEdB,

(1)

Gelfand problem xE

B,

xEaB,

K(X) = 0,

(2)

where B1 = {x E R”: 1x1< l}, 6 > 0 is the Frank-Kamenetski parameter, E > 0 is the reciprocal of the activation energy, and u(x) is the temperature. Both of these models play a fundamental role in the mathematical theory of thermal explosions for finite rigid and gaseous systems (Kassoy [l], Kassoy and Poland [2], Kapila [3], Bebernes and Kassoy [4], Bebernes and Bressan [5]). For rigid systems the physical processes are determined by a pointwise balance between chemical heat addition and heat loss by conduction. During the inductive period, with a duration measured by the conduction time scale of the bounding container, the heat released by the chemical reaction is redistributed by thermal conduction. As the temperature of the container increases, the reaction rate grows dramatically. Eventually the characteristic time for heat release becomes significantly smaller than the conduction time in a well-defined hot spot embedded in the system. Then the heat released is used almost entirely to increase the hot spot temperature. The purpose of this paper is to show that both models detect this hot spot development in a very precise manner. This hot spot development had previously been detected only numerically (Kassoy [2], Gustafson and Eaton [6]). * Research supported by U.S. Army Research Office under contract number DAAG 29-82-0069. 165

_I. BEBERNES. D. EBERLY and b’. FCLKS

166

For (1) with an arbitrary smooth bounded domain Q, it is known that there is a critical value 6” > 0 such that (1) has at least one positive solution for 0 < 6 < 6* and no solution for 6 > 6*. When R = Br, then very precise multiplicity results are known. These are reviewed in Section 2. Our main result for the Gelfand problem (1) is theorem 2 of Section 3. For R = B,, we prove that, for dimension n 2 3, the minimal solution will be bell-shaped, that is, exhibit a hot spot, for a range of parameter values 6.8 < 6 < 6*. For 0 c 6 < 5, we prove that the physically significant minimal solution is concave down and all other solutions are bell-shaped. For the perturbed Gelfand problem (2), we prove that: (a_)fern = 2, the solutions are either bell-shaped or concave down; (b) for n 3 3, there exists 6 = a(~) such that for 6 > 6(~) all solutions are not concave down; (c) for n 2 3, for each 6 C d(&), the minimal solution is concave down for E sufficiently close to 0; (d) the large solutions (notation used is that of Dancer [lo]) are not concave down and have an odd number of points of inflection and (e) for n Z 3, there exist solutions with at least two points of inflection. 2. KNOWN

RESULTS

FOR THE

GELFXND

PR0BLE.M

For 52 = B, C KY, we summarize in this section how to find all solutions U(X) E C’(B,, R) of the Gelfand problem (1). These results are then used in the next sections. Problem (1) is referred to as the Gelfand problem because of the multiplicity results obtained by Gelfand [ 1l] for n = 3. But the equation (1) has a long history and can be traced back to Liouville in 1853 (see Bandle [7]). A key paper giving many earlier references is that of Joseph and Lundgren [8] where (1) is fully analyzed for all dimensions n P 3 via a phase plane approach. As in most earlier references the solutions were assumed radially symmetric. By the maximum principle, any solution U(X) E C?(B,.Z) of (1) is positive on B,. By the result of Gidas, Ni and Nirenberg [9], all solutions of (1) and (2) are radially symmetric, that is, u = u(r) where r = 1x1.Consequently, for S2 = II,. the Gelfand problem (1) is equivalent to n-l u” + -r4’+6eU=0, r

O
U’(0) = 0, U(1) = 0 or

(rnT1~‘)’ + drnmleu = 0, 0 < r < 1 (4)

U(0) = (Y,u(1) = 0 or n-l u” + -u'+be"=O, r

OCrCl

(5)

u(l) = 0, u’(1) = c where (Yand c are related to 6 through the relationship u’(O) = 0. The results on existence are given below. THEOREM

1. (a) n = 1: There exists 6* > 0 such that: (i) for 6 E (0, S*), there exists 2 solutions; (ii) for 6 = 6*, there exists 1 solution; (iii) for 6 > 6*, there are no solutions.

Solution profiles for some simple combustion models

167

(b) n (i) for (ii) for (iii) for (iv) for

= 2: Let 6* = 2. Then: 6 E (0, 6”), there exist 2 solutions; S = 6*, t here exists 1 solution; S > 6*, t here are no solutions; c = c((u, 6) =-K’(l), C’ + 4c + 26 = 0. (c) 3 s n 5 9: Let 6 = 2(” - 2). There exists 6* > 8 such that: (i) for d = 6*, there exists 1 solution; (ii) for S > 6*, there are no solutions; (iii) for 6 = 8, there exjsts a countable infinity of solutions; (iv) for 6 E (0, 6”) - {6}, t h ere exists a finite number of solutions. (d) n h 10: Let 6* = 2(n - 2). Then: (i) for d 1 6*, there are no solutions; (ii) for b E (0, S*), there exists 1 solution. Proof. Let (Y= u(O), p = -c = -u’(l). For n = 1, (3) can be solved by integration to obtain ,B = (p’ + 26)*/* tanh[+@ 1 + (IS=3eValn u(r)

+ 26)]

e-Q)l/’

[ 1 - (1 - ,-y/2

2 1

(6)

= a - 2 In cosh[4(26 en)‘i’zr],

For n = 2, (3) can also be solved by making the change of variables r = e-‘, w(r) = u(r) - 2t to obtain ti + 6e” = 0. Then p’ - 4p + 26 = 0 6 = 8(e- @a - e-a) u(r)

(7)

= a - 2 ln(1 + 66e”e*).

For n z 3, let

and u(r) = LY+ 2r + z(t), Then (4) becomes i -n-2

i+2ez

-2=O,

z(x)=-xc,

with compatability

tl
i(x)=-2

condition z(tt) = - CY- 2t,. Let y(t) = i(t) + 2 and I

(8) = 2(n - 2)e-(‘), then

x = x(y - 2) t*
j=(n-2)y-x, x(r)

= y(x)

= 0

(9)

J.

168

with compatability

BEBERNES, D. EBERLY and W. FCLKS

condition I, +!ln(VJ,

and thus, 6 = x(f,), p = y(ti). The two-dimensional system (9) has critical points at (0, 0) and (2(n - 2), 2). If 3 5 n S 9, then (2(n - 2), 2) is an unstable spiral and (0,O) is a saddle. If n 2 10, then (2(n - 2), 2) is an unstable node and (0,O) is a saddle point. One can prove that there exists a heteroclinic orbit joining these two critical points. We summarize our observations in terms of the following (6, ,B) bifurcation diagrams (Fig. 1). 3.

KNOWN RESULTS

For the perturbed Gelfand proved the following. THEOREM

FOR THE PERTURBED

problem

GELFAND

PROBLEtM

(2) with special domain R = B, C R”, Dancer

[lo]

2. For any E > 0, 6 > 0, there exists at least one and at most finitely many solutions.

Fig. 1

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It (w) J! (u)

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aJe suo!inIos11~‘1 = u Jod (e) 'f wIHo3HJ

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J. BEBERNES. D. EBERLYand W. Furu

170

/3 > 6/(n - l), then u”(1) > 0 and u”(0) < 0 imply there exists R E (0, 1) such that u”(R) = 0 and (R, u(R)) is a point of inflection. By the uniqueness of inflection points, the solution u(r) corresponding to (6, p) is therefore bell-shaped. If (6, p) E D satisfies p < S/(n - l), then u”(1) < 0 and u”(0) < 0 imply no inflection points or more than one. Uniqueness rules out the latter case. For n = 2, since D = ((6, p): 6 > 0, ,6’ - 4p + 26 = 0) the result is immediate assuming uniqueness of points of inflection. To show that D intersects p = S/(n - 1) uniquely for n Z 3, we prove a sequence of lemmas. LEMMA

1. Let K(X) be a solution of (1). Then 6K(X) e

where w, is the surface area of the unit sphere in R”. As a consequence, for n > 2.

,6’ - 2nfi + 26 < 0

Proof. Define ui = K,! and uil = I(,[ K~, = uxix,. Then A (r$)

= A (~xP,) = i$ [XiAKi + ~(CX,). (VU,) f

K~Ax,]

a

=r~--(AK)+2AU

using equation (1). Let u =

S(K

- 2)e".

Then r(du/dr) =

I

au

r-dx El ar

6(u

=

l)e”r(du/ar)

and u = -26 for x E 6Br. Since

(Vu) . x du

= i =

-

(Vu).

(V4r') dx

El

I

dBI

= -26

u

5 (Vtr')

ds -

i

uA($r’) dx El

Solution

we

profiles

for some simple combustion

models

171

have

where w,, is the surface area of B1. Define

Then, using the first part of the proof, jE,[~ue"(r~+2)-SeYr~]~

I=

=

I[ I[

6(u-

l)e”r~+26ue”

BI

=

I

dx

I

dx

rg+no+(2-n)u+46eY

El

= -26w,

+

I El

[(2 - n)u + 2n]f5 eU dx.

But by Green’s theorem,

ds

sincert=O

on

aB,,r=l

and

&=”

ar

on

f3B1.

So I = /3*wn and hence, [(2 - n)u + 2n]6 eU dx = (/3* + 240,.

i El Also, by Green’s theorem,

Je, 6 ey dx = PO,,. This and the previous equation yields

I

81

&.4(x) e u(x)

&(

=

(P2+ 24 + 24 w,. 2-n

Since the integral is positive and since n L 3, we necessarily have /3* - 2n/3 + 26 < 0.

172

J. BEBERNES,D. EBERLY and W. FCLKS

Remarks. The idea for the proof comes from Weinberger

[22]. For p = 2, the critical value 6* must satisfy 6* < 2(n - 1) via the preceding inequality. This bound on 6’ is used later. It is also possible to compute an upper bound on 6*, namely, 6” < Si/e < 2(n - l), where 6i is the first eigenvalue of -A~+!J= 6~ on B,, I+I= 0 on dB1. See, for example, Kassoy [l]. LEXN;\ 2. The heteroclinic orbit D and the graph of L(6) = S/(n - 1) intersect in at least one point where n - 1 < 6 < 2(n - l), 1 < /3 < 2. Proof. Let /3(S) be the arc of D which originates at (0,O) and terminates of intersection (6* ,2) on the line /? = 2. From (9), we have /J’(S)

=

at the first point

(n - 2)P- 6

w -

2)

The graph of p(S) has a vertical tangent at (6*, 2). From lemma 1, it must be the case that 6 < 2(n - 1) for any pair (ZJ?,B) E D. Thus, the graph of p(S) must reach p = 2 before 6 reaches 2(n - 1) and as a result, since L(2(n - 1)) = 2, the graph of /3(S) must intersect that of L(S) for 6 < 2(n - 1). See Fig. 3. Moreover, if p(S,J = L(6,) = PO for any PO E (0, I], a0 E (0, n - 11, then ,B’(S,J = [(n - 1) (2 - ,$,)]-’ < (n - 1)-l = L’(6,). Thus. if there are any points of intersection for PO E (0, 11, then there can only be one. This also implies that p’(O) 2 (n - 1)-i. However, p’(0) = n-l < (n - 1)-i, a contradiction. So there can be no points of intersection for p 5 1. 3

1

*L.___-__----.___

,/

\

Iv//

9:;

--

/,il I

-.

b

2,/?-I)

6

Fig. 3

The uniqueness of the point of intersection of the graphs of ,6(S) and L(a) at (aa, ,&J where n - 1 < 6a < 2(n - l), PO > 1 follows from the fact that p’(S,) = [(n - 1) (2 - PO)]-’ > (n - 1>-‘. Remark. For n 2 10, it is clear from the geometry

that this point of intersection is unique. It remains to be shown that for 3 5 n 5 9 there are no more points of intersection than the one just constructed. LEMMA

3. Let 3 5 n Z 9. The heteroclinic

orbit D intersects the graph of L(6) in a unique

point. Proof. There is at least one point of intersection (which was constructed in lemma 2). It is conceivable that as the heteroclinic orbit spirals into (2(n - 2), 2), it intersects the graph of

Solution profiles for some simple combustion models

173

L(6) again. To show this cannot happen, consider the region R, bounded by: L1 = ((6, 6): 6 = 4, 2 5 p d 3) for n = 3. LI = ((6, p): 6 = 2(n - l), 2 s /3 $ 2((n - l)/(n - 2))) for n z 4; Ll = {(S, p>: p = 3, 3 5 6 5 4) for n = 3, L, = ((6, p>: p = 2((n - l)/(n - 2)), n d 6 ~5 2(n - l)} for n E 4; L3 = ((6, 0): p = -(l/4)6’ + (3/2)6 + (3/4), 1 d 6 5 3) for n = 3, L3 = {(6, PI: P = (6/b - 2)) + 1, n - 2 5 6 5 n} for n Z 4; L, = ((6, p): 6 = n - 2, 1 5 p 5 3); Lj = ((6, p): p = 1, Il - 2 $ 6 s n - 1); Lb = ((6, @)I fi = S/(n - l), n - 1 5 6 5 2(n - 1)). See Fig. 4.

2-

.I

!I

n-2

L5 III

n-l

I

Zh-II

”

B

2ln-2) Fig. 4.

With the observation that p’(S) < 0 in S = ((6, j3): 6 > 2(n - 2), 2 < /3 C 6/(n - 2)) U {(S, /3): 0 < 6 < 2(n - 2), 6/(n - 2) < /3 < 2) and p’(S) > 0 in ((8, p): p > 0, 6 > 0) - S, we see that the heteroclinic orbit cannot leave R through L, or L2. The orbit cannot leave R through L3, L4, Lj, or L6 since the slope at such a crossing would not agree with p’(6) evaluated on these sets. Thus, R is an invariant region for the heteroclinic orbit and so the first point of intersection of the orbit with L6 is the only point of intersection.

LEMMA 4.Consider the problem n-l u” + -u'+6e"=O, r

u’(0) = 0,

u(1) = 0

OCrCl

(u(0) = Ly,u’(1) = -p>

(10) (11)

(n - 1>p - 6 = 0. There exists a unique pair (8, ,@which satisfies (11) and, consequently, U(T) in the sense that this solution is the only one with u”(1) = 0.

yields a unique solution

Proof. For the case n = 2, we had p? - 4/3 + 26 = 0. This condition along with j3 - 6 = 0 yields the unique solution p = 2, 6 = 2. For the cases n L 3, the point of intersection of D with L(6) yields the pair (6, ,6) asked for in the conclusion of the lemma. By our previous work, this is the only pair which satisfies the condition (11).

171

J. BEBERNES, D. EBERLY and W. FULKS

We now show that points of inflection (when they exist) are unique. LEMMA 5. ,Let u(r) be a solution to (10) for n Z 2. If u”(1) > 0, then u” = 0 has exactly one solution R E (0, 1)

Proof. Let R E (0, 1) be the smallest value of r such that u” = 0. For this value R, define m = u’(R). Then

In (lo)-(ll), we have

make the change of variables r = sR and U(S) = U(T) - u(R). Restrictings

n-1 u” + s

v’ + deu =o,

E [0, 11,

O
v’(0) = 0, u(1) = 0 [u’(l) = - p]

(n -

l);g-

s=o

(13)

where 6 = -(n - 1)mR > 0 and p = -mR > 0. By lemma 4, there is a unique pair (8, /) and a unique solution u(s) satisfying (12)-(13). Note that u”(1) = 0 is implied by condition (13). Since u”(r) < 0 for 0 5 r < R, we have that o”(s) < 0 for 0 Z s < 1. Suppose there is a second value of r where u” = 0. Let P be this second value. Then 0 < R < P d 1 and u”(R) = u”(P) = 0. Define I= u’(P). Then u(P)=ln(-l(nSpl)). Make the change of variables r = sf’ and u(s) = u(r) - u(P): Restricting s E [0, 11, we have that u(s) satisfies (12)-(13) where 6 = -(n - 1)IP > 0 and p = -1P 7 0. By our uniqueness claim to (lo)-(ll), we must have that u”(s) < 0 for s E [0, 11. However, u”(R/P) = P2u”(R) = 0. This is a contradiction, so it must be the case that there is at most one value R where u” = 0. Theorem 3 must follow as a result of the comments made earlier in this section. The results of this section are part of the thesis of Eberly [13]. 5. SOLUTION

PROFILES

FOR

THE

PERTURBED

GELFAND

PROBLEM

Information about solution profiles can be obtained for the perturbed Gelfand problem. This information is not as precise as for the Gelfand problem, but it appears that such precision is attainable with more detailed work. By the maximum principle, any solution to (2) must be positive on B,. Since all positive solutions are radially symmetric, the following system equivalent to (2):

Solution profiles for some simple combustion models n-l u” + -

u’ + 6 exp

r

u’(0) = 0,

U

( )

=O, 1 + EU

175

O
u(1) = 0.

Using the same notation as in the Gelfand problem, let (Y= u(0) and p = -u’(l). PROPOSITION 1. For n = 2, points of inflection of solutions of (14) are unique. Consequently,

all solutions of (14) are either bell-shaped or concave down. Proof.

Let f(u) = exp(u/(l

+ EU)). Differentiating

u”’

+

(n

-

1)

( y

(14) gives

1+

Gf’(u)u’

= 0.

Suppose (R, u(R)) is a point of inflection. Then Ru”(R)

+ &u(R)]’ since u”(R) = 0. The function ru’(r) is decreasing and the function [l + au(r)]-’ is increasing. Suppose (P, u(P)) 1s . another point of inflection where r < P < 1, and where u”(r) < 0 for r E [0, R), u”(r) > 0 for R < r < P. Then Pu’(P)

Ru’(R)

’ + [l + EU(P)]~ < 1 + [l + EU(R)]~ < ’

where the last inequality is valid by (15) since u”‘(R) > 0. This set of inequalities forces u”‘(P) > So (R, u(R)) is the only point of inflection.

0, a contradiction.

LEMMA 6. Let n Z 3. For E > 0 sufficiently small, the graphs of the bifurcation curve and the line (n - l)p - 6 = 0 intersect in at least three points. The point of intersection with the smallsmall branch is unique, there are at least two points of intersection with the large-small branch. Proof. By Section 4 we have a unique point of intersection, (6, p) = (6(O), p(O)) (where E = 0), for the bifurcation curve D and the line L(6). At this point of intersection, the angle of intersection is positive so that a small perturbation of the bifurcation curve still yields a unique point of intersection (S(E), &E)). By our earlier remark on the closeness of the largesmall branch to the small-small branch, the large-small branch and the straight line must intersect in at least two more points. For E sufficiently close to 0, one of these points must occur near (B(E), P(E)); the other must occur near (0,O). See Fig. 5. At these points of intersection, u”( 1) = 0 for a solution to (14). From <15), un( 1) = -/_?(1 y p). Thus, at (at, PI) near (O,O), u”‘(l) < 0 and at (&, /3*) and (QE), P(E)) near (6(O), P(O)), u”‘(l) > 0 for solutions u(r) corresponding to these pairs of 6 and j3. With these observations, we have the following result. PROPOSITION 2. For E > 0 sufficiently small, there exist solutions to (14) with at least two points

of inflection.

176

J. BEBERVES,D. EBERLY and W. FULKS

Fig. 5

Proof. The solution to (14) corresponding to the pair (S,, /3,) must have an even number of points of inflection since u”(1) = 0 and u”‘(1) < 0 imply that u”(r) > 0 for r < 1 (1 - r small). Since u”(O) < 0, the intermediate value theorem guarantees that there is at least one value of R E (0, 1) such that u”(R) = 0. The graph of u(r) has at least two points of inflection (which by the geometry must occur in pairs). THEOREM

4, Let (S(E), fi(~)) be the pair constructed in lemma 6. The minimal solution to (14) where 6 > b(~) must have at least one point of inflection.

Proof. For 6 > &(E), pairs (6, p) on the bifurcation curve D, satisfy the relationship u”(1) = (n - 1)p - 6 > 0 (by the construction of (d(~), P(E))). By the intermediate value theorem, there must be at least one value R E (0, 1) such that u”(R) = 0 and (R, u(R)) is a point of inflection. PROPCNTION

3. For a fixed 6 < d(0) and E sufficiently close to 0, the minimal solution of (14) is concave down.

Proof. The minimal solution to (14) can be written as u = u(r, 6, E). The minimal solution to the Gelfand problem is u(r, 6,O). By continuous dependence on parameters, and by the C1 closeness result of Dancer [lo], we can also conclude that the second derivatives of solutions to (14) depend continuously on the parameters 6 and E.-Thus, since the minimal solution of the Galfand problem is concave down, for a fixed 6 < 6(O) and for E sufficiently close to 0, the minimal solution to the perturbed Gelfand problem is concave down. In the notation of Dancer [lo], the pair (6, /3) yields a largesolution u(r) if there is a constant k and a number u (close to 1) such that u(r) 2 ke’!’ for r E [0, p]. We have the following result. PROPOSITION 0.

4.

For E sufficiently close to zero, any large solution of (14)

must

satisfy

u”(1)

>

Solution profiles for some simple combustion models Proof.

Integration p = 6

of the differential

j’ rn-‘f(u(r))

equation (11) yields

dr 2 6 jy rn-*f(kel’E)

0

177

dr = d $f(kel’E)

> A

0

for E sufficiently close to 0. Thus, n”(l) = (n - l)p - 6 > 0 for E sufficiently close to 0. Finally, we have been unable to prove but conjecture that the number of points of intersection of D, with L(6) is exactly three and, thus, there are at most two points of inflection for any solution of (14) for E small. REFERENCES

in combustion, ignition and explosion, SIAM-A.LfS Trans. 10, 61-72 (1976). KASSOYD. & POLMD J., The thermal explosion confined by a constant temperature boundary II-the extremely rapid transient, SIAM J. appl. Math. 41, 231-246 (1981). KAPILAA., SIAM J. appl. Math. 39, 476-484 (1980). BEBERNESJ. & KASSOY D., A mathematical analysis of blow-up for thermal reactions-the spatially nonhomogeneous case, SIAM J. appl. Marh. 40, 476-484 (1981). BEBERNESJ. & BRESS~NA., Thermal behavior for a confined reactive gas, J. diff, Eqns 44, 118-133 (1982). GUSTAFSONK. & EATON B., Exact solutions and ignition parameters in the Arrhenius conduction theory of gaseous thermal explosion, J. appl. Marh. Phys. (ZAMP) 33, 2392401 (1982). BUNDLEC., Isoperimetric Inequnliries and Their Applications, Pitman. London (1980). JOSEPH D. & LUNDGRENT., Quasilinear Dirichlet problems driven by positive sources, Archs rarion. mech.

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2. 3. 4. ;: 7. 8.

Analysis

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(1979).

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