We consider the problem
u00(x) + f"(u) = 0, 1 < x < 1, u( 1) = u(1) = 0,
f"(u) = "u3+ u2 u + , ; " > 0. (5.1)
Problem (5.1) was …rst systematically studied by a celebrated paper by Smoller and Wasserman [11]. In particular, they consider (5.1) with " = 1 and that cubic nonlinearity
f"=1(u) has three real zeros a < b < c. In this section we discuss the general case with
" > 0 and ; ; 2 R, so that f"(u) may have exactly one positive zero, two distinct positive zeros or three distinct positive zeros. First, note that, if 0 or 0, we can show that the structure of bifurcation curve S" of positive solutions for (5.1) is one of the following cases:
(i) The bifurcation curve S" of (5.1) is an empty set (that is, (5.1) has no positive solution for all > 0).
(ii) The bifurcation curve S" of (5.1) is a monotone curve on the ( ; kuk1)-plane.
(iii) The bifurcation curve S" of (5.1) has exactly one turning point where the curve turns to the right on the ( ; kuk1)-plane.
Thus problem (5.1) has at most two positive solutions if 0 or 0. See [5] for the details of the above results.
If > 0and > 0, then (5.1) reduces to (1.1). It is more di¢ cult to determine precisely the exact multiplicity of (1.1) since problem (1.1) may have three positive solutions for some positive numbers "; ; . We analyze (1.1) more precisely in this section. First, if
p ;
the exact multiplicity results of positive solutions for problem (1.1) was determine precisely by Theorem 2.1 and [3, Theorem 2.1]. By some numerical simulations, we give next three conjectures on the shape of bifurcation curves ^S" of positive solutions of (1.1) with
> p , de…ned by
S^" f( ; ku k1) : > 0 and u is a positive solution of (1.1) with >p
g . Conjecture 5.1. Consider (1.1) where
p < p
3 :
Then there exists a positive number ~" = ~"( ; ; ) satisfying satisfying (25
32(
3
27 ))1=2 < ~" < (
3
27 )1=2 such that all results in Theorem 2.1(i)–(iii) hold.
While
>p
3 ; (5.2)
we remark that there exists some " > 0 such that cubic nonlinearity f"(u) has three positive zeros 0 < a < b < c andRc
a f"(t)dt > 0 (see Fig. 8(i).) For these f"(u), it is easy
to check that a + c > 2b and there exists 2 (b; c) such thatR
a f"(t)dt = 0. So problem
(1.1), (5.2) can be written as
u00(x) + "(u a)(u b)(c u) = 0, 1 < x < 1, u( 1) = u(1) = 0;
; " > 0, 0 < a < b < c, a + c > 2b. (5.3)
Fig. 8. (i) The graph of f"(u) satisfying (5.3). (ii) The conjectured bifurcation curve of problem (5.3).
It was conjectured that the bifurcation curve of positive solution of problem (5.3) is broken S-shaped (see Fig. 8(ii)) on the ( ; kuk1)-plane. A proof was claimed by Smoller and Wasserman [11, Theorem 2.1], but their proof has a gap. Assuming di¤erent conditions on constants a, b and c, Wang [12] and Korman, Li and Ouyang [6] gave partial proof of the above conjecture independently. For this conjecture, Korman, Li and Ouyang [7]
gave a computer-assisted proof. Further investigation on this conjecture is needed. We give two conjectures about problem (1.1), (5.2) which is more general than problem (5.3).
Conjecture 5.2. Consider (1.1) where
p3 < < 2p : (5.4)
Then there exist two positive numbers ~"0 = ~"0( ; ; ) < "0 = "0( ; ; ) such that the following assertions (i)–(iii) hold:
(i) (See Fig. 2(i).) If 0 < " < ~"0, then the bifurcation curve ^S" is S-shaped on the ( ;kuk1)-plane. Moreover, the exact multiplicity results of positive solutions in Theorem 2.1(i) hold.
(ii) (See Fig. 8(ii).) If ~"0 " < "0, then the bifurcation curve ^S" is broken S-shaped on the ( ; kuk1)-plane. Moreover, there exist > 0 such that problem (1.1), (5.4) has exactly three positive solutions for > , exactly two positive solutions for
= , and exactly one positive solution for 0 < < .
(iii) (See Fig. 2(iii).) If " "0, then the bifurcation curve ^S"is a monotone curve on the ( ;jjujj1)-plane. Moreover, problem (1.1), (5.4) has exactly one positive solution for all > 0.
Conjecture 5.3. Consider (1.1) where
2p : (5.5)
Then there exists a positive number "0 = "0( ; ; ) such that the following assertions (i) and (ii) hold:
(i) (See Fig. 8(ii).) If 0 < " < "0, then the bifurcation curve ^S" is broken S-shaped on the ( ; kuk1)-plane. Moreover, there exist > 0 such that problem (1.1), (5.5) has exactly three positive solutions for > , exactly two positive solutions for
= , and exactly one positive solution for 0 < < .
(ii) (See Fig. 2(iii).) If " "0, then the bifurcation curve ^S"is a monotone curve on the ( ;jjujj1)-plane. Moreover, problem (1.1), (5.5) has exactly one positive solution for all > 0.
Acknowledgments. Most of the computation in this paper has been checked using the symbolic manipulator Mathematica 7.0.
References
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6. Appendix
Proof of Lemma 3.4.
The proof of Lemma 3.4 is rather long and technical, we divide the proof into next Steps 1–5.
Step 1. We compute G0"( ).
By (3.3)–(3.5), we compute that
G"( ) = 8p
where
K"( ; u) = 20(4F")2(4f") 22(4F")2(4 ~f") 4(4F")2(4 ^f")
+33(4F")(4f")2+ 18(4F")(4f")(4 ~f") 15(4f")3, (6.2)
4F" = F"( ) F"(u), (6.3)
4f"= f"( ) uf"(u), (6.4)
4 ~f"= 2f"0( ) u2f"0(u), (6.5)
4 ^f"= 3f"00( ) u3f"00(u). (6.6)
Since f"(u) = "u3+ u2 u + , we have that
F"(u) = "u4=4 + u3=3 u2=2 + u, (6.7)
uf"(u) = "u4+ u3 u2 + u, (6.8)
u2f"0(u) = 3"u4+ 2 u3 u2, (6.9)
u3f"00(u) = 6"u4+ 2 u3. (6.10)
For 0 < u < , we let A "( 4 u4), B ( 3 u3), C ( 2 u2), D ( u).
Then A; B; C; D > 0. By (6.3)–(6.10), we obtain that
4F" = A=4 + B=3 C=2 + D, (6.11)
4f" = A + B C + D, (6.12)
4 ~f"= 3A + 2B C, (6.13)
4 ^f"= 6A + 2B. (6.14)
Substitute (6.11)–(6.14) into (6.2), we have
K"( ; u) = 1
72(168ABC 1356ABD 504ACD 168BCD
+9A3 144D3 2AB2+ 12A2B + 90AC2 207A2C 60B2C + 2646AD2+ 1134A2D
1248BD2+ 560B2D + 468CD2 + 72C2D). (6.15) So Lemma 3.4 holds if we can prove that K"( ; u) > 0for any …xed " 2 [(107(273 ))1=2; (273 )1=2],
2 [ "; ") and 0 < u < .
Step 2. We make a transformation for K"( ; u).
Although both T"( ) and G"( ) are only de…ned for 2 (0; "), K"( ; u) is well de…ned for all 2 R. So Lemma 3.4 holds if we can prove K"( ; u) > 0 for any …xed
" 2 [(107(273 ))1=2; (273 )1=2], ", 0 p and 0 < u < . Since " = 3", we consider K"( ; u) when ", 0 p
3" 2", 107" 3" " 3", and 0 < u < . Let
= (r + 1) ", r 2 [0; 1),
= s" 2", s 2 [0;p 3],
= t" 3", t 2 [ 7 10; 1], u = y ", y 2 (0; r + 1).
Thus
A = "( 4 u4) = " 4"[(r + 1)4 y4], (6.16) B = ( 3 u3) = 3" 4"[(r + 1)3 y3], (6.17) C = ( 2 u2) = s" 4"[(r + 1)2 y2], (6.18) D = ( u) = t" 4"(r + 1 y). (6.19) Substitute (6.16)–(6.19) into (6.15), we obtain
K"( ; u) = 1
8"3 12" (r + 1 y)3Ke"(r; s; t; y); (6.20) where
Ke"(r; s; t; y) = P9 j=0
kj(r; s; t)yj; (6.21)
k0(r; s; t) = (3 122t2 + 10s2 16t3+ 8s2t + 234t 27s + 52st2 112st) +( 392st + 16s2t + 50t2+ 736t + 50s2+ 27 125s + 52st2)r +(100s2+ 466t2+ 8s2t 243s + 730t 504st + 106)r2+ ( 280st +238 + 294t2+ 240t + 100s2 285s)r3+ ( 265s 56st + 50s2 +336 + 190t)r4+ ( 207s + 304t + 308 + 10s2)r5+ (126t 105s +182)r6 + (66 23s)r7+ 13r8+ r9,
k1(r; s; t) = (9 + 52st2+ 468t + 30s2 81s 224st 122t2+ 16s2t) + (16s2t +172t2 560st + 72 294s + 1004t + 120s2)r + (294t2+ 180s2
435s 448st + 456t + 246)r2+ ( 112st + 24t + 468 + 120s2 420s)r3+ (30s2+ 356t + 540 375s)r4 + (252t + 384 246s)r5 +(162 69s)r6+ 36r7+ 3r8,
k2(r; s; t) = (18 + 40s2 135s + 702t 224st + 8s2t 122t2) + (126 + 294t2 +804t + 120s2 355s 336st)r + ( 370s 120t + 366 112st +120s2)r2 + (40s2+ 156t 330s + 570)r3+ ( 295s + 378t + 510)r4 +(258 115s)r5 + 66r6+ 6r7,
k3(r; s; t) = ( 125s + 40s2+ 268t 168st + 30 + 294t2) + ( 236s 80t + 80s2 +176 112st)r + ( 258s + 156t + 40s2+ 414)r2+ (496 308s +504t)r3+ (314 161s)r4+ 96r5+ 10r6,
k4(r; s; t) = (36 61s 56st + 34t + 30s2) + (172 148t 103s + 30s2)r is strictly decreasing with respect to s on [0;p
3].
By (6.16)–(6.20), we compute that
@2Ke"
By (6.16)–(6.19) and (6.22), we compute that
~ decreasing with respect to s on [0;p
3].
Step 4. We show eK"(r; s; t; y) > 0 for any …xed y 2 (0; r + 1), (r; t) 2 [0;1)
In order to prove (6.28), we claim that for any …xed y 2 (0; r + 1) and (r; t) 2 ,
eh6(r; t) = 54r9+ 662r8+ (3282 713p
As polynomials of r, it is easy to check that the coe¢ cients of ehn(r; t) are all positive for n2 f0; 1; 2; ; 9gn f1g, where t 2 [107; 1]. So for any …xed y 2 (0; r + 1) and (r; t) 2 ,
ehn(r; t) > 0; n2 f0; 1; 2; ; 9gn f1g :
Note that for n = 1 and t 2 [107; 1], it is easy to check that the coe¢ cients of eh1(r; t) are all positive besides that of r3. By (6.31), we obtain that
eh1(r; t)
and
Step 5. Finally, we prove the lemma by the above analyses.
By Step 4 and (6.20) in Step 2, we have K"( ; u) > 0for any …xed " 2 [(107(273))1=2; (273)1=2],
So Steps 1–5 complete the proof of Lemma 3.4.