For a class of semilinear parabolic equations under nonlinear dynamical boundary conditions in a bounded domain, we obtain finite time blow-up solutions when the initial data varies in the phase space H10(Ω) at positive initial energy level and get global solutions with the initial data at low and critical energy level. Our main tools are potential well method and concavity method.
Citation: Mingyou Zhang, Qingsong Zhao, Yu Liu, Wenke Li. Finite time blow-up and global existence of solutions for semilinear parabolic equations with nonlinear dynamical boundary condition[J]. Electronic Research Archive, 2020, 28(1): 369-381. doi: 10.3934/era.2020021
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For a class of semilinear parabolic equations under nonlinear dynamical boundary conditions in a bounded domain, we obtain finite time blow-up solutions when the initial data varies in the phase space H10(Ω) at positive initial energy level and get global solutions with the initial data at low and critical energy level. Our main tools are potential well method and concavity method.
Let p be a prime, f be a polynomial with k variable and Fp=Z/(p) be the finite field, where Z is the integer ring, and let
N(f;p)=#{(x1,x2,⋯,xk)∈Fkp|f(x1,x2,⋯,xk)=0}. |
Many scholars studied the exact formula (including upper bound and lower bound) for N(f;p) for many years, it is one of the main topics in the finite field theory, the most elementary upper bounds was given as follows (see [14])
N(f;p)≤pk−1degf. |
Let ordp denote the p-adic additive valuation normalized such that ordpp=1. The famous Chevalley-Warning theorem shows that ordpN(f;p)>0 if n>degf. Let [x] denote the least integer ≥x and e denote the extension degree of Fq/Fp. Ax (see [2]) showed that
ordpN(f;q)≥e[n−degfdegf]. |
In 1977, S. Chowla et al. (see [7]) investigated a problem about the number of solutions of a equation in finite field Fp as follow,
x31+x32+⋅⋅⋅+x3k≡0, |
where p is a prime with p≡1mod 3 and xi∈Fp, 1≤i≤k.
Let Mk denotes the number of solutions of the above equation. They proved that
M3=p2+d(p−1),M4=p2+6(p2−p),∞∑s=1Msxs=x1−px+x2(p−1)(2+dx)1−3px2−pdx3, |
where d is uniquely determined by 4p=d2+27y2 and d≡1 mod 3.
Myerson [12] extended the result in [2] to the field Fq and first studied the following equation over Fq,
x31+x32+⋅⋅⋅+x3k≡0. |
Recently J. Zhao et al. (see [17]) investigated the following equations over field Fp,
f1=x41+x42+x43,f2=x41+x42+x43+x44. |
And they give exact value of N(f1;p) and N(f2;p). For more general problem about this issue interested reader can see [6,9,10,11].
In this paper, let A(k,p) denotes the number of solutions of the following equation in Fp,
x61+x62+⋅⋅⋅+x6k≡0, |
where p is a prime with p≡1mod 3 and xi∈Fp, 1≤i≤k, and for simplicity, in the rest of this paper, we assume there exists an integer z such that z3≡2mod p, we use analytic methods to give a recurrence formula for the number of solutions of the above equation. And our method is based on the properties of Gauss sum. It is worth noting that we used a novel method to simplify the steps and avoid a lot of complicated calculations. We proved the following:
Theorem 1. For any positive integer k≥1, we have the recurrence formula
A(k+6,p)=5pA(k+4,p)+10dpA(k+3,p)+(46p2+5d2p+dp)A(k+2,p)+(2p2+120dp2+3d3p+d2p+dp)A(k+1,p)+(−4p5+2d2p4+64p7/2+381p3+2d2p5/2+(129d2+11d+6)p2+d4p)A(k,p)+pk+5−pk+4−(10dp+2d2)pk+3−64pk+5/2−(429+121d+5d2)pk+2−2d2pk+3/2−(3d3+130d2+12d+6)pk+1−d4pk, |
with the initial condition
A(1,p)=1,A(2,p)=4(p−1)+p,A(3,p)=10d(p−1)+p2,A(4,p)=56p(p−1)+10d2(p−1)+p3,A(5,p)=188dp(p−1)+5d3(p−1)+16dC(p)(p−1)+p4,A(6,p)=p5+1400p2(p−1)+(388d2+8d−576)p(p−1)+d2p−d2, |
where d is uniquely determined by 4p=d2+27y2 and d≡1 mod 3, and C(p)=p∑a=1ep(a3).
Remark. Our method is suitable to calculus the number of solutions of the following equation in Fp,
xt1+xt2+⋅⋅⋅+xtk≡0, |
where p satisfied a certain congruence conditions, and t is any nature number.
Our Theorem 2 can be deduced from Theorem 1 and the theory of the Difference equations.
Theorem 2. Let ti (1≤i≤k) be the real root of the below equation with multiplicity si (1≤i≤k) respectively, and ρje±iwj (1≤j≤h) be the complex root of the below equation with multiplicity rj (1≤j≤h) respectively,
x6=5px4+10dpx3+(46p2+5d2p+dp)x2+(2p2+120dp2+3d3p+d2p+dp)x+(−4p5+2d2p4+64p7/2+381p3+2d2p5/2+(129d2+11d+6)p2+d4p). |
We have
A(n,p)=k∑i=1si∑a=1Ciansi−atni+h∑j=1rj∑b=1Djbnrj−bρnjcosnwj+h∑j=1rj∑b=1Ejbnrj−bρnjsinnwj, |
where Cia,Djb,Ejb, are determined by
A(6,p)=k∑i=1si∑a=1Cia6si−at6i+h∑j=1rj∑b=1Djb6rj−bρ6jcos6wj+h∑j=1rj∑b=1Ejb6rj−bρ6jsin6wj,A(5,p)=k∑i=1si∑a=1Cia5si−at5i+h∑j=1rj∑b=1Djb5rj−bρ5jcos5wj+h∑j=1rj∑b=1Ejb5rj−bρ5jsin5wj,A(4,p)=k∑i=1si∑a=1Cia4si−at4i+h∑j=1rj∑b=1Djb4rj−bρ4jcos4wj+h∑j=1rj∑b=1Ejb4rj−bρ4jsin4wj,A(3,p)=k∑i=1si∑a=1Cia3si−at3i+h∑j=1rj∑b=1Djb3rj−bρ3jcos3wj+h∑j=1rj∑b=1Ejb3rj−bρ3jsin3wj,A(2,p)=k∑i=1si∑a=1Cia2si−at2i+h∑j=1rj∑b=1Djb2rj−bρ2jcos2wj+h∑j=1rj∑b=1Ejb2rj−bρ2jsin2wj,A(1,p)=k∑i=1si∑a=1Ciati+h∑j=1rj∑b=1Djbρjcoswj+h∑j=1rj∑b=1Ejbρjsinwj. | (1.1) |
Before we prove these lemmas, we give some notations, χ2 denotes the second-order character of Fp, χ denotes the third-order character of Fp, ψ denotes the sixth order character of Fp.
ep(x)=e2πixp,τ(χ)=p∑m=1χ(m)ep(m),G(χ,m)=p∑a=1χ(a)ep(am). |
We call G(χ,m) the Gauss sum, and we have the following:
G(χ,m)=τ(χ)¯χ(m),(m,p)=1. | (2.1) |
And also we have
|τ(χ)|=√p, | (2.2) |
where χ is a primitive character of Fp. And let G(m,6;p)=p−1∑a=0ep(ma6). For the property of the exponential sum and the general Gauss sum, interested readers can see [1,4,5,8,13,15].
Lemma 1. Let p be a prime with p≡1mod 3. Then for any third-order character χ of Fp, we have the identity
τ3(χ)+τ3(¯χ)=dp, |
where d is uniquely determined by 4p=d2+27y2 and d≡1mod 3.
Proof. For the proof of this lemma see [3].
Lemma 2. Let χ be a third-order character of Fp with p≡1mod 3, and C(p)=τ(χ)+τ(¯χ), then C(p)=p∑a=1ep(a3).
Proof.
A=τ(χ)+τ(¯χ)=p∑a=1(1+χ(a)+¯χ(a))e(ap)=p∑a=1e(a3p). |
Lemma 3. Let p≡1mod 6, 2≡z3mod p for some z, and let χ be a third-order character of Fp, ψ be a sixth-order character of Fp, then we have the identity
τ(ψ)=τ2(χ)√p. |
Proof. This is Lemma 3 in [16].
Lemma 4. As the definition above, we have the identity
G(m,6;p)=√pχ2(m)+¯x2√pψ(m)+x2√p¯ψ(m)+¯xχ(m)+x¯χ(m), |
where (m,p)=1 and x=τ(χ).
Proof. Firstly we have the identity
1+χ2(m)+χ(m)+¯χ(m)+ψ(m)+¯ψ(m)={6, if m≡a6 mod p;0,otherwise. |
So we have
G(m,6;p)=p−1∑a=0(1+χ2(a)+χ(a)+¯χ(a)+ψ(a)+¯ψ(a))ep(ma)=G(χ2,m)+G(ψ,m)+G(¯ψ,m)+G(χ,m)+G(¯χ,m) |
By (2.1) and Lemma 3, we have
G(m,6;p)=τ(χ2)χ2(m)+τ(¯ψ)ψ(m)+τ(ψ)¯ψ(m)+τ(¯χ)χ(m)+τ(χ)¯χ(m)=√pχ2(m)+¯x2√pψ(m)+x2√p¯ψ(m)+¯xχ(m)+x¯χ(m). | (2.3) |
By (2.3), we complete the proof of our lemma.
Next we let,
Gn(m,6;p)=an+bnχ2(m)+cnψ(m)+dn¯ψ(m)+enχ(m)+fn¯χ(m). | (2.4) |
Then we have following Lemma 5.
Lemma 5. Let an,bn,cn,dn,en,fn are defined as above, then we have that an,bn,cn,dn,en,fn are uniquely determined by n, where n≥1.
Proof. By the orthogonality of characters of Fp, we have
p−1∑a=1χ(a)={p−1, if χ=χ0;0,otherwise. | (2.5) |
By (2.4) and (2.5) we have
p−1∑m=1Gn(m,6;p)=(p−1)an+bnp−1∑m=1χ2(m)+cnp−1∑m=1ψ(m)+dnp−1∑m=1¯ψ(m)+enp−1∑m=1χ(m)+fnp−1∑m=1¯χ(m)=(p−1)an. |
So we have
an=1p−1p−1∑m=1Gn(m,6;p). | (2.6) |
By the same method, we have
bn=1p−1p−1∑m=1χ2(m)Gn(m,6;p),cn=1p−1p−1∑m=1¯ψ(m)Gn(m,6;p),dn=1p−1p−1∑m=1ψ(m)Gn(m,6;p),en=1p−1p−1∑m=1¯χ(m)Gn(m,6;p),fn=1p−1p−1∑m=1χ(m)Gn(m,6;p). |
So now it is easy to see the conclusion of the lemma.
Lemma 6. The sequences {an}, {bn}, {cn}, {dn}, {en}, {fn} are defined above, then they satisfied the following recurrence formulae (n≥0):
an+1=√pbn+¯x2√pdn+x2√pcn+xen+¯xfn, | (2.7) |
bn+1=√pan+¯x2√pen+x2√pfn+xdn+¯xcn, | (2.8) |
cn+1=√pfn+¯x2√pan+x2√pen+xbn+¯xdn, | (2.9) |
dn+1=√pen+¯x2√pfn+x2√pan+xcn+¯xbn, | (2.10) |
en+1=√pdn+¯x2√pcn+x2√pbn+xfn+¯xan, | (2.11) |
fn+1=√pcn+¯x2√pbn+x2√pdn+xan+¯xen, | (2.12) |
with the initial condition
a0=1,b0=c0=d0=e0=f0=0. |
Proof. We only prove (2.7), the rest can be proved in the same way. By Lemma 5, we know an is unique determined by n. We can compare the coefficient of the equation
Gn+1(m,6;p)=Gn(m,6;p)G(m,6;p). |
We have
an+1=√pbn+¯x2√pdn+x2√pcn+xen+¯xfn. |
So we complete the proof of the lemma.
Lemma 7. Let an is defined as above, then we have
a0=1,a1=0,a2=5p,a3=10dp,a4=56p2+10d2p,a5=188dp2+5d3p+16dpC(p). |
Proof. By Lemma 4 and after some elementary calculations we have
G2(m,6;p)=5p+2dp1/2χ2(m)+4p1/2xψ(m)+4p1/2¯x¯ψ(m)+(p−1¯x4+3x2)χ(m)+(p−1x4+3¯x2)¯χ(m),G3(m,6;p)=10dp+(16p3/2+dp1/2)χ2(m)+(15p¯x+2dx2+p−1x5)χ(m)+(15px+2d¯x2+p−1¯x5)¯χ(m)+(4p−1/2x4+12p1/2¯x2+2dp1/2x)ψ(m)+(4p−1/2¯x4+12p1/2x2+2dp1/2¯x)¯ψ(m),G4(m,6;p)=60p2+9d2p+dp+48dp3/2χ2(m)+(p−2x8+17¯x4+46px2+16dp)χ(m)+(p−2¯x8+17x4+46p¯x2+16dp)¯χ(m)+(56p3/2x+4dp−1/2x4+12dp1/2¯x2+8p−1/2¯x5)ψ(m)+(56p3/2¯x+4dp−1/2¯x4+12dp1/2x2+8p−1/2x5)¯ψ(m),G5(m,6;p)=188dp2+5d3p+16dpC(p)+(52d2p3/2+208p5/2+16dp1/2(x2+¯x2))χ2(m)+(p−2/5x10+p−3/2¯x8+4dp−1/2¯x5+71p1/2x4+(46p3/2+16p1/2)x2+(129p3/2+10d2p1/2)¯x2+60dp3/2x+16dp3/2)ψ(m)+(p−2/5¯x10+p−3/2x8+4dp−1/2x5+71p1/2¯x4+(46p3/2+16p1/2)¯x2+(129p3/2+10d2p1/2)x2+60dp3/2¯x+16dp3/2)¯ψ(m)+(8p−1¯x7+p−1x7+25x5+52dpx2+(28dp+46p2)x+16d¯x4+112p2¯x)χ(m)+(8p−1x7+p−1¯x7+25¯x5+52dp¯x2+(28dp+46p2)¯x+16dx4+112p2x)¯χ(m), |
and comparing the above formulae with (2.6), we have
a0=1,a1=0,a2=5p,a3=10dp,a4=60p2+9d2p+dp,a5=188dp2+5d3p+16dpC(p). |
Lemma 8. Let an, bn, cn, dn, en, fn are defined as above, then we have
a6=5pa4+10dpa3+(46p2+5d2p+dp)a2+(2p2+120dp2+3d3p+d2p+dp)a1+(−4p5+2d2p4+64p7/2+381p3+2d2p5/2+(129d2+11d+6)p2+d4p)a0b6=5pb4+10dpb3+(46p2+5d2p+dp)b2+(2p2+120dp2+3d3p+d2p+dp)b1+(−4p5+2d2p4+64p7/2+381p3+2d2p5/2+(129d2+11d+6)p2+d4p)b0c6=5pc4+10dpc3+(46p2+5d2p+dp)c2+(2p2+120dp2+3d3p+d2p+dp)c1+(−4p5+2d2p4+64p7/2+381p3+2d2p5/2+(129d2+11d+6)p2+d4p)c0d6=5pd4+10dpd3+(46p2+5d2p+dp)d2+(2p2+120dp2+3d3p+d2p+dp)d1+(−4p5+2d2p4+64p7/2+381p3+2d2p5/2+(129d2+11d+6)p2+d4p)d0e6=5pe4+10dpe3+(46p2+5d2p+dp)e2+(2p2+120dp2+3d3p+d2p+dp)e1+(−4p5+2d2p4+64p7/2+381p3+2d2p5/2+(129d2+11d+6)p2+d4p)e0f6=5pf4+10dpf3+(46p2+5d2p+dp)f2+(2p2+120dp2+3d3p+d2p+dp)f1+(−4p5+2d2p4+64p7/2+381p3+2d2p5/2+(129d2+11d+6)p2+d4p)f0 |
Proof. We only proof the first formula, the rest can be proof in the same way. By Lemma 6, we have
a6=√pb5+¯x2√pd5+x2√pc5+xe5+¯xf5=5pa4+2dp1/2b4+4p1/2¯xc4+4p1/2xd4+(3¯x2+p−1x4)e4+(3x2+p−1¯x4)f4=5pa4+10dpa3+(d2p1/2+12p3/2)b3+(2dp1/2¯x+8p1/2x2+p−1/2¯x4)c3+(2dp1/2x+8p1/2¯x2+p−1/2x4)d3+(11px+¯x2+p−1¯x5)e3+(11p¯x+x2+p−1x5)f3=5pa4+10dpa3+(46p2+5d2p+dp)a2+(25dp3/2+2p3/2)b2+(p−3/2¯x7+2p−1/2x5+p−1/2¯x4+42p3/2¯x+2dp1/2x2+(d2+1)p1/2¯x)c2+(p−3/2x7+2p−1/2¯x5+p−1/2x4+42p3/2x+2dp1/2¯x2+(d2+1)p1/2x)d2+(10x4+(32p+d2)¯x2+(4dp+p)x)e2+(10¯x4+(32p+d2)x2+(4dp+p)¯x)f2=5pa4+10dpa3+(46p2+5d2p+dp)a2+(2p2+120dp2+3d3p+d2p+dp)a1+(−4p5+2d2p4+64p7/2+381p3+2d2p5/2+(129d2+11d+6)p2+d4p)a0. |
So we complete the proof of this lemma.
Lemma 9. Let an is defined as above, then for any integer n≥0, we have
an+6=5pan+4+10dpan+3+(46p2+5d2p+dp)an+2+(2p2+120dp2+3d3p+d2p+dp)an+1+(−4p5+2d2p4+64p7/2+381p3+2d2p5/2+(129d2+11d+6)p2+d4p)an. |
Proof. By (2.4) and Lemma 8, we have
G6(m,6;p)=5pG4(m,6;p)+10dpG3(m,6;p)+(46p2+5d2p+dp)G2(m,6;p)+(2p2+120dp2+3d3p+d2p+dp)G(m,6;p)+(−4p5+2d2p4+64p7/2+381p3+2d2p5/2+(129d2+11d+6)p2+d4p). |
We multiple Gn(m,6;p) to the both side of the above formula, we have
Gn+6(m,6;p)=5pGn+4(m,6;p)+10dpGn+3(m,6;p)+(46p2+5d2p+dp)Gn+2(m,6;p)+(2p2+120dp2+3d3p+d2p+dp)Gn+1(m,6;p)+(−4p5+2d2p4+64p7/2+381p3+2d2p5/2+(129d2+11d+6)p2+d4p)Gn(m,6;p). |
By Lemma 5, we can compare the coefficient of the above equation, we have
an+6=5pan+4+10dpan+3+(46p2+5d2p+dp)an+2+(2p2+120dp2+3d3p+d2p+dp)an+1+(−4p5+2d2p4+64p7/2+381p3+2d2p5/2+(129d2+11d+6)p2+d4p)an. |
In the formula below, we always let k≥1. By the following formula,
p−1∑a=0ep(ma)={p, if p∣m;0, otherwise, |
we have
A(k,p)=1pp−1∑m=0p−1∑x1=0,x2=0,⋅⋅⋅,xk=0ep(m(x61+x62+⋅⋅⋅+x6k))=1pp−1∑m=0Gk(m,6;p). | (3.1) |
By (8), we have
A(k,p)=1pp−1∑m=0Gk(m,6;p)=1p(p−1∑m=1Gk(m,6;p)+pk)=1p((p−1)ak+pk)=p−1pak+pk−1. | (3.2) |
So by Lemma 9, we have
A(k+6,p)−pk+5=5p(A(k+4,p)−pk+3)+10dp(A(k+3,p)−pk+2)+(46p2+5d2p+dp)(A(k+2,p)−pk+1)+(2p2+120dp2+3d3p+d2p+dp)(A(k+1,p)−pk)+(−4p5+2d2p4+64p7/2+381p3+2d2p5/2+(129d2+11d+6)p2+d4p)(A(k,p)−pk−1). |
So we have
A(k+6,p)=5pA(k+4,p)+10dpA(k+3,p)+(46p2+5d2p+dp)A(k+2,p)+(2p2+120dp2+3d3p+d2p+dp)A(k+1,p)+(−4p5+2d2p4+64p7/2+381p3+2d2p5/2+(129d2+11d+6)p2+d4p)A(k,p)+pk+5−pk+4−(10dp+2d2)pk+3−64pk+5/2−(429+121d+5d2)pk+2−2d2pk+3/2−(3d3+130d2+12d+6)pk+1−d4pk. |
And by Lemma 7 and (3.2), we have the initial conditions
A(1,p)=1,A(2,p)=4(p−1)+p,A(3,p)=10d(p−1)+p2,A(4,p)=56p(p−1)+10d2(p−1)+p3,A(5,p)=188dp(p−1)+5d3(p−1)+16dC(p)(p−1)+p4.A(6,p)=p5+1400p2(p−1)+(388d2+8d−576)p(p−1)+d2p−d2. |
So we complete the proof of the theorem.
The main purpose of this paper is using analytic methods to give a recurrence formula of the number of solutions of an equation over finite field. And we give an expression of the number of solutions of the above equation by the root of sixth degree polynomial. We use analytic methods to give a recurrence formula for the number of solutions of the above equation. And our method is based on the properties of the Gauss sum. It is worth noting that we used a novel method to simplify the steps and avoid complicated calculations.
The author thanks to referees for very important recommendations and warnings which improved the paper.
The author declares that there is no competing interest.
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