# Nonexistence of Global Weak Solutions of Nonlinear Keldysh Type Equation with One Derivative Term

Authors: Kangqun Zhang
Journal: Advances in Mathematical Physics
 (2018)

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## Abstract

We focus on the nonexistence of global weak solutions of nonlinear Keldysh type equation with one derivative term. In terms of the analysis of the first Fourier coefficient, we show the solution of singular initial value problem and singular initial-boundary value problem of the nonlinear equation with positive initial data blow-up in some finite time interval.


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## Body

## 1. Introduction

In this paper we consider the nonexistence of global weak solution of nonlinear Keldysh type equation (1) with modified initial data (2) and boundary value (3); that is,(1)t∂t2u-△u+b∂tu=fu,in0,T×Ω(2)u0,x=φ1x,limt→0tb∂tut,x=φ2xinΩand(3)ut,x=0on0,T×∂Ωwhere △=∑i=1n∂xi2 is the Laplace differential operator and b∈(0,1) is a parameter. Here (1)-(2) is a singular initial value problem for Ω=Rn. And (1)-(3) is a singular initial-boundary value problem when Ω⊆Rn is a bounded domain with a piecewise smooth boundary ∂Ω. The singularity of the given second datum was analyzed by T. Hristov in [1]. In order to formulate the mechanism of blow-up, we assume that, for s>s0 with a positive constant s0, the nonlinear source term f(s) is a convex function such that(4)fs-μs≥cs1+p,p>0with some positive constants μ and c.The Keldysh type equations are known in some specific applications in plasma physics, optics, and analysis on projective spaces [2]. In one-dimensional space, homogeneous equation of (1) turns into the classic Keldysh equation which is closely connected with flow dynamics [3, 4]. In this case, Chen S.X. [5] established the fundamental solutions to the corresponding differential operator for b<1/2 by using the finite part of divergence integrals in theory of distribution. For n=3, the existence of singular solution to Protter problem and Protter-Morawetz problem was considered in [6, 7].In the upper half space, (1) is a hyperbolic equation with variable propagation speed and characteristics degenerate on t=0. For nonlinear wave equation with constant coefficients, there exist many well-known results including existence and blow-up of solutions. For the existence of solutions, one can see [8–11]. Here we emphasize the nonexistence of global solution; then we will enumerate some classical results. In [12] for n=3, Keller used a comparison theorem and Riemann function showed the solution is blow-up in finite time. In [13], Jörgens extended Keller’s idea to the case for nonlinear wave equation with n>3. For more results, one can see [14–18]. Similar idea of the proof also has been used to establish the blow-up of solution of nonlinear parabolic equations by Kaplan [19] and Fujita [20, 21].In this paper, we use the first Fourier coefficient of the solution to show the blow-up of solution. This method was first introduced by Kaplan [19] to study a nonlinear second-order parabolic equation, later modified by Glassey [14] for classical solutions of nonlinear wave equations and extended by Levine [22, 23] to weak solutions of parabolic and wave equations. In order to show the first Fourier coefficient of the solution is positive and approaches infinity at finite time, we derive a nonlinear second-order singular differential inequality:(5)ddttbF′t≥ctb-1F1+pt.Then, based on the cautious calculation of singular differential inequality (5), we can deal with the nonexistence of global solution to the nonlinear problem and the proof is independent of the weak conservation of energy law and concavity of argument.This paper is organized as follows: In Section2, the blow-up result for singular initial-boundary value problem is obtained. In Section 3, based on a similar method used in Section 2 but with some differences in calculus, we show the blow-up of solution to singular initial value problem.


## 2. Singular Initial-Boundary Value Problem

In this section, we consider the blow-up of solution of problem (1)-(3) on a bounded domain Ω. The method of proof is independent of the spatial dimension and the Riemann function.Letψ(x) denote the first eigenfunction of Dirichlet problem:(6)△ψ+λψ=0,inΩψx=0on∂Ωwith the first eigenvalue λ. According to a classical result ([24], Vol.I, pp. 451-455), then there is only one normalized eigenfunction ψ(x)>0 satisfying(7)∫Ωψxdx=1.Now we consider problem (1)-(3) in bounded domain Ω with smooth boundary ∂Ω.Assume that(a)
φ1(x)≥0, φ2(x)≥0 for all x∈Ω; there exist two points xi∈Ω, i=1,2, such that φ1(x1)>0, φ2(x2)>0;(b)
∫Ωψ(x)φ1(x)dx=α>0, ∫Ωψ(x)φ2(x)dx=β>0.Set(8)ϕt,·,ut,·=∫Ωϕt,xut,xdx.Definition 1.
The functionu(t,x) is called a weak solution of initial-boundary value problem (1)-(3) if u(t,·) and tb∂tu(t,·) are continuous with respect to t, if u satisfies (2) and (3), and if for all ϕ:(0,T)×Ω→R which are twice continuously differentiable and satisfies(9)tϕt,·,∂tut,·=∫0t1-bϕτ,·,∂τuτ,·+τ∂τϕτ,·,∂τuτ,·+△ϕτ,·,uτ,·+ϕτ,·,fuτ,·dτ.Theorem 2.
Assume that a weak solution of problem (1)-(2) with μ≥λ in (4) under assumptions (a)-(b) in the sense of Definition 1 exists; then there is some positive constant T such that(10)limt→T-ut,·L∞Ω=+∞,(11)limt→T-ut,·LqΩ=+∞,q∈1,+∞.Proof.
Forψ(x) defined in (6)-(7), set(12)Ft=ψx,ut,x=∫Ωψxut,xdx,and then F(0)=α>0 and(13)F′t=ψx,∂tut,x=∫Ωψx∂tut,xdxwith limt→0tbF′(t)=β>0. Since there is no differentiability of second order on the weak solution u with respect to t in Definition 1, then we will derive the second order of F(t) by use of (9). Multiplying t on (13), in terms of (8) and (9), we obtain(14)F′′t=t-1tF′t′=-t-1F′t+t-1tF′t′=t-1-bψx,∂tut,x+ψx,fut,x-λψx,ut,x.According to the convexity of f(·) and Jensen’s inequality, one has(15)ψx,fut,x≥fψx,ut,x=fFt.Then substitute (15) into (14), together with (4) for λ=μ; we obtain(16)F′′t≥t-1-bψx,∂tut,x+fψx,ut,x-λψx,ut,x≥-bt-1F′t+ct-1F1+ptwhich is equivalent to(17)ddttbF′t≥ctb-1F1+p.
Note thatF(0)=α>0 and limt→0tbF′(t)=β>0; then there exists a positive number η such that F(t)>α for t∈(0,η). Hence (17) means that tbF′(t) is strictly increasing on the interval (0,η). Multiply tbF′(t) on (15) and then integrate it over (ε,t) such that (ε,t)∈(0,η),(18)tbF′t2≥εbF′ε2+2c∫εtτ2b-1F1+pτF′τdτ.It follows that F′(t)>βt-b and F(t)>α on the entire existence interval (0,η).
Now, forε>0, we set(19)Ht=∫εtτ2b-1F1+pτF′τdτ,(20)Gt=2+p-1F2+pt.
On one hand, it is easy to verify thatH′(t)=t2b-1G′(t) holds and H(t)>0, G(t)>0. For b≥1/2, integrating by parts, it follows that(21)Ht≤t2b-1Gt-ε2b-1Gε+2b-1∫εtτ2b-1Gτdτ≤t2b-1Gt.On the other hand, in terms of (17)-(20), we derive that(22)H′t2=t2b-2F2+2pttbF′t2≥2+p2+2pt2b-1εbF′ε2+2cHtG1+p/2+pt.Then, substituting (21) into (22), we obtain(23)H′t2≥2+p2+2pt-1+p1-2b/2+pεbF′ε2+2cHtH1+p/2+pt,which is equivalent to(24)εbF′ε2+2cHtH1+p/2+pt-1/2H′t≥2+p1+pt-1/2+p1-2b/22+p.
Forb∈(1/2,1), integrate inequality (24) over (ε,t); then(25)+∞>∫HεHtεbF′ε2+2cHτ-1/2H-1/2-p/22+pτdHτ≥∫εt2+p1+pτ-1/2+p1-2b/22+pdτ≥12+p1-2b22+p-12+p1+pt1/2+p1-2b/22+p-ε1/2+p1-2b/22+p.Fix ε; inequality (25) implies that there is a singularity of H(t) at some point T≤T0, where(26)T0=ε1/2+p1-2b/22+p+12+p1-2b22+p2+p-1-p×∫Hε+∞εbF′ε2+2ct-1/2t-1/2-p/22+pdt1/1/2+p1-2b/22+p.Then, for some t0≤T0, we derive(27)limt→T-Ht=+∞.Moreover, by a direct computation on (21), we have(28)limt→T-Ht≤T2b-1limt→T-Gt.Hence, combining (20) through (27) and (28), we confirm that(29)limt→T-Ft=+∞.
Finally, because ofψ(x)>0, in terms of Hölder’s inequality, we have(30)Ft=Ft=ψx,ut,x=ψxut,xL1Ω≤ψxLpΩut,xLqΩwhich implies (11). In particular, since ψ(x) satisfies (7), then, from (30), we obtain(31)Ft<ut,xL∞Ω.This derives (10).
Next, we consider the case forb∈[0,1/2). Therefore 2b-1<0. Then, for 0<ε≤τ≤t, we have t2b-1≤τ2b-1≤ε2b-1. Since c>0, p>0, F(t)>0, and F′(t)>0 on the entire existence interval from (18), it follows that(32)tbF′t2≥t2b-1εF′ε2+2c∫εtF1+pτF′τdτ.This is equivalent to(33)F′t2≥t-1εF′ε2+2c2+pF2+pt-F2+pε.Note F(t) is increasing; then(34)εF′ε2+2c2+pF2+pt-F2+pε-1/2F′t≥t-1/2.
Fixε>0 and for t>ε and integrate (34) on [ε,t]; it follows that(35)∫FεFtεF′ε2+2c2+py2+p-F2+pε-1/2dy≥2t1/2-ε1/2>0.
SetT1=∫α+∞[ε(F′(ε))2+2c/2+p(y2+p-F2+p(ε))]-1/2dy; then (35) implies that there exists some point T<T1 such that(36)limt→T-Ft=+∞.Hence, there is T∈(0,T1) such that (30) holds for b∈(0,1/2). Then (10) and (11) are yielded by use of Ho¨lder’s inequality.
Finally, we complete the proof of Theorem2.


## 3. Singular Initial Value Problem

In this section we show the blow-up of solution of singular initial value problem (1)-(2) with the idea of proof is similar to that used in Section 2, but it is more complicated in calculus.Letψ(x) denote the first eigenfunction of Dirichlet problem (6); then there is only one eigenfunction ψ(x)>0 satisfying (7). As shown in [24], set Sr={x:x<r}=Ω in (6) and denote ψr(x) as the eigenfunction corresponding to the first eigenvalue λr; then ψr(x) is also the function of |x| only and λr decrease with r.Assume that(c)
nonnegative functionsφ1(x), φ2(x) have support in SR-δ for some positive constant R and δ∈(0,R); there exists xi∈SR-δ,i=1,2 such that φ1(x1)>0 and φ2(x2)>0;(d)
∫SR-δψr(x)φ1(x)dx=α>0, ∫SR-δψr(x)φ2(x)dx=β>0.Set(37)ϕt,·,ut,·=∫Rnϕt,xut,xdx.Taking a similar procedure of the proof of(36) and (46) in [25] with the principle of superposition, we can obtain a solution of problem (1)-(2) which is the representation at a point (t,x) as a solution of the homogeneous linear equation plus a Volterra integral (formed by confluent hypergeometric function; see [26]) of the inhomogeneous term against an appropriate kernel over the retrograde cone through (t,x). Then we give the following definition which is similar to definition 3 in [23].Definition 3.
One says that a solution of problem (1)-(2) has a cone compact support property if (for any R) whenever φi(x),i=1,2 has support in {x∈Rn∣x<R}, u(t,·) has support in {x∈Rn∣|x|<R+γ(t)} for all t∈(0,T), where(38)γt=2t.Now fixr>R and the value will be given later. Let t≤γ-1(r-R); then u(t,x) given in Definition 3 has support in SR-δ+γ(t)×{t≤γ-1(r-R)}, which is contained in Sr-δ×{t≤γ-1(r-R)}.Definition 4.
The functionu(t,x) is called a weak solution of problem (1)-(2) which enjoys the cone compact support property, if u(t,·) and tb∂tu(t,·) are continuous with respect to t, if u satisfies (2), and if for all ϕ:SR-δ×(0,T)→R which are twice continuously differentiable and satisfies(39)tϕt,·,∂tut,·=∫0t1-bϕτ,·,∂τuτ,·+τ∂τϕτ,·,∂τuτ,·+△ϕτ,·,uτ,·+ϕτ,·,fuτ,·dτ.Theorem 5.
Assume that a weak solution of problem (1)-(2) with μ≥λ in (4) under conditions (c)-(d) in the sense of Definition 4 exists; then there is some positive constant T such that(40)limt→T-ut,·L∞Rn=+∞.(41)limt→T-ut,·LqRn=+∞,q∈1,+∞.Proof.
Define a functionΨr(x); that is,(42)Ψrx=ψrx,x≤r-δ2ψ0x,r-δ2≤x≤r0,x≥rwhere ψr is first eigenfunction of Dirichlet problem (6); ψ0 is a twice continuously differentiable function of ρ=x which satisfies ψ0i(r-δ/2)=ψri(r-δ/2) and ψ0i(r)=0 for i=0,1,2. When t<γ-1(r-R), substitute ϕ(t,x) by Ψr(t,x) in (39); then it is easy to verify that(43)tψr·,∂tut,·=∫0t1-bψr·,∂τuτ,·+△ψr·,uτ,·+ψr·,fuτ,·dτ.since the support of u(t,x) is contained in Sr-δ×[0,γ-1(r-R)).
Fort≤γ-1(r-R), set(44)Frt=ψrx,ut,x=∫Rnψrxut,xdx,and then, by the assumption,(45)Frt=∫Br-δψrxut,xdx,(46)Fr′t=ψrx,∂tut,x=∫Br-δψrx∂tut,xdx.In terms of (43) and (46), it follows that(47)tFr′t=∫0t1-bψr·,∂τuτ,·+△ψr·,uτ,·+ψr·,fuτ,·dτ.Therefore, we obtain(48)Fr′′t=t-1tFr′t′=t-1-bψrx,∂tut,x+ψrx,fut,x-λrψrx,ut,x,which is equivalent to(49)ddttbF′t≥ctb-1ψrx,fut,x-λrψrx,ut,x.
In terms of the convexity off(·) and Jensen’s inequality, one has(50)ψrx,fut,x=∫Rnψrxfut,xdx≥f∫Rnψrxut,xdx=fFrt.Then substituting (50) into (49), together with (4) for λr=μ, we obtain(51)ddttbF′t≥ctb-1Fr1+pt.In fact, if set f(s)=s1+p, then we can choose R such that λr<αp for Dirichlet problem (6) in Sr, and then (51) holds which is independent of (4).
Note thatFr(0)=α>0 and limt→0tbFr′(0)=β>0; then there exists an interval (0,η) such that Fr(t)>α for t∈(0,η). Hence, on (0,η),(52)ddttbFr′t≥ctb-1Fr1+pt>0.This means that tbFr′(t) is strictly increasing on (0,η). Then we obtain Fr′(t)>βt-b>0 on (0,η). Multiply tbFr′(t) on (51) and then integrate it over (ε,t) where (ε,t)∈(0,η):(53)tbFr′t2≥εbFr′ε2+2c∫εtτ2b-1Fr1+pτFr′τdτ.It follows that Fr′(t)>βt-b and Fr(t)>α on the entire existence interval.
Now, forε>0, we set(54)Hrt=∫εtτ2b-1Fr1+pτFr′τdτ,(55)Grt=2+p-1Fr2+pt.Then, there, Hr′(t)=t2b-1Gr′(t) holds and Hr(t)>0, Gr(t)>0. For b∈[1/2,1), integrating by parts, it follows that(56)Hrt≤t2b-1Grt-ε2b-1Grε+2b-1∫εtτ2b-1Grτdτ≤t2b-1Grt.
Combining to (51)-(56), we derive that(57)Hr′t2≥2+p2+2pt-1+p1-2b/2+pεbFr′ε2+2cHrtHr1+p/2+pt,which is equivalent to(58)εbFr′ε2+2cHtHr1+p/2+pt-1/2Hr′t≥2+p1+pt-1/2+p1-2b/22+p.
Then, forb∈[1/2,1), integrating (58) over (ε,t), we obtain(59)∫HrεHrtεbFr′ε2+2cHrτ-1/2Hr-1/2-p/22+pτdHrτ≥∫εt2+p1+pτ-1/2+p1-2b/22+pdτ≥2+p1+pt1/2+p1-2b/22+p-ε1/2+p1-2b/22+p.According to the previous analysis, we have εbFr′(ε)>β>0 independent of r; then(60)∫HrεHrtεbFr′ε2+2cHrτ-1/2Hr-1/2-p/22+pτdHrτ≤∫HrεHrtεbFr′ε2+2cτ-1/2τ-1/2-p/22+pdτ<∫β+∞εbFr′ε2+2ct-1/2t-1/2-p/22+pdt.Set K0=∫β+∞(2ct)-1/2t-1/2-p/2(2+p)dt, in terms of (59)-(60); then K0 is a finite positive number independent of r and ε, and(61)2+p1+pt1/2+p1-2b/22+p-ε1/2+p1-2b/22+p<K0.
Now fixε; we choose r=r0,(62)r0=K02+p-1-p+ε1/2+p1-2b/22+p1/1/2+p1-2b/22+p+2R,and then we obtain(63)t<r0-2R<r0-Rfor all t∈(ε,η)∩(0,r0-R), where (0,η) is the existence interval of solution (1)-(2). Hence there is a T≤r0-2R, such that(64)limt→T-Hr0t=+∞.From (55)-(56) and (64), we obtain(65)limt→T-Fr0t=+∞.At last, by use of Hölder inequality on (65), we derive (40) and (41) for b∈[1/2,1).
Forb∈(0,1/2), we also obtain (35) with the term F being substituted by Fr by a procedure as mentioned in Section 2; that is,(66)∫FrεFrtεFr′ε2+2c2+py2+p-Fr2+pε-1/2dy≥2t1/2-ε1/2>0.Since we do not know the sign of the term ε(Fr′(ε))2-2c/2+pFr2+p(ε), we calculate(67)∫FrεFrtεFr′ε2+2c2+py2+p-Fr2+pε-1/2dy≤2c2+p-1/2∫Frε+∞y2+p-Fr2+pε-1/2dy=2c2+p-1/2Fr-p/2ε∫1+∞z2+p-1-1/2dz<2c2+p-1/2α-p/2∫1+∞z2+p-1-1/2dz<+∞where we use p>0 and Fr(ε)>α.
LetK1≔(2c/2+p)-1/2α-p/2∫1+∞(z2+p-1)-1/2dz which is a finite positive constant independent of r and ε; then(68)K1>2t1/2-ε1/2.Fixing ε, we choose r=r1:(69)r1=ε1/2+12K12+2R,and then it follows from (68) that(70)t<r1-2R<r1-Rfor all t∈(ε,η)∩(0,r1-R), where (0,η) is the existence interval of solution (1)-(2). Hence there is T≤r1-2R, such that(71)limt→T-Hr1t=+∞.By a computation with (55)-(56) and (71), we obtain(72)limt→T-Fr1t=+∞.Hence, in terms of Hölder inequality, the last formula implies (40) and (41) for b∈(0,1/2).
Then we conclude that Theorem5 holds.



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