In this section, we will give some conditions to guarantee the finite time blowup. Motivated by Mckean’s deep observation for the Camassa-Holm equation [7], we can consider the similar particle trajectory as
(5)qt=bu(q,t), 0<t<T, x∈ℝ,q(x,0)=x, x∈ℝ,
where T is the lifespan of the solution: then q is a diffeomorphism of the line. Taking derivative (5) with respect to x, we obtain
(6)dqtdx=qtx=bux(q,t)qx, t∈(0,T).
Therefore
(7)qx(x,t)=exp{b∫0tux(q,s)ds}, qx(x,0)=1.
Hence, from (2), the following identity can be proved:
(8)y(q)qxa/b=y0(x)e-λt.
In fact, direct calculation yields
(9)ddt(y(q)qxa/b)=[yt(q)+bu(q,t)yx(q) +aux(q,t)y(q)]qxa/b=-λyqxa/b.

Proof.
Suppose that the solution exists globally. From (8) and initial condition (10), we have y(q(x0,t),t)=0 and
(12)y(q(x,t),t)≥0(≢0), for x∈(-∞,q(x0,t)),y(q(x,t),t)≤0(≢0), for x∈(q(x0,t),∞),
for all t≥0. Due to u(x,t)=G*y(x,t), we can write u(x,t) and ux(x,t) as
(13)u(x,t)=12e-x∫-∞xeξy(ξ,t)dξ+12ex∫x∞e-ξy(ξ,t)dξ,ux(x,t)=-12e-x∫-∞xeξy(ξ,t)dξ+12ex∫x∞e-ξy(ξ,t)dξ.
Therefore,
(14)ux2(x,t)-u2(x,t)=-∫-∞xeξy(ξ,t)dξ∫x∞e-ξy(ξ,t)dξ,
for all t>0.

By direct calculation, for x≤q(x0,t), we have
(15)ux2(x,t)-u2(x,t) =-∫-∞xeξy(ξ,t)dξ∫x∞e-ξy(ξ,t)dξ =-(∫-∞q(x0,t)eξy(ξ,t)dξ-∫xq(x0,t)eξy(ξ,t)dξ) ×(∫q(x0,t)∞e-ξy(ξ,t)dξ+∫xq(x0,t)e-ξy(ξ,t)dξ) =ux2(q(x0,t),t)-u2(q(x0,t),t) -∫-∞xeξy(ξ,t)dξ∫xq(x0,t)e-ξy(ξ,t)dξ +∫xq(x0,t)eξy(ξ,t)dξ∫q(x0,t)∞e-ξy(ξ,t)dξ ≤ux2(q(x0,t),t)-u2(q(x0,t),t).
Similarly, for x≥q(x0,t), we have
(16)ux2(x,t)-u2(x,t)≤ux2(q(x0,t),t)-u2(q(x0,t),t).
So for any fixed t, combination of (15) and (16), we obtain
(17)ux2(x,t)-u2(x,t)≤ux2(q(x0,t),t)-u2(q(x0,t),t),
for all x∈ℝ.

From the expression of ux(x,t) in terms of y(x,t), differentiating ux(q(x0,t),t) with respect to t, we have
(18)∂tux(q(x0,t),t) =uxt(q(x0,t),t)+uxx(q(x0,t),t)qt(q(x0,t),t) =a2u2(q(x0,t),t)+a-b2ux2(q(x0,t),t) -λux(q(x0,t),t)-G*(a2u2(x,t)+3b-a2ux2(x,t)) =G*(a2u2(q(x0,t),t)+a-b2ux2(q(x0,t),t) -a2u2(x,t)-3b-a2ux2(x,t))-λux(q(x0,t),t) =G*(a-2b2(u2(q(x0,t),t)-ux2(q(x0,t),t) -u2(x,t)+ux2(x,t))a-2b2) +G*(bu2(q(x0,t),t)-b2ux2(q(x0,t),t) -bu2(x,t)-b2ux2(x,t))-λux(q(x0,t),t) ≤b2u2(q(x0,t),t)-b2ux2(q(x0,t),t)-λux(q(x0,t),t),
where we have used (17), and the inequality G*(u2(x,t)+(1/2)ux2(x,t))≥(1/2)u2. In addition, we also used the equation utx+uuxx-(a/2)u2-((b-a)/2)ux2+G*((a/2)u2+((3b-a)/2)ux2)+λux=0, which is obtained by differentiating equation (3).

For (11), we know that
(19)(u0x(x0)+λb)2-(u0(x0)+λb)2 =-e-x0∫-∞x0eξy0(ξ)dξ×(ex0∫x0∞e-ξy0(ξ)dξ+2λb)>0,(u0x(x0)+λb)2-(u0(x0)-λb)2 =-ex0∫x0∞e-ξy0(ξ)dξ×(e-x0∫-∞x0eξy0(ξ)dξ-2λb)>0.

Claim.
u
x
(
q
(
x
0
,
t
)
,
t
)
<
0
is decreasing. (u(q(x0,t),t)+λ/b)2<(ux(q(x0,t),t)+λ/b)2 and (u(q(x0,t),t)-λ/b)2<(ux(q(x0,t),t)+λ/b)2, for all t≥0.

Suppose that there exists a t0 such that (u(q(x0,t),t)+λ/b)2<(ux(q(x0,t),t)+λ/b)2 and (u(q(x0,t),t)-λ/b)2<(ux(q(x0,t),t)+λ/b)2 on [0,t0); then (u(q(x0,t0),t0)+λ/b)2=(ux(q(x0,t0),t0)+λ/b)2 or (u(q(x0,t0),t0)-λ/b)2=(ux(q(x0,t0),t0)+λ/b)2.

Now, let
(20)I(t)∶=12e-q(x0,t)∫-∞q(x0,t)eξy(ξ,t)dξ,II(t)∶=12eq(x0,t)∫q(x0,t)∞e-ξy(ξ,t)dξ.
Firstly, differentiating I(t), we have
(21)dI(t)dt=-b2u(q(x0,t),t)e-q(x0,t)∫-∞q(x0,t)eξy(ξ,t)dξ +12e-q(x0,t)∫-∞q(x0,t)eξyt(ξ,t)dξ=b2u(ux-u)(q(x0,t),t)-12e-q(x0,t) ×∫-∞q(x0,t)eξ(a-2b2buyx+2buxy +a-2b2(u2-ux2)x+λy)dξ≥b2u(ux-u)(q(x0,t),t)+b4(u2+ux2-2uux) ×(q(x0,t),t)-λ2(u-ux)(q(x0,t),t)=b4(ux2-u2)(q(x0,t),t)-λ2(u-ux)(q(x0,t),t)=b4(ux(q(x0,t),t)+λb)2 -b4(u(q(x0,t),t)+λb)2>0, on[0,t0).
Secondly, by the same argument, we obtain
(22)dII(t)dt=b2u(q(x0,t),t)eq(x0,t)∫q(x0,t)∞e-ξy(ξ,t)dξ +12eq(x0,t)∫q(x0,t)∞e-ξyt(ξ,t)dξ=b2u(ux+u)(q(x0,t),t)-12eq(x0,t) ×∫q(x0,t)∞e-ξ(a-2b2buyx+2buxy + a-2b2(u2-ux2)x+λy)dξ≤b2u(ux+u)(q(x0,t),t) -b4(u2+ux2+2uux)(q(x0,t),t) -λ2(ux+u)(q(x0,t),t)=-b4(ux2-u2)(q(x0,t),t) -λ2(ux+u)(q(x0,t),t)=-b4(ux(q(x0,t),t)+λb)2 +b4(u(q(x0,t),t)-λb)2<0, on[0,t0).
Therefore, it follows from (21), (22), and the continuity property of ODEs that
(23)(ux(q(x0,t),t)+λb)2-(u(q(x0,t),t)+λb)2 =-4I(t)(II(t)+λb)>-4I(0)(II(0)+λb)>0,(ux(q(x0,t),t)+λb)2-(u(q(x0,t),t)-λb)2 =-4(I(t)-λb)II(t)>-4(I(0)-λb)II(0)>0,
for all t>0. This implies that t0 can be extended to the infinity.

Moreover, using (21) and (22) again, we have the following equation for [2(ux+λ/b)2-(u+λ/b)2-(u-λ/b)2](q(x0,t),t):
(24)ddt[2(ux+λb)2-(u+λb)2-(u-λb)2](q(x0,t),t) = -4ddt[I(t)(II(t)+λb)]-4ddt[(I(t)-λb)II(t)] ≥-b[(ux+λb)2-(u+λb)2](q(x0,t),t)(II(t)+λb) +b[(ux+λb)2-(u-λb)2](q(x0,t),t)I(t) -b[(ux+λb)2-(u+λb)2](q(x0,t),t)II(t) +b[(ux+λb)2-(u-λb)2](q(x0,t),t)(I(t)-λb) =b(-λb[2(ux+λb)2-(u+λb)2-(u-λb)2] ×(q(x0,t),t)-ux(q(x0,t),t) ×(2(ux(q(x0,t),t)+λb)2) +2(u+λb)2II(t)-2(u-λb)2I(t)[2(ux+λb)2-(u+λb)2-(u-λb)2]) =b((u-λb)2-(ux(q(x0,t),t)+λb) ×[2(ux+λb)2-(u+λb)2-(u-λb)2] ×(q(x0,t),t)-ux(q(x0,t),t) ×[(u+λb)2+(u-λb)2](q(x0,t),t) +2(u+λb)2II(t)-2(u-λb)2I(t)) ≥-b(ux(q(x0,t),t)+λb) ×[2(ux+λb)2-(u+λb)2-(u-λb)2](q(x0,t),t),
where we use ux(q(x0,t),t)=-I(t)+II(t).

Now, recalling (18), we have
(25)∂tux(q(x0,t),t)(q(x0,t),t) ≤b2u2(q(x0,t),t)-b2ux2(q(x0,t),t)-λux =b4[(u+λb)2+(u-λb)2-2(ux+λb)2] ×(q(x0,t),t).

Putting (25) into (24), it yields
(26)ddt[2(ux+λb)2-(u+λb)2-(u-λb)2](q(x0,t),t) ≥b24[2(ux+λb)2-(u+λb)2-(u-λb)2] ×(q(x0,t),t) ×(∫0t[2(ux+λb)2-(u+λb)2-(u-λb)2] ×(q(x0,τ),τ)dτ-4u0x(x0)-4λb∫0t[2(ux+λb)2-(u+λb)2-(u-λb)2]).
Before finishing the proof, we need the following technical lemma.

Lemma 3 (see [15]). Suppose that Ψ(t) is twice continuously differential satisfying(27)Ψ′′(t)≥C0Ψ′(t)Ψ(t), t>0, C0>0,Ψ(t)>0, Ψ′(t)>0.Then ψ(t) blows up in finite time. Moreover the blow-up time can be estimated in terms of the initial datum as(28)T≤max{2C0Ψ(0),Ψ(0)Ψ′(0)}.

Let Ψ(t)=∫0t[2(ux+λ/b)2-(u+λ/b)2-(u-λ/b)2](q(x0,τ),τ)dτ-4u0x(x0)-4λ/b; then (26) is an equation of type (27) with C0=b2/4. The proof is complete by applying Lemma 3.