By Oleinik's line method, we study the existence and the uniqueness of the classical
solution of the Cauchy problem for the following equation in [0,T]×R2: ∂xxu+u∂yu−∂tu=f(⋅,u), provided that T is suitable small. Results of numerical experiments
are reported to demonstrate that the strong solutions of the above
equation may blow up in finite time.

1. Introduction

We consider the following Cauchy problem:
∂xxu+u∂yu-∂tu=f(·,u),(t,x,y)∈(0,T]×R2,u(0,·)=u0(x,y),(x,y)∈R2.
This problem arises in financial mathematics recently; more and more mathematicians have been interested in it. In [1], Antonelli et al. introduced a new model for agents' decision under risk, in which the utility function is the solution to (1.1)-(1.2); they also proved, by means of probability methods, the existence of a continuous viscosity solution of (1.1)-(1.2), which satisfies
|u(x,y,t)-u(ξ,η,τ)|≤CT(|x-ξ|+|y-η|)
for every (x,y),(ξ,η)∈R2, t∈[0,T), under the assumption that f is uniformly Lipschitz continuous function. In [2], Citti et al. studied the interior regularity properties of this problem; they proved that the viscosity solutions are indeed classical solutions. On the other hand, Antonelli and Pascucci [3] showed that the solution u found in [1] can be also considered as a distributional solution.

However, all the above results are obtained when T is suitably small; say, the solution is local. The global weak solutions of the Cauchy problem for a more general class of equations, that contains (1.1), are obtained in [4–7], and so forth. This kind of solutions, however, is few regular and does not satisfy condition (1.3) in general.

In this paper, we will solve the Cauchy problem (1.1)-(1.2) in another simpler way and get the result as [2] again. Moreover, some examples are provided by numerical computation. The results of computation show that the strong solutions of the above equation may blow-up in finite time, though there exist the global weak solutions.

2. Line Method

In order to describe our method, we have to quote the well-known Prandtl system for a nonstationary boundary layer arising in an axially symmetric incompressible flow past a solid body, it has the form
∂tu+u∂xu+v∂yu=∂tU+U∂xU+∂y2u,∂x(ru)+∂y(rv)=0
in a domain D={0<t<T,0<x<X,0<y<∞}, where U(t,x) and r(x) are given functions. If we introduce the Crocco variables:
τ=t,ξ=x,η=u(t,x)U(t,x),
we obtain the following equation for w(τ,ξ,η)=∂yu/U:
w2wηη-wτ-ηUwξ+Awη+Bw=0.
Oleinik and Samokhin [8] had done excellent work in the boundary theory by the line method. Comparing this equation with (1.1), it is natural to conjecture that we are able to solve problem (1.1)-(1.2) by Qleinik's method.

Consider the following initial boundary problem:
wηη-wτ+wwξ=f(τ,ξ,η,w),w(0,ξ,η)=u0(ξ,η),
where u0∈C2(R2); its first-order derivatives and u0ηη are all bounded.

Definition 2.1.

A function w(τ,ξ,η) is said to be a solution of problem (2.4)-(2.5) if w has first-order derivatives in (2.4) which is continuous in (0,T]×R2, and its derivative wηη is continuous; w satisfies (2.4) in (0,T]×R2, together with condition (2.5).

The solution of problem (2.4)-(2.5) will be constructed as the limit of a sequence wn, n→∞, which consists of solutions of the equations
Ln(wn)=wηηn-wτn+wn-1wξn-f(·,wn)=0,wn(0,ξ,η)=w0(ξ,η).
As w0(τ,ξ,η) we take a function which is smooth in [0,T]×R2.

Suppose that for some nonnegative number p|f(·,v)|≤c(1+|v|p),
and when v1-v2≥0,
c1(v1-v2)≥f(·,v1)-f(·,v2)≥c2(v1-v2),max{|∂f∂τ|,|∂f∂ξ|,|∂f∂η||∂2f∂v2|}≤c.

Lemma 2.2.

Let V be a smooth function such that Ln(V)≥0 in (0,T)×R2, V≤wn for τ=0. Then V≤wn everywhere (0,T)×R2. Let V1 be a smooth function such that Ln(V1)≤0 in (0,T)×R2, V≥wn for τ=0. Then V1≥wn everywhere in (0,T)×R2.

Proof.

Let us prove the first statement of Lemma 2.2. The difference zn=wn-V satisfies the inequality
0≥Ln(zn)=Ln(wn)-Ln(V)=zηηn-zτn+wn-1zξn-(f(·,wn)-f(·,V)).

Let z1n=e-ατzn. Then
0≥z1ηηn-z1τn+αz1n+wn-1z1ξn-e-ατ(f(·,wn)-f(·,V))≥z1ηηn-z1τn-αz1n+wn-1z1ξn-c1z1n.
If we choose α large enough, by the maximal principle, we know V≤wn everywhere in (0,T)×R2.

Let us construct functions satisfying the conditions of Lemma 2.2. To this end, we define a twice continuously differentiable even function such that V1=(1-e-β|η|)eβτ for |η|>1, V1=φ(η)eβτ for |η|≤1, where φ(η) is a C2 function, |φηη|≤c.

When |η|>1,
Ln(V1)=V1ηηn-V1τ-wn-1V1ξn-f(·,V1)=-β12e-β1|η|eβτ-β(1-e-β1|η|)eβτ-f(·,V1)≤-β12e-β1|η|eβτ-β(1-e-β1|η|)eβτ+c(1-e-β1|η|)pepβτ+c<0
if we chose β large enough and βτ≤T0 small enough.

When |η|≤1,
Ln(V1)=φηηeβτ-βφeβτ-f(·,V1)≤φηηeβτ-βφeβτ+c(1+φpeβτp)<0
by the same reason.

Let V=ψ(η)e-ατ, α1>ψ(η)≥α0>0, |ψηη|≤c. Then
Ln(V)=ψηηe-ατ+αψeατ-f(·,V)≥ψηηe-ατ+αψeατ-c(1+ψpeατp)≥0
if we chose α large enough and ατ≤T0 small enough.

Similarly, we are able to prove the second statement of Lemma 2.2.

Thus we have the following.

Lemma 2.3.

Suppose that f satisfies (2.9) and V(0,ξ,η)≤w0≤V1(0,ξ,η), then
V≤wn≤V1.

The smooth functions V, V1 can be constructed as in [8], and we omit details here.

Let
Φn=Φ=(uτn)2+(uξn)2+(uηn)2,
where un=wn. We will show that there exist positive constants M and T such that the conditions Φμ≤M for τ≤T, μ≤n-1, imply that Φn≤M for τ≤T.

First, we rewrite (2.6) as
uηηn-uτ+un-1uξn-f(·,un)=0,(τ,ξ,η)∈(0,T]×R2.

Applying the operator 2uτn(∂/∂τ)+2uξn(∂/∂ξ)+2uηn(∂/∂η) to (2.17),
2uτnuτηηn+2uτn(uτn-1uξn+un-1uξτn)-2uτnuττn-2∂f∂u(uτn)2-2uτn∂f∂τ,2uξnuξηηn+2uξn(uξn-1uξn+un-1uξξn)-2uξnuτξn-2∂f∂u(uξn)2-2uξn∂f∂ξ,2uηnuηηηn+2uηn(wηn-1uξn+wn-1uξηn)-2uηnuτηn-2∂f∂u(uηn)2-2uηn∂f∂ξ=0,
then
un-1Φξ=(2uτnuτξn+2uξnuξξn+2uηnuηξn)un-1,-Φτ=-2uτnuττn-2uξnuξτn-2uηnuητn,Φηη=2(uτη)2+2uτnuτηηn+2(uξη)2+2uξnuξηηn+2(uηη)2+2uηnuηηηn,Φηη+un-1Φξ-Φτ-2∂f∂uΦ+2uτnuξnuξn-1+2(uξn)2uξn-1+2uηnuξnuηn-1-2uτn∂f∂τ-2uξn∂f∂ξ-2uηn∂f∂ξ=0.

By (2.9), (2.15), and Cauchy inequality, we are able to get
Φηη+un-1Φξ-Φτ+RnΦ≥0,
where Rn depends on un-1 and its derivatives are up to the second. Let Φ1=Φe-γτ with a positive constant γ to be chosen later. Then
Φ1ηη+un-1Φ1ξ-Φ1τ+(Rn-γ)Φ≥0
if we choose γ according to M such that Rn-γ≤-1. If Φ1 attains its positive maximum at τ=0, then
Φ1|τ=0=Φe-γτ|τ=0=Φ|τ=0≤c,
where the constant c does not depend on n. At the same time, the positive maximum of Φ1 in (0,T]×R2 cannot be attained by maximal principle. Thus we have
Φ1≤c.

So, if we let T1≤T small enough such that eγT1=2 and set M=2c, then
Φ≤ceγT1=M.

In order to estimate the second derivatives of un in [0,T1]×R2, consider the function
F=(uττn)2+(uξξn)2+(uηηn)2+(uτξn)2+(uξηn)2+(uτηn)2.
Applying the operator
P=2uττn∂2∂τ2+2uξξn∂2∂ξ2+2uηηn∂2∂η2+2uτξn∂2∂τ∂ξ+2uτηn∂2∂τ∂η+2uξηn∂2∂ξ∂η
to both sides of (2.17), we find that
0=2uττnuηηττn+2uττn(uττn-1uξn+2uτn-1uξτn+un-1uξττn)-2uττnuτττn+-2uττn(∂2f∂u2(uτn)2+∂f∂uuττn)+2uξξnuηηξξn+2uξξn(uξξn-1uξn+2uξn-1uξξn+un-1uξξξn)-2uξξnuτξξn-2uξξn(∂2f∂u2(uξn)2+∂f∂uuξξn)+2uηηnuηηηηn+2uηηn(uηηn-1uξn+2uηn-1uξηn+un-1uξηηn)-2uηηnuτξξn-2uηηn(∂2f∂u2(uηn)2-∂f∂uuηηn)+2uτξnuηητξn+2uτξn(uτξn-1uξn+uτn-1uξξn+uξn-1uξτn+un-1uξξτn)-2uτξnuττξn-2uτξn(∂2f∂u2uτnuξn+∂f∂uuτξn)+2uξηnuηηξηn+2uξηn(uξηn-1uξn+uξn-1uξηn+uηn-1uξξn+un-1uξξηn)-2uξηnuττξn-2uξηn(∂2f∂u2uηnuξn+∂f∂uuξηn)+2uτηnuηητηn+2uτηn(uτηn-1uξn+uτn-1uξηn+uηn-1uξτn+un-1uξτηn)-2uτηnuττηn-2uτηn(∂2f∂u2uηnuτn+∂f∂uuτηn).

At the same time, we can calculate that
Fη=2uττnuττηn+2uξξnuξξηn+2uηηnuηηηn+2uτξnuτξηn+2uξηnuξηηn+2uτηnuτηηn,Fηη=2(uττηn)2+2uττnuττηηn+2(uξξηn)2+2uξξnuηηξξn+2(uηηηn)2+2uηηnuηηηηn+2(uτξηn)2+2uτξnuτξηηn+2(uηξξn)2+2uξηnuξηηηn+2(uτηηn)2+2uτηnuτηηηn,un-1Fξ=un-1(2uττnuττξn+2uξξnuξξξn+2uηηnuηηξn+2uτξnuτξξn+2uξηnuξηξn+2uτηnuτηξn),-Fτ=-(2uττnuτττn+2uξξnuξξτn+2uηηnuηητn+2uτξnuτξτn+2uξηnuξητn+2uτηnuτητn),
and so we have
Fηη+un-1Fξ-Fτ-2∂f∂uF-2uττn(wττn-1uξn+2wτn-1uξτn)-2uξξn(uξξn-1uξn+2uξn-1uξξn)-2uηηn(uηηn-1uξn+2wηn-1uξηn)-2uτξn(uτξn-1uξn+uτn-1uξξn+uξn-1uξτn)-2uξηn(uξηn-1uξn+uξn-1uξηn+uηn-1uξξn)-2uτηn(uτηn-1uξn+uτn-1uξηn+uηn-1uξτn)-2uττn∂2f∂u2(uτn)2-2uξξn∂2f∂u2(uξn)2-2uηηn∂2f∂u2(uηn)2-2uτξn∂2f∂u2uτnuξn-2uξηn∂2f∂u2uηnuξn-2uτηn∂2f∂u2uηnuτn-2∂f∂u[(uτn)2+(uξn)2+(uηn)2]=0.
By the introduced assumption that the first-order and second-order derivatives of un-1, ∂f/∂u, and ∂2f/∂u2 are all bounded and using Cauchy inequality, we can get from (2.29) that
Fηη-2αFη-un-1Fξ-Fτ+R1nF≥0.
By the transformation F1=Fe-γτ, if we chose γ large enough, we are able to show that there exist positive constants M and T such that the conditions Fμ≤M for τ≤T, μ≤n-1, imply that Fn≤M for τ≤T. Thus we have the following.

Theorem 2.4.

Let wn be the solutions of problems (2.4)-(2.5), then the derivatives of wn up to the second-order are uniformly bounded with respect to n in the domain (0,T]×R2 with a small positive number T.

Now let us establish uniform convergence of wn=un in [0,T]×R2. For vn=wn-wn-1 we obtain the following equation from (2.6):
vηηn-vτn+wn-1vξn-vn-1wξn-1-(f(·,wn)-f(·,wn-1))=0,vn(0,ξ,η)=0.

Let vn=eατv1η. Then
v1ηηn-v1τn+wn-1v1ξn-v1n-1wξn-1-αv1n-e-ατ(f(·,wn)-f(·,wn-1))=0,v1ηηn-v1τn+wn-1v1ξn-v1n-1wξn-1=αv1n+e-ατ(f(·,wn)-f(·,wn-1))=αv1n+e-ατ∂f∂wv1n≥(α-c)v1n,
where we have chosen τ≤T small enough such that e-ατ=2, and 2(∂f/∂w)≥-c.

If v1 attains its positive maximal value in (0,T]×R2, we can choose α large enough such that
|wξn-1α-c|<1,
and then at the maximal point we have
(α-c)v1n≤-v1n-1wξn-1.
If v1n attains its negative minimal value in (0,T]×R2, we have
(α-c)(-v1n)≤-v1n-1wξn-1.
Notice that at τ=0,v1n=vn=0. By (2.34) and (2.35),
max|v1n|≤qmax|v1n-1|,q<1,
which means that the series v11+v12+⋯+v1n+⋯, whose sum has the form wne-ατ, is majorized by a geometrical progression and, therefore, is uniformly convergent. The fact that wn and its derivatives up to the second-order are bounded implies that the first derivatives of wn are uniformly convergent as n→∞.

It follows from (2.6) that wηηn are also uniformly convergent as n→∞.

Now, we can take w-1=w0=w0; then by the above discussion, we have the following theorem.

Theorem 2.5.

Suppose that V(0,ξ,η)≤w0≤V1(0,ξ,η) and f satisfies (2.9) and is suitable smooth, then there exists a small positive number T such that the Cauchy problem (2.4) has a classical solution.

By the way, it is easy to prove the uniqueness of the solution for the Cauchy problem (2.4), and we omit the details here.

3. Computational Examples

In this section, a numerical simulate is made for the equations by differential method. Numerical computation of these examples shows that the strong solutions for the corresponding Cauchy problem of (1.1)-(1.2) will blow-up in finite time.

Let Ω=[0,Lx]×[0,Ly] and u(x,y,0)=u0(x,y), (x,y)∈Ω, but u(x,y,0)=0, (x,y)∈R2/Ω. Then instead of studying the Cauchy Problem (1.1)-(1.2), we can study the following initial boundary problem:
∂xxu+u∂yu-∂tu=f(·,u),(x,y,t)∈Ω×(0,T],u(x,y,0)=u0(x,y),(x,y)∈Ω,u|∂Ω=0.
If f(·,0)=0, it is clear that if u(x,y,t) is a classical solution of (3.1), then u(x,y,t) is a strong solution of the Cauchy problem (1.1)-(1.2).

To dissect domain Ω, suppose that Lx=Ly=2π and hx=2π/N, hy=2π/M stands for the space step-length in the axis x and axis y, and k=T/J stands for the time step-length. Let Ωh={(ihx,jhy)∣0≤i≤N;0≤j≤M} and define uijn=u(ihx,jhy,nk). The differential scheme of the original equation is (to ensure numerical stability, here we apply arithmetic averages in order to avoid “oscillation’’ and “shifting’’ of the numerical solution)
ui+1,jn-2ui,jn+ui-1,jnhx2+ui+1,jn+ui,jn+ui-1,jn+ui,j+1n+ui,jn+ui,j-1n6ui,j+1n-ui,j-1n2hy-ui,jn+1-(1/4)(ui-1,jn+ui+1,jn+ui,j+1n+ui,j-1n)k=f(ihx,jhy,nk,ui+1,jn+ui,jn+ui-1,jn+ui,j+1n+ui,jn+ui,j-1n6),un|∂Ωh=0,(n=1,2,…),ui,j0=u0(ihx,jhy).

So we get the following explicit formula:
ui,jn+1=14(ui-1,jn+ui+1,jn+ui,j+1n+ui,j-1n)+khx2(ui+1,jn-2ui,jn+ui-1,jn)+k12hy(ui+1,jn+2ui,jn+ui-1,jn+ui,j+1n+ui,j-1n)(ui,j+1n-ui,j-1n)-kf(ihx,jhy,nk,ui+1,jn+2ui,jn+ui-1,jn+ui,j+1n+ui,j-1n6).

Experiment 1.

Suppose Ω=[0,2π]×[0,2π], hx=hy=2π/40, k=0.001, u0(x,y)=sinxsiny which itself does not satisfy (1.1); we get the graphs (see Figures 1–3) where u(x,y,t) changes according to the changes of t when different functions are given to f(·,u).

f(·,u)=u.

f(·,u)=sinu.

The numerical results in (a) at t=0 and in (b) at t=0.046 when f(·,u)=u2.

Figure 1 shows that when f(·,u)=u, at t=0.04, the numerical solutions become oscillatory; at t=0.042, the bifurcation of solutions occurs; when t>0.042, the solutions will blow-up. Similarly Figure 2 shows that when f(·,u)=sinu, at t=0.6, the bifurcation of solutions occurs; when t>0.6, the solutions will blow-up. Figure 3 is the spatiotemporal graphs of solutions when f(·,u)=u2 at t=0 (initial value) and t=0.0046. When t>0.0046, the solutions will blow-up.

Experiment 2.

The initial value is unknown in the general situation; so we use random numbers ([-0.01,0.01]) as the initial value and draw graphs (see Figures 4 and 5) where u(x,y,t) changes as t changes when different functions are given to f(·,u).

f(·,u)=u.

f(·,u)=1-sinu.

Figures 4 and 5 show that even though the initial value is sufficiently small, the blow-up will appear in finite time for the different functions.

The numerical result shows that there is a locality solution of the equation. When t becomes larger, the bifurcation of solutions occurs in finite time and blow-up appears. For this problem, it is essential to have a further research.

Acknowledgments

The research was supported by the Fujian National Science Foundation of China Grant 2008J0198, 2009J1009. The authors would like to thank Professor Zhao Junning for insightful discussions and Professor Xu Chuanju for helpful suggestions and comments.

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