# On the Cauchy Problem of a Quasilinear Degenerate Parabolic Equation

Authors: Zongqi Liang; Huashui Zhan
Journal: Discrete Dynamics in Nature and Society
 (2009)

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## Abstract

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.


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## Body

## 1. Introduction

We consider the following Cauchy problem:(1.1)∂xxu+u∂yu-∂tu=f(·,u),(t,x,y)∈(0,T]×R2,(1.2)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
(1.3)|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 whenT 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(2.1)∂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:
(2.2)τ=t,ξ=x,η=u(t,x)U(t,x),
we obtain the following equation for w(τ,ξ,η)=∂yu/U:
(2.3)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:(2.4)wηη-wτ+wwξ=f(τ,ξ,η,w),(2.5)w(0,ξ,η)=u0(ξ,η),
where u0∈C2(R2); its first-order derivatives and u0ηη are all bounded.Definition 2.1.
A functionw(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
(2.6)Ln(wn)=wηηn-wτn+wn-1wξn-f(·,wn)=0,(2.7)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 numberp(2.8)|f(·,v)|≤c(1+|v|p),
and when v1-v2≥0,
(2.9)c1(v1-v2)≥f(·,v1)-f(·,v2)≥c2(v1-v2),max{|∂f∂τ|,|∂f∂ξ|,|∂f∂η||∂2f∂v2|}≤c.Lemma 2.2.
LetV 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 Lemma2.2. The difference zn=wn-V satisfies the inequality
(2.10)0≥Ln(zn)=Ln(wn)-Ln(V)=zηηn-zτn+wn-1zξn-(f(·,wn)-f(·,V)).
Letz1n=e-ατzn. Then
(2.11)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 Lemma2.2. 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,
(2.12)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,
(2.13)Ln(V1)=φηηeβτ-βφeβτ-f(·,V1)≤φηηeβτ-βφeβτ+c(1+φpeβτp)<0
by the same reason.
LetV=ψ(η)e-ατ, α1>ψ(η)≥α0>0, |ψηη|≤c. Then
(2.14)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 Lemma2.2.Thus we have the following.Lemma 2.3.
Suppose thatf satisfies (2.9) and V(0,ξ,η)≤w0≤V1(0,ξ,η), then
(2.15)V≤wn≤V1.
The smooth functionsV, V1 can be constructed as in [8], and we omit details here.Let(2.16)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
(2.17)uηηn-uτ+un-1uξn-f(·,un)=0,(τ,ξ,η)∈(0,T]×R2.Applying the operator2uτn(∂/∂τ)+2uξn(∂/∂ξ)+2uηn(∂/∂η) to (2.17),
(2.18)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
(2.19)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
(2.20)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
(2.21)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
(2.22)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
(2.23)Φ1≤c.So, if we letT1≤T small enough such that eγT1=2 and set M=2c, then
(2.24)Φ≤ceγT1=M.In order to estimate the second derivatives ofun in [0,T1]×R2, consider the function
(2.25)F=(uττn)2+(uξξn)2+(uηηn)2+(uτξn)2+(uξηn)2+(uτηn)2.
Applying the operator
(2.26)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
(2.27)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(2.28)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
(2.29)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
(2.30)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.
Letwn 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 ofwn=un in [0,T]×R2. For vn=wn-wn-1 we obtain the following equation from (2.6):
(2.31)vηηn-vτn+wn-1vξn-vn-1wξn-1-(f(·,wn)-f(·,wn-1))=0,vn(0,ξ,η)=0.Letvn=eατv1η. Then
(2.32)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.Ifv1 attains its positive maximal value in (0,T]×R2, we can choose α large enough such that
(2.33)|wξn-1α-c|<1,
and then at the maximal point we have
(2.34)(α-c)v1n≤-v1n-1wξn-1.
If v1n attains its negative minimal value in (0,T]×R2, we have
(2.35)(α-c)(-v1n)≤-v1n-1wξn-1.
Notice that at τ=0,v1n=vn=0. By (2.34) and (2.35),
(2.36)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 takew-1=w0=w0; then by the above discussion, we have the following theorem.Theorem 2.5.
Suppose thatV(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.LetΩ=[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:
(3.1)∂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 domainΩ, 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)
(3.2)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:(3.3)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.
SupposeΩ=[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).Figure 1
f(·,u)=u.Figure 2
f(·,u)=sinu.The numerical results in (a) att=0 and in (b) at t=0.046 when f(·,u)=u2.
(a)(b)Figure1 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).Figure 4
f(·,u)=u.Figure 5
f(·,u)=1-sinu.Figures4 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. Whent 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.



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