MATHEMATICS
S. K. GODUNOV
Submitted 1962-01-01 | RussiaRxiv: ru-196201.54452 | Translated from Russian

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MATHEMATICS

S. K. GODUNOV

ON NONUNIQUENESS FOR PARABOLIC SYSTEMS

(Presented by Academician I. G. Petrovskii, February 26, 1962)

In the work \((^1)\) I attempted to show that a nonlinear parabolic system of the form

\[ \frac{\partial u_i}{\partial t} = \frac{\partial}{\partial x} \left[ \sum_k a_{ik}\frac{\partial u_k}{\partial x} \right]. \tag{1} \]

need not have a unique solution in the case of discontinuous initial data. S. N. Kruzhkov drew my attention to an error that I had made in my calculations. He also proved that for the system considered in \((^1)\),

\[ \frac{\partial u}{\partial t} = \frac{\partial}{\partial x} \left[ a(u,v)\frac{\partial u}{\partial x} \right], \]

\[ \frac{\partial v}{\partial t} = \frac{\partial^2 v}{\partial x^2}, \]

the solution is unique also for discontinuous initial data. Nevertheless, the assertion that a nonunique solution is possible for systems of type (1) is correct. We shall now show this for a system of three equations, based on the same idea as in \((^1)\).

First let us note that the solution of the equation

\[ \frac{\partial u}{\partial t} = \frac{\partial}{\partial x} \left[ K(\xi)\frac{\partial u}{\partial x} \right] \qquad \left( \xi=\frac{x}{\sqrt{t}} \right) \]

for

\[ K(\xi)= \begin{cases} 1, & \text{for } -\infty<\xi<-2,\\ f(\xi), & \text{for } -2\leq \xi\leq -1,\\ 1, & \text{for } -1<\xi<+\infty \end{cases} \]

with the initial condition \(u|_{t=0}=\operatorname{sign} x\) is written by the formulas

\[ u(\xi) = -1+2\int_{-\infty}^{\xi}\frac{z(\eta)}{2K(\eta)}\,d\eta; \]

\[ z(\xi)= \begin{cases} A_1 \exp\left[-\dfrac{\xi^2}{4}\right], & \text{for } \xi\leq -2,\\[1.2em] A_2 \exp\left[-\displaystyle\int_{-2}^{\xi}\frac{\eta\,d\eta}{2f(\eta)}\right], & \text{for } -2<\xi\leq -1,\\[1.2em] A_3 \exp\left[-\dfrac{\xi^2}{4}\right], & \text{for } -1<\xi. \end{cases} \]

Of the expressions for the constants \(A_1, A_2, A_3\) we shall give only one:

\[ A_3=\left\{\int_{-\infty}^{+\infty} K^{-1}(\xi)\exp\left[-\int_0^\xi \frac{\eta\,d\eta}{2K(\eta)}\right]d\xi\right\}^{-1}. \]

The solution of the equation

\[ \frac{\partial w}{\partial t}=\frac{\partial^2 w}{\partial x^2} \qquad \left(w\big|_{t=0}=\operatorname{sign}x\right) \]

is defined as follows:

\[ w(\xi)=-1+2\int_0^\xi \bar z(\eta)\,d\eta, \]

where

\[ \bar z(\xi)=\left[\int_{-\infty}^{+\infty}\exp\left[-\frac{\xi^2}{4}\right]d\xi\right]^{-1} \exp\left[-\frac{\xi^2}{4}\right]. \]

It is clear from these formulas that in the plane \(u,w\) the curve defined by the parametric equations

\[ u=u(\xi),\qquad w=w(\xi), \]

connects the points \((-1,-1)\), \((+1,+1)\) and contains two rectilinear segments. One of these segments corresponds to the interval \(-1<\xi<+\infty\), and on it

\[ \frac{du}{dw}:\frac{z(\xi)}{\bar z(\xi)} = A_3\int_{-\infty}^{+\infty}\exp\left[-\frac{\xi^2}{4}\right]d\xi = \operatorname{const}=B. \]

We shall now show that \(B<1\). Since

\[ B= \left\{ \int_{-\infty}^{0}\exp\left[-\frac{\xi^2}{4}\right]d\xi + \int_0^{+\infty}\exp\left[-\frac{\xi^2}{4}\right]d\xi \right\} \times \]

\[ \times \left\{ \int_{-\infty}^{0}\frac{1}{K(\xi)} \exp\left[-\int_0^\xi \frac{\eta\,d\eta}{2K(\eta)}\right]d\xi + \int_0^\infty \exp\left[-\frac{\xi^2}{4}\right]d\xi \right\}^{-1}, \]

it suffices to verify that

\[ \int_{-\infty}^{0}\exp\left[-\frac{\xi^2}{4}\right]d\xi < \int_{-\infty}^{0}\frac{1}{K(\xi)} \exp\left[-\int_0^\xi \frac{\eta\,d\eta}{2K(\eta)}\right]d\xi. \tag{2} \]

Denote

\[ s(\xi)=-\left[2\int_0^\xi \frac{\eta\,d\eta}{K(\eta)}\right]^{1/2}. \]

The inverse function satisfies the equation

\[ \xi(s)=-\left[2\int_0^s K|\xi(\gamma)|\,\gamma\,d\gamma\right]^{1/2}. \]

Since \(K(\xi)\le 1\) and \(K(\xi)\not\equiv 1\), we have \(\xi(s)/s\le 1\) and \(\xi(s)/s\not\equiv 1\). The integral appearing on the right-hand side of inequality (3) can be rewritten as

\[ \int_{-\infty}^{0}\frac{1}{K(\xi)} \exp\left[-\int_0^\xi \frac{\eta\,d\eta}{2K(\eta)}\right]d\xi = \int_{-\infty}^{0}\frac{s}{\xi}\exp\left[-\frac{s^2}{4}\right]ds > \]

\[ > \int_{-\infty}^{0}\exp\left[-\frac{s^2}{4}\right]ds = \int_{-\infty}^{0}\exp\left[-\frac{\xi^2}{4}\right]d\xi; \]

thus it is proved that \(B<1\).

It is seen from this that the curve \(u = u(\xi), w = w(\xi)\) can have common points with the line \(u=w\), distinct from \((+1,+1)\), only for \(\xi < -1\).

Defining the function \(v(\xi)\) as a solution of the equation

\[ \frac{dv}{dt}=\frac{\partial}{\partial x}\left[K(-\xi)\frac{\partial v}{\partial x}\right] \]

with initial data \(v|_{t=0}=\operatorname{sign} x\), we find that \(v(\xi)=-u(-\xi)\), and thereby establish that the curve \(v=v(\xi), w=w(\xi)\) can intersect the line \(v=w\) only at points with \(\xi>1\) and at the point \(v=w=-1\). Hence it is clear that in the space \((u,v,w)\) the curve \(u=u(\xi), v=v(\xi), w=w(\xi)\) intersects the line \(u=v=w\) only when \(u=v=w=\pm 1\). Let us also note that \(u=u(\xi), v=v(\xi), w=w(\xi)\) are strictly monotone functions.

Define along the curve \(u=u(\xi), v=v(\xi), w=w(\xi)\) the coefficients

\[ \begin{aligned} a(u,v,w)&=K(\xi),\\ b(u,v,w)&=K(-\xi),\\ c(u,v,w)&=1. \end{aligned} \]

On the line \(u=v=w\) put \(a=b=c=1\). After this we extend \(a,b,c\) to the whole space \((u,v,w)\). These extended \(a,b,c\) may be regarded as greater than \(1/2\), not exceeding \(1\), and sufficiently smooth. The possibility of a smooth extension is ensured by the smoothness of the curve \(u=u(\xi), v=v(\xi), w=w(\xi)\). This smoothness can be achieved by choosing a smooth \(f(\xi)\).

The system of equations

\[ \begin{aligned} \frac{\partial u}{\partial t}&=\frac{\partial}{\partial x}\left[a\frac{\partial u}{\partial x}\right],\\ \frac{\partial v}{\partial t}&=\frac{\partial}{\partial x}\left[b\frac{\partial v}{\partial x}\right],\\ \frac{\partial w}{\partial t}&=\frac{\partial}{\partial x}\left[c\frac{\partial w}{\partial x}\right] \end{aligned} \]

with initial data

\[ u|_{t=0}=v|_{t=0}=w|_{t=0}=\operatorname{sign} x \]

has the solutions:

\[ \begin{aligned} 1)\quad \begin{cases} u=u(\xi),\\ v=v(\xi),\\ w=w(\xi); \end{cases} \qquad 2)\quad \begin{cases} u=u(\xi),\\ v=u(\xi),\\ w=u(\xi). \end{cases} \end{aligned} \]

Received
15 II 1962

CITED LITERATURE

  1. S. K. Godunov, DAN, 136, No. 6, 1281 (1961).

Submission history

MATHEMATICS