MATHEMATICAL PHYSICS

R. M. ZAIDEL and K. A. SEMENDYAEV

LIMITING SOLUTIONS OF A NONLINEAR EQUATION OF PARABOLIC TYPE

(Presented by Academician A. D. Sakharov on 16 III 1960)

1. We consider a nonlinear equation of parabolic type in the form (1)

[
\frac{1}{a}\frac{\partial F}{\partial t}=F\Delta F+\frac{1}{k}(\nabla F)^2.
\tag{1}
]

For the one-dimensional case equation (1) takes the form

[
\frac{1}{a}\frac{\partial F}{\partial t}
=
F\left(\frac{\partial^2 F}{\partial r^2}
+
\frac{\nu-1}{r}\frac{\partial F}{\partial r}\right)
+
\frac{1}{k}\left(\frac{\partial F}{\partial r}\right)^2,
\tag{2}
]

where, as usual, the parameter (\nu) assumes the values (1, 2, 3), respectively, for plane, cylindrical, and spherical solutions. It is possible, however, to assign meaning to any other values of (\nu), including negative ones, if in the quasi-one-dimensional approximation one considers motion in a tube of variable cross section (S(r)\sim r^{\nu-1}).

1. Let us consider a solution of the type of a wave converging to the singular point (r=0) of equation (1). We take the focusing time to be zero, so that the quantity (\tau=-t) varies in the range (0\leq \tau<\infty). Instead of (2) we obtain the equation

[
\frac{1}{a}\frac{\partial F}{\partial \tau}
+
F\left(\frac{\partial^2 F}{\partial r^2}
+
\frac{\nu-1}{r}\frac{\partial F}{\partial r}\right)
+
\frac{1}{k}\left(\frac{\partial F}{\partial r}\right)^2
=0.
\tag{3}
]

We seek the solution in self-similar form

[
F(r,\tau)=\frac{r^2}{a\tau}\varphi(\xi),
\tag{4}
]

where (\xi=r/A\tau^\alpha), (A) is a constant, and (\alpha) is the exponent to be determined.

For the function (\varphi(\xi)), according to (3), we obtain the equation in total derivatives

[
\varphi\left[
\xi^2\frac{d^2\varphi}{d\xi^2}
+
(\nu+3)\xi\frac{d\varphi}{d\xi}
+
2\nu\varphi
\right]
+
\frac{1}{k}
\left(
\xi\frac{d\varphi}{d\xi}
+
2\varphi
\right)^2
-
\left(
\alpha\xi\frac{d\varphi}{d\xi}
+
\varphi
\right)
=0,
\tag{5}
]

which, by the substitution (p=\xi\,d\varphi/d\xi), is reduced to the first-order equation

[
\frac{dp}{d\varphi}
=
-\frac{1}{p\varphi}
\left[
2\left(\nu+\frac{2}{k}\right)\varphi^2
+
\left(\nu+2+\frac{4}{k}\right)p\varphi
+
\frac{1}{k}p^2
-
\alpha p
-
\varphi
\right].
\tag{6}
]

The independent variable is recovered by the quadrature

[
\ln \xi=\int_0^\varphi \frac{d\varphi}{p(\varphi)}.
\tag{7}
]

The lower limit in this integral is chosen on the basis that at the wave front there must be (\varphi=0), and by changing the scale of the coordinates and time one can arrange that the wave front correspond to (\xi=1). In this case the inequality (1\leq \xi\leq \infty) is satisfied.

The condition that the flux (q=\operatorname{const}\cdot F^{1/k}\dfrac{\partial F}{\partial r}) vanish at the wave front gives

[
\varphi^{1/k}\xi\frac{d\varphi}{d\xi}=0
\quad \text{for } \xi=1 .
\tag{8}
]

We shall seek the expansion of (\varphi(x)) as (x=\ln \xi\to 0) in the form (\varphi(x)=x^\lambda(C_0+C_1x+\cdots)), where (\lambda\ne0) and (C_0\ne0). Substituting this expansion into (5), and in the quadratic and linear groups of terms retaining the terms containing (x) to the lowest power, we then obtain

[
\lambda C_0\left{\left(\lambda\frac{k+1}{k}-1\right)C_0x^{\lambda-1}-\alpha\right}=0 .
\tag{9}
]

Moreover, from (8) it follows that
[
\lim_{x\to0}\lambda C_0^{\,1+\frac1k}x^{\lambda\left(1+\frac1k\right)-1}=0,
]
i.e. (\lambda>\dfrac{k}{k+1}).

The two possibilities, (\lambda>1) and (\lambda<1), according to (9), lead respectively to the conditions (\lambda C_0^2=0) and (\lambda C_0\alpha=0), which contradict the initial assumptions. Hence it follows that (\lambda=1) and (C_0=\alpha k). Instead of (8) we now have

[
p=\xi\frac{d\varphi}{d\xi}=\alpha k
\quad \text{for } \xi=1 .
\tag{10}
]

We seek a solution which at the moment (\tau=0) would give a finite value for (F(r,0)) when (r\ne\infty). For (r\ne0) and (\tau\to0), the quantity (\xi=r/A\tau^\alpha\to\infty). From (4) we conclude that

[
\varphi(\xi)=\operatorname{const}\cdot \xi^{-1/\alpha}
\quad \text{as } \xi\to\infty .
\tag{11}
]

Then (F(r,0)=\operatorname{const}\cdot r^{2-1/\alpha}), and since (F(0,0)=0), it follows that (\alpha>1/2). From physical considerations it is clear that the required solutions correspond to accelerated motion of the wave front, which gives (\alpha<1).

From condition (11) it follows that as (\xi\to\infty), (\varphi\to0) and (p=\xi\,d\varphi/d\xi=-\varphi/\alpha). Thus, the sought integral curve of equation (6) must leave the point ((\varphi=0;\ p=\alpha k)) and enter the origin ((\varphi=0;\ p=0)) along the straight line (\alpha p+\varphi=0).

Fig. 1

The character of the integral curves of equation (6) is shown in Fig. 1. This equation has three singular points: (L(\varphi=0,\ p=0)), (M(\varphi=0;\ p=\alpha k)), (N\left(\varphi=\dfrac{1}{2\nu+4/k};\ p=0\right)). The coordinate axes (except for the singular points) are isoclines of (\infty). The ordinate axis, although it is an integral curve, is of no interest.

The singular point (M) is of saddle type; its second separatrix (the first is the line (\varphi=0)) near (M) has the form:

[
p=\alpha k+A_1\varphi+A_2\varphi^2+A_3\varphi^3+\cdots,
\tag{12}
]

where

[
A_1=-\frac{1}{\alpha(k+1)}
\left[\alpha k\left(\nu+2+\frac{4}{k}\right)-1\right],
]

[
A_2=-\frac{1}{\alpha(2k+1)}
\left[2\left(\nu+\frac{2}{k}\right)+\frac{A_1}{\alpha k}\right],
]

[
A_3=-\frac{A_2}{\alpha(3k+1)}
\left[\frac{1}{\alpha k}+\left(2+\frac{1}{k}\right)A_1\right].
]

The character of the singular point (N) depends on (\alpha). For (1/2<\alpha<1) it is a focus. The origin of coordinates (the point (L)) is a complicated singular point (?). Along the direction (\alpha p+\varphi=0), for (\varphi>0), only one integral curve (a separatrix) enters the point (L), for which we have the expansion

[
p=B_1\varphi+B_2\varphi^2+B_3\varphi^3+\ldots,
\tag{13}
]

where

[
B_1=-\frac{1}{\alpha},\qquad
B_2=\frac{2}{\alpha^3}\left(\nu+\frac{2}{k}\right)
\left(\alpha-\frac{1}{2}\right)
\left(\alpha-\frac{k+1}{k\nu+2}\right)>0,
]

[
B_3=\frac{B_2}{\alpha}
\left[\nu+2+\frac{4}{k}-\frac{1}{\alpha}\left(3+\frac{2}{k}\right)\right].
]

For (\varphi<0), a multitude of integral curves enters along the direction (\alpha p+\varphi=0). Along the direction (\varphi=0), one integral curve enters the point (L). Since (F(r,\tau)) is an essentially positive quantity, the behavior of the integral curves in the half-plane (\varphi>0) is of practical interest.

The method of numerical computation consisted in starting from the points (M) and (L) with the aid of the expansions (12) and (13) and, by selecting the value of (\alpha), achieving the joining of the two branches of the integral curve on the ray (p=0,\ \varphi>0). It turned out that small changes in the exponent (\alpha) lead to a considerable divergence of the two branches. Therefore, with a small number of trials, for given (\nu) and (k) one can obtain the exponent (\alpha) with sufficient accuracy. Table 1 gives the values of (\alpha) for (\nu=2,3) and (k=1,2,\ldots,10).

Table 1

In Figs. 2 and 3, respectively for (\nu=2) and (\nu=3) and the same values of (k), graphs are given of the function (\varphi(\eta)), where (\eta=\xi^{-1/\alpha}), constructed according to formula (7).

1. The self-similar solutions constructed are limiting in the sense that, if at some moment (\tau>0) there is a region with center at the point (r=0) such that throughout the region (F(r,\tau)=0), then, provided that dependence on other spatial coordinates is absent, the solution in a neighborhood of the point (r=0,\ \tau=0) will necessarily enter a self-similar regime with the corresponding exponent (\alpha).

One can indicate two values of the exponent (\alpha) for which the solution in the variables ((p,\varphi)) has the form of a straight line passing through the points (M) and (N):

[
p=\alpha k\left[1-2\left(\nu+\frac{2}{k}\right)\varphi\right]
=\xi\frac{d\varphi}{d\xi},
]

which gives

[
\varphi(\xi)=\frac{1}{2(\nu+2/k)}\left[1-\xi^{-2\alpha(k\nu+2)}\right].
]

For the value (\alpha_1=\dfrac{1}{k\nu+2}) we obtain the inverse of the known solution (3).

The second value (\alpha_2=\dfrac{1}{2(k+1)}) gives an analogue of solution (4). For (\nu=2) both these solutions coincide. The two indicated solutions do not satisfy the boundedness condition for (F(r,0)) as (r\ne\infty), and therefore are not limiting solutions.

Fig. 2

Fig. 3

Since at the instant (\tau=0), (F(r,0)=\mathrm{const}\cdot r^{2-1/\alpha}) depends on the coordinate in a power-law manner, the function (F(r,t)) for (t>0) is found by a method analogous to that indicated in (5).

Received
28 XII 1959

REFERENCES

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3. Ya. B. Zeldovich, A. S. Kompaneets, in: Collection Dedicated to the Seventieth Birthday of Academician A. F. Ioffe, Acad. Sci. USSR Press, 1950; G. I. Barenblatt, Prikl. matem. i mekh., 16, issue 1 (1952).
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