UDC 517.911

SOLUTION OF A LIMITING BOUNDARY VALUE PROBLEM FOR THE FILTRATION EQUATION TAKING EVAPORATION INTO ACCOUNT

V. G. MELAMED

In steady-state filtration, the case of supported filtration from a single channel, taking into account losses only due to evaporation, is of interest. Under real conditions [2, 3] this leads to a nonlinear limiting*) boundary value problem for the equation

\[ \frac{d^2 y}{dx^2}+w(y)=0 \tag{1} \]

under the conditions

\[ y(0)=0, \tag{2} \]

\[ y(\infty)=1. \tag{3} \]

The evaporation function \(w(y)\) satisfies the following conditions: \(w(y)\) is defined and continuous for \(y \ge 0\), with \(w(y)>0\) for \(0 \le y<1\) and \(w(y)\equiv 0\) for \(y \ge 1\). The solution of (1)—(3) is sought in the class of twice continuously differentiable positive functions.

It follows from [1] that fulfillment of the Lipschitz condition for \(w(y)\) guarantees the existence and uniqueness of the solution of (1)—(3), and the Lipschitz condition plays an essential role in the proof of uniqueness. There an algorithm is also proposed for solving the limiting boundary value problem for an autonomous second-order equation by the method of finite differences, which at the same time gives the existence of a solution.

In the present work, the existence and uniqueness of the solution of (1)—(3) are proved without imposing the Lipschitz condition, and the solution is found in quadratures.

Let us prove that if \(y\) satisfies (1)—(3), then \(y'(x)>0\) for \(0 \le y<1\) and \(x \ge 0\). First of all, let us prove that \(y'(0)>0\). Indeed, if \(y'(0)<0\), then in some right half-neighborhood of the point \(x=0\) one must have \(y(x)<0\), which is impossible. If, however, \(y'(0)=0\), then let us consider \(y''(0)\). Since \(w(y)>0\) for \(0 \le y<1\), it follows from (1) that \(y''(0)<0\), and in some right half-neighborhood of the point \(x=0\) we have \(y(x)<0\), which is impossible.

Finally, suppose that at some point \(x_0>0\), \(y(x_0)<1\), one has \(y'(x_0)\le 0\).

If \(y'(x_0)<0\), then, since by (1) \(y''(x_0)<0\) and

\[ y(x)=y(x_0)+y'(x_0)(x-x_0)+\frac12 y''[x_0+\Theta(x-x_0)](x-x_0)^2,\quad 0<\Theta<1, \]

we obtain \(y(x)<y(x_0)<1\) for \(x>x_0\), which contradicts (3).

If, on the other hand, \(y'(x_0)=0\), then, since \(y''(x_0)<0\), in some right half-neighborhood of the point \(x_0\) we shall have \(y'(x)<0\), which, as has been proved, is impossible.

*) In the sense of the condition at \(\pm\infty\) [1].

Thus it has been proved that the desired solution of (1)—(3) must be monotonically increasing on the semi-interval \(0 \leq y < 1\).

This circumstance makes it possible to regard \(y\) in (1) as the independent variable, with \(x=x(y)\) a single-valued function.

Then (1)—(3) takes the form

\[ x''-(x')^3 w(y)=0, \tag{4} \]

\[ x(0)=0, \tag{5} \]

\[ x(1)=\infty . \tag{6} \]

Integrating (4) and using (5), we obtain

\[ x'(y)=\left\{[x'(0)]^{-2}-2\int_0^y w(\alpha)\,d\alpha\right\}^{-1/2}, \tag{7} \]

\[ x(y)=\int_0^y \frac{ds}{\sqrt{[x'(0)]^{-2}-2\int_0^s w(\alpha)\,d\alpha}} . \tag{8} \]

Depending on the choice of \(x'(0)\), three cases are possible.

1.

\[ [x'(0)]^{-2}<2\int_0^1 w(\alpha)\,d\alpha . \]

Then there exists \(y_1\), \(0<y_1<1\), such that

\[ [x'(0)]^{-2}=2\int_0^{y_1} w(\alpha)\,d\alpha, \]

and, consequently, \(x(y)\) does not exist for \(y_1<y<1\), i.e., it cannot be a solution of (4)—(6).

2.

\[ [x'(0)]^{-2}>2\int_0^1 w(\alpha)\,d\alpha . \]

Then the function under the sign of the outer integral on the right-hand side of (8) is continuous for \(0\leq y<1\) and has a finite limit as \(y\to 1\). Consequently, the integral in (8) as \(y\to 1\) is proper and, therefore, \(x(1)\) and \(x'(1)\) are finite, with \(x'(1)>0\).

Returning to the independent variable \(x\), we obtain \(y[x(1)]\equiv 1\), \(0<y'[x(1)]<\infty\). Since from (1) \(y''(x)=0\) for \(y>1\), the solution under consideration for \(x>x(1)\) is the straight line

\[ y-y[x(1)]=y'[x(1)][x-x(1)], \]

which contradicts (3).

3.

\[ [x'(0)]^{-2}=2\int_0^1 w(\alpha)\,d\alpha . \tag{9} \]

Then it follows from (8) that \(x=x(y)\) exists for all \(y\), \(0\leq y<1\), and for \(y=1\) the integral on the right-hand side of (8) is improper.

If it diverges, then \(x(y)\) is a solution of (1)—(3). If, however, it converges, then we construct a solution of (1)—(3) in the following way.

For \(0 < x < x(1)\) we determine from (8) \(y\) as an implicit function of \(x\), and for \(x \geqslant x(1)\) we set \(y \equiv 1\). By direct verification we see that the solution “glued together” in this way satisfies conditions (2) and (3), and, for \(x \ne x(1)\), also equation (1).

Let us show that the solution under consideration satisfies (1) also for \(x = x(1)\). Indeed, it is continuous. Moreover,
\[ y'[x(1)+0]=0,\qquad y'[x(1)-0]=[x'(1)]^{-1}=0. \]
Thus, \(y'(x)\) is continuous at \(x=x(1)\). By construction,
\[ y''[x(1)+0]=0. \]
To determine \(y''[x(1)-0]\), take an arbitrary \(x_1,\ 0<x_1<x(1)\), and apply Lagrange’s theorem to \(y'(x)\) on the interval \((x_1, x(1))\). Then
\[ \frac{y'(x_1)-y'[x(1)]}{x_1-x(1)}=y''(\xi)=-w(\xi), \]
where \(x_1<\xi<x(1)\). Hence
\[ y''[x(1)-0]=\lim_{x\to x(1)-0} \frac{y'(x_1)-y'[x(1)]}{x_1-x(1)}=0. \]
Thus, \(y''[x(1)]\) exists and is equal to zero, which satisfies (1). Thereby the existence of a solution of (1)—(3) has been proved; it may finally be written in the form
\[ x(y)=\int_0^y \frac{ds}{\sqrt{\displaystyle\int_s^1 w(t)\,dt}} . \tag{10} \]

Along the way, the uniqueness of the choice of \(x'(0)\), and therefore of \(y'(0)\), has been proved. In the case of divergence of the improper integral on the right-hand side of (8) for \(y=1\), the uniqueness of the solution of (1)—(3) follows from the uniqueness of \(y'(0)\), according to the general theory of differential equations.

In the case of convergence of the above-mentioned integral, the same considerations imply the uniqueness of the solution of (1)—(3) on \([0,x(1)]\).

For \(x>x(1)\), consider an arbitrary \(x_2>x(1)\). If \(y(x_2)<1\), then, since \(y''(x_2)<0\), there exists \(x_3,\ x(1)<x_3<x_2\), such that \(y'(x_3)<0\), which, as was proved above, is impossible. If, however, \(y(x_2)>1\), then on the interval \((x(1),x_2)\) the solution is a straight line not parallel to the \(x\)-axis, which is impossible in view of \(y'[x(1)]=0\).

Thus, the solution \(y\equiv 1\) found for \(x>x(1)\) is also unique. Thereby it has been proved that the solution of (1)—(3) satisfies the inequality \(0\leqslant y(x)\leqslant 1\).

The problem under consideration of backed-up filtration with allowance for evaporation in the classical formulation [4] has the form
\[ \frac{d}{dx}\left(y\frac{dy}{dx}\right)+w(y)=0,\qquad y(0)=0,\quad y(\infty)=1. \]
With the substitution \(y=\sqrt{z}\) we obtain
\[ z''=-2w(\sqrt{z}),\qquad z(0)=0,\quad z(\infty)=1. \tag{11} \]

In this case \(w'(0)\) may become infinite. However, solution (11) is obtained in a manner entirely analogous to (1)—(3), since the finiteness of \(w'\) in the solution of (1)—(3) was not used.

References

1. Klokov Yu. A. Mathematical Collection. New series, 53, 1961.
2. Averyanov S. F. Hydraulic Engineering and Land Reclamation, 9, No. 10, 1950.
3. Averyanov S. F. Collection The Influence of Irrigation Systems on the Regime of Groundwater. Edited by P. Ya. Polubarinova-Kochina. Publishing House of the Academy of Sciences of the USSR, 1956.
4. Polubarinova-Kochina P. Ya. Theory of the Motion of Groundwater. Tekhteorizdat, 1952.

Received by the editors
October 12, 1965

Moscow State University
named after M. V. Lomonosov