MATHEMATICS

G. V. SHCHERBINA

ON A BOUNDARY-VALUE PROBLEM FOR THE BLASIUS EQUATION

(Presented by Academician I. G. Petrovskii, 28 XI 1960)

In the present paper the existence of a solution of the following boundary-value problem will be proved:

\[ y''' + 2yy' = 0; \tag{1} \]

\[ y(\pm 0)=0,\quad y'(+0)=y'(-0),\quad y''(+0)=cy''(-0),\quad \lim_{x\to+\infty} y'(x)=a, \]

\[ \lim_{x\to-\infty} y'(x)=b \quad (c>0,\ a\geq 0,\ b\geq 0,\ a\ne b). \tag{1'} \]

The boundary-value problem (1), (1′) for \(a=1,\ b=1+\lambda\) \((-1\leq \lambda\leq 1),\ c=1\) arises in the problem of the mixing of two gas jets \((^{1})\).

Lemma 1. For any solution of equation (1), one of the following two alternatives holds: either \(y''(x)\ne 0\) \((-\infty<x<\infty)\), or \(y''(x)\equiv 0\).

The proof follows easily from the uniqueness of the solution of equation (1) with initial conditions \(y(x_0)=y_0,\ y'(x_0)=y'_0,\ y''(x_0)=0\).

Theorem 1. If \(y'(x)>0\) \((-\infty<x<\infty)\), then both limits \(\lim_{x\to+\infty} y'(x)\), \(\lim_{x\to-\infty} y'(x)\) exist and are finite.

Proof. We shall prove the theorem by contradiction. Since, under the substitution \(x\) by \(-x\), \(y\) by \(-y\), equation (1) does not change its form, it is sufficient in Theorems 1 and 2 to consider the case \(x\geq 0\). Integrating equation (1) twice, we obtain

\[ y'(x)=y'(0)+y''(0)x+\frac{1}{2}\left[\int_0^x y'(x)\,dx\right]^2 -\int_0^x\int_0^t y'^2(x)\,dx\,dt, \]

whence it is seen that either \(x\to c,\ y\to+\infty,\ y'\to+\infty\), or \(y\to+\infty,\ x\to+\infty,\ y'\to+\infty\). Make the change of variables \(y'=p(y)\). Then (1) becomes the equation

\[ p\frac{d^2p}{dy^2}+\left(\frac{dp}{dy}\right)^2+2y\frac{dp}{dy}=0 \tag{2} \]

and \(\lim_{y\to\infty} p(y)=\infty\). It is easy to see that \(d^2p/dy^2\), beginning with some \(y\), is monotone, since if \(d^3p/dy^3=0\), then, differentiating (2), we obtain

\[ \frac{d^2p}{dy^2}=\frac{2\,dp/dy}{3\,dp/dy+2y} =-p\left[\left(\frac{dp}{dy}\right)^2+2\frac{dp}{dy}\right], \quad \frac{p}{y^2}=\left(\frac{(\partial p/\partial y)}{y}+2\right) \left(\frac{(3dp/dy)}{y}+2\right). \tag{3} \]

But \(dp/dy>0\), hence \(d^2p/dy^2<0\) and \(dp/dy=o(y),\ p=o(y^2)\), and equality (3) is impossible for sufficiently large \(y>0\). From the preceding considerations it follows that

\[ \lim_{y\to\infty} \frac{d^2p}{dy^2}=0. \]

In this case either \(dp/dy\to c>0\),

or \(\lim\limits_{y\to\infty} dp/dy=0\). The equality \(dp/dy\to c>0\) is impossible, since in equation (2) the last term would be of higher order than the others. Thus, finally,

\[ \frac{d^2p}{dy^2}\to 0,\qquad \frac{dp}{dy}\to 0,\qquad p\to\infty,\qquad \text{if } y\to\infty . \tag{4} \]

Make the change of variables \(\rho=uy^2,\ y=e^t\). Equation (2) becomes the equation

\[ u\,[u''+3u'+2u]+[u'+2u]^2+u'+2u=0 . \tag{5} \]

From (4) we have, as \(t\to\infty\): \(du/dt=e^tdu/dy=(yp'-2p)/y^2\to0,\ u\to0,\ d^2u/dt^2\to0\). But then from (5) it follows that \(du/dt\sim -2u,\ u\sim ce^{-2t},\ du/dt\to -2ce^{-2t}\) and \(du/dy\sim c_1/y^3\). Taking into account that \(\lim\limits_{t\to\infty}u=0\), we obtain \(u\sim c_1/y^2\) and \(p\sim c_2\).

The contradiction obtained proves the theorem.

Theorem 2. Let \(y(x)\) be a solution of equation (1) and \(\lim\limits_{x\to+\infty} y'(x)=a>0,\ \lim\limits_{x\to-\infty} y'(x)=b>0,\ y(0)=c_0,\ y'(0)=c_1,\ y''(0)=c_2\). Then for every \(\varepsilon>0\) one can indicate a \(\delta>0\) such that as soon as \(|y_0-c_0|<\delta,\ |y'_0-c_1|<\delta,\ |y''_0-c_2|<\delta\), then

\[ \max_{-\infty<x<+\infty}\left|\,y'(x)-y'(x,y_0,y'_0,y''_0)\,\right|<\varepsilon . \]

Proof. Pass to equation (2). Let \(\varepsilon>0\) be given. Make the substitution \(p^2=u\). Then \(u(x)\) satisfies the equation

\[ \frac{\sqrt{u}}{2}\frac{d^2u}{dy^2}+y\frac{du}{dy}=0 . \tag{6} \]

Denote by \(\bar y\) such a \(y\) that

\[ e^{4y^2/a}\int_y^\infty e^{-4y^2/a}\,dy<\max\left\{\frac{\varepsilon}{4},\,\frac{a}{4}\right\},\qquad \frac{du}{dy}<\frac12\quad (y>\bar y),\qquad \frac{a}{2}<\sqrt{u(\bar y)}, \tag{7} \]

and by \(G\) a neighborhood of the point \(\bar y\) in which (7) is true and \(\sqrt{u(y)}>a/4\); write (6) in the form
\[ -\frac{d^2u/dy^2}{du/dy}=\frac{2y}{\sqrt{u}}. \]
If \(y\in G\), then
\[ -\frac{d^2u/dy^2}{du/dy}<\frac{8y}{a}. \]
Integrating twice, we obtain

\[ |\Delta u|<\left|\frac{du}{dy}\right|_{y=\bar y}\, e^{4\bar y^2/a}\int_{\bar y}^{y} e^{-4y^2/a}\,dy . \tag{8} \]

From inequalities (7) and (8) it follows that, first, \(|\Delta u|<a/4\), i.e. \(G\supset(\bar y,\infty)\), and also \(|\Delta u|<\varepsilon/4\) \((\bar y<y<\infty)\). Since on a finite segment the solution is a continuous function of the initial data, the theorem is proved.

Theorem 3. Denote \(y'(0)=\alpha,\ y''(0)=\beta\). For any \(a\ge0\) and any \(\alpha\ge0\) there exists such a \(\beta(\alpha)\) that \(y(x)\) satisfies equation (1) and \(\lim\limits_{x\to+\infty} y'(x)=a\). Moreover: if \(a-\alpha\ge0\), then \(\beta(\alpha)/\alpha\) is a continuous monotonically decreasing function of \(\alpha\); if \(a-\alpha<0\), then, defining \(\bar\beta(\alpha)\) as the exact upper bound of those \(y''(0)\) for which \(\lim\limits_{x\to\infty} y'(x)+a\), we obtain that \(\bar\beta(\alpha)\alpha^{-1}\) is a nonincreasing function of \(\alpha\).

The proof of Theorem 3 in the case \(a-\alpha\ge0\) follows easily from Theorems 1, 2 and from the strict monotonicity, proved in the works of Iglisch \((2,3)\), of \(\lim\limits_{x\to+\infty}y'(x)\) with respect to \(y'(0)\) for fixed \(y''(0)>0\). In the case \(a-\alpha<0\), Iglisch’s arguments, in which the positivity of \(y''(x)\) is essentially used, do not go through, since for \(a-\alpha<0\) we have \(y''(x)<0\). Let us make

make the change of variables \(y' = p(y)\) and then \(p = y^2/u,\ y = e^t\). For \(u(x)\) we obtain the equation

\[ u\left(2u+\frac{du}{dt}-\frac{d^2u}{dt^2}\right) -2\frac{du}{dt}\left(2u-\frac{du}{dt}\right) +\left(2u-\frac{du}{dt}\right)^2 +2u^2\left(2u-\frac{du}{dt}\right)=0. \]

If \(du/dt = 0\), then \(d^2u/dt^2 = 6u+4u^2\), i.e. \(du/dt = 0\) no more than once, but
\(\lim_{t\to-\infty}u=0,\ \lim_{t\to+\infty}u=\infty,\ u>0\); therefore \(du/dy>0\) and \(du/dy=z(u)\). We obtain the equation

\[ \frac{dz}{du}=6\frac{u}{z}+3\frac{z}{u}+4\frac{u^2}{z}-2u-7. \tag{9} \]

Since for \(u\ne0\) we have \(z\ne0\), and for \(u>0\) the uniqueness theorem for solutions of equation (9) is valid, if for some \(u>0\) the inequality \(z_1(u)>z_2(u)\) holds, then this inequality is true for all \(u>0\). Substituting in (9) \(z=y\,du/dy=(2y^2p-yp')/p^2\) and expanding the resulting fractions in a Taylor series, we obtain

\[ \frac{dz}{du} =2-\frac{3}{2}\frac{p'(0)}{p(0)}y+O(y^2) =2-\frac{3}{2}\frac{p'(0)}{\sqrt{p(0)}}\sqrt{u}+o(\sqrt{u}). \]

Let now \(p'_1(0)=p'_2(0)<0,\ p_1(0)>p_2(0)\). Then

\[ z_1(0)=z_2(0),\qquad \frac{dz_1}{du}<\frac{dz_2}{du}, \tag{10} \]

if \(u\) is small. From the fact that \(p_1(0)>p_2(0)\) and from the equality \(p=y^2/u\), for sufficiently small \(y>0\) we have \(u_1(y)<u_2(y)\). Let \(y_0>0\) be the first point at which \(u_1=u_2\). Since \(u_1(y)<u_2(y)\) if \(0<y<y_0\), then \(du_1/dy\big|_{y=y_0}>du_2/dy\big|_{y=y_0}\), i.e. \(z_1(u_0)>z_2(u')\), and this entails \(z_1(u)>z_2(u)\) for all \(u>0\), which contradicts (10). The case \(p_1(0)=p_2(0),\ 0>dp_1/dy\big|_{y=0}>dp_2/dy\big|_{y=0}\) is considered analogously.

We shall show that \(\bar{\beta}(\alpha)\alpha^{-1}\) does not increase when \(a=\alpha<0\). Suppose this is not so and \(\alpha_1>\alpha_2,\ \alpha_1^{-1}\bar{\beta}(\alpha_1,\alpha)=p'_1(0)>p'_2(0)=\alpha_2^{-1}\bar{\beta}(\alpha_2,\alpha)\). By the definition of \(\bar{\beta}(\alpha)\)

\[ \lim_{y\to\infty}p(y,\alpha_1,p'_1(0)) = \lim_{y\to\infty}p(y,\alpha_2,p'_2(0)) =a,\qquad \lim_{y\to\infty}p(y,\alpha_2,p'_2(0)) < \lim_{y\to\infty}p(y,\alpha_2,p'_2(0)). \]

By what has been proved,

\[ \lim_{y\to\infty}p(y,\alpha_2,p'_1(0)) \le \lim_{y\to\infty}p(y,\alpha_1,p'_2(0)) \]

and finally,

\[ \lim_{y\to\infty}p(y,\alpha_2,p'_2(0)) < \lim_{y\to\infty}p(y,\alpha_1,p'_1(0)), \]

which is impossible. The theorem is proved.

We now prove the existence of a solution of the boundary-value problem (1), \((1')\). To this end we formulate it somewhat differently. Namely, since under the substitution \(x\) by \(-x\), \(y\) by \(-y\), equation (1) is not changed, the boundary-value problem (1), \((1')\) is equivalent to the following problem.

Find two solutions of equation (1) satisfying the following conditions:

\[ y_1(0)=y_2(0)=0,\qquad y'_1(0)=y'_2(0),\qquad y''_2(0)=c y''_2(0), \]

\[ \lim_{x\to+\infty}y'_1(x)=a,\qquad \lim_{x\to+\infty}y'_2(x)=b,\qquad a<b,\quad a\ge0,\quad b>0. \]

Denote
\(\varphi_1(\alpha)=\bar{\beta}(a,\beta)\alpha^{-1}\),
\(\varphi_2(\alpha)=-c\alpha^{-1}\bar{\beta}(a,\alpha)\)
\((a\le\alpha\le b)\). As we have shown, \(\varphi_1(\alpha)\) decreases and is continuous, \(\varphi_2(\alpha)\) does not decrease. If there exists \(a<\alpha_0<b\) such that \(\varphi_1(\alpha_0)=\varphi_2(\alpha_0)\), the problem is solved. Suppose this is not so. Then there exist two sequences of numbers \((a_n,b_n)\) such that \(a_n<b_n\), \(\varphi_2(a_n)<\varphi_1(a_n)\), \(\varphi_2(b_n)>\varphi_1(b_n)\), \(b_n-a_n<1/2^n\). If \(\alpha_0=\lim_{n\to\infty}a_n=\lim_{n\to\infty}b_n\), then \(\lim \varphi_2(a_n)=A\le a_1(\alpha_0),\ \lim \varphi_2(b_n)=B\ge \varphi_1(\alpha_0)\). By the defini-

By the definition of \(\varphi_1(a)\) and \(\varphi_2(a)\), we have \(\lim_{x\to\infty} y'(x,\alpha_0-A/c\alpha_0)=\lim_{x\to\infty} y'(x,\alpha_0,-B/c\alpha_0)=a\), and, by theorem (2), \(\lim_{x\to\infty} y'(x,\alpha_0,-\varphi_1(\alpha_0)/c\alpha_0)=a\), which, together with \(\lim_{x\to\infty} y'(x,\alpha',\varphi_1(\alpha_0)/\alpha_0)=b\), solves the problem.

In conclusion I consider it my duty to express my gratitude to Prof. A. D. Myshkis, who drew the author’s attention to this problem.

Physical-Technical Institute of Low Temperatures
Academy of Sciences of the Ukrainian SSR

Received
24 XI 1960

REFERENCES

\({}^{1}\) L. G. Napolitano, Quart. Appl. Math., 16, No. 4, 397 (1959).
\({}^{2}\) R. Iglisch, ZAMM, 33, 143 (1953).
\({}^{3}\) R. Iglisch, ZAMM, 34, No. 12, 441 (1954).