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UDC 517.946.9:517.947.42
ON THE UNCONDITIONAL SOLVABILITY OF THE OBLIQUE DERIVATIVE PROBLEM
A. JANUŠAUSKAS
In this paper we establish one criterion for the unconditional solvability of the oblique derivative problem for harmonic functions.
1. Consider, in three-dimensional Euclidean space \(R^3\), the domain \(D\) bounded by the torus
\[ (x^2+y^2+z^2+R^2-r^2)^2=4R^2(x^2+y^2), \qquad R>r. \]
Putting \(z=\eta,\ x=(R+\xi)\cos\varphi,\ y=(R+\xi)\sin\varphi\), we obtain
\[ \xi^2+\eta^2=r^2,\qquad \xi=\sqrt{x^2+y^2}-R. \]
The domain \(D\) is obtained by rotating the circle \(C:\{x=0,\ z^2+(y-R)^2<r^2\}\) about the axis \(Oz\). As coordinates of a point \(P\in D\) we take the following triple of numbers: \((y,z,\varphi)\), where \((y,z)\) are the coordinates of the point \(P_1\in C\) from which the point \(P\) is obtained by rotating the circle \(C\) about the axis \(Oz\), and \(\varphi\) is the angle between the planes \(yOz\) and \(POz\). If the point \(P\in S\), then it is uniquely determined by the pair \((\xi,\varphi)\), where \(\xi\) is a point of the circumference \(K\) bounding the circle \(C\).
Consider the following oblique derivative problem for harmonic functions: find a harmonic function \(u(x,y,z)\), regular in the domain \(D\), and satisfying on the surface \(S\) the condition
\[ a(\xi,\varphi)\left(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}\right) +b(\xi,\varphi)\frac{\partial u}{\partial z} =h(\xi,\varphi), \tag{1} \]
where \(h(\xi,\varphi)\) is a Hölder-continuous function, and \(a\) and \(b\) have continuous first derivatives with respect to \(\xi\) and are analytic with respect to \(\varphi\).
We reduce this problem, in exactly the same way as in [1], to a dynamical system. For this purpose extend the functions \(a(\xi,\varphi)\) and \(b(\xi,\varphi)\) into the domain \(D\) so that the extended functions \(\alpha_1(y,z,\varphi)\) and \(\beta_1(y,z,\varphi)\) vanish simultaneously only on one circumference \(L_0:\{y=y_0,\ z=z_0\}\), and so that \(\alpha_1(y,z,\varphi)\) and \(\beta_1(y,z,\varphi)\) are analytic in \(D\). The harmonic function \(u(x,y,z)\), regular in the domain \(D\), is analytic in this domain [2]; therefore the function \(h(\xi,\varphi)\) also extends analytically into the domain \(D\). From the boundary condition (1) we obtain the following first-order equation:
\[ \alpha_1(y,z,\varphi)\left(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}\right) +\beta_1(y,z,\varphi)\frac{\partial u}{\partial z} =g_1(y,z,\varphi). \tag{2} \]
Under conditional or unconditional solvability, problem (1) is equivalent to the following problem: find all functions \(g_1\), analytic in \(D\), satisfying the condition \(g_1|_S=h(\xi,\varphi)\), such that equation (2) has a harmo-
... harmonic solution \(w(x,y,z)\), regular in the domain \(D\). In order to show the solvability of the latter problem, it suffices to find a function \(g_1\), analytic in \(D\) and satisfying the condition \(g_1|_S=h\), such that equation (2) has a solution \(u(x,y,z)\) satisfying the condition
\[ \left.\frac{\partial u}{\partial n}\right|_S = \left.\frac{\partial w}{\partial n}\right|_S, \]
where \(w\) is a harmonic function, regular in the domain and assuming on \(S\) the same values as \(u\).
We proceed to study equation (2). Introduce cylindrical coordinates
\[ x=\rho\cos\varphi,\qquad y=\rho\sin\varphi,\qquad z=z . \]
Equation (2) assumes the following form:
\[ \alpha(\rho,z,\varphi)\rho\frac{\partial u}{\partial \rho} + \beta(\rho,z,\varphi)\frac{\partial u}{\partial z} = g(\rho,z,\varphi), \tag{3} \]
where
\[ \alpha(\rho,z,\varphi)=\alpha_1(\rho\sin\varphi,z,\varphi),\qquad \beta(\rho,z,\varphi)=\beta_1(\rho\sin\varphi,z,\varphi), \]
\[ g(\rho,z,\varphi)=g_1(\rho\sin\varphi,z,\varphi). \]
Introduce into equation (3) new independent variables \(\xi\) and \(\eta\), which are first integrals of the following equations:
\[ \frac{d\rho}{\rho\alpha(\rho,z,\varphi)} = \frac{dz}{\beta(\rho,z,\varphi)}, \tag{4} \]
\[ \frac{d\rho}{-\beta(\rho,z,\varphi)} = \frac{dz}{\rho\alpha(\rho,z,\varphi)}. \tag{5} \]
Assume that the first integrals \(\xi\) and \(\eta\) of equations (4) and (5) are holomorphic in the domain \(D\). Equation (3), in the new variables, is rewritten as follows:
\[ T(u)=\{[\rho\alpha(\rho,z,\varphi)]^2+[\beta(\rho,z,\varphi)]^2\}\frac{\partial u}{\partial\eta} = f(\xi,\eta). \]
It follows from this equality that the operator \(T(u)\) maps the set of all functions analytic in the domain \(D\) into the set \(J\) of functions analytic in \(D\) and having the following representation: \(f=(\rho^2\alpha^2+\beta^2)g\), where \(g\) is an arbitrary function analytic in the domain \(D\). Functions from the set \(J\) can assume on the boundary \(S\) of the domain \(D\) any values prescribed in advance.
As was noted above, the equation of the torus has the form
\[ \omega\equiv (x^2+y^2+z^2+R^2-r^2)^2-4R^2(x^2+y^2)=0. \]
Equation (3) is solvable when the function \(f\) has the form
\[ f=(\rho^2\alpha^2+\beta^2)\omega g_0, \]
where \(g_0\) is an arbitrary function analytic in the domain \(D\).
Every function \(u\) analytic in the domain \(D\) can be represented near the boundary \(S\) in the following way:
\[ u=\sum_{n=0}^{\infty}\omega^n u_n, \tag{6} \]
where \(u_n\) are harmonic functions regular in the domain \(D\). In particular, the solution of the equation can also be represented in this form
\[ T(u)=(\rho^2\alpha^2+\beta^2)\omega g_0 . \tag{7} \]
Substituting (6) into (7), we obtain, for \(\omega=0\),
\[ T(u_0)+T(\omega)u_1=0, \]
\[ T(u_1)+2T(\omega)u_2=(\rho^2\alpha^2+\beta^2)g_0, \]
\[ T(u_n)+(n+1)T(\omega)u_{n+1}=0,\quad n\geq 2. \]
Putting \(u_n=0\) for \(n\geq 2\), we obtain
\[ (\rho^2\alpha^2+\beta^2)g_0=T(u_1),\quad T(u_0)+T(\omega)u_1=0. \tag{8} \]
It follows from the equalities (8) that there always exists a harmonic function \(u_0\), regular in the domain \(D\), satisfying the condition \(T(u_0)=T(\omega)h\), where \(h\) is an arbitrary function continuous, in the sense of Hölder, on the surface \(S\). If \(u_1\) is regarded as an arbitrary analytic function, then the second equality (8) is an equation with respect to \(u_0\).
Thus, when both equations (4), (5) have holomorphic first integrals, and the function \(h\) has the form \(Ah_1\), equation (3) has a solution \(u(x,y,z)\) such that the value of its normal derivative on \(S\) coincides with the value of the normal derivative of the harmonic function \(w(x,y,z)\), regular in the domain \(D\) and taking on \(S\) the same values as \(u\). Consequently, problem (1) in this case always has a solution.
Let now \(h(\zeta,\varphi)\) be an arbitrary Hölder-continuous function. Let \(u(x,y,z)\) be a solution of equation (3); then the harmonic function \(w_1\), taking on \(S\) the same values as \(u\), satisfies on \(S\) the boundary condition
\[ \alpha\rho \frac{\partial w_1}{\partial \rho}+\beta \frac{\partial w_1}{\partial z} = h-T(u_1). \tag{9} \]
This follows from equality (7). Consider the harmonic function \(w_0\), regular in the domain \(D\) and satisfying on the surface \(S\) the condition \(T(w_0)=T(\omega)u_1\). The existence of such a harmonic function has already been established. The harmonic function \(w=w_1-w_0\) is a solution of problem (1).
If the first integrals of equations (4) and (5) are holomorphic in the domain \(D\), then problem (1) is unconditionally solvable. Conditions for the holomorphy of the first integrals have been studied in [3].
The existence of holomorphic first integrals in equations (4), (5) is completely determined by the type of singularities of the vector field \(\{x\alpha_1,\ y\alpha_1,\ \beta_1\}\). As was noted above, for fixed \(\varphi=\varphi_0\) this field in the disk \(C_\varphi:\{\rho_1\leq r,\ \varphi=\varphi_0\}\) has one singular point \(Q_\varphi=(\rho_0,z_0)\). If this singular point \(Q_\varphi\) is a saddle, then the first integrals of equations (4) and (5) are holomorphic [3]. It is known [4] that the singular point \(Q_\varphi\) is a saddle only in the case when the rotation of the field \(A:\{x a,\ y a,\ b\}\) on the boundary \(K_\varphi\) of the disk \(C_\varphi\) is negative.
We shall show that the rotation \(n\) of the field \(A\) on \(K_\varphi\) does not depend on \(\varphi\). As is known [4],
\[ n=\frac{1}{2\pi}\int_{K_\varphi} \frac{\rho a(\zeta,\varphi)\,db(\zeta,\varphi)-b(\zeta,\varphi)\,d\rho\,a(\zeta,\varphi)} {\rho^2 a^2+b^2}. \]
If \(a^2+b^2\ne 0\), then, since \(a\) and \(b\) are analytic functions of \(\varphi\), \(n\) is also an analytic function of the variable \(\varphi\). On the other hand, \(n\) is an integer [4]; therefore \(n\) does not depend on \(\varphi\).
The case \(n=0\) can be considered analogously. In this case the field \(A:\{xa, ya, b\}\) can be extended into the domain \(D\) without singular points [4]. For \(n=0\), equations (4) and (5) also have holomorphic first integrals [3].
Thus we have the following theorem.
Theorem 1. If the rotation \(n\) of the vector field \(A:\{xa, ya, b\}\) on the meridian \(K_{\varphi}\) of the torus \(S\) is negative or equal to zero, then problem (1) is unconditionally solvable.
Let us consider one particular case of problem (1). Suppose the coefficients \(a(\xi)\) and \(b(\xi)\) do not depend on \(\varphi\). The boundary condition (1) takes the following form:
\[ a(\xi)\left(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}\right)+b(\xi)\frac{\partial u}{\partial z}=h(\xi,\varphi). \tag{10} \]
Since the harmonic function \(u\) is, by Hölder’s theorem, a continuous function of \(\varphi\) in the closed domain \(\overline D\), it can be represented by a uniformly convergent series [5]
\[ u=\sum_{l=0}^{\infty}\left[A_l(\rho^2,z)\cos l\varphi+B_l(\rho^2,z)\sin l\varphi\right], \tag{11} \]
where the functions \(A_l(t,z)\) and \(B_l(t,z)\) satisfy the equation
\[ 4t\frac{\partial^2 v}{\partial t^2}+\frac{\partial^2 v}{\partial z^2}+4\frac{\partial v}{\partial t}-\frac{l}{t}v=0. \tag{12} \]
Substituting (11) into (10) and equating the coefficients of \(\cos l\varphi\) and \(\sin l\varphi\) to the corresponding Fourier coefficients of the function \(h\), we obtain the following boundary condition
\[ a(\xi)t\frac{\partial v}{\partial t}+b(\xi)\frac{\partial v}{\partial z}=h_l(\xi) \quad \text{for} \quad (t+z^2+R^2-r^2)^2=4R^2t. \tag{13} \]
Thus problem (10) is reduced to the following problem: find a solution \(v\) of equation (12), regular in the domain \(E:\{t>0,\ (t+z^2+R^2-r^2)^2-4R^2t<0\}\), satisfying condition (13) on the boundary of the domain \(E\).
As is known [6], the difference between the number of linearly independent solutions of the homogeneous problem corresponding to problem (10) and the number of necessary and sufficient orthogonality conditions imposed on the function \(h_l\) for solvability of problem (13) is equal to \(-2n+2\). It was shown above that problem (10) is unconditionally solvable; therefore problem (13), for \(n\le 0\), is also unconditionally solvable. Consequently, for \(n\le 0\) problem (10) is unconditionally solvable, and the corresponding homogeneous problem has infinitely many linearly independent solutions.
Remark 1. In exactly the same way as above the oblique derivative problem was considered for harmonic functions regular in the domain \(D\) bounded by the torus \(S\), one can study problem (1) for any domain \(G\) formed by rotating some plane simply connected domain \(G_0\) about the axis \(Oz\). Moreover, the domain \(G_0\) has no common points with the axis \(Oz\). A theorem analogous to Theorem 1 also holds in this case.
2. Let now \(D\) be a domain symmetric with respect to the axis \(Oz\), homeomorphic to a ball, and let the surface \(S\) bounding the domain \(D\),
has continuous curvature. Let \(Q\) denote the section of the domain \(D\) by the half-plane \(M:\{y=0,\ x \geqslant 0\}\), and let \(L\) denote the curve \(M\cap S\). The domain \(D\) is obtained by rotating the plane domain \(Q\) about the axis \(Oz\). By \(\xi\) we shall denote a point of the curve \(L\).
Consider the following oblique derivative problem: find a harmonic function \(u\), regular in the domain \(D\), satisfying on the surface \(S\) the condition
\[ a(\xi,\varphi)\left(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}\right) +b(\xi,\varphi)\frac{\partial u}{\partial z} =h(\xi,\varphi), \tag{14} \]
where \(h\) is Hölder-continuous, and the functions \(a(\xi,\varphi)\) and \(b(\xi,\varphi)\) have continuous first derivatives with respect to \(\xi\) and are analytic with respect to \(\varphi\). In addition, \(a\) and \(b\) satisfy the following inequalities:
\[ [a(\xi,\varphi)]^2(x^2+y^2)+[b(\xi,\varphi)]^2\ne 0,\qquad b(\xi_0,\varphi)\cdot b(\xi_1,\varphi)>0, \tag{15} \]
where \(\xi_0\) and \(\xi_1\) are the points of intersection of the curve \(L\) with the axis \(Oz\).
As in the preceding case, we extend the coefficients \(a\) and \(b\) inside the domain \(D\). In consequence of the second inequality (15), the coefficients can be extended so that the extended coefficient \(b(\beta(0,z,\varphi))\) does not vanish at any point of the axis \(Oz\). Consequently, the vector field
\(A:\{xa,ya,b\}\) can be extended without singularities to a segment \(J\) of the axis \(Oz\) contained in the domain \(D\), in such a way that the rotation of the extended field on this segment is equal to zero. Further, problem (14) is studied exactly as problem (1) was studied above, except that instead of the circles \(C_\varphi\) one here considers the domains
\(Q_\varphi=D\cap M_\varphi\), \(M_\varphi:\{x\geqslant 0,\ px+qy=0\}\).
Introducing cylindrical coordinates, we reduce problem (14) to the following dynamical system:
\[ \frac{d\rho}{\rho\alpha(\rho,z,\varphi)} = \frac{dz}{\beta(\rho,z,\varphi)} = \frac{du}{g(\rho,z,\varphi)}, \tag{16} \]
where \(\alpha,\beta\), and \(g\) are the extensions into the domain \(D\), respectively, of the functions \(a,b\), and \(h\).
Since the rotation of the field \(B:\{x\alpha,y\alpha,\beta\}\) extended in \(D\) on the segment \(J\) of the axis \(Oz\) is equal to zero, the rotation of this field on the boundary of the plane domain \(Q_\varphi\) is equal to the rotation on the curve \(L_\varphi=S\cap M_\varphi\) of the field \(A:\{xa,ya,b\}\). This rotation is an integer and does not depend on \(\varphi\).
As in the preceding case, the validity of the following theorem is established.
Theorem 2. If the coefficients of the boundary condition (14) satisfy the inequalities (15) and the rotation \(n\) of the vector field \(A:\{xa,ya,b\}\) on the curve \(L_\varphi=S\cap T_\varphi\) is negative or equal to zero, then problem (14) is unconditionally solvable.
Remark 2. Above, the oblique derivative problem was considered for the Laplace equation. In exactly the same way one can study the oblique derivative problem with boundary condition (1) or (14) for any elliptic equation in the domain \(D\) of the form
\[ \sum_{j,i=1}^{3} a_{ij}\frac{\partial^2 u}{\partial x_i\partial x_j} +\sum_{i=1}^{3} b_i\frac{\partial u}{\partial x_i} =0, \tag{17} \]
where \(a_{ij}, b_i\) \((i=1,2,3;\ j=1,2,3)\) are functions analytic in the domain \(D\). In this case a theorem analogous to Theorem 2 is also valid.
Suppose that the second inequality (15) is not satisfied; then, by virtue of the first inequality (15), we have
\[ b(\xi_0,\varphi)b(\xi_1,\varphi)<0. \]
In this case the function \(b(\zeta,\varphi)\) cannot be extended to the segment \(J\) of the \(Oz\) axis in such a way that it does not vanish. However, in this case it can be extended to \(J\) in the form of a product \(z\beta_1(z)\), where \(\beta_1(z)\ne 0\) on the segment \(J\) of the \(Oz\) axis. The vector field \(B:\{xa,ya,\beta\}\), obtained by extending the field \(A:\{xa,ya,b\}\) into the domain \(D\), has a singular point \((0,0,0)\) on the \(Oz\) axis. This singular point may be either a node or a saddle. The presence of this singular point, together with the singular curve \(N\), complicates the study of the oblique derivative problem.
- Consider the domain \(D\) bounded by the torus \(S\),
\[ (x^2+y^2+z^2+R^2-r^2)^2=4R^2(x^2+y^2) \]
and the oblique derivative problem with the following boundary condition:
\[ a(\zeta,\varphi)\left(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}\right)+b(\zeta,\varphi)\frac{\partial u}{\partial z}=g(\zeta,\varphi), \tag{18} \]
where the function \(g(\zeta,\varphi)>0\) everywhere on the surface \(S\), and the coefficients \(a(\zeta,\varphi)\) and \(b(\zeta,\varphi)\) satisfy the same conditions as in the boundary condition (1). Denote by \(C_\varphi\) the intersection of the surface \(S\) with the plane \(T_\varphi:px+qy=0\), where \(\varphi=\arctg \dfrac{q}{p}\).
It follows from the boundary condition (18) that, for the solution \(u\) of problem (18), the inequality
\[ s(x,y,z)\equiv \left(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}\right)^2+\left(\frac{\partial u}{\partial z}\right)^2\ne 0 \tag{19} \]
holds everywhere on the surface \(S\).
Inequality (19) means that the gradient of the function \(u\) is not orthogonal at any point of the surface \(S\) to the plane \(T_\varphi\) passing through that point. Consider the vector field \(\operatorname{grad} u\) on the plane \(T_\varphi\), and denote by \(H_\varphi\) its projection onto the plane \(T_\varphi\). It follows from the boundary condition (18) that the vectors of the fields \(H_\varphi\) and \(P_\varphi:\{ax,ay,b\}\) are not directed oppositely to each other at any point of the circle \(C_\varphi\), and hence it follows that these fields are homotopic and that their rotations on the circle \(C_\varphi\) coincide [4].
We also note that it follows from inequality (19) that the function \(u\) cannot become a constant on the plane \(T_\varphi\) for any \(\varphi\).
Consider the function \(u_\varphi\)—the value of the function \(u\) on the plane \(T_\varphi\). The vector field \(\operatorname{grad} u_\varphi\) coincides with the field \(H_\varphi\). The set of singular points of the field \(\operatorname{grad} u_\varphi\) is the set \(N_\varphi\) of points lying on \(T_\varphi\) at which the analytic function \(s\) vanishes. Therefore the set \(N_\varphi\) is a real-analytic set of dimension not exceeding one [7] and is entirely contained in some compact subset \(A_\varphi\) of the disk \(K_\varphi=D\cap T_\varphi\). As is known [7], the set \(N_\varphi\) decomposes into a finite number of isolated points and a finite number of purely one-dimensional real-analytic sets \(L_i\). In consequence of the fact that the set \(N_\varphi\) is contained inside the compact set \(A_\varphi\), all the sets \(L_i\) are closed level lines of the function \(u_\varphi\). Consider the connected subset of the set \(K_\varphi\) bounded by the closed level line \(L_0\) of the function \(u_\varphi\). As is known [4], this domain contains at least one extremum point of the function \(u_\varphi\), and this point is a singular point of node type for the field \(\operatorname{grad} u_\varphi\).
Denote by \(G_\varphi\) the field of vectors tangent to the level lines of the function \(u_\varphi\). The vectors of the fields \(G_\varphi\) and \(\operatorname{grad} u_\varphi\) are orthogonal to one another at every point of the circle \(K_\varphi\); therefore all isolated singular points of the field \(\operatorname{grad} u_\varphi\) coincide with the isolated singular points of the field \(G_\varphi\). Since \(u_\varphi\) is a continuous function, the field \(G_\varphi\) can have isolated singular points of only two types: saddles and centers, and a singular point of center type is an extremum point of the function \(u_\varphi\). As \(\varphi\) varies, the isolated singular points of the field \(G_\varphi\) describe a certain finite number of closed curves \(L_i\). Let the curve \(L_1\) be a curve on which all the functions \(u_\varphi\) attain a maximum. Denote by \(\mathfrak{w}_1\) the value of the harmonic function \(u\) on the curve \(L_1\). Since the curve \(L_1\) is closed, the function \(\mathfrak{w}_1\) attains a maximum at some point \(X\). This point \(X\) is also a point of maximum for the function \(u\). By the maximum principle for harmonic functions, \(u \equiv \operatorname{const}\).
Thus, if \(u_\varphi\) is the value on the plane \(T_\varphi\) of a solution \(u\) of problem (18), then the vector field \(\operatorname{grad} u_\varphi\) in the circle \(K_\varphi\) can have only isolated singular points of saddle type.
If the rotation \(n\) of the vector field \(P_\varphi : \{xa, ya, b\}\) on the circumference \(C_\varphi\) satisfies the inequality \(n<0\), then the gradient of a solution \(u\) of problem (18) in the domain \(D\) vanishes on no fewer than one analytic curve and on no more than \(-n\) analytic curves. Moreover, these curves are closed curves and intersect one another in no more than a finite number of points. Denote by \(B\) the set of points of the domain \(D\) at which \(\operatorname{grad} u=0\), and by \(M\) denote the set of points of the domain \(D\) lying on level surfaces of the function \(u\) that pass through points of the set \(B\). If the set \(B\) is removed from the domain \(D\), then in the resulting domain \(D_1\) the set \(M\) decomposes into no more than \(-4n\) connected components \(M_i\), each of which is a two-dimensional real-analytic manifold.
Consider the homogeneous problem corresponding to problem (1). Let \(u_1\) and \(u_2\) be solutions of problem (18); then the function \(v=u_2-u_1\) is a solution of the homogeneous problem corresponding to problem (1). Obviously, in this form every solution of the homogeneous problem corresponding to problem (1) can be represented. If at no point of the surface \(S\) is the gradient of the solution \(v\) of the homogeneous problem corresponding to problem (1) orthogonal to the plane \(T_\varphi\) passing through this point, then the vector field \(\operatorname{grad} v\) has all the properties that were established above for the gradient field of a solution of problem (18), since the rotations of the fields \(\operatorname{grad} v\) and \(P\) on the circumference \(C_\varphi\) coincide.
From all that has been said above there follows the following
Theorem 3. Let \(u\) be a harmonic function regular in the domain \(D\), whose gradient at no point of the surface \(S\) is orthogonal to the plane \(T_\varphi\) passing through this point; let the rotation of the field \(\operatorname{grad} u_\varphi\) on the circumference \(C_\varphi\) be equal to \(k\). If the rotation \(n\) of the field \(P_\varphi : \{xa, ya, b\}\) on the circumference \(C_\varphi\) is negative, then the function
\[ \mu=\left[ a(\zeta,\varphi)\left(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}\right) +b(\zeta,\varphi)\frac{\partial u}{\partial z} \right]_{S} \]
assumes both positive and negative values when \(k\ne n\).
Proof. Since \(k\ne n\), the vector fields \(P_\varphi\) and \(\operatorname{grad} u_\varphi\) are not homotopic, and consequently at least at one point of the circumference \(C_\varphi\) the vectors of these fields are directed in the same way and at least at one point have opposite directions [4]. The assertion of the theorem follows from this.
Remark 3. An analogous theorem holds when, as the domain \(D\), one considers an axisymmetric domain homeomorphic to a ball and conditions (15) are satisfied, and also for the oblique-derivative problem for equation (17) with boundary condition (1).
Let \(u_0\) denote the solution of problem (18) for \(g(\xi,\varphi)\equiv 1\), and let \(B_0\) denote the set of points of the domain \(D\) at which the equality \(\operatorname{grad} u_0=0\) holds. Next, let \(M_0\) denote the set of points of the domain \(D\) lying on the level surfaces of the function \(u_0\) that pass through points of the set \(B_0\), with the points of the set \(B_0\) themselves not included in the set \(M_0\). As was noted above, the set \(M_0\) decomposes into \(l\leq -4n\) connected components.
Let \(u_i\) denote a solution of the homogeneous problem corresponding to problem (18), whose gradient vanishes on the set \(B_0\), and let the set \(M_i\) (which is constructed from the function \(u_i\) in the same way as the set \(M_0\) from the function \(u_0\)) consist of \(l\) connected components. We shall show that there can be no more than two linearly independent solutions \(u_i\) of the homogeneous oblique-derivative problem having this property.
Suppose that there exist at least three such solutions \(u_1,u_2,u_3\). Consider the function
\[ w=C_1(u_0+u_1)+C_2(u_0+u_2)+C_3(u_0+u_3), \]
where \(C_1,C_2,C_3\) are arbitrary constants. These constants can always be chosen so that \(C_0=C_1+C_2+C_3\ne 0\) and \(\operatorname{grad} w\) vanishes, in addition to the set \(B_0\), on at least one closed curve \(A_0\). It follows from this that the rotation of the field \(\operatorname{grad} w\) on the circle \(C_\varphi\) is equal to \(n-1\) or is less than this number. The function \(w\) satisfies boundary condition (18) for \(g(\xi,\varphi)\equiv C_0\). As was noted above, the rotation of the field \(\operatorname{grad} w\) on the circle \(C_\varphi\) is equal to \(n\). The contradiction obtained shows that there cannot be three such functions.
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Received by the editors
March 2, 1966
Institute of Mathematics, Siberian Branch of the Academy of Sciences of the USSR