ON A CERTAIN CLASS OF PROBLEMS IN INTEGRAL GEOMETRY
MATHEMATICS
Submitted 1967-01-01 | SovietRxiv: ru-196701.93595 | Translated from Russian

Abstract Generated abstract

This note studies an inverse problem in integral geometry for a two-parameter family of curves in the upper half-plane: to recover a continuous function from its integrals over the curves. Under assumptions of uniqueness of curves through point pairs, nesting of bounded regions, and smooth dependence on parameters, the authors state a uniqueness theorem showing that the function is determined inside a region bounded by one curve from the integral data on the corresponding parameter square. The proof is outlined by introducing partial curve integrals, deriving a second-order differential equation involving the curvature of the curve family, transforming variables to reduce the problem to a Cauchy problem, and applying Hörmander’s uniqueness theorem for linear partial differential operators.

Full Text

UDC 517.946

MATHEMATICS

M. M. LAVRENT’EV, Yu. E. ANIKONOV

ON A CERTAIN CLASS OF PROBLEMS IN INTEGRAL GEOMETRY

(Presented by Academician I. M. Vinogradov on 10 VI 1967)

Let \(\Gamma(x_1,x_2)\) be a two-parameter family of curves in the half-plane \(y>0\), satisfying the following conditions:

1) Through any pair of points of the half-plane with coordinates \((x_1,y_1)\), \((x_2,y_2)\), \(x_1\ne x_2\), there passes one and only one curve of the family. Moreover, through the pair of points \((x_1,0)\), \((x_2,0)\) there passes the curve \(\Gamma(x_1,x_2)\).

2) If \(x_1' < x_1 < x_2 < x_2'\), then the part of the half-plane bounded by the curve \(\Gamma(x_1,x_2)\) lies strictly inside the part of the half-plane bounded by the curve \(\Gamma(x_1',x_2')\).

3) The functions defining the equations of the curves \(\Gamma\) in normal form are three times continuously differentiable with respect to arc length and to the parameters \(x_1,x_2\).

Further, let \(u(x,y)\) be a function continuous in the half-plane, and let

\[ v(x_1,x_2)=\int_{\Gamma(x_1,x_2)} u(x,y)\,ds . \]

Consider the following problem of integral geometry: given the function \(v(x_1,x_2)\), it is required to determine the function \(u(x,y)\).

A number of inverse problems for partial differential equations reduce to the formulated problem of integral geometry (see \((^3,^4)\)). In the work of V. G. Romanov \((^4)\), analogous problems of integral geometry were investigated by the method of moments.

In the present note we give a formulation and a scheme of proof of the uniqueness theorem for the posed problem, without those additional restrictions on the families of curves \(\Gamma\) that are contained in \((^5)\).

Theorem. The function \(u(x,y)\) is uniquely determined in the part of the half-plane \(y>0\) bounded by the curve \(\Gamma(x_1^0,x_2^0)\), \(x_1^0<x_2^0\), by the values of the function \(v(x_1,x_2)\) in the square \(x_1^0\le x_1\le x_2^0\), \(x_1^0\le x_2\le x_2^0\).

We give the scheme of the proof of the theorem. Consider some point \((x,y)\) lying on the curve \(\Gamma(x_1,x_2)\), and denote by \(\theta\) the angle formed by the tangent to the curve \(\Gamma(x_1,x_2)\) at the point \((x,y)\) with the \(x\)-axis, \(x_1<x_2\). Denote by \(\widetilde{\Gamma}(x,y,\theta)\) the part of the curve \(\Gamma(x_1,x_2)\) lying between the points \((x_1,0)\), \((x,y)\), and consider the function

\[ w(x,y,\theta)=\int_{\widetilde{\Gamma}(x,y,\theta)} u(\xi,\eta)\,ds . \]

It is not difficult to verify that the function \(w\) satisfies the following second-order differential equation:

\[ \frac{\partial}{\partial\theta}\left(\cos\theta\,\frac{\partial}{\partial x}+\sin\theta\,\frac{\partial}{\partial y}-K\,\frac{\partial}{\partial\theta}\right)w=0, \tag{1} \]

where \(K(x,y,\theta)\) is the curvature of the curve \(\Gamma(x_1,x_2)\) at the point \((x,y)\).

The differential operator in equation (1) can be brought to the form

\[ \begin{gathered} \left(K\partial/\partial\theta-\frac12\cos\theta\,\partial/\partial x-\frac12\sin\theta\,\partial/\partial y\right)^2-\\ -\frac14\left(\cos\theta\,\partial/\partial x+\sin\theta\,\partial/\partial y\right)^2+\\ +K\left(\sin\theta\,\partial/\partial x-\cos\theta\,\partial/\partial y\right) +\left(\cos\theta K_x+\sin\theta K_y\right)\partial/\partial\theta . \end{gathered} \tag{2} \]

Let us consider, instead of the variables \(x,y,\theta\), new variables \(t,\alpha,\beta\) such that \(t,\alpha,\beta\), as functions of \(x,y,\theta\), satisfy the equations

\[ K\,\partial t/\partial\theta-\frac12\cos\theta\,\partial t/\partial x-\frac12\sin\theta\,\partial t/\partial y=0, \]

\[ \cos\theta\,\partial\alpha/\partial x+\sin\theta\,\partial\alpha/\partial y=0, \]

\[ \cos\theta\,\partial\beta/\partial x+\sin\theta\,\partial\beta/\partial y=0. \]

In the variables \(t,\alpha,\beta\), equation (1) takes the form:

\[ a_0\partial^2 w/\partial t^2-\left(a_1\partial/\partial\alpha+a_2\partial/\partial\beta\right)^2 w +a_3\,\partial w/\partial t+ \]

\[ +a_4\,\partial w/\partial\alpha+a_5\,\partial w/\partial\beta=0, \tag{3} \]

where the \(a_j\) are certain functions.

With a suitable choice of the variables \(t,\alpha,\beta\), the integral-geometry problem under consideration is reduced to the Cauchy problem for equation (3). The Cauchy data are prescribed on the surface

\[ t=\varphi(\alpha,\beta), \]

where \(\varphi\) is a function with positive second differential.

The formulated uniqueness theorem follows, after the transformations indicated above, from Hörmander’s general theorem (see \((^2)\)).

Computing Center
of the Siberian Branch of the Academy of Sciences of the USSR

Received
9 VI 1967

REFERENCES

\(^1\) I. M. Gelfand, M. I. Graev, N. Ya. Vilenkin, Integral Geometry and Related Questions of Representation Theory, Moscow, 1967.
\(^2\) L. Hörmander, Linear Partial Differential Operators, Moscow, 1965.
\(^3\) M. M. Lavrent’ev, V. G. Romanov, DAN, 171, No. 6 (1966).
\(^4\) V. G. Romanov, DAN, 173, No. 4 (1967).
\(^5\) V. G. Romanov, Siberian Mathematical Journal, 8 (1967) (in press).

Submission history

ON A CERTAIN CLASS OF PROBLEMS IN INTEGRAL GEOMETRY