MATHEMATICS

Corresponding Member of the Academy of Sciences of the USSR A. N. TIKHONOV

ON NONLINEAR EQUATIONS OF THE FIRST KIND

The purpose of the present article is to develop methods for solving equations of the first kind

\[ A[z]=u \qquad (z\in Z,\ u\in U), \]

where \(A[z]\) is a continuous operator from \(Z\) into \(U\), satisfying the uniqueness condition: \(A[z_1]\ne A[z_2]\), if \(z_1\ne z_2\), and \(Z\) and \(U\) are metric spaces satisfying the conditions formulated below. These methods are a development of the ideas set forth in \((^{1-3})\).

1. We shall assume that in \(Z\) there exists a set \(\overline Z\), which admits a new metrization \(\rho_1(\overline z_1,\overline z_2)\), majorant with respect to the metrization \(\rho_0(z_1,z_2)\) of the space \(Z\):

\[ \rho_0(\overline z_1,\overline z_2)\leqslant \rho_1(\overline z_1,\overline z_2), \]

and, moreover, the sphere \(\overline S_C(\overline z_0)=\{\overline z:\rho_1(\overline z,\overline z_0)\leqslant C\}\) is compact in \(Z\) in the sense of the metric \(\rho_0\). We shall say that the space \(\overline Z_1\) (the set \(\overline Z\) in the metric \(\rho_1\)) is \(s\)-compactly embedded in \(Z\).

Let the space \(\overline Z_1\), \(s\)-compactly embedded in \(Z\), be a convex closed set of some Hilbert space, so that
\(\rho_1(\overline z_1,\overline z_2)=(\overline z_1-\overline z_2,\overline z_1-\overline z_2)_1=\|\overline z_1-\overline z_2\|_1\), and let the concept of a segment be defined

\[ [\overline z_1,\overline z_2]=\{\overline z=\alpha \overline z_1+\beta \overline z_2\subset \overline Z,\ \alpha>0,\ \beta>0,\ \alpha+\beta=1\}. \]

We shall assume that the diameter of the segment \([\overline z_1,\overline z_2]\) in the norm \(\rho_0\) tends to zero if \(\rho_0(\overline z_1,\overline z_2)\to 0\), which is always fulfilled if the space \(Z\) is normed.

The spaces \(Z=C(a,b)\) and \(\overline Z=W_2^1(a,b)\), or

\[ \overline Z=\{\overline z(s):\overline z\in W_2^1,\ q(s)\leqslant \overline z(s)\leqslant p(s)\}, \]

where \(q(s)\) and \(p(s)\) are prescribed bounded functions, satisfy these conditions.

Consider the smoothing functional

\[ M^\alpha[\overline z,\widetilde u]=\rho_2^2(A[\overline z],\widetilde u)+\alpha\Omega[\overline z]\qquad(\overline z\in\overline Z), \]

where \(\widetilde u\) is an arbitrary element of \(U\); \(\rho_2(u_1,u_2)\) is the metric of \(U\), \(\Omega[\overline z]=\|\overline z\|^2\), and \(\alpha>0\) is a numerical parameter.

Theorem 1. If \(\overline Z\) is a closed convex set of a Hilbert space, then for every \(\widetilde u\) and \(\alpha>0\) there exists an element \(z^\alpha\in\overline Z\) realizing the minimum \(M^\alpha[\overline z,\widetilde u]\).

Let

\[ M_0=\inf M^\alpha[\overline z,\widetilde u], \qquad \overline z\in\overline Z, \]

and let \(\overline z_n\) be a minimizing sequence

\[ M_n=M^\alpha[\overline z_n,\widetilde u]\to M_0 \qquad (\text{as } n\to\infty). \]

Without loss of generality one may assume that \(M_{n-1}\leqslant M_n\). Then we have

\[ \Omega(\overline z_n)\leqslant C_1^2 \qquad \left(C_1^2=\frac{1}{\alpha}M_1\right), \]

i.e. \(\overline z_n\in\overline S_{C_1}\) for every \(n\). Without changing the notation, we shall assume that the sequence \(\overline z_n\) converges to \(z^\alpha\) in the sense of the metric \(\rho_0\). It is not difficult to see

show that \(z^\alpha \in \overline Z\). Indeed, for any \(\varepsilon\) we have \(\|\bar z_n-\bar z_{n+p}\|_1 \leq \varepsilon\), if \(n \geq n(\varepsilon)\) and \(p>0\). Suppose that this is not so, i.e., that there exists an \(\varepsilon_0\) such that

\[ \|\bar z_n-\bar z_{n+p_n}\|_1 \geq \varepsilon_0 \]

for arbitrarily large indices \(n\). Put \(\xi_n=\bar z_n-\bar z_{n+p_n}\). Obviously,

\[ \rho_0(\bar z_n,\bar z_{n+p_n})\to 0 \quad (n\to\infty). \]

Consider the element \(\xi_n=\frac12(z_n+z_{n+p_n})=z_n-\frac12\xi_n=\bar z_{n+p_n}+\frac12\xi_n\) and

\[ M^\alpha[\xi_n,u] = \rho_2^2\bigl(A[\bar z_n-\tfrac12\xi_n],\tilde u\bigr) +\alpha\bigl(\|\bar z_n\|_1^2-(\bar z_n,\xi_n)_1+\tfrac14\|\xi_n\|_1^2\bigr) \geq M_0. \]

Since \(\rho_0(\xi_n,z^\alpha)\to 0\) and \(\rho_0(\xi_n,\bar z_n)\to 0\), we have

\[ \rho_2^2(A[\xi_n],\tilde u)=\rho_2^2(A[\bar z_n],\tilde u)+\Delta_n^{(1)} \quad (\Delta_n^{(1)}\to 0 \text{ as } n\to\infty), \]

whence it follows that

\[ \alpha\bigl(-(\dot z_n,\xi_n)_1+\tfrac14\|\xi_n\|_1^2\bigr) \geq M_0-M_n-|\Delta_n^{(1)}|=-\Delta_n', \]

\[ (\Delta_n'>0 \text{ and } \Delta_n'\to 0,\ n\to\infty). \]

Similarly, representing \(\xi_n\) as \(\bar z_{n+p_n}+\frac12\xi_n\), we obtain

\[ \alpha\bigl((\bar z_{n+p_n},\xi_n)_1+\tfrac14\|\xi_n\|_1^2\bigr) \geq -\Delta_n'' \quad (\Delta_n''>0 \text{ and } \Delta_n''\to 0,\ n\to\infty), \]

and also

\[ \alpha\bigl(-(\bar z_n-\bar z_{n+p_n},\xi_n)_1+\tfrac12\|\xi_n\|_1^2\bigr) = -\tfrac12\alpha\|\xi_n\|_1^2 \geq -(\Delta_n'+\Delta_n'') \]

or

\[ \|\xi_n\|_1=\|\bar z_n-\bar z_{n+p_n}\|_1 \leq \varepsilon_0/2 \quad \text{for } n\geq n(\varepsilon), \]

which contradicts the assumption. By the completeness of the space \(\overline Z\), the element \(z^\alpha\in\overline Z\).

Theorem 2. Let \(\bar u^{(0)}=A[\bar z^{(0)}]\), where \(\bar z^{(0)}\in\overline Z\) (\(\overline Z\) satisfies the conditions of Theorem 1), and let \(\rho_2(\tilde u,\bar u^{(0)})\leq\delta\). Then for any \(\varepsilon>0\) and any function \(\alpha_0(\delta)\): \(\alpha_0(\delta)\to 0\) as \(\delta\to 0\) and \(\alpha_0(\delta)\geq q_0\delta^2\), there exists a \(\delta_0(\varepsilon\|\bar z^{(0)}\|_1,\alpha_0,q_0)\) such that any element \(z^\alpha\) realizing the minimum of \(M^\alpha[\bar z,\tilde u]\) in \(\overline Z\) satisfies the condition

\[ \rho_0(z^\alpha,\bar z^{(0)})\leq\varepsilon, \quad \text{if } \quad q_0\delta^2\leq\alpha\leq\alpha_0(\delta). \]

Indeed,

\[ M^\alpha[z^\alpha,\tilde u]\leq M^\alpha[\bar z^{(0)},\tilde u]\leq \delta^2+\alpha\|\bar z^{(0)}\|_1^2, \]

whence:

\[ 1^\circ.\quad \|z^\alpha\|^2\leq \frac1\alpha\bigl(\delta^2+\alpha\|\bar z^{(0)}\|_1^2\bigr) \leq \frac1{q_0}+\|\bar z^{(0)}\|_1^2 = C^2 \quad (\text{i.e. } z^\alpha\in\overline S_C). \]

\[ 2^\circ.\quad \rho_2(A[z^\alpha],A[\bar z^{(0)}]) \leq \rho_2(A[z^\alpha],\tilde u)+\delta \leq \delta+\sqrt{\delta^2+\alpha_0(\delta)\|\bar z^{(0)}\|_1^2} =\eta(\delta). \]

Since all \(z^\alpha\) and \(\bar z^{(0)}\) belong to the compact set \(K_C\)—the closure of \(\overline S_C\) in \(Z\)—there exists an \(\bar\eta(\varepsilon)\) such that if \(\rho_2(u^\alpha,\bar u^{(0)})\leq \bar\eta(\varepsilon)\), then \(\rho_0(z^\alpha,\bar z^{(0)})\leq\varepsilon\). Choosing \(\delta_0\) so that \(\eta(\delta)\leq\bar\eta(\varepsilon)\) for \(\delta\leq\delta_0\), we shall have

\[ \rho_0(z^\alpha,\bar z^{(0)})\leq\varepsilon \quad \text{for } \delta\leq\delta_0,\ \text{if } q\delta^2\leq\alpha\leq\alpha_0(\delta). \]

II. 2. If one does not make the assumption of completeness of the space \(\bar Z\), then the assertion that \(z^\alpha \in \bar Z\) is false. However, if \(\hat z^\alpha\) denotes some element for which

\[ M^\alpha[\hat z^\alpha,\tilde u]\leq M_0+\alpha C\qquad (C=\mathrm{const}), \]

then the following holds.

Theorem 3. Let \(\bar Z\) be \(s\)-compactly embedded in \(Z\), \(\bar u^{(0)}=A[\bar z^{(0)}]\), where \(\bar z^{(0)}\in \bar Z\). Then, for any \(\varepsilon\) and any function \(\alpha_0(\delta): \alpha_0(\delta)\to 0\) as \(\delta\to 0\), and \(\alpha_0(\delta)\geq q_0\delta^2\), there exists a \(\delta_0(\varepsilon,\|\bar z\|,\alpha_0,q_0,C)\) such that any element \(\hat z^\alpha\) for which

\[ M^\alpha[\hat z^\alpha,\tilde u]\leq M_0+\alpha C \qquad (\rho_2(\tilde u,\bar u^{(0)})\leq \delta), \]

satisfies the condition

\[ \rho_0(\hat z^\alpha,\bar z^{(0)})\leq \varepsilon, \qquad \text{if } q_0\delta^2\leq \alpha\leq \alpha_0(\delta). \]

The proof of this theorem is close in idea to the proof of Theorem 2.

II. 3. In the present subsection we shall give a condition under which \(z^\alpha\) is determined uniquely.

Theorem 4. If \(Z\) is normed, \(\bar Z\) and \(U\) are Hilbert spaces, \(A[\bar z^{(0)}]=\bar u^{(0)}\), and the operator \(A(z)=u\) has a first differential \(A[z,\xi]\) satisfying the conditions:

1) \(\|A_1[z_1,\xi]-A_1[z_2,\xi]\|_2\leq A_2[\bar z^{(0)}]\|z_1-z_2\|_0\|\xi\|_0\), if \(\|z_1-\bar z^{(0)}\|_0\leq \varepsilon\), \(\|z_2-\bar z^{(0)}\|_0\leq \varepsilon\);

2) \(A_1[\bar z^{(0)},\xi]\equiv 0\) only for \(\xi\equiv 0\),

and if \(q_0\delta^2\leq \alpha\leq \alpha_0(\delta)\), then there exists a \(\delta_1(\|\bar z^{(0)}\|_1,q_0,\alpha_0)\) such that for every function \(\tilde u\) such that \(\|\tilde u-\bar u^{(0)}\|_2\leq \delta_1\), the function \(z^\alpha\) is determined uniquely.

Let \(z^\alpha\) be some element realizing the minimum

\[ M^\alpha[\bar z,\tilde u]=\|A[\bar z]-\tilde u\|_2^2+\alpha\|\bar z\|_1^2. \]

In this case

\[ \delta M^\alpha[z^\alpha,\xi,\tilde u] =(A[z^\alpha]-\tilde u;\,A_1[z^\alpha,\xi])_2+\alpha(z^\alpha,\xi)=0. \]

If there exist two elements \(z_1^\alpha\) and \(z_2^\alpha\) for which \(\delta M^\alpha\), then, taking their difference \(v=z_1^\alpha-z_2^\alpha\), we obtain

\[ R=\delta M^\alpha[z_1^\alpha,\xi,\tilde u]-\delta M^\alpha[z_2^\alpha,\xi,\tilde u] =(A[z_1^\alpha]-A[z_2^\alpha];\,A_1[z_1^\alpha,\xi])_2+ \]

\[ +(A[z_2^\alpha]-\tilde u;\,A_1[z_1^\alpha,\xi]-A_1[z_2^\alpha,\xi])_2+\alpha(v,\xi)_1=0. \]

We shall use the fact that

\(1^\circ.\) \(A[z_1^\alpha]-A[z_2^\alpha]=A_1[\bar z^{(0)},v]+\hat u,\quad \|\hat u\|_2\leq \eta(\delta)\|v\|_0\) \((\eta(\delta)\to 0\) as \(\delta\to 0)\), since \(z_1^\alpha\) and \(z_2^\alpha\), by Theorem 2, for sufficiently small \(\delta\), lie in an \(\varepsilon\)-neighborhood of \(\bar z^{(0)}\).

\(2^\circ.\) \(\|A_1[z_1^\alpha,\xi]-A_1[z_2^\alpha,\xi]\|_2\leq A_2\|v\|_0\|\xi\|_0.\)

\(3^\circ.\) \(A_1[z_1^\alpha,\xi]=A_1[\bar z^{(0)},\xi]+\hat u,\quad \|\hat u\|_2\leq \eta(\delta)\|v\|_0.\)

\(4^\circ.\) \(\|A[z_2^\alpha]-\tilde u\|_2\leq \eta(\delta).\)

Putting \(\xi=v\), we shall have

\[ R=\|A_1[\bar z^{(0)},v]\|_2^2+r+\alpha\|v\|_1^2=0, \qquad |r|\leq \eta(\delta)\|v\|_0^2. \]

Let us verify that

\[ \|A_1[\bar z^{(0)},v]\|_2^2\geq s_0^2\|v\|_0^2, \]

from which it will follow that \(v\equiv 0\).

Indeed, in the proof of Theorem 2 it was established that \(\|z^\alpha\|_1\leq C\); thus \(\|v\|_1\leq 2C\). Consider

\[ s_0=\inf \|A_1[\bar z^{(0)},\xi]\|_2,\qquad \xi\in \overline{Z}\{\xi:\ \|\xi\|_0=1,\ \|\xi\|_1\leq 2C\}. \]

We shall show that \(s_0 > 0\). Let \(\zeta_n\) be a minimizing sequence:
\(\|A_1[\bar z^{(0)}, \zeta_n]\|_2 = s_n \to s_0\) (as \(n \to \infty\)). In view of the fact that \(\bar Z\) is compact in \(Z\), we may assume that \(\zeta_n\) converges in the metric \(\|\ \|_0\) to some element \(\zeta_0\), and, by the continuity of \(A_1[\bar z^{(0)}, \zeta]\) with respect to \(\zeta\), in the metric \(\|\ \|_0\) we shall have \(\|A_1[\bar z^{(0)}, \zeta_0]\| = s_0\), whence we conclude that \(s_0 \ne 0\), since \(\|\zeta_0\| = 1\).

The proof given is based on the fact that the functional \(M^\alpha[z,\tilde u]\) has only a single stationary point for sufficiently small \(\alpha\). With the help of this proposition, the method of steepest descent is justified for sufficiently small \(\alpha\).

REFERENCES

\({}^{1}\) A. N. Tikhonov, DAN, 151, No. 3, 501 (1963).
\({}^{2}\) A. N. Tikhonov, DAN, 156, No. 6, 1296 (1964).
\({}^{3}\) A. N. Tikhonov, S. B. Glasko, Zhurn. vychisl. matem. i matem. fiz., 4, 564 (1964).