MATHEMATICS

I. S. Arshon and M. A. Evgrafov

ESTIMATE OF THE GROWTH OF A SOLUTION OF A SYSTEM WITH NONHOMOGENEOUS BOUNDARY CONDITIONS AND THE PHRAGMÉN—LINDELÖF THEOREMS

(Presented by Academician M. V. Keldysh, May 3, 1960)

We shall consider the system

\[ \frac{\partial u}{\partial y} = A\frac{\partial u}{\partial x}+Pu,\qquad B_1u(x,0)+B_2u(x,1)=f(x), \tag{1} \]

where \(A, P, B_1, B_2\) are matrices of order \(n\); \(u\) and \(f\) are vectors.

We impose the following restrictions on the matrices \(A, B_1, B_2\):

1) Among the eigenvalues of the matrix \(A\) there are no purely real ones.

2) Denote by \(E^+\), respectively \(E^-\), the projection operator onto the sum of the invariant subspaces of the matrix \(A\) corresponding to the eigenvalues \(a_k\) for which \(\operatorname{Im} a_k>0\), respectively \(\operatorname{Im} a_k<0\), and require that

\[ \det(B_1E^+ + B_2E^-)\ne 0,\qquad \det(B_1E^- + B_2E^+)\ne 0. \tag{2} \]

Conditions 1) and 2), generally speaking, do not ensure completeness of the eigenfunctions of the system

\[ \frac{du}{dy}=(zA+P)u,\qquad B_1u(0)+B_2u(1)=0. \tag{3} \]

With respect to the vector \(f(x)\), assume that

\[ 3)\qquad \|f(x)-f(s)\|<|x-s|^\alpha(\varphi(x)+\varphi(s)),\qquad \alpha>0, \tag{4} \]

where

\[ \varphi(x)>0,\qquad \lim_{x\to+\infty}\frac{\varphi'(x)}{\varphi(x)}=0. \]

Theorem 1. If \(\det(B_1+B_2e^{P+zA})\) has no purely imaginary zeros, then there exists a solution \(u_0(x,y)\) of system (1) for which

\[ u_0(x,y)=O(\varphi(x)+\|f(x)\|). \]

(It should be noted that the zeros of \(\det(B_1+B_2e^{P+zA})\) are eigenvalues of system (3).)

Theorem 2. If the greatest multiplicity of the purely imaginary zeros of \(\det(B_1+B_2e^{P+zA})\) is equal to \(p\), then there exists a solution of system (1) for which

\[ u_0(x,y)=O\left(\varphi(x)+\|f(x)\|+x^{p-1}\int_0^x \|f(t)\|\,dt\right). \]

Theorem 3. If \(\det(B_1+B_2e^{P+zA})\) has one simple purely imaginary zero \(z=i\lambda_0\), then there exists a solution of system (1) of the form

\[ u_0(x,y)=e^{Py+i\lambda_0(x+Ay)}G_0\int_0^x e^{-i\lambda_0 t}f(t)\,dt +O(\varphi(x)+\|f(x)\|), \]

where \(G_0\) is a certain constant matrix of rank one.

The example given at the end of the article shows that in Theorems 1–3 one cannot dispense with restriction 3), even in the classical case of the system

Cauchy--Riemann (boundedness of a harmonic function does not imply boundedness of its conjugate).

Combining Theorems 1--3 with the results of the note [1], we obtain Phragmén--Lindelöf theorems for system (1). For example:

Theorem 4. If \(\det(B_1+B_2e^{P+zA})\) has no zeros in the strip \(0\le \operatorname{Re} z<\beta\), \(u(x,y)\) is a solution of system (1) satisfying the condition \(u(x,y)=o(e^{\beta x})\), \(x\to+\infty\), then
\[ u(x,y)=O(\varphi(x)+\|f(x)\|). \]

We shall divide the proof of Theorems 1--3 into four lemmas.

Lemma 1. Put
\[ M(z,y)=e^{(P+zA)y}(B_1+B_2e^{P+zA})^{-1}. \]
For
\[ |\arg z\pm \pi/2|\le \eta,\qquad \eta=\frac12\min_k\{|\arg(\pm a_k)|\}, \]
\[ \|M(z,y)\|<Ce^{-\tau y(1-y)|z|},\qquad 0\le y\le 1. \]

Proof. Put \(A(z)=A+\dfrac1zP\). It is clear that, for sufficiently large \(z\), among the eigenvalues of the matrix \(A(z)\) there are no purely real ones. If \(E^+(z)\) and \(E^-(z)\) denote for \(A(z)\) the same as \(E^+\) and \(E^-\) for \(A\), then, obviously, \(E^\pm(z)\to E^\pm\) as \(z\to\infty\), and therefore
\[ \det(B_1E^\pm(z)+B_2E^\mp(z))\to \det(B_1E^\pm+B_2E^\mp). \]

Let, for definiteness, \(z\to\infty\), \(|\arg z-\pi/2|<\eta\). Then
\[ E^-(z)e^{zA(z)}=O(e^{-\tau|z|}),\qquad E^+(z)e^{-zA(z)}=O(e^{-\tau|z|}),\qquad \tau>0. \tag{5} \]
Since \(M(z,y)=(B_1e^{-yzA(z)}+B_2e^{(1-y)zA(z)})^{-1}\), putting
\[ D=B_1E^-(z)e^{-yzA(z)}+B_2E^+(z)e^{(1-y)zA(z)}, \]
\[ \varepsilon=B_1E^+(z)e^{-yzA(z)}+B_2E^-(z)e^{(1-y)zA(z)}, \]
we can, by virtue of the identity \(E^+(z)+E^-(z)=E\), write
\[ M(z,y)=(D+\varepsilon)^{-1}=D^{-1}(E+\varepsilon D^{-1})^{-1} =D^{-1}-D^{-1}\varepsilon D^{-1}(E+\varepsilon D^{-1})^{-1}. \tag{6} \]
Using the identities \(E^\pm(z)E^\mp(z)=0\), \((E^\pm(z))^2=E^\pm(z)\), and the commutativity of \(E^\pm(z)\) and \(A(z)\), we easily find
\[ D=(B_1E^-(z)+B_2E^+(z))(E^-(z)e^{-yzA(z)}+E^+(z)e^{(1-y)zA(z)}), \]
\[ D^{-1}=(E^-(z)e^{yzA(z)}+E^+(z)e^{-(1-y)zA(z)})(B_1E^-(z)+B_2E^+(z))^{-1}, \]
\[ \varepsilon D^{-1}=(B_1E^+(z)e^{-zA(z)}+B_2E^-(z)e^{zA(z)})(B_1E^-(z)+B_2E^+(z))^{-1}. \]
By (5) this gives us \(D^{-1}=O(e^{-\tau y(1-y)|z|})\), \(\varepsilon D^{-1}=O(e^{-\tau|z|})\), whence, with the aid of (6), we obtain the assertion of the lemma.

Lemma 2. The residue of \(M(z,y)e^{zx}\) in the pole \(z=z_n\) of multiplicity \(p\) has the form \(e^{z_nx}H_n(x,y)\), where \(H_n(x,y)\) is a polynomial in \(x\) of degree \(p-1\) and
\[ B_1H_n(x,0)+B_2H_n(x,1)=0. \]

Proof. We have
\[ \operatorname*{res}_{z=z_n} M(z,y)e^{zx} =\frac{1}{(p-1)!}\frac{d^{p-1}}{dz^{p-1}}(z-z_n)^p e^{zx}M(z,y)\bigg|_{z=z_n} =e^{z_nx}\sum_{k=0}^{p-1}x^kH_{n,k}(y), \]
and, since by definition \(B_1M(z,0)+B_2M(z,1)=E\), it follows that
\[ B_1H_n(x,0)+B_2H(x,1) =\frac{e^{-z_nx}}{(p-1)!}\frac{d^{p-1}}{dz^{p-1}}(z-z_n)^p e^{zx}E\bigg|_{z=z_n}=0, \]
which proves the lemma.

Denote by \(L_\delta^\pm\) the boundary of the angle \(|\arg(\pm z-\delta)|<\pi/2-\eta_1\), where \(\delta>0\) and \(0<\eta_1<\eta\) (see Lemma 1) are chosen so that between \(L_\delta^+\) and \(L_\delta^-\) lie ...

have only purely imaginary poles of \(M(z,y)\). By \(L_0\) we denote an arbitrary closed contour lying between \(L_\delta^+\) and \(L_\delta^-\) and containing inside itself the point \(z=0\) and all purely imaginary poles of \(M(z,y)\).

Lemma 3. Denote
\[ K(x,y)=\frac{1}{2\pi i}\int_{L_\delta^-} M(z,y)e^{zx}\,dz,\qquad x>0, \]
\[ K(x,y)=\frac{1}{2\pi i}\int_{L_\delta^+} M(z,y)e^{zx}\,dz,\qquad x<0. \]

The inequality
\[ \|K(x,y)\|<\frac{Ce^{-\delta |x|}}{|x|+\tau_1y(1-y)},\qquad 0\le y\le 1 \]
holds.

Proof. Let, for definiteness, \(x<0\). From Lemma 1 we obtain
\[ \|K(x,y)\|<\frac{1}{2\pi}\int_{L_\delta^+}|e^{zx}|\,\|M(z,y)\|\,|dz| <C_1\int_0^\infty e^{-\delta |x|-r|x|\sin\eta_1}e^{-r\tau y(1-y)}\,dr= \]
\[ =\frac{C_1e^{-\delta |x|}}{\tau y(1-y)+|x|\sin\eta_1} =\frac{Ce^{-\delta |x|}}{|x|+\tau_1y(1-y)}, \]
as was required.

We define the desired solution \(u_0(x,y)\) by the formula
\[ u_0(x,y)=\int_{-\infty}^{\infty}K(x-t,y)\bigl(f_1(t)-f_1(x)\bigr)\,dt +\sum_{\operatorname{Re}z_n=0}\int_{-\infty}^{x} e^{z_n(x-t)}H_n(x-t,y)\times \]
\[ {}\times f_1(t)\,dt+\frac{1}{2\pi i}\int_{L_0}\frac{M(z,y)}{z}\,dz\, f_1(x) =u_1(x,y)+u_2(x,y)+u_3(x,y), \]
where \(f_1(x)=f(x)\) for \(x\ge 0\), \(f_1(x)=0\) for \(x\le -1\), and it satisfies condition (4).

Lemma 4. The function \(u_0(x,y)\) is continuous for \(x\ge 0,\ 0\le y\le 1\) and satisfies system (1). Moreover, \(u_1(x,y)=O(\varphi(x))\).

Proof. We have
\[ \|u_1(x,y)\|= \left\|\int_0^\infty K(u,y)\bigl(f_1(x-u)-f_1(x)\bigr)\,du +\int_0^\infty K(-u,y)\bigl(f_1(x+u)-f_1(x)\bigr)\,du\right\| \]
\[ < C\int_0^\infty u^{\alpha-1}e^{-\delta u}\,[\varphi(x-u)+2\varphi(x)+\varphi(x+u)]\,du< \]
\[ < C_1\varphi(x)\int_0^\infty u^{\alpha-1}e^{-(\delta-\delta_1)u}\,du =O(\varphi(x)), \]
since from \(\displaystyle \lim_{x\to+\infty}\frac{\varphi'(x)}{\varphi(x)}=0\) it follows that, for any \(\delta_1>0\),
\[ \frac{\varphi(x+u)}{\varphi(x)}<C_{\delta_1}e^{\delta_1|u|}. \]
Hence we also obtain the continuity of \(u_1\). The continuity of \(u_2\) and \(u_3\) is obvious.

To verify the boundary conditions, note that for \(x\ne t\)
\[ B_1K(x-t,0)+B_2K(x-t,1)=0 \]

Thus, \(B_1u_1(x,0)+B_2u_1(x,1)=0\). Further, by Lemma 2,

\[ B_1u_2(x,0)+B_2u_2(x,1)=0, \]

\[ B_1u_3(x,0)+B_2u_3(x,1) =\operatorname*{res}_{z=0}\frac{B_1M(z,0)+B_2M(z,1)}{z}\,f(x) =\operatorname*{res}_{z=0}\frac{E}{z}f(z)=f(x). \]

It remains to verify that for \(x>0,\ 0<y<1\) the derivatives \(\partial u_0/\partial x,\ \partial u_0/\partial y\) exist and that
\(\partial u_0/\partial y=A\,\partial u_0/\partial x+Pu_0\). For \(0<y<1\), in view of the estimates of Lemma 3, we have

\[ \int_{-\infty}^{\infty} K(x-t,y)[f_1(t)-f_1(x)]\,dt =\int_{-\infty}^{\infty} K(x-t,y)f_1(t)\,dt -\int_{-\infty}^{\infty} K(u,y)\,du\, f_1(x). \]

Since \(M(z,y)=O(e^{-\tau y(1-y)|z|})\), in the definition of \(K(x,y)\) for \(0<y<1\) integration over the contours \(L_\delta^+\) and \(L_\delta^-\) may be replaced by integration along the straight lines \(\operatorname{Re}z=\delta\) and \(\operatorname{Re}z=-\delta\). Therefore

\[ \int_{-\infty}^{\infty} K(u,y)\,du =\frac{1}{2\pi i}\left\{ \int_{-\infty}^{0}\int_{\delta-i\infty}^{\delta+i\infty} M(z,y)e^{zu}\,dz\,du +\int_{0}^{\infty}\int_{-\delta-i\infty}^{-\delta+i\infty} M(z,y)e^{zu}\,dz\,du \right\} = \]

\[ =\frac{1}{2\pi i}\left\{ \int_{\delta-i\infty}^{\delta+i\infty} M(z,y)\frac{dz}{z} -\int_{-\delta-i\infty}^{-\delta+i\infty} M(z,y)\frac{dz}{z} \right\} =\frac{1}{2\pi i}\int_{L_0} M(z,y)\frac{dz}{z}. \]

Thus,

\[ u_0(x,y)=\int_{-\infty}^{\infty} K(x-t,y)f_1(t)\,dt +\sum_{\operatorname{Re}z_n=0}\int_{-\infty}^{x} e^{z_n(x-t)}H_n(x-t,y)f_1(t)\,dt = \]

\[ =\int_{-\infty}^{\infty} K_1(x-t,y)f_1(t)\,dt, \]

where

\[ K_1(x,y)=\frac{1}{2\pi i}\int_{\delta-i\infty}^{\delta+i\infty} M(z,y)e^{zx}\,dz,\qquad 0<y<1. \]

As in Lemma 3, we easily obtain the inequalities

\[ \|K_1(x,y)\|<C_y e^{\delta x},\qquad \left\|\frac{\partial}{\partial x}K_1(x,y)\right\|<C_y e^{\delta x},\qquad x<0. \]

From these inequalities and from 4) it is clear that differentiation under the integral sign is admissible, and the lemma is completely proved.

Applying Lemma 4 to each of the cases, we obtain Theorems 1–3.

The following example shows that in Theorems 1–3 one cannot make do with conditions only on the growth of \(\|f(x)\|\). We take the Cauchy–Riemann system with boundary conditions

\[ B_1=\begin{pmatrix}1&0\\0&0\end{pmatrix},\qquad B_2=\begin{pmatrix}0&0\\1&0\end{pmatrix}, \]

and construct a bounded and differentiable \(f(x)\) such that nothing can be said about the growth of the particular solution.

Namely, put

\[ f(x)= \begin{pmatrix} \alpha(x,0)\\ \alpha(x,1) \end{pmatrix}, \qquad \text{where }\alpha(x,y)=\operatorname{Re}g(x+iy), \]

\[ g(z)=i\sum_{n=1}^{\infty}\frac{1}{(e^{\pi z}-n)^2+1} \ln\frac{1-ip_n(e^{\pi z}-n)}{p_n+1}, \qquad p_n\to\infty. \]

The solution of least growth will be

\[ u_0(x,y)= \begin{pmatrix} \operatorname{Re}g(x+iy)\\ \operatorname{Im}g(x+iy) \end{pmatrix}, \]

and it is not difficult to check that, by choosing \(p_n\), the function \(u_0(x,y)\) can be made to tend to infinity arbitrarily fast.

Received
28 IV 1960

REFERENCES

1. M. A. Evgrafov, DAN, 126, No. 3 (1959).