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UDC 517.944.8
ESTIMATING THE LIMITING RATE OF DECREASE OF SOLUTIONS OF AN ELLIPTIC SYSTEM
I. S. ARSHON, G. A. JAFARLI
§ 1. STATEMENT OF THE PROBLEM
The present note, like the notes [1, 2], is devoted to estimating the limiting rate of decrease of solutions of an elliptic equation or system in a cylinder (strip).
In note [1] the question of harmonic functions of three variables was considered, while in [2] a system with constant coefficients was investigated,
\[ \frac{\partial \hat u}{\partial y} = A \frac{\partial \hat u}{\partial x} + P\hat u, \]
where \(A\) and \(P\) are matrices of arbitrary order \(n\), and \(\hat u(x,y)\) is an \(n\)-dimensional vector.
We shall show that the method of reasoning adopted in [2] is also applicable in a somewhat more general case, namely when the coefficients depend on one of the variables.
It should be noted that the very existence of a limiting rate of decrease, as well as the order of this quantity for the case of solutions of a second-order elliptic equation with any number of variables and with coefficients depending on all these variables, was established in [3] by E. M. Landis. Although the present work concerns the special case of a system with two variables, it is of some interest in connection with the absence of the maximum principle and the explicit expression of the limiting rate of decrease in terms of the coefficients of the system.
We formulate the result obtained by us.
Theorem. Let \(A(y)\), \(P(y)\) be matrices of arbitrary order \(n\), with \(A(y)\) continuously differentiable on the interval \([0,1]\), and \(P(y)\) continuous on this interval. Suppose that for \(0 \le y \le 1\) the eigenvalues \(a_k(y)\) of the matrix \(A(y)\) have nonzero imaginary part. If \(\hat u(x,y)\) is a solution of the system
\[ \frac{\partial \hat u}{\partial y} = A(y)\frac{\partial \hat u}{\partial x} + P(y)\hat u, \tag{1} \]
continuous together with its first partial derivatives in the strip \(-\infty < x < +\infty,\ 0 \le y \le 1\), and satisfying in this strip the conditions
\[ \|\hat u(x,y)\| < C_1 \exp\{\alpha x\},\quad x<0, \tag{2} \]
\[ \|\hat u(x,y)\| < C_2 \exp\left\{-\exp \frac{\pi+\eta}{h}\,x\right\},\quad x>0, \tag{2'} \]
where \(a\) is arbitrary, \(\eta>0\) is fixed and arbitrarily small, and
\[ h=\int_0^1 \min_k |\operatorname{Im} a_k(s)|\,ds, \tag{3} \]
then
\[ \hat u(x,y)\equiv 0. \]
Let us make the following remark concerning system (1). Let
\(A(y)=T^{-1}(y)R(y)T(y)\), where \(R(y)\) is the Jordan normal form of the matrix \(A(y)\). Then the function \(\hat v(x,y)=T(y)\hat u\) is a solution of the system
\[ \frac{\partial \hat v}{\partial y} = R(y)\frac{\partial \hat v}{\partial x} + P_1(y)\hat v,\qquad P_1(y)=T(y)P(y)T^{-1}(y)+T'_y(y)T^{-1}(y). \]
Therefore the matrix \(A(y)\) in (1) may be assumed to be given in Jordan normal form. For simplicity of exposition we shall assume it to be a diagonal matrix.
§ 2. OUTLINE OF THE PROOF
Consider the function
\[ \tilde u(z,y)=\int_{-\infty}^{+\infty}\hat u(x,y)e^{xz}\,dx, \]
which is regular in \(z\), for every fixed \(y\) on \([0,1]\), in the half-plane \(\operatorname{Re} z>0\).
According to a well-known theorem from the theory of analytic functions, if \(f(z)\) is regular in the half-plane \(\operatorname{Re} z>0\), and on the real axis \(z=\sigma>0\)
\[
|f(\sigma)|<C_3\Gamma\left(\frac{\sigma+1}{\rho}\right),
\]
while on the imaginary axis \(z=i\tau\)
\[
|f(i\tau)|<C_4\exp\{-a|\tau|\},
\]
and in the whole half-plane
\[
|f(z)|=\exp\{o|z|^2\},\qquad z\to\infty,
\]
and if the inequality
\[
\rho>\frac{\pi}{2a}
\]
holds, then \(f(z)\equiv 0\). We shall use this result to prove the theorem.
Let us estimate the growth of \(\hat u(z,y)\) on the positive part of the real axis \(z=\sigma>0\). From the assumptions of the theorem we obtain
\[ \|\tilde u(\sigma,y)\| \le C_5+C_6\int_{-\infty}^{+\infty}\exp\{-\exp \rho x+\sigma x\}\,dx < C_7\Gamma\left(\frac{\sigma}{\rho}\right). \]
Applying Stirling’s formula to estimate the \(\Gamma\)-function, we finally obtain
\[ \|\tilde u(\sigma,y)\| < C_8\exp\left\{\frac{\sigma}{\rho}\ln\frac{\sigma}{\rho}\right\}. \]
Let us note that this estimate with \(\sigma=\operatorname{Re} z\) is valid for \(\operatorname{Re} z>0\). Hence, in particular, it follows that
\[ \|\tilde u(z,y)\|=\exp\{o|z|^2\},\qquad z\to\infty. \]
It remains to prove that for at least one \(y=y_0\) the inequality
\[
\|\widetilde u(i\tau,y_0)\|<C_9\exp\left\{-\,\frac{h-\varepsilon}{2}|\tau|\right\}
\tag{4}
\]
holds with arbitrarily small \(\varepsilon>0\). Then we shall have \(\widetilde u(z,y_0)\equiv0\), whence, by the inversion formula for the Laplace transform, \(\widehat u(z,y)\equiv0\).
After this, an appeal to Carleman’s theorem on the uniqueness of the solution of the Cauchy problem, in view of the absence of real characteristics for system (1), completes the proof.
Thus the problem consists in proving inequality (4), i.e., in estimating the function \(\widetilde u(z,y)\) on the imaginary axis \(z=i\tau\). We shall use a slight modification of the method proposed in [4].
§ 3. THE METHOD OF SUCCESSIVE APPROXIMATIONS FOR THE ESTIMATE
Let us introduce some notation. Put
\[
\Lambda(y)=-i\int_0^y A(s)\,ds.
\]
Further, let \(E^{\pm}\) be matrices whose main diagonal contains ones in those positions where the matrix \(A(y)\) has eigenvalues \(a_k(y)\) with positive (respectively negative) imaginary parts. The remaining elements of \(E^{\pm}\) are zeros.
Now define the matrix
\[
Q(i\tau,y,t)=E^{\pm}\exp\{\tau[\Lambda(y)-\Lambda(t)]\}\operatorname{sign}(y-t),
\]
\[
\tau\operatorname{sign}(y-t)>0.
\]
Finally, denote
\[
\delta(y)=\int_0^y \min_k |\operatorname{Im} a_k(s)|\,ds.
\]
Lemma 3.1. The estimate
\[
\|Q(i\tau,y,t)\|<q_0\exp\{-|\tau|\,|\delta(y)-\delta(t)|\}
\]
holds.
Proof. Consider, for example, the case \(\tau>0,\ y>t\). Then
\[
Q(i\tau,y,t)=E^{-}\exp\{\tau[\Lambda(y)-\Lambda(t)]\}.
\]
This is a diagonal matrix. Its nonzero elements have the form
\[
\exp\left\{-i\tau\int_t^y a_k(s)\,ds\right\},\qquad \operatorname{Im}a_k(s)<0.
\]
Therefore,
\[
\|Q(i\tau,y,t)\|<C\max_k \exp\left\{-|\tau|\int_t^y |\operatorname{Im}a_k(s)|\,ds\right\}\le
\]
\[
\le q_0\exp\{-|\tau|\,|\delta(y)-\delta(t)|\},
\]
which was to be proved.
Lemma 3.2. The function \(\widetilde u(i\tau,y)\) is a solution of the system of integral equations
\[
\tilde u(i\tau,y)=Q(i\tau,y,0)\tilde u(i\tau,0)-Q(i\tau,y,1)\tilde u(i\tau,1)+
\]
\[
+\int_0^1 Q(i\tau,y,t)P(t)\tilde u(i\tau,t)\,dt.
\tag{5}
\]
Proof. Indeed, since the function \(\hat u(x,y)\) satisfies system (1) in the strip \(0\le y\le 1\), its Fourier transform \(\tilde u(i\tau,y)\) is a solution of the system of ordinary differential equations
\[ \frac{d\tilde u}{dy}=-i\tau A(y)\tilde u+P(y)\tilde u, \]
which in turn is equivalent to either of the two systems of integral equations
\[
\tilde u(i\tau,y)=\exp\{\tau\Lambda(y)\}\tilde u(i\tau,0)+
\]
\[
+\int_0^y \exp\{\tau[\Lambda(y)-\Lambda(t)]\}P(t)\tilde u(i\tau,t)\,dt.
\tag{6}
\]
\[
\tilde u(i\tau,y)=\exp\{\tau[\Lambda(y)-\Lambda(1)]\}\tilde u(i\tau,1)+
\]
\[
+\int_1^y \exp\{\tau[\Lambda(y)-\Lambda(t)]\}P(t)\tilde u(i\tau,t)\,dt.
\tag{6'}
\]
If, for \(\tau\ge 0\), one multiplies both sides of (6) by \(E^{\mp}\), both sides of \((6')\) by \(E^{\pm}\), and adds the resulting equalities, then one obtains the assertion of the lemma.
We shall solve system (5) by the usual method of successive approximations. In other words, introduce the operator
\[ Mf(i\tau,y)=\int_0^1 Q(i\tau,y,t)P(t)f(i\tau,t)\,dt. \]
Then, denoting for brevity
\[ \tilde f_0(i\tau,y)=Q(i\tau,y,0)\tilde u(i\tau,0);\qquad \tilde f_1(i\tau,y)=-Q(i\tau,y,1)\tilde u(i\tau,1), \]
we find the following representation for the function \(\tilde u(i\tau,y)\):
\[ \tilde u(i\tau,y)=\sum_{n=0}^{\infty}M^n\tilde f_0(i\tau,y)+ \sum_{n=0}^{\infty}M^n\tilde f_1(i\tau,y) =\tilde u_0(i\tau,y)+\tilde u_1(i\tau,y). \]
§ 4. AUXILIARY PROPOSITIONS
For the estimate of the successive approximations we shall need the following propositions.
Lemma 4.1. Let the numbers \(a_{n,k}\), with integral nonnegative indices \(n,k\), satisfy the conditions
\[ a_{n,k}=\sum_{s=k-1}^{n-1}a_{n-1,s},\qquad a_{n,k}=0\quad(k>n), \]
\(a_{0,0}=1\) (if at least one of the indices is negative, then we set \(a_{n,k}=0\)). Then \(0<a_{n,k}<4^n\).
Proof. Put \(a_{n,k}=C_{2n-k,n-k}\). Then
\[ C_{2n-k,n-k}=\sum_{s=-k-1}^{n-1} C_{2n-2-s,n-1-s}. \]
Taking \(s=k-1+t\), we rewrite the preceding equality in the form
\[ C_{2n-k,n-k}=\sum_{t=0}^{n-k} C_{2n-k-1-t,n-k-t}. \]
Next, replacing \(2n-k=\alpha,\ n-k=\beta\), we find
\[ C_{\alpha,\beta}=\sum_{t=0}^{\beta} C_{\alpha-1-t,\beta-t}. \]
Consequently,
\[ C_{\alpha,\beta}=C_{\alpha-1,\beta}+\sum_{t=0}^{\beta-1} C_{\alpha-2-t,\beta-1-t}, \]
i.e. \(C_{\alpha,\beta}=C_{\alpha-1,\beta}+C_{\alpha-1,\beta-1}\).
But if \(k<0\) there was \(a_{n,k}=0\), which means that for \(2\beta>\alpha\) it must be that \(C_{\alpha,\beta}=0\). Similarly, from \(a_{n,k}=0\) for \(k>n\) it follows that \(C_{\alpha,\beta}=0\) for \(\beta<0\). Finally, \(C_{0,0}=1\). Comparing the numerical triangle \(\{C_{\alpha,\beta}\}\) with the known Pascal triangle for the binomial coefficients \(\binom{\beta}{\alpha}\), we obtain
\[ C_{\alpha,\beta}\leq \binom{\beta}{\alpha}. \]
Thus,
\[ a_{n,k}\leq a_{n,0}=C_{2n,n}\leq \binom{n}{2n}<4^n, \]
as was required to prove.
Lemma 4.2. Let
\[ h_{n-1}(i\tau,y)=\exp\{-|\tau|\delta(y)\}\,\frac{\delta^{\,n-1}(y)}{(n-1)!}. \]
Then
\[ \|Mh_{n-1}(i\tau,y)\|<C_0\exp\{-|\tau|\delta(y)\}\sum_{k=0}^{n}\frac{\delta^k(y)}{k!(2|\tau|)^{\,n-k}}. \]
Proof. Obviously,
\[ \|Mh_{n-1}(i\tau,y)\|\leq p_0\int_0^1 \|Q(i\tau,y,t)\|\,\|h_{n-1}(i\tau,t)\|\,dt, \]
where \(p_0=\max_t\|P(t)\|\) for \(0\leq t\leq 1\). Hence, by Lemma 3.1,
\[ \|Mh_{n-1}(i\tau,y)\|\leq p_0q_0\left\{\exp[-|\tau|\delta(y)]\times \right. \]
\[ \left. {}\times \int_0^y \frac{\delta^{\,n-1}(t)}{(n-1)!}\,dt +\exp[|\tau|\delta(y)]\int_y^1 \exp[-2|\tau|\delta(t)]\, \frac{\delta^{\,n-1}(t)\,dt}{(n-1)!}\right\}. \]
According to the definition of \(\delta(y)\), we have
\[ \min_{0\leq y\leq 1}\delta'(y)=\min_y\max_k|\operatorname{Im}a_k(y)|=\Delta_0>0. \]
Hence, for any function \(f(t)\) positive on \([0,1]\),
\[ \int_0^1 f(t)\,dt\leq \frac{1}{\Delta_0}\int_0^1 f(t)\delta'(t)\,dt. \]
Using this inequality and making in the obtained integrals the substitution
\(\delta(t)=s\), \(\delta'(t)\,dt=ds\), we shall have
\[ \|Mh_{n-1}(i\tau,y)\|\leq \frac{p_0q_0}{\Delta_0} \left\{\exp[-|\tau|\delta(y)]\times\right. \]
\[ \left. {}\times \int_0^{\delta(y)}\frac{s^{n-1}}{(n-1)!}\,ds +\int_{\delta(y)}^{+\infty}\exp[|\tau|\delta(y)-2|\tau|s]\, \frac{s^{n-1}\,ds}{(n-1)!}\right\}, \]
which after integration gives the required result.
§ 5. ESTIMATE OF SUCCESSIVE APPROXIMATIONS
Solving the integral equation (5), we obtained the following representation for
\(\tilde u(i\tau,y)\)
\[ \tilde u(i\tau,y)=\tilde u_0(i\tau,y)+\tilde u_1(i\tau,y) =\sum_{n=0}^{\infty} M^n\tilde f_0(i\tau,y) +\sum_{n=0}^{\infty} M^n\tilde f_1(i\tau,y), \]
where
\[ \tilde f_0(i\tau,y)=Q(i\tau,y,0)\tilde u(i\tau,0); \qquad \tilde f_1(i\tau,y)=-Q(i\tau,y,1)\tilde u(i\tau,1). \]
We shall show that the first of the series on the right converges and estimate its sum. Since from the definition of \(\tilde u(z,y)\) there obviously follows the inequality
\(\|\tilde u(i\tau,y)\|\leq \mathrm{const}\), Lemma 3.1 gives
\[ \|\tilde f_0(i\tau,y)\|\leq q_0\exp\{-|\tau|\delta(y)\}. \]
Obviously,
\[ \|M\tilde f_0(i\tau,y)\|\leq C_0\exp\{-|\tau|\delta(y)\} \left(\delta(y)+\frac{1}{2|\tau|}\right), \]
\[ \|M^2\tilde f_0(i\tau,y)\|=\|M(M\tilde f_0(i\tau,y))\| \leq C\exp\{-|\tau|\delta(y)\}\times \]
\[ {}\times \sum_{k=0}^{2} c_{2,k}\, \frac{\delta^k(y)}{k!(2|\tau|)^{2-k}}, \]
where \(c_{2,0}=1,\ c_{2,1}=2,\ c_{2,2}=2\). Similarly we obtain
\[ \|M^3\tilde f_0(i\tau,y)\|\le C\exp\{-|\tau|\delta(y)\}\sum_{k=0}^{3} c_{3,k}\, \frac{\delta^k(y)}{k!(2|\tau|)^{3-k}}. \]
Repeating this process, we obtain
\[ \|M^{n-1}\tilde f_0(i\tau,y)\|< C\exp\{-|\tau|\delta(y)\}\sum_{k=0}^{n-1} c_{n-1,k}\, \frac{\delta^k(y)}{k!(2|\tau|)^{\,n-1-k}}. \]
Applying the operator \(M\) once more and applying Lemma 3.1, we obtain
\[ \|M^n\tilde f_0(i\tau,y)\|\le C\exp\{-|\tau|\delta(y)\}\sum_{k=0}^{n-1} \frac{c_{n-1,k}}{(2|\tau|)^{\,n-1-k}}\times \]
\[ {}\times \sum_{j=0}^{k+1} \frac{\delta^j(y)}{j!(2|\tau|)^{\,k+1-j}} = \exp\{-|\tau|\delta(y)\}\sum_{j=0}^{n} \frac{\delta^j(y)}{j!(2|\tau|)^{\,n+1-j}} \sum_{k=j-1}^{n-1} c_{n-1,k}, \]
where \(c_{n,k}\) are certain positive numbers. Then, by Lemma 4.1, we have
\[ \|M^n\tilde f_0(i\tau,y)\|< C^n\exp\{-|\tau|\delta(y)\}\sum_{k=0}^{n} \frac{\delta^k(y)}{k!(2|\tau|)^{\,n-k}}. \]
Now, for all sufficiently large \(|\tau|\), we have
\[ \|\tilde u'(i\tau,y)\|\le \sum_{n=0}^{\infty}\|M^n\tilde f_0(i\tau,y)\|\le \exp\{-|\tau|\delta(y)\}\times \]
\[ {}\times \sum_{n=0}^{\infty}\sum_{k=0}^{n} \frac{C^n\delta^k(y)}{k!(2|\tau|)^{\,n-k}} = \exp\{-|\tau|\delta(y)\}\times \]
\[ {}\times \sum_{k=0}^{\infty}\frac{\delta^k(y)}{k!} \sum_{n=0}^{\infty}\frac{C^n}{(2|\tau|)^n} = C\exp\{-|\tau|\delta(y)\}. \]
In an entirely analogous way we verify that
\[ \|\tilde u_1(i\tau,y)\|\le C\exp\{-|\tau|[\delta(1)-\delta(y)]\}. \]
Defining \(y_0\) from the condition \(\delta(y_0)=\delta(1)-\delta(y_0)\), i.e., taking \(\delta(y_0)=\dfrac{h}{2}\), we obtain
\[ \|\tilde u(i\tau,y)\|\le C\exp\left\{-\frac{h}{2}|\tau|\right\}. \]
Inequality (4) has been proved; hence our theorem has also been proved.
References
- Arshon I. S., Iglitskii M. A. DAN SSSR, 152, No. 4, 775–778, 1963.
- Dzhafarli G. A. On the decrease of solutions of elliptic equations and systems in a strip. Scientific Notes of the Azerbaijan State University named after S. M. Kirov, No. 1, 1965.
- Landis E. M. Uspekhi Mat. Nauk, vol. XVIII, issue 1 (109), 3–61, 1963.
- Arshon I. S. Matem. Sb., 61 (103), issue 3, 1963, pp. 362–376.
Received by the editors
April 28, 1965
Moscow Institute of
Electronic Machine Building
Azerbaijan Polytechnic Institute