By Jacek Szarski

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**Sample text**

Local existence of a solution of the Cauchy problem "along a given charac- teristic". Consider the equation (1, 4) with the initial condition (2. 4) and introduce the following Assumptions B. 4) Y = Y(x), z = z(x), Q = Q(x) be a characteristic strip of (1. 4), i. e. a solution of the system (5. 4) Y(x) = Y, z(x) - w(Y), Q(x) = wy(Y) We shall solve the following problem. Problem 1. 4. Establish the necessary and sufficient conditions for a solution of (1. 4) and (2. 4) al < x < a2 i. e. 4). We will prove first the following Lemma 1.

Y(x),z(x,Y(x)),zy(x,Y(x)) (x,Y(x)+H1) - zy3(x,Y(x))] fq](x,Y(x), z(x,Y(x)), zy(x,Y(x)))l . Now, setting for 0 < t < 1, (x, Y) E D Pi(x, Y, t, h) _ (x, Y+tH1, z(x,Y)+t[z(x,Y+H1)- z(x, Y)], zy(x, Y) + + t[zy(x, Y+Hi) - y(x, H)]), we have obviously (4. 5) kim Pi(x,Y(x), t, h) _ (x,y(x), z(x,Y(x)), zY(x,Y(x))) = (x,Y(x), z{x), Q(x)) -38- uniformly with respect to x and t. Applying Hadamard's mean value theorem to the extreme right-hand member of (3. 5) 1 1 J fy (Pi (x,Y(x), t, h))dt + J fz(Pi(x,Y(x),t,h)) dt qi(x, h) 0 0 1 1 zyj(x,Y(x))]h-1 fq(x, Y (x), z (x), Since zy(x, Y) satisfy a Lipschitz condition with respect to Y the expression i IzyJx, Y(x) +Hi)- zyJ(x,Y(x))] h 1 J is bounded and hence by (2.

X, Y) yi satisfy a Lipschitz condition with respect to Y. Under these assumptions z(x, Y) has the property P. Proof. Let Y(x) be an aribtrary solution of (8. , 0) h standing on the i-th place. We have obviously by (1. 5) (2. 5) lim ql(x, h) = z (x, Y(x)) = qi(x) yi h0 uniformly with respect to x. Differentiating qi(x, h) with respect to x for fixed h we obtain by (1. 4) and (8. Y(x),z(x,Y(x)),zy(x,Y(x)) (x,Y(x)+H1) - zy3(x,Y(x))] fq](x,Y(x), z(x,Y(x)), zy(x,Y(x)))l . Now, setting for 0 < t < 1, (x, Y) E D Pi(x, Y, t, h) _ (x, Y+tH1, z(x,Y)+t[z(x,Y+H1)- z(x, Y)], zy(x, Y) + + t[zy(x, Y+Hi) - y(x, H)]), we have obviously (4.