By Yang D., Zhou Y.
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Additional resources for A Boundedness Criterion via Atoms for Linear Operators in Hardy Spaces
To find the orthogonal trajectories, we must solve this differential equation. To this end, note that this equation is separable and thus 1 1 dx = dy ⇒ x y ⇒ eln |x|−C = eln |y| ln |x| = ln |y| + C ⇒ where k = ±e−C . y = kx, Therefore, the orthogonal trajectories are lines through the origin. (c) Let F (x, y) = xy. Then we have Fx (x, y) = y and Fy (x, y) = x. Plugging these expressions into the final result of part (a) gives x dx − y dy = 0. To find the orthogonal trajectories, we must solve this differential equation.
Qdx + y0 µ(x0 ) x0 x=x0 (d) We assume that y(x) is a solution to the initial value problem (15). Since µ(x) is a continuous positive function on (a, b), the equation (5) is equivalent to (4). Since, from the part (a), the left-hand side of (5) is the derivative of the product µ(x)y(x), this function must be an antiderivative of the right-hand side, which is µ(x)Q(x). Thus, we come up with (8), where the integral means one of the antiderivatives, for example, the one suggested in the part (c) (which has zero value at x0 ).
To use the method described in Problem 32, we rewrite the equation x2 + y 2 = kx in the form x + x−1 y 2 = k. Thus, F (x, y) = x + x−1 y 2 , ∂F = 1 − x−2 y 2 , ∂x 54 ∂F = 2x−1 y. 4 Substituting these derivatives in the equation given in Problem 32(b), we get the required. Multiplying the equation by xn y m , we obtain 2xn−1 y m+1 dx + xn−2 y m+2 − xn y m dy = 0. Therefore, ∂M = 2(m + 1)xn−1 y m , ∂y ∂N = (n − 2)xn−3 y m+2 − nxn−1 y m . ∂x Thus, to have an exact equation, n and m must satisfy n−2=0 2(m + 1) = −n .