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CALCULUS OF VARIATIONS b) Given a point (x0 , y0 ) within D 2 , and a direction through it, show that the equation you derived in part a) determines a unique geodesic curve passing through (x0 , y0 ) in the given direction, but does not determine the parametrization of the curve. c) Show that there exists a solution to the equation in part a) in the form x(t) = R cos t + x0 y(t) = R sin t. Find a relation between x0 and R, and from it deduce that the geodesics are circular arcs that cut the bounding unit circle (which plays the role of the line at infinity in the Lobachevski plane) at right angles.
113) This will be conserved if J is time independent. If J = 0, it is the total field energy. Suppose J is neither zero nor time independent. Then, looking back at the derivation of the time-independence of the first integral, we see that if L does depend on time, we instead have dE ∂L =− . 115) so that − dE d J˙ · A d3x = = (Field Energy) − dt dt ˙ + J˙ · A J·A d3 x. 3. LAGRANGIAN MECHANICS 27 ˙ we find Thus, cancelling the duplicated term and using E = −A, d (Field Energy) = − dt J · E d3 x. 117) Now J · (−E) d3x is the rate at which the power source driving the current is doing work against the field.
If you are the Urbana city engineer worrying about the capacity of the sewer system to cope with a downpour, you are primarily concerned with the maximum value of R(t). For you a big rain has a big “sup |R(t)|”1 . 1 Norms and convergence We can seldom write down an exact solution function to a real-world problem. We are usually forced to use numerical methods, or to expand as a power series in some small parameter. The result is a sequence of approximate solutions fn (x), which we hope will converge to the desired exact solution f (x) as we make the numerical grid smaller, or take more terms in the power series.