By David Bleecker, George Csordas (auth.)

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**Example text**

25) Indeed, if the wire loop is nearly planar, and is held nearly level, then the minimal surface formed will be close (in a sense which is rather difficult to make precise) to the graph of the corresponding solution of Laplace's equation. Troubles arise when the supposition u~ + u~ < < 1 turns out to be incorrect. As an illustration, we compare the solutions of (24) and (25) in the case . l where u IS assumed to have the form u = f( r) , r = [x + y ]>. By the computation done in Example 1, it is found that (24) and (25) become rfll(r) + f/(r)(1 + [f/(r)]2) = 0 (26) rf"(r) + f' (r) (27) = 0, respectively.

It is a standard fact that u E Ck implies u E Ck- 1 for k > O. For a C2 function u, recall that uxy = uyx ' More generally, the order in which one takes k or fewer partial derivatives of a Ck function is immaterial. Remark. We have assumed above that the function u is defined for all values of the independent variables. The function might only be defined for (x,y,z, ... ) in a certain region D. , rectangles, strips, discs). If such a region includes some point p of its boundary, then technically the notion of partial derivative of u at p is not defined, unless one wishes to deal with one-sided derivatives.

We expect that the rod's temperature will approach the temperature (zero) of its icy environment. Indeed, the factor exp[-7r2ktJ in (8) tells us that, as t -+ 00 , the temperature of the rod approaches 0, and it does so more rapidly for larger values of the heat conductivity k. More generally, choosing f(x) = sin(nme) for an arbitrary positive integer n, we get the solution as t -+ 00 u(x,t) = exp[-n 27r2ktjsin(nme) . Note that the rate at which this solution approaches is faster for larger values of n.