Mathematical Methods for Physics and Engineering - Matematica.NET

20.3 GENERAL AND PARTICULAR SOLUTIONSwhich, when substituted into the PDE (20.9), give[A(x, y) ∂p]∂p df(p)+ B(x, y) =0.∂x ∂y dpThis removes all reference to the actual form of the function f(p) since fornon-trivial p we must haveA(x, y) ∂p ∂p+ B(x, y) =0. (20.10)∂x ∂yLet us now consider the necessary condition for f(p) to remain constant as xand y vary; this is that p itself remains constant. Thus for f to remain constantimplies that x and y must vary in such a way thatdp = ∂p ∂pdx + dy =0. (20.11)∂x ∂yThe forms of (20.10) and (20.11) are very alike and become the same if werequire thatdxA(x, y) = dyB(x, y) . (20.12)By integrating this expression the form of p can be found.◮Forx ∂u ∂u− 2y =0, (20.13)∂x ∂yfind (i) the solution that takes the value 2y +1 on the line x =1, and (ii) a solution thathas the value 4 at the point (1, 1).If we seek a solution of the form u(x, y) =f(p), we deduce from (20.12) that u(x, y) willbe constant along lines of (x, y) thatsatisfydxx = dy−2y ,which on integrating gives x = cy −1/2 . Identifying the constant of integration c with p 1/2(to avoid fractional powers), we conclude that p = x 2 y. Thus the general solution of thePDE (20.13) isu(x, y) =f(x 2 y),where f is an arbitrary function.We must now find the particular solutions that obey each of the imposed boundaryconditions. For boundary condition (i) a little thought shows that the particular solutionrequired isu(x, y) =2(x 2 y)+1=2x 2 y +1. (20.14)For boundary condition (ii) some obviously acceptable solutions areu(x, y) =x 2 y +3,u(x, y) =4x 2 y,u(x, y) =4.683