Mathematical Methods for Physics and Engineering - Matematica.NET

20.3 GENERAL AND PARTICULAR SOLUTIONSIf we take, for example, h(x, y) =expy, which clearly satisfies (20.15), then the generalsolution isu(x, y) =(expy)f(x exp(− 1 y)). 2Alternatively, h(x, y) =x 2 also satisfies (20.15) and so the general solution to the equationcan also be writtenu(x, y) =x 2 g(x exp(− 1 y)), 2where g is an arbitrary function of p; clearly g(p) =f(p)/p 2 . ◭20.3.2 Inhomogeneous equations and problemsLet us discuss in a more general form the particular solutions of (20.13) foundin the second example of the previous subsection. It is clear that, so far as thisequation is concerned, if u(x, y) is a solution then so is any multiple of u(x, y) orany linear sum of separate solutions u 1 (x, y)+u 2 (x, y). However, when it comesto fitting the boundary conditions this is not so.For example, although u(x, y) in (20.14) satisfies the PDE and the boundarycondition u(1,y)=2y + 1, the function u 1 (x, y) =4u(x, y) =8xy + 4, whilstsatisfying the PDE, takes the value 8y+4 on the line x = 1 and so does not satisfythe required boundary condition. Likewise the function u 2 (x, y) =u(x, y)+f 1 (x 2 y),for arbitrary f 1 , satisfies (20.13) but takes the value u 2 (1,y)=2y +1+f 1 (y) onthe line x = 1, and so is not of the required form unless f 1 is identically zero.Thus we see that when treating the superposition of solutions of PDEs twoconsiderations arise, one concerning the equation itself and the other connectedto the boundary conditions. The equation is said to be homogeneous if the factthat u(x, y) is a solution implies that λu(x, y), for any constant λ, is also a solution.However, the problem is said to be homogeneous if, in addition, the boundaryconditions are such that if they are satisfied by u(x, y) then they are also satisfiedby λu(x, y). The last requirement itself is referred to as that of homogeneousboundary conditions.For example, the PDE (20.13) is homogeneous but the general first-orderequation (20.9) would not be homogeneous unless R(x, y) = 0. Furthermore,the boundary condition (i) imposed on the solution of (20.13) in the previoussubsection is not homogeneous though, in this case, the boundary conditionu(x, y) = 0on the line y =4x −2would be, since u(x, y) =λ(x 2 y − 4) satisfies this condition for any λ and, being afunction of x 2 y, satisfies (20.13).The reason for discussing the homogeneity of PDEs and their boundary conditionsis that in linear PDEs there is a close parallel to the complementary-functionand particular-integral property of ODEs. The general solution of an inhomogeneousproblem can be written as the sum of any particular solution of theproblem and the general solution of the corresponding homogeneous problem (as685