Mathematical Methods for Physics and Engineering - Matematica.NET

20.3 GENERAL AND PARTICULAR SOLUTIONSWe could, of course, have taken h(x, y) =y, but this only leads to a solution thatis already contained in (20.25).◮Solve∂ 2 u∂x +2 ∂2 u2 ∂x∂y + ∂2 u∂y =0, 2subject to the boundary conditions u(0,y)=0and u(x, 1) = x 2 .From our general result, functions of p = x + λy will be solutions provided1+2λ + λ 2 =0,i.e. λ = −1 and the equation is parabolic. The general solution is thereforeu(x, y) =f(x − y)+xg(x − y).The boundary condition u(0,y) = 0 implies f(p) ≡ 0, whilst u(x, 1) = x 2 yieldsxg(x − 1) = x 2 ,which gives g(p) =p + 1, Therefore the particular solution required isu(x, y) =x(p +1)=x(x − y +1). ◭To reinforce the material discussed above we will now give alternative derivationsof the general solutions (20.22) and (20.25) by expressing the original PDEin terms of new variables before solving it. The actual solution will then becomealmost trivial; but, of course, it will be recognised that suitable new variablescould hardly have been guessed if it were not for the work already done. Thisdoes not detract from the validity of the derivation to be described, only fromthe likelihood that it would be discovered by inspection.We start again with (20.20) and change to new variablesζ = x + λ 1 y, η = x + λ 2 y.With this change of variables, we have from the chain rule thatUsing these and the fact that∂∂x = ∂∂ζ + ∂∂η ,∂∂y = λ ∂1∂ζ + λ ∂2∂η .equation (20.20) becomesA + Bλ i + Cλ 2 i =0 fori =1, 2,[2A + B(λ 1 + λ 2 )+2Cλ 1 λ 2 ] ∂2 u∂ζ∂η =0.691