Mathematical Methods for Physics and Engineering - Matematica.NET

PDES: GENERAL AND PARTICULAR SOLUTIONSin which κ is a constant with the dimensions length 2 × time −1 . The physicalconstants that go to make up κ in a particular case depend upon the nature ofthe process (e.g. solute diffusion, heat flow, etc.) and the material being described.With (20.34) we cannot hope to repeat successfully the method of subsection20.3.3, since now u(x, t) is differentiated a different number of times on the twosides of the equation; any attempted solution in the form u(x, t) =f(p) withp = ax + bt will lead only to an equation in which the form of f cannot becancelled out. Clearly we must try other methods.Solutions may be obtained by using the standard method of separation ofvariables discussed in the next chapter. Alternatively, a simple solution is alsogiven if both sides of (20.34), as it stands, are separately set equal to a constantα (say), so that∂ 2 u∂x 2 = α κ ,These equations have the general solutions∂u∂t = α.u(x, t) = α2κ x2 + xg(t)+h(t) and u(x, t) =αt + m(x)respectively and may be made compatible with each other if g(t) is taken asconstant, g(t) =g (where g could be zero), h(t) =αt and m(x) =(α/2κ)x 2 + gx.An acceptable solution is thusu(x, t) = α2κ x2 + gx + αt + constant. (20.35)Let us now return to seeking solutions of equations by combining the independentvariables in particular ways. Having seen that a linear combination ofx and t will be of no value, we must search for other possible combinations. Ithas been noted already that κ has the dimensions length 2 × time −1 and so thecombination of variablesη = x2κtwill be dimensionless. Let us see if we can satisfy (20.34) with a solution of theform u(x, t) =f(η). Evaluating the necessary derivatives we have∂u∂x = df(η) ∂ηdη ∂x = 2x df(η)κt dη ,∂ 2 u∂x 2 = 2 (df(η) 2xκt dη+ κt) 2d 2 f(η)dη 2 ,∂u∂t = − x2 df(η)κt 2 dη .Substituting these expressions into (20.34) we find that the new equation can be696