Mathematical Physics Vol 1

7.5 A formal procedure for solving LDE

363

Example 239 Heat conduction equation

∂ 2 u ∂ x 2 1

∂ 2 u ∂ x 2 2

∂ 2 u ∂ x 2 3

∂ u ∂ t

= a 2

(7.147)

+

+

is, according to 4 ◦ and u = u ( x 1 , x 2 , x 3 , t ) , of parabolic type.

Example 240 Laplace equation

∂ 2 u ∂ x 2 1

∂ 2 u ∂ x 2 2

∂ 2 u ∂ x 2 3

∆ u ≡

= 0

(7.148)

+

+

is, according to 1 ◦ and u = u ( x 1 , x 2 , x 3 ) , of elliptic type.

7.5 A formal procedure for solving LDE Observe the equation of the form

a 11 u xx + 2 a 12 u xy + a 22 u yy + b 1 u x + b 2 u y + cu = 0 .

(7.149)

Further, assume that there exists a solution of the form u = C · e α x + β y ,

(7.150)

where α and β are, for now, undetermined constants. Given that, by assumption, u is a solution, this function must identically satisfy the initial equation, and thus, by differentiating (7.150), followed by substitution of obtained derivatives into (7.149) and dividing by u , we obtain a 11 α 2 + 2 a 12 α · β + a 22 β 2 + 2 b 1 α + 2 b 2 β + c = 0 . (7.151) As we have two arbitrary constants, let us assume that α is an integer, i.e. α = k ( k = 0 , 1 ,... ) . (7.152) From (7.151) we now obtain for β : β = ak + b ± p ck 2 + dk + f , (7.153) where a , b , c , d and f are constants that depend of a i j , b i and c . Given that k is an integer, the value under the square root is not equal to zero, and we obtain, for each k , two solutions C k · e kx +( ak + b + √ ck 2 + dk + f ) y , D k · e kx +( ak + b − √ ck 2 + dk + f ) y . (7.154) If C k and D k are constants, for the solution we obtain

kx +( ak + b + √ ck 2 + dk + f ) y

kx +( ak + b − √ ck 2 + dk + f ) y .

∞ ∑ k = 0

∞ ∑ k = 0

u =

C k · e

D k · e

(7.155)

+