Mathematical Physics Vol 1

7.4 Linear second order PDE

355

The last expression represents a linear first order partial equation of an unknown function g . As we have shown earlier, solving this equation comes down to solving a system of ordinary equations: d x ∂ f ∂ p = d y ∂ f ∂ q = d z ∂ f ∂ p p + ∂ f ∂ q q = d p − ∂ f ∂ x − ∂ f ∂ z p = d q − ∂ f ∂ y − ∂ f ∂ z q ,

or, if we introduce the notation

∂ f ∂ x

∂ f ∂ y

∂ f ∂ z

∂ f ∂ p

∂ f ∂ q

f x =

f y =

f z =

f p =

f q =

,

,

,

,

,

the previous expression can be written in the form

d x f p

d y f q

d z pf p + qf q

d p − f x − pf z

d q − f y − qf z

(7.80)

=

=

=

=

.

This system yields f ( x , y , z , p , q )= 0 as one of the first integrals. Thus, at least another first integral of the same system must be found, in order to determine the values of p and q andobtain the total differential (7.69).

7.4 Linear second order PDE

A linear non-homogeneous second order partial differential equation is an equation of the form L ( u ) ≡ n ∑ i , j = 1 a i j ∂ 2 u ∂ x i ∂ x j + n ∑ i = 1 a i ∂ u ∂ x i + bu = c (7.81) where L denotes the linear operator, u = u ( x 1 ,..., x n ) is the unknown function, and the coefficients are functions of the form

a i j = a i j ( x 1 ,..., x n ) ; a i = a i ( x 1 ,..., x n ) ; b = b ( x 1 ,..., x n ) ; c = c ( x 1 ,..., x n ) , where a i j = a ji .

(7.82)

R Note that the assumption on the symmetry of coefficients a i j is not a constraint, because for continuous functions ∂ 2 u ∂ x i ∂ x j = ∂ 2 u ∂ x j ∂ x i .

A linear homogeneous second order partial differential equation is an equation of the form

n ∑ i , j = 1

n ∑ i = 1

∂ 2 u ∂ x i ∂ x j

∂ u ∂ x i

L ( u ) ≡

a i j

a i

+ bu = 0 .

(7.83)

+

We can write these equations in a shorter form

L ( u )= c , that is, L ( u )= 0 .

(7.84)