Mathematical Physics Vol 1

7.3 Linear and quasilinear first order PDE

349

Adding the left hand and right hand sides of these relations, respectively, we further obtain P ∂ R ∂ y − ∂ Q ∂ z + Q ∂ P ∂ z − ∂ R ∂ x + R ∂ Q ∂ x − ∂ P ∂ y = P ∂ Q ∂ z − ∂ R ∂ y + Q ∂ R ∂ x − ∂ P ∂ z + R ∂ P ∂ y − ∂ Q ∂ x . From the last equation we obtain P ∂ R ∂ y − ∂ Q ∂ z + Q ∂ P ∂ z − ∂ R ∂ x + R ∂ Q ∂ x − ∂ P ∂ y = 0 , (7.46) namely, the condition of integrability . Thus, there exists a function v , by which (7.36) needs to be multypelied in order to be represented as a total differential The condition of integrability of the Pfaffian equation (7.46) can be represented in a more suitable form. Namely, the expressions next to P , Q and R , in equation (7.46), represent components of a rotor of the vector v = P i + Q j + R k (see definition (4.46), on p. 92). The expression (7.46) itself, represents now a scalar product (see (1.44), p. 31) v · rot v = 0 . (7.47) d u = v ( P d x + Q d y + R d z )= 0 .

R Note. Given that P , Q and R are not identically equal to zero (otherwise we would have the identity 0=0), it follows that this condition (7.47) is satisfied either when rot v = 0, or when vectors v and rot v are orthogonal.

R Note. If the conditions for integrability (7.47) are satisfied, then the function v can be determined form one of the three equations (7.43) – (7.45).

Let us now outline a procedure for determining the solution of the equation (7.36), when the condition (7.47) is satisfied, and rot v̸ = 0. Assume that one variable, say z , is constant. The the observed equation becomes

P ( x , y , z ) d x + Q ( x , y , z ) d y = 0 ,

(7.48)

namely, an ordinary differential equation, in which z has the role of a parameter. The solution of this equation is a function of the form

u ( x , y , z )= c ( z ) .

(7.49)

The arbitrary "constant" is a function of parameter z . The function c ( z ) is selected so that the equation (7.36) is satisfied. By differentiating the equation (7.49) we obtain

∂ u ∂ x

∂ u ∂ y

∂ u ∂ z

d c d z

d x +

d y +

d z −

d z = 0 ,

that is

d y +

d c d z

∂ u ∂ x

∂ u ∂ y

∂ u ∂ z −

d x +

d z = 0 .

(7.50)