Show that if z is abse nt from the e quation F (x, y, z, p, q)= 0,Charpit's
method coincides with Jacobi's method.
Solution:-
we know that f(x,y,z,p,q)=0 and g(x,y,z,p,q)=0 (1) are compatible if
∂ ( f , g ) ∂ ( x , p ) + p ∂ ( f , g ) ∂ ( z , p ) + ∂ ( f , g ) ∂ ( y , q ) + q ∂ ( f , g ) ∂ ( z , q ) = 0 { \partial (f,g)\above{2pt}\partial (x,p) }+p{ \partial(f,g)\above{2pt} \partial (z,p)}+{\partial (f,g) \above{2pt} \partial (y,q)}+q{ \partial (f,g)\above{2pt} \partial (z,q)}=0 ∂ ( x , p ) ∂ ( f , g ) + p ∂ ( z , p ) ∂ ( f , g ) + ∂ ( y , q ) ∂ ( f , g ) + q ∂ ( z , q ) ∂ ( f , g ) = 0 (2)
First part:- Comparing the given equations f(x,y,zp,q)=0 and g(x,y,z,p,q)=0 with (1) , we find that z is absent in given equations and so
∂ f ∂ z = 0 {\partial f \above{2pt}\partial z } =0 ∂ z ∂ f = 0 and ∂ g ∂ z = 0 {\partial g \above{2pt} \partial z}=0 ∂ z ∂ g = 0 (3)
Now,
∂ ( f , g ) ∂ ( z , p ) = ∣ ∂ f ∂ z ∂ f ∂ p ∂ g ∂ z ∂ g ∂ p ∣ {\partial (f,g) \above{2pt}\partial (z,p) }=\begin{vmatrix}
{ \partial f\above{2pt} \partial z} & {\partial f \above{2pt}\partial p } \\
{\partial g\above{2pt}\partial z } & { \partial g\above{2pt}\partial p }
\end{vmatrix} ∂ ( z , p ) ∂ ( f , g ) = ∣ ∣ ∂ z ∂ f ∂ z ∂ g ∂ p ∂ f ∂ p ∂ g ∣ ∣ = = = ∣ 0 ∂ f ∂ p 0 ∂ g ∂ p ∣ \begin{vmatrix}
0 & {\partial f \above{2pt}\partial p } \\
0 & { \partial g\above{2pt}\partial p }
\end{vmatrix} ∣ ∣ 0 0 ∂ p ∂ f ∂ p ∂ g ∣ ∣ = 0 =0 = 0 and
∂ ( f , g ) ∂ ( z , q ) = {\partial (f,g) \above{2pt}\partial (z,q) }= ∂ ( z , q ) ∂ ( f , g ) = ∣ ∂ f ∂ z ∂ f ∂ q ∂ g ∂ z ∂ g ∂ q ∣ \begin{vmatrix}
{ \partial f\above{2pt} \partial z} & {\partial f \above{2pt}\partial q } \\
{\partial g\above{2pt}\partial z } & { \partial g\above{2pt}\partial q }
\end{vmatrix} ∣ ∣ ∂ z ∂ f ∂ z ∂ g ∂ q ∂ f ∂ q ∂ g ∣ ∣ = ∣ 0 ∂ f ∂ q 0 ∂ g ∂ a ∣ =\begin{vmatrix}
0 & {\partial f \above{2pt}\partial q } \\
0 & { \partial g\above{2pt}\partial a }
\end{vmatrix} = ∣ ∣ 0 0 ∂ q ∂ f ∂ a ∂ g ∣ ∣ = 0 =0 = 0
Substituting these value in equation (2) the required solution is
∂ ( f , g ) ∂ ( x , p ) + ∂ ( f , g ) ∂ ( y , q ) = 0 { \partial (f,g)\above{2pt}\partial (x,p) }+{\partial (f,g) \above{2pt} \partial (y,q)}=0 ∂ ( x , p ) ∂ ( f , g ) + ∂ ( y , q ) ∂ ( f , g ) = 0
Charpit’s method:-
It is a general method for finding the general solution of a nonlinear PDE of first-order of the form
f ( x , y , z , p , q ) = 0 f(x,y,z,p,q)=0 f ( x , y , z , p , q ) = 0 (1)
Basic Idea: To introduce another partial differential equation of the first order
g ( x , y , z , p , q , a ) = 0 g(x,y,z,p,q,a)=0 g ( x , y , z , p , q , a ) = 0 (2)
which contains an arbitrary constant a and is such that
(i) equations (1) and (2) can be solved for p and q to obtain
p = p ( x , y , u , a ) p=p(x,y,u,a) p = p ( x , y , u , a ) , q = q ( x , y , u , a ) q=q(x,y,u,a) q = q ( x , y , u , a )
(ii) the equation
d u = p ( x , y , u , a ) d x + q ( x , y , u , a ) d y du=p(x,y,u,a)dx+q(x,y,u,a)dy d u = p ( x , y , u , a ) d x + q ( x , y , u , a ) d y (3)
is integrable.
When such a function g is found, the solution
F ( x , y , u , a , b ) = 0 F(x,y,u,a,b)=0 F ( x , y , u , a , b ) = 0
of (3) containing two arbitrary constants a and b will be the solution of (6). The compatibility of equations (1) and (2) yields
The compatibility of equations (1) and (2) yields
[ f , g ] = [f,g]= [ f , g ] = ∂ ( f , g ) ∂ ( x , p ) + p ∂ ( f , g ) ∂ ( z , p ) + ∂ ( f , g ) ∂ ( y , q ) + q ∂ ( f , g ) ∂ ( z , q ) = 0 { \partial (f,g)\above{2pt}\partial (x,p) }+p{ \partial(f,g)\above{2pt} \partial (z,p)}+{\partial (f,g) \above{2pt} \partial (y,q)}+q{ \partial (f,g)\above{2pt} \partial (z,q)}=0 ∂ ( x , p ) ∂ ( f , g ) + p ∂ ( z , p ) ∂ ( f , g ) + ∂ ( y , q ) ∂ ( f , g ) + q ∂ ( z , q ) ∂ ( f , g ) = 0
Expanding it, we are led to the following linear PDE in g(x, y, u, p, q):
f p ∂ g ∂ x + f q ∂ g ∂ y + ( p f p + a f q ) ∂ g ∂ u − ( f x + p f u ) ∂ g ∂ p − ( f y + q f u ) ∂ g ∂ q = 0 f_p{ \partial g\above{2pt} \partial x }+f_q{ \partial g\above{2pt} \partial y}+(pf_p+af_q){ \partial g\above{2pt}\partial u }-(f_x+pf_u){ \partial g\above{2pt}\partial p }-(f_y+qf_u){ \partial g\above{2pt}\partial q }=0 f p ∂ x ∂ g + f q ∂ y ∂ g + ( p f p + a f q ) ∂ u ∂ g − ( f x + p f u ) ∂ p ∂ g − ( f y + q f u ) ∂ q ∂ g = 0 (4)
Now solve (9) to determine g by finding the integrals of the following auxiliary equations:
d x f p = d y f q = d u p f p + q f q = d p − ( f x + p f u ) = d q − ( f y + q f u ) {dx \above{2pt} f_p}={dy \above{2pt}f_q }={du \above{2pt} pf_p+qf_q}={dp \above{2pt} -(f_x+pf_u)}={dq\above{2pt} -(f_y+qf_u)} f p d x = f q d y = p f p + q f q d u = − ( f x + p f u ) d p = − ( f y + q f u ) d q (5)
These equations are known as Charpit’s equations. Once an integral g(x, y, u, p, q, a) of this kind has been found, the problem reduces to solving for p and q, and then integrating equation (3).
Comments
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solve by jacobi method q-px-p^2=0