Answer to Question #104128 in Differential Equations for maria

Question

1. 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.


2. how how to solve, by Jacobi's method, a partial differential equation of

the type f( x, delu/delx, delu/delz ) = g(y, delu/dely, delu/delz )

Answer 1

 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

${ \partial (f,g)\above{2pt}\partial (x,p) }+p{ \partial(f,g)\above{2pt} \partial (z,p)}+{\partial (f,g) \above{2pt} \partial (y,q)}+q{ \partial (f,g)\above{2pt} \partial (z,q)}=0$ (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

${\partial f \above{2pt}\partial z } =0$ and ${\partial g \above{2pt} \partial z}=0$ (3)

Now,

${\partial (f,g) \above{2pt}\partial (z,p) }=\begin{vmatrix} { \partial f\above{2pt} \partial z} & {\partial f \above{2pt}\partial p } \\ {\partial g\above{2pt}\partial z } & { \partial g\above{2pt}\partial p } \end{vmatrix}$ $=$ $\begin{vmatrix} 0 & {\partial f \above{2pt}\partial p } \\ 0 & { \partial g\above{2pt}\partial p } \end{vmatrix}$ $=0$ and

${\partial (f,g) \above{2pt}\partial (z,q) }=$ $\begin{vmatrix} { \partial f\above{2pt} \partial z} & {\partial f \above{2pt}\partial q } \\ {\partial g\above{2pt}\partial z } & { \partial g\above{2pt}\partial q } \end{vmatrix}$ $=\begin{vmatrix} 0 & {\partial f \above{2pt}\partial q } \\ 0 & { \partial g\above{2pt}\partial a } \end{vmatrix}$ $=0$

Substituting these value in equation (2) the required solution is

${ \partial (f,g)\above{2pt}\partial (x,p) }+{\partial (f,g) \above{2pt} \partial (y,q)}=0$

Charpit’s method:-

It is a general method for finding the general solution of a nonlinear PDE of first-order of the form 

$f(x,y,z,p,q)=0$ (1)

Basic Idea: To introduce another partial differential equation of the first order

$g(x,y,z,p,q,a)=0$ (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

$p=p(x,y,u,a)$ , $q=q(x,y,u,a)$

(ii) the equation 

$du=p(x,y,u,a)dx+q(x,y,u,a)dy$ (3)

is integrable. 

When such a function g is found, the solution

$F(x,y,u,a,b)=0$

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

$[f,g]=$ ${ \partial (f,g)\above{2pt}\partial (x,p) }+p{ \partial(f,g)\above{2pt} \partial (z,p)}+{\partial (f,g) \above{2pt} \partial (y,q)}+q{ \partial (f,g)\above{2pt} \partial (z,q)}=0$

Expanding it, we are led to the following linear PDE in g(x, y, u, p, q):

$f_p{ \partial g\above{2pt} \partial x }+f_q{ \partial g\above{2pt} \partial y}+(pf_p+af_q){ \partial g\above{2pt}\partial u }-(f_x+pf_u){ \partial g\above{2pt}\partial p }-(f_y+qf_u){ \partial g\above{2pt}\partial q }=0$ (4)

Now solve (9) to determine g by finding the integrals of the following auxiliary equations: 

${dx \above{2pt} f_p}={dy \above{2pt}f_q }={du \above{2pt} pf_p+qf_q}={dp \above{2pt} -(f_x+pf_u)}={dq\above{2pt} -(f_y+qf_u)}$ (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).