Question #104128
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 )
1
Expert's answer
2020-03-03T10:57:43-0500

 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 (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

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

Now,

(f,g)(z,p)=fzfpgzgp{\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} == 0fp0gp\begin{vmatrix} 0 & {\partial f \above{2pt}\partial p } \\ 0 & { \partial g\above{2pt}\partial p } \end{vmatrix} =0=0 and

(f,g)(z,q)={\partial (f,g) \above{2pt}\partial (z,q) }= fzfqgzgq\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} =0fq0ga=\begin{vmatrix} 0 & {\partial f \above{2pt}\partial q } \\ 0 & { \partial g\above{2pt}\partial a } \end{vmatrix} =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

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)=0f(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)=0g(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) , q=q(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)dydu=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)=0F(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)(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


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

fpgx+fqgy+(pfp+afq)gu(fx+pfu)gp(fy+qfu)gq=0f_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: 

dxfp=dyfq=dupfp+qfq=dp(fx+pfu)=dq(fy+qfu){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). 



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Comments

Assignment Expert
02.02.21, 00:05

Dear YUVASRI RAMAKRISHNAN, please use the panel for submitting new questions.

YUVASRI RAMAKRISHNAN
29.01.21, 23:58

solve by jacobi method q-px-p^2=0

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