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Question While solving a non-linear partial differential equation, i am stuck to a problem $$z^2(x^2p^2+q^2)=1$$ And let $$F=z^2(x^2p^2+q^2)-1$$ Where $p=\frac{\partial z}{\partial x}$ And $q= \frac{\partial z}{\partial y}$

my approach From Charpit method $$\frac{dx}{F_p}=\frac{dy}{F_q}=\frac{dz}{pF_p+qF_q}= \frac{-dp}{F_x+pF_z}=\frac{-dq}{F_y+qF_q}$$

$$\frac{dx}{2px^2z^2}=\frac{dy}{2qz^2}=\frac{dz}{2p^2x^2z^2+2q^2z^2}=\frac{-dp}{2xp^2z^2+2pz(x^2p^2+q^2)}=\frac{-dq}{2zq(x^2p^2+q^2)}$$

The solution of the pde is given by $$dz=pdx+qdy$$ My problem is, i am unable to seperate the functions for further solution. Any subtle hints are highly apperciated.

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  • $\begingroup$ What is the boundary condition ? $\endgroup$
    – JJacquelin
    Sep 22 '19 at 7:50
  • $\begingroup$ The boundary conditions are not given $\endgroup$ Sep 22 '19 at 8:55
  • $\begingroup$ Note that $dz=pdx+qdy$ doesn't give the solution of the PDE because it is a general property (total derivative) : $$dz(x,y)=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$$ . $\endgroup$
    – JJacquelin
    Sep 22 '19 at 13:48
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$$z^2x^2\left(\frac{\partial z}{\partial x} \right)^2+z^2\left(\frac{\partial z}{\partial y}\right)^2=1$$

Below, this is not a direct answer to the question, but an hint, too long to be edited in the comments section.

The change of variables :$\quad\begin{cases} Z=\frac12 z^2\\ X=\frac12 x^2\\ Y=y \end{cases}\quad$ transforms the PDE into : $$\left(\frac{\partial Z}{\partial X} \right)^2+\left(\frac{\partial Z}{\partial Y}\right)^2=1$$ In the littérature one can find examples of solving the Eikonal equation with various boundary conditions :

Example 11, p.21 in http://www.ehu.eus/luis.escauriaza/apuntes_problemas_y_examene/method-of-characteristics.pdf

Example 2.26, p.21 in https://www.math.ualberta.ca/~xinweiyu/436.A1.12f/PDE_Meth_Characteristics.pdf

And on Math.StackExchange : Finding a complete integral solution for the Eikonal equation $u_x^2+u_y^2=1$

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  • $\begingroup$ Can i use any boundry condition provided above for solution? $\endgroup$ Sep 22 '19 at 8:58
  • $\begingroup$ You can find analytic solution in some cases of boundary conditions. I doubt in other cases. Look at the extensive littérature about eikonal equation. $\endgroup$
    – JJacquelin
    Sep 22 '19 at 9:21
  • $\begingroup$ Thankyou for support $\endgroup$ Sep 22 '19 at 9:22

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