Charpit's method

Suppose one wants to solve a first order nonlinear PDE

$$f(x, y, u, u_x, u_y) = 0$$ (1.22)

As mentioned earlier, the fundamental idea in Charpit's method is to introduce a compatible PDE of the first order

$$g(x, y, u, u_x, u_y, a) = 0,$$ (1.23)

which contains an arbitrary constant $a$. Then Eq.() and Eq.() can be solved to give

$$u_x = u_x(x, y, u, a), \hspace{1cm} u_y = u_y(x, y, u, a).$$

This gives integrable equation

$$du = u_x(x, y, u, a) dx + u_y(x, y, u, a) dy.$$ (1.24)

When such a function $g$ has been found, the solution of Eq.().

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

containing two arbitrary constants $a$ and $b$ will be the solution of Eq.(). The conditions for such a compatible PDE $g=0$ are

$$J=\frac{\partial(f, g)}{\partial(u_x, u_y)} \neq 0,$$    and $$[f, g]=0.$$

Expanding $[f,g]=0$, one get

$$\hspace{-1cm}f_{u_x} \frac{\partial g}{\partial x} + f_{u_y} \fra... ...ac{\partial g}{\partial u_x} - (f_y + u_yf_u) \frac{\partial g}{\partial u_y}=0$$ (1.25)

This is linear PDE in $g.$ The auxiliary equations for it are

$$\hspace{-0.5cm} \frac{dx}{f_{u_x}} = \frac{dy}{f_{u_y}} = \frac{d... ... + u_y f_{u_y}} = \frac{du_x}{-(f_x + u_x f_u)} = \frac{du_y}{- (f_y + u_yf_u)}$$ (1.26)

These equation is known as Charpit's equations and are equivalent to the characteristic equations. Any integral of Eq.() involving $u_{x}$ or $u_{y}$ or both can be taken as the compatible PDE Eq.(). Then integration of Eq.() gives the desired complete integral of Eq.().

Not all of Charpit's equations () need to be used.
It is enough to choose the simplest of them.

Example 1.6.1   Find the complete integral of $u u_{x}^{2} + u_{y}^{2} = 4$.

Solution This is a nonlinear PDE. One will solve it by Charpit's method.

Here $f=u u_{x}^{2} + u_{y}^{2} -4.$ To find compatible PDE, the auxiliary equations are

$$\frac{dx}{f_{u_x}} = \frac{dy}{f_{u_y}} = \frac{du}{u_x f_{u_x} + u_y f_{u_y}} = \frac{du_x}{-(f_x + u_x f_u)} = \frac{du_y}{- (f_y + u_yf_u)}$$

$$\therefore~ \frac{dx}{2 u u_x} = \fr... ..._{x}^{2} + 2 u_{y}^{2}} = \frac{du_x}{-u_{x}^{3} = \frac{du_y}{-u_y u_{x}^{2}}}$$

From the last two equations $\dfrac{du_x}{u_x} = \dfrac{du_y}{u_y}$ , which gives $u_x=a u_y$ ($a$ is constant.) The required compatible equation is $u_x=a u_y.$ Now one solve the given equation and this eq. for $u_x$ and $u_y$. Put $u_x=a u_y$ in given PDE.

$$u a^2 u_{y}^{2} + u_{y}^{2} = 4$$

$$\therefore u_y = \frac{\pm 2}{\sqrt{1+ a^2u}}, \hspace{1cm} u_x = \frac{\pm 2a}{\sqrt{1+ a^2u}}$$

Substitute this in equation $du = u_x dx + u_y dy$, one get

$$du = \frac{\pm 2a dx \pm 2 dy}{\sqrt{1+ a^2u}},$$

$$\sqrt{1+ a^2u}~du = \pm 2a dx \pm 2 dy$$

Integrating it,

$$\frac{(1 + a^2 u)^{\frac{3}{2}}}{\frac{3}{2}a^2} = \pm 2a x \pm 2 y + 2b ~\hspace{1cm} (2b~ ),$$

$$\implies~(1 + a^2 u)^{\frac{3}{2}} = \pm 3a^{3} x \pm 3 a^2 y + 3a^2b,$$

$$\implies~(1 + a^2 u) = (\pm 3a^{3} x \pm 3 a^2 y + 3a^2b)^{\frac{2}{3}}.$$

This is the required complete integral. Check that it satisfies the given PDE and the compatible PDE.