User:Egm6321.f12.team1.pb/report4

Solution:
=Problem 4.5* - Equivalence of two exactness criterion =

Given:
The equation providing the second exactness condition of the second order linear differential equation $$G(x,y,y') = 0$$ as

Where

Prove:
Prove that ($$) is equivalent to the following conditions stated earlier

Where
 * $$p=y'$$

and$$,\; f_{xy} = \frac {\partial^2}{\partial x \partial y} f(x,y,p)$$

Solution:
We have


 * $$g_0 = \frac{\partial G(x,y,p)}{\partial y}$$


 * $$g_1 = \frac{\partial G(x,y,p)}{\partial y'} = \frac{\partial G(x,y,p)}{\partial p}$$
 * $$\Rightarrow g_1 = g_{p}+f_{p}p'$$
 * $$\Rightarrow \frac {\mathrm{d}g_1}{\mathrm{d}x} = \frac {\mathrm{d}}{\mathrm{dx}} \left(g_{p}+f_{p}p'\right) = \frac {\mathrm{d}}{\mathrm{dx}} \left(g_{p}\right) +\frac {\mathrm{d}}{\mathrm{dx}} \left(f_{p}p'\right)$$
 * $$\Rightarrow \frac {\mathrm{d}g_1}{\mathrm{d}x} = \left (g_{px} + g_{py}y' + g_{pp}p'\right) +\frac {\mathrm{d}}{\mathrm{dx}} \left(f_{p}p'\right)$$
 * $$\Rightarrow \frac {\mathrm{d}g_1}{\mathrm{d}x} = \left (g_{px} + g_{py}y' + g_{pp}p'\right) +\frac {\mathrm{d}}{\mathrm{dx}} \left(f_{p}p'\right)$$
 * $$\Rightarrow \frac {\mathrm{d}g_1}{\mathrm{d}x} = \left (g_{px} + g_{py}y' + g_{pp}p'\right) +\frac {\mathrm{d}}{\mathrm{dx}} \left(f_{p}p'\right)$$


 * $$g_2 = \frac{\partial G(x,y,p)}{\partial y''} = \frac{\partial G(x,y,p)}{\partial p'}$$
 * $$\Rightarrow g_2 = f$$
 * $$\Rightarrow \frac {\mathrm{d}g_2}{\mathrm{d}x} = \frac {\mathrm{d}f}{\mathrm{dx}} = f_{x} + f_{y} y' + f_{p} p'$$
 * $$\Rightarrow \frac {\mathrm{d^2}g_2}{\mathrm{d^2}x} = \frac {\mathrm{d}}{\mathrm{dx}} \left( \frac {\mathrm{d}g_2}{\mathrm{d}x} \right) = \frac {\mathrm{d}}{\mathrm{dx}} \left( f_{x} + f_{y} y' + f_{p} p' \right)$$
 * $$\Rightarrow \frac {\mathrm{d^2}g_2}{\mathrm{d^2}x} = \frac {\mathrm{d}}{\mathrm{dx}} \left( f_{x} \right)+ \frac {\mathrm{d}}{\mathrm{dx}} \left(  f_{y} y'  \right)+ \frac {\mathrm{d}}{\mathrm{dx}} \left(  f_{p} p' \right)$$
 * $$\Rightarrow \frac {\mathrm{d^2}g_2}{\mathrm{d^2}x} = \left( f_{xx} + f_{xy}y' + f_{xp}p'\right)+ \frac {\mathrm{d}}{\mathrm{dx}} \left( f_{y} \right) y' + f_{y} y'' +\frac {\mathrm{d}}{\mathrm{dx}} \left(f_{p}p'\right)$$
 * $$\Rightarrow \frac {\mathrm{d^2}g_2}{\mathrm{d^2}x} = \left( f_{xx} + f_{xy}y' + f_{xp}p'\right)+ \left(  f_{yx} + f_{yy}y' + f_{yp}p'\right) y' + f_{y} y'' + \frac {\mathrm{d}}{\mathrm{dx}} \left(f_{p}p'\right)$$
 * $$\Rightarrow \frac {\mathrm{d^2}g_2}{\mathrm{d^2}x} = \left( f_{xx} + f_{xy}p + f_{xp}p'\right)+ \left(  f_{yx} + f_{yy}p + f_{yp}p'\right) p + f_{y} p' + \frac {\mathrm{d}}{\mathrm{dx}} \left(f_{p}p'\right)$$

Substituting the terms obtained in ($$), ($$) and ($$) into ($$)
 * $$ \left [ g_{y}+f_{y}p' \right ] - \left [ \left (g_{px} + g_{py}p + g_{pp}p'\right) +\cancel{\frac {\mathrm{d}}{\mathrm{dx}} \left(f_{p}p'\right)} \right ] + \left [ \left( f_{xx} + f_{xy}p + f_{xp}p'\right)+ \left(  f_{yx}p + f_{yy}p^2 + f_{yp}pp'\right) + f_{y} p' + \cancel{\frac {\mathrm{d}}{\mathrm{dx}} \left(f_{p}p'\right)} \right ] = 0$$

Rearranging terms to segregate into the form $$a(x,y,p)p'+b(x,y,p) = 0$$
 * $$ \left ( f_{y} - g_{pp} + f_{xp} + f_{yp}p + f_{y}\right ) p' + \left ( g_{y} - g_{px} - g_{py}p + f_{xx} + f_{xy}p + f_{yx}p + f_{yy}p^2 \right ) = 0$$

Assuming $$f_{xy}=f_{yx}$$ and $$f$$ and $$g$$ are invariant to the order of differentiation with respect to other variables

Since ($$) is an identity, it should hold for all values of $$p'$$. So each of the two coefficients should be $$0$$ in order to satisfy this identity, i.e. in order to satisfy ($$). So
 * $$ 2f_{y} - g_{pp} + f_{xp} + pf_{yp} = 0$$

And,$$\; g_{y} - g_{xp} - pg_{yp}p + f_{xx} + 2pf_{xy} + p^2f_{yy} = 0$$

($$) and ($$) are identical to ($$) and ($$) respectively. Which implies that ($$) is equivalent to ($$) and ($$) together$$.\qquad \blacksquare$$

=References=