User:Egm6321.f11.team1.zheng.zx/personal place/HW4

= R*4.6 Show the equivalence of two forms of 2nd exactness condition of N2-ODE =

Given
$$ \displaystyle g_0 - \frac{d}{dx}g_1+\frac{d^2}{dx^2}g_2=0 $$

$$ \displaystyle g_i:=\frac{\partial G}{\partial y^{(i)}} \, \quad for \ i=0,1,2 $$

Find
It's equal to the 2nd exactness condition in form:

$$ \displaystyle \phi_{xy}=\phi_{yx}\ ,\quad \phi_{px}=\phi_{xp}\ ,\quad \phi_{py}=\phi_{yp} $$

Solution
As defined:

$$ \displaystyle G=\frac{d \phi}{dx} (x,y,p)=0 $$

$$ \displaystyle G=\phi_x+\phi_y \cdot p +\phi_p \cdot q $$

$$ \displaystyle for \quad p=y' \, \ q=y'' $$

$$ \displaystyle g_0 = \frac{\partial G}{\partial y} = \frac{\partial}{\partial y}(\phi_x+\phi_yp+\phi_pq) = \frac{\partial}{\partial y}(\frac{d \phi}{dx}) $$

$$ \displaystyle \frac{d}{dx}g_1 = \frac{d}{dx}(\frac{\partial G}{\partial p}) = \frac{d}{dx}(\phi_{xp}+\phi_{yp}p+\phi_y+\phi_{pp}q) $$

$$ \displaystyle \begin{align} \frac{d^2}{dx^2} g_2& = \frac{d^2}{dx^2}(\frac{\partial G}{\partial q})\\ &=\frac{d^2}{dx^2} [\frac{\partial}{\partial q}(\phi_x + \phi_yp +\phi_pq)] \\ &= \frac{d^2}{dx^2} \phi_p\\ &= \frac{d}{dx}(\phi_{px}+\phi_{py}p+\phi_{pp}q) \end{align} $$

For we have:

$$ \displaystyle g_0-\frac{d}{dx}g_1+\frac{d^2}{dx^2}g_2=0 $$

$$ \displaystyle \frac{\partial}{\partial y}(\frac{d \phi}{dx}) - \frac{d}{dx}(\phi_{xp}+\phi_{yp}p+\phi_y+\phi_{pp}q) + \frac{d}{dx}(\phi_{px}+\phi_{py}p+\phi_{pp}q) = 0 $$

Rearranging:
 * {| style="width:100%" border="0" align="left"

\displaystyle [\frac{\partial}{\partial y} (\frac{d\phi}{dx})-\frac{d}{dx}(\frac{\partial \phi}{\partial y})] + \frac{d}{dx} [(\phi_{px}-\phi_{xp})+(\phi_{py}-\phi_{yp})p+(\phi_{pp}-\phi_{pp})q] = 0 $$ $$
 * $$\displaystyle (Equation\;4.6.1 )
 * }
 * }

So, both two parts in the left side of equation 4.6.1 are equal to 0 separately: For first part:

$$ \displaystyle \frac{\partial}{\partial y} (\frac{d\phi}{dx})-\frac{d}{dx}(\frac{\partial \phi}{\partial y})=0 $$

$$ \displaystyle (\phi_{xy}-\phi_{yx})+(\phi_{yy}-\phi_{yy})p+(\phi_{py}-\phi_{yp}q)=0 $$

so:

$$ \displaystyle \phi_{xy}=\phi_{yx} $$

$$ \displaystyle \phi_{py}=\phi_{yp} $$

And, second part:

$$ \displaystyle \frac{d}{dx} [(\phi_{px}-\phi_{xp})+(\phi_{py}-\phi_{yp})p+(\phi_{pp}-\phi_{pp})q] = 0 $$

$$ \displaystyle (\phi_{px}-\phi_{xp})+(\phi_{py}-\phi_{yp})p+(\phi_{pp}-\phi_{pp})q = 0 $$

thus we get:

$$ \displaystyle \phi_{px}=\phi_{xp} $$

= References =