Given 3 functions $u(v,p)$,$g(v,p)$ and $f(v,p)$ is related by the condition $\frac{dp}{dv} =f(v,p)=\frac{g-u_v}{u_p}$ and initial condition $p(v_0)=p_0$. Show that the solution of $p_1$ that fit in in this equation
$u(v_0,p_0)+g(v_0,p_0)h=u(v_0+h,p_1)$
converge to an Euler method when h converge to $0$ which mean
$p_1\approx p_0+hf(v_0,p_0)$
I'm expand it using taylor series but somehow is it true if the function u have a independent variable $v$ and dependant variable $p$, it doesn't make sense to use such approach below to proof?
Rearrange the formula $u(v_0,p_0)+g(v_0,p_0)h=u(v_0+h,p_1)$
to
$u(v_0+h,p_1)-u(v_0,p_0)=g(v_0,p_0)h$
By taylor expansion on the above LHS formula
$u(v_0+h,p_1)-u(v_0,p_0)$
$=u_v(v_0+h-v_0)+u_p(p_1-p_0)+\frac{1}{2}[u_{vv}(v_0+h-v_0)^2+2u_{pv}(v_0+h-v_0)(p_1-p_0)+u_{pp}(p_1-p_0)^2+...]$
On the RHS have $g(v_0,p_0)h$ where $g=u_pf+u_v$ which is a condition impose on the function of $g$, and $f=\frac{dp}{dv}$
Hence
$u_v(v_0+h-v_0)+u_p(p_1-p_0)+\frac{1}{2}[u_{vv}(v_0+h-v_0)^2+2u_{pv}(v_0+h-v_0)(p_1-p_0)+u_{pp}(p_1-p_0)^2+...]=(u_v+u_pf_0)h$
the above equation is use to solve for the identity of $p_1$
2 things know about $p_1$
1)When h change, $p_1$ change, it is a function dependency on h which can be represent by power series
2)When h=0,$p_1=p_0$ which means the power series is center around $0$
so
$p_1=a_0+a_1h+a_2h^2+...$
when $h=0,p_1=p_0$ so $a_0=p_0$, substitute this to above taylor series give
$u_v(h)+u_p(a_1h+a_2h^2+...)+\frac{1}{2}[u_{vv}(h)^2+2u_{pv}(h)(a_1h+a_2h^2+...)+u_{pp}(a_1h+a_2h^2+...)^2]+...=(u_v+u_pf_0)h$
Notice the coefficient of h on the RHS equation is $u_pf_0+u_v$
$u_vh+u_p(a_1h)=(u_pf_0+u_v)h$ get $a_1=f_0$
In order to find $a_2$, need to group all the coefficient of $h^2$ on the left hand side taylor series and consider the sum of coefficient of $h^2=0$ since on the RHS equation can be treated as $g_0h+0h^2+0h^3+...$, which is
$u_p(a_2h^2)+\frac{1}{2}[u_{vv}h^2+2u_{pv}f_0h^2+u_{pp}(hf)^2]=0h^2$
$a_2=-\frac{[u_{vv}+2u_{pv}f_0+u_{pp}f^2]}{2u_p}$
which can use same method to find $a_n$ and so on
This show that why when stepsize h tend to 0, the solution $p_1$ by directly solve from this formula $u(v_0,p_0)+g(v_0,p_0)h=u(v_0+h,p_1)$ , tend to become an Euler method