To compute asymptotic approximations of solutions $u(x,y),v(x,y)$ of the system of equations \begin{cases} u^a+v^b+x+y= 4\\ u+v+x+y= 4 \end{cases} near $(u,v)=(1,1),(x,y)=(1,1)$.
I replace $u^a$ by $1+a(u-1)$, similarly for $v^b$, I get a linear equation \begin{cases} 1+a(u-1)+1+b(v-1)+x+y=4\\ u+v+x+y=4 \end{cases} Solving this, I get the first order asymptotic $$u_1=1+\frac{(x-1) (b-1)}{a-b}+\frac{(y-1) (b-1)}{a-b}\\ v_1=1+\frac{(x-1)(1-a)}{a-b}+\frac{(y-1) (1-a)}{a-b}$$ How to find the second order asymptotic?
Mathematica (code) gives the result $$u_2=u_1-\frac{(a-1) (b-1) (a (2 b-1)-b) (x+y-2)^2}{2 (a-b)^3}\\ v_2=v_1+\frac{(a-1) (b-1) (a (2 b-1)-b) (x+y-2)^2}{2 (a-b)^3} $$
To compute asymptotic approximations of solutions $u(x,y),v(x,y)$ of the system of equations \begin{cases} u^a+v+x^c+y= 4\\ u+v+x+y= 4 \end{cases} near $(u,v)=(1,1),(x,y)=(1,1)$.
I replace $u^a$ by $1+a(u-1)$, similarly for $x^c$, I get a linear equation \begin{cases} 1+a(u-1)+b+1+c(x-1)+y=4\\ u+v+x+y=4 \end{cases} Solving this, I get the first order asymptotic $$u_1=1+\frac{(1-c) (x-1)}{a-1}\\ v_1=2-y+\frac{(x-1) (c-a)}{a-1}$$ How to find the second order asymptotic?
Mathematica (code) gives the result $$u_2=u_1-\frac{(c-1) (x-1)^2 (a (2 c-1)-c)}{2 (a-1)^2}\\ v_2=v_1+\frac{(c-1) (x-1)^2 (a (2 c-1)-c)}{2 (a-1)^2} $$