Help me!
Using the theorem (Existence and uniqueness of the Brouwer degree). There is a unique map $\deg_{B}: \Sigma \rightarrow \mathbb{Z}$ (Brouwer degree), where $\mathbb{Z}$ represents the set of integers, such that the homotopy invariance holds.
I must prove the existence of the solution of the following system, using the topological degree (Brouwer degree)
$$\begin{cases}
8x+6y+ \dfrac{\log (x^{2}+2)}{x^{2}+y^{2}+1}-\sin (y-7)+3=0 \\
3x-y+ e^{(-y^{2}-4)}+ \cos(xy)+2=0
\end{cases}\label{eq1}\tag{1}$$
I've done this:
Step (a)
Take the linear part of the system and propose a function
$f: \mathbb{R}^2 \longrightarrow \mathbb{R}^2$
$$(x,y) \longrightarrow (8x+6y,3x-y)= \begin{bmatrix}
8 & 6 \\
3 & -1
\end{bmatrix} \begin{bmatrix}
x\\
y
\end{bmatrix} $$
i.e the linear system can be written in a matrix form where $A= \begin{bmatrix}
8 & 6 \\
3 & -1
\end{bmatrix}$
We know that if $\deg_{B} (f, \Omega, 0) \neq 0$ the equation $f(x)=0$ has a solution, and $\deg_{B} (f, \Omega, 0) = \mathop{\rm sgn}(\det (A))$, but $\det (A)=-26$, then:
$\deg_{B} (f, B_{\mathbb{R}^{2}}(0,r), 0) = \mathop{\rm sgn}(\det (A)) = -1 \neq 0$, in the case $\Omega = B_{\mathbb{R}^{2}}(0,r)$
Step (b)
Take the nonlinear part of the system and propose a function
$g: \mathbb{R}^{2} \longrightarrow \mathbb{R}^{2}$
$$(x,y) \longrightarrow \left( \dfrac{\log (x^{2}+2)}{x^{2}+y^{2}+1}-\sin (y-7)+3, e^{(-y^{2}-4)}+ \cos(xy)+2 \right)$$
Then $(\ref{eq1}) \Leftrightarrow (f+g)(x,y)=(0,0)$.
So I need to prove that $\deg_{B} (f+g, B_{\mathbb{R}^{2}}(0,r) , 0) \neq 0 $, so I propose a homotopy from $f$ to $f + g$ of the form $f + tg$, $\forall t \in [0,1]$:
$$\begin{aligned}
H:[0,1]\times B_{\mathbb{R}^{2}}(0,r) &\longrightarrow \mathbb{R}^{2} \\
(t,x,y) &\longrightarrow f(x,y) + tg(x,y) \\
(t, \chi) &\longrightarrow f(\chi) + tg(\chi)
\end{aligned}$$
For this homotopy to be valid, it must be true that: $\forall \chi=(x,y): \Vert \chi \Vert =r, \forall t \in [0,1] \Rightarrow H(t, \chi) \neq 0 $
I test this using "Methods of a priori dimensions", that is, if $H(t, \chi) = 0 $
$ \forall t \in [0,1]$, $\chi \in \mathbb{R}^{2} \Rightarrow \Vert \chi \Vert < M$
Since we take $ \Omega = B_{\mathbb{R}^{2}} (0, r) $, it must be true that $ M \leq r $, then $\Vert \chi \Vert < M$, and it is true, since the system can be written as:
$$ \begin{aligned}
A &\begin{bmatrix}
x\\
y
\end{bmatrix} + t \begin{bmatrix}
\dfrac{ \log (x^{2}+2)}{x^{2}+y^{2}+1}-\sin (y-7)+3\\
e^{(-y^{2}-4)}+ \cos(xy)+2
\end{bmatrix}\\ \Rightarrow &\begin{bmatrix}
x\\
y
\end{bmatrix} = A^{-1}\begin{bmatrix}
-t \dfrac{ \log (x^{2}+2)}{x^{2}+y^{2}+1}+t\sin (y-7)-3t \\
-te^{(-y^{2}-4)} -t\cos(xy)-2t
\end{bmatrix}\\ \Rightarrow &\begin{bmatrix}
x\\
y
\end{bmatrix} \leq M
\end{aligned} $$
Is my solution correct? or can you give me an idea how to do it.
Thanks!
Best Answer
The solution I mention is correct and complete.