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My aim is to prove that the system $$ \begin{cases} 2x+y+\sin (x+y) = 0 \\ x-2y + \cos (x+y)=0\end{cases}$$ has, at least, one solution in the ball $B_r(0)$ where $r>1/\sqrt{5}$. This problem is in the book Nonlinear functional Analysis (Ch. 1) by K. Deimling. By some basic algebra I know that if the system has a solution $(x_0,y_0)$, then $x_0^2+y_0^2 = 1/5$. But How to prove that the system actually has a solution?

I want to apply Brouwer degree theory to it. Any hint would be appreciate.

Thanks in advance.

Senna
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1 Answers1

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Denote $z=\binom{x}{y}$. Let $$ F(z,\lambda)=Az+\lambda g(z),\lambda\in[0,1] $$ with $$ A=\bigg(\begin{matrix}2&1\\1&-2\end{matrix}\bigg),g(z)=\binom{\sin(x+y)}{\cos(x+y)}. $$ On the circle $\Omega: |z|=r$ with $r>\frac1{\sqrt5}$, $$ \|F(z,\lambda)\|\ge \|f(x,y)\|-\lambda\|g(x,y)\|=5r-\lambda>0. $$ So by the degree theory, $$ d(F(\cdot,1), \Omega, 1)=d(F(\cdot,1), \Omega, 0)=\text{sign}\det A=-1\neq0. $$ Therefore $F(z,1)=0$ has at least one solution in $B_r(0)$.

xpaul
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