Solve nonlinear system of equations and show it has infinite solutions

Question

So we have this system of nonlinear equations \begin{align*} \sin(x+u) - e^y + 1 = 0\\ x^2 + y + e^u = 1 \end{align*} and we want to show that it has infinitely many solutions $(x,y,u)$.

I tried starting by finding the Jacobian matrix (so I can use the theorem of implicit functions) $$ J_{(x,y,u)} = \left[\begin{array}{ccc} \cos(x+u) & -e^y & \cos(x+u) \\ 2x & 1 & e^u \end{array}\right] $$ but obviously this is not a square matrix so the determinant is not defined. I was thinking we could add another equation, I was told "$H(x,y,u) = u$ (or $x$ or $y$)" could work but how is that "allowed"?

Answer 1

You might notice that $(x,y,u) = (0,0,0)$ is a solution. In a neighbourhood of that, consider looking for a solution of the form $x = x(u), y = y(u)$. Then by differentiating the equations with respect to $u$, you get

$$ \eqalign{(1 + x') \cos(x+u) - y' e^y &= 0\cr 2 x x' + y' + e^u &= 0\cr}$$ and therefore the system of differential equations $$ \eqalign{x' &= - \frac{e^{u+y} + \cos(x+u)}{2 x e^y + \cos(x+u)} \cr y' &= - \frac{\cos(x+u) (e^u - 2 x)}{2 x e^y + \cos(x+u)}\cr}$$ A solution of this system with initial conditions $x(0)=0, y(0)=0$ exists in some interval around $u=0$, and will be a solution of your system.

Answer 2

What you need to apply the implicit function theorem is merely that some $\ 2\times2\ $ submatrix of the Jacobian be invertible at some solution of the equations. If you take the solution $\ x=y=u=0\ $ suggested Robert Israel in his answer, your Jacobian matrix at that solution is

\begin{align} J_{(x,y,u)}(0,0,0)&=\pmatrix{\cos(0+0)&-e^{0}&\cos(0+0)\\ 2\times0&1&e^0}\\ &=\pmatrix{1&-1&1\\ 0&1&1}\ . \end{align} Since the matrix $$ \pmatrix{1&-1\\ 0&1} $$ formed by the first two columns of the Jacobian is invertible, the implicit function theorem tells you that there's a neighbourhood $\ U\ $ of $\ 0\ $ and a function $\ \big(\bar{x},\bar{y}\big):U\rightarrow\mathbb{R}\ $ such that $\ x=\bar{x}(v),y=\bar{y}(v),u=v\ $ will be a solution of the equations for all $\ v\in U\ .$

Relation to Robert Israel's answer

The first set of equations in Robert Israel's answer come from using the chain rule to differentiate the equations \begin{align} \cos\big(x(u)+u\big)-e^{y(u)}+1&=0\\ x(u)^2+y(u)+e^u&=1 \end{align} with respect to $\ u\ ,$ and can be written in matrix form as \begin{align} {\scriptstyle J_{(x,y,u)}\big(x(u),y(u),u\big)}&{\scriptstyle\pmatrix{x'(u)\\y'(u)\\1}}\\ &{\scriptstyle=\pmatrix{\cos\big(x(u)+u\big)&-e^{y(u)}&\cos\big(x(u)+u\big)\\ 2x(u)&1&e^{y(u)}}\pmatrix{x'(u)\\y'(u)\\1}}\\ &{\scriptstyle=\pmatrix{\cos\big(x(u)+u\big)&-e^{y(u)}\big)\\ 2x(u)&1}\pmatrix{x'(u)\\y'(u)}+\pmatrix{\cos\big(x(u)+u\big)\\e^{y(u)}}}\\ &{\scriptstyle=\pmatrix{0\\0}}\ . \end{align} The last of this series of equations can be written $$ \scriptstyle\pmatrix{x'(u)\\y'(u)}=-\pmatrix{\cos\big(x(u)+u\big)&-e^{y(u)}\big)\\ 2x(u)&1}^{-1}\pmatrix{\cos\big(x(u)+u\big)\\e^{y(u)}}\ , $$ provided the inverse of the matrix $\ \scriptstyle\pmatrix{\cos\big(x(u)+u\big)&-e^{y(u)}\big)\\ 2x(u)&1}\ $ exists. This is Robert Israel's system of differential equations written as a matrix equation. What he has done, therefore, is give an explicit proof of the implicit function theorem for this particular example. In fact, one standard way of proving the general case of that theorem is just a more general version of the one Robert Israel has given here.