Let \(u,v:\mathbb R^2\to\mathbb R\) be \(C^1\). Assume \(\nabla u\) is nowhere zero and that \(\nabla v\) and \(\nabla u\) are linearly dependent at every point. Given \(p_0\in\mathbb R^2\), show that there is a \(C^1\) function \(F\) of one variable such that \[ v(x,y)=F(u(x,y)) \] in some neighborhood of \(p_0\).
The source prints “linearly independent” in place of “linearly dependent.” That printed statement is false: for \[ u(x,y)=x, \qquad v(x,y)=y, \] the gradients are everywhere linearly independent, but \(v\) cannot be a function of \(u\) on any neighborhood. Moreover, any identity \(v=F(u)\) implies \[ \nabla v=F'(u)\nabla u, \] so linear dependence is necessary.
Solution
1. There is a \(C^1\) local coordinate system near \(p_0\) whose first coordinate is \(u\).
Write \[ p_0=(x_0,y_0). \] Since \(\nabla u(p_0)\neq0\), either \[ u_x(p_0)\neq0 \] or \[ u_y(p_0)\neq0. \]
If \(u_x(p_0)\neq0\), define \[ \Phi(x,y)=(u(x,y),y). \] Its Jacobian determinant at \(p_0\) is \(u_x(p_0)\neq0\). If instead \(u_y(p_0)\neq0\), define \[ \Phi(x,y)=(u(x,y),x), \] whose Jacobian determinant at \(p_0\) is \(-u_y(p_0)\neq0\).
In either case, the inverse function theorem gives neighborhoods \[ U\ni p_0, \qquad W\ni\Phi(p_0) \] such that \[ \Phi:U\longrightarrow W \] is a \(C^1\) diffeomorphism. After shrinking \(W\), take it to be an open rectangle \[ W=I\times J. \] Its first coordinate is \(u\).
2. In the coordinates \((s,t)=\Phi(x,y)\), the function\[ \widetilde v(s,t)=v\bigl(\Phi^{-1}(s,t)\bigr) \]satisfies\[ \frac{\partial\widetilde v}{\partial t}(s,t)=0 \]throughout \(I\times J\).
Fix \((s,t)\in I\times J\) and put \[ q=\Phi^{-1}(s,t). \] Let \[ X=\frac{\partial\Phi^{-1}}{\partial t}(s,t)\in T_q\mathbb R^2. \] Because the first coordinate of \(\Phi\) is \(u\), the identity \[ u\bigl(\Phi^{-1}(s,t)\bigr)=s \] gives, after differentiating with respect to \(t\), \[ du_q(X)=0. \]
The hypothesis says that \(du_q\) and \(dv_q\) are linearly dependent. Since \(du_q\neq0\), there is a scalar \(\lambda_q\) such that \[ dv_q=\lambda_q\,du_q. \] Hence \[ dv_q(X)=0. \] By the chain rule, \[ \frac{\partial\widetilde v}{\partial t}(s,t) = dv_q(X) = 0. \]
3. There is a \(C^1\) function \(F:I\to\mathbb R\) such that\[ \widetilde v(s,t)=F(s) \]for every \((s,t)\in I\times J\).
Choose any \(t_0\in J\) and define \[ F(s)=\widetilde v(s,t_0). \] Since \(\widetilde v\) is \(C^1\), so is \(F\).
For fixed \(s\in I\), step 2 says that the one-variable function \[ t\longmapsto\widetilde v(s,t) \] has derivative zero throughout the interval \(J\). By the mean value theorem it is constant on \(J\). Therefore \[ \widetilde v(s,t)=\widetilde v(s,t_0)=F(s) \] for all \(t\in J\).
4. On the neighborhood \(U\) of \(p_0\),\[ \boxed{v(x,y)=F(u(x,y))}. \]
For \((x,y)\in U\), write \[ \Phi(x,y)=(s,t). \] By construction, \[ s=u(x,y). \] Step 3 gives \[ v(x,y) = \widetilde v(s,t) = F(s) = F(u(x,y)). \]
5. Q.E.D.
Step 4 supplies the required \(C^1\) function and neighborhood.