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Let $f:\mathbb{R}^2\rightarrow \mathbb{R}$ be a differentiable function such that for all $x\in\mathbb{R}^2$, $$ \frac{\partial f}{\partial x_2}(x)=2\cdot\frac{\partial f}{\partial x_1}(x). $$ Prove that for every $c\in\mathbb{R}$ function $f$ is constant on $$M_c=\left\{x\in\mathbb{R}^2: 2x_2+x_1=c\right\}.$$

I don't know how to approach.

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is there a minus sign missing somewhere? – user20266 Jun 18 '12 at 14:14
@Thomas, I checked. The task is well-prescribed. – ray Jun 18 '12 at 14:15
Approach : Over the set $M_c$, show that the partial derivative w.r.t $x_1$ is zero. Hint : Find a relationship between partial derivative w.r.t $x_1$ and partial derivative w.r.t $x_2$ that is satisfied over the set $M_c$. – TenaliRaman Jun 18 '12 at 14:16
up vote 5 down vote accepted

Consider the map $\phi \colon \mathbb{R} \to \mathbb{R}^2$ defined by $$ \phi(t) = \begin{pmatrix} c-2t \\ t \end{pmatrix}. $$ Now try to compute $$ \frac{d}{dt}\left( f\circ \phi \right) = \frac{\partial f}{\partial x_1} \cdot (-2) + \frac{\partial f}{\partial x_2} $$ and use the assumption. What can you conclude from this?

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Ok, finally I see it. With this parameterization we have that $\phi: \mathbb{R}\rightarrow M_c$ so we need to prove that $(f\circ\phi)$ is a constant function. But by chain rule $\frac{d}{dt}(f\circ\phi)=-2\frac{\partial f}{\partial x_1}+\frac{\partial f}{\partial x_2}=0$ (assumptions) so the derivative is equal to zero, so the function is constant. Beautiful, thank you so much again! – ray Jun 18 '12 at 14:48

Note that by the given assumption the gradient of $f$ is orthogonal to $\mathbf{d}=(2,-1)$ at any point $\mathbf{x}$ in the domain of $f$, i.e. $\nabla{f(\mathbf{x})}\cdot{\mathbf{d}}=0$ all $\mathbf{x}\in\mathbb{R^n}$ Now use the mean value theorem in several variables which states that for a differentiable function $f:\mathbb{R}^n\mapsto\mathbb{R} $ for every $\mathbf{a},\mathbf{b}\in \mathbb{R^n}$ there is a $\mathbf{c} \in [\mathbf{a},\mathbf{b}]$, where $$[\mathbf{a},\mathbf{b}]:=\{\mathbf{x}\in \mathbb{R^n}| \mathbf{x}=t\mathbf{a}+(1-t)\mathbf{b}, t \in [0,1]\subset \mathbb{R}\}$$ such that $f(\mathbf{a})-f(\mathbf{b})=\nabla{f(\mathbf{c})\cdot{}(\mathbf{a}-\mathbf{b})}$.

Now pick a point $\mathbf{a}=(a_1,a_2) \in M_c$, and let $\mathbf{b}=(b_1,b_2)\in M_c$ be arbitrary. Then clearly $\mathbf{a}=(c-2a_2,a_2)$ and $\mathbf{b}=(c-2b_2,b_2)$. From the mean value theorem above it follows that $f(\mathbf{a})-f(\mathbf{b})=\nabla{f(\mathbf{c})}\cdot(\mathbf{a}-\mathbf{b})=\nabla{f(\mathbf{c})}\cdot{}(2,-1)(b_2-a_2)=(b_2-a_2)\nabla{f(\mathbf{c})}\cdot \mathbf{d}=0$.

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This method is instructive, but it requires an additional assumption, namely: $f$ has $C^2$ class (existence and continuity of the partial derivatives of second order) – Mohamed Jun 18 '12 at 15:55
@Mohamed I think that it is enough that f is differentiable, have a look at these notes for example : – Henrik Jun 18 '12 at 16:03
I am sorry I had a stupid confusion between the symbols $\nabla$ and $\Delta$ (Laplacian). I hope that a moderator removes the last part of my comment above : "but it requires ..." till the end and replace by : " I keep it as an excellent alternative " – Mohamed Jun 18 '12 at 18:41

Well! we can generalize this : $f$ is constant on all subset $M_{u,v,I}$ of $\mathbb R^2$ where :$$M_{u,v,I}=\{(u(t),v(t)) \mid t \in I \}$$ where : $I$ is an open intervall and $u,v$ having $C^1$ class on $I$ to $\mathbb R$ such that : $u' + 2 v'=0$

In our example we have : $I=\mathbb R$ and $u(t)=t$ and $v(t)=\frac{c-t}2$

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