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Thomas Richard
Thomas Richard
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Set $V$ to be the subbundle of $TM$ whose fiber at $p$ is the span of $\nabla\phi_1(p)$ and $\nabla\phi_2(p)$. Your equation on $u$ is equivalent to requiring that $du$ vanishes on $V$. Frobenius integrability theorem tells you it has (EDIT: non constant) local solutions if and only if $V$ is integrable (i.e. it is stable under Lie bracket). It is not too hard to find functions $\phi_1$ and $\phi_2$ such that $[\nabla\phi_1(p),\nabla\phi_2(p)]\notin V$. For instance on $\mathbb{R}^3$, take $\phi_1(x,y,z)=x$ and $\phi_2(x,y,z)=xyz$.

EDIT: This example is actually false, the one provided by Deane Yang works.
For instance on $\mathbb{R}^3$, take $\phi_1(x,y,z)=x$ and $\phi_2(x,y,z)=xyz$.

Set $V$ to be the subbundle of $TM$ whose fiber at $p$ is the span of $\nabla\phi_1(p)$ and $\nabla\phi_2(p)$. Your equation on $u$ is equivalent to requiring that $du$ vanishes on $V$. Frobenius integrability theorem tells you it has (EDIT: non constant) local solutions if and only if $V$ is integrable (i.e. it is stable under Lie bracket). It is not too hard to find functions $\phi_1$ and $\phi_2$ such that $[\nabla\phi_1(p),\nabla\phi_2(p)]\notin V$. For instance on $\mathbb{R}^3$, take $\phi_1(x,y,z)=x$ and $\phi_2(x,y,z)=xyz$.

Set $V$ to be the subbundle of $TM$ whose fiber at $p$ is the span of $\nabla\phi_1(p)$ and $\nabla\phi_2(p)$. Your equation on $u$ is equivalent to requiring that $du$ vanishes on $V$. Frobenius integrability theorem tells you it has (EDIT: non constant) local solutions if and only if $V$ is integrable (i.e. it is stable under Lie bracket). It is not too hard to find functions $\phi_1$ and $\phi_2$ such that $[\nabla\phi_1(p),\nabla\phi_2(p)]\notin V$.

EDIT: This example is actually false, the one provided by Deane Yang works.
For instance on $\mathbb{R}^3$, take $\phi_1(x,y,z)=x$ and $\phi_2(x,y,z)=xyz$.

Added the fact that the solution needs to be non constant.
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Thomas Richard
Thomas Richard
  • 4.1k
  • 1
  • 24
  • 39

Set $V$ to be the subbundle of $TM$ whose fiber at $p$ is the span of $\nabla\phi_1(p)$ and $\nabla\phi_2(p)$. Your equation on $u$ is equivalent to requiring that $du$ vanishes on $V$. Frobenius integrability theorem tells you it has (EDIT: non constant) local solutions if and only if $V$ is integrable (i.e. it is stable under Lie bracket). It is not too hard to find functions $\phi_1$ and $\phi_2$ such that $[\nabla\phi_1(p),\nabla\phi_2(p)]\notin V$. For instance on $\mathbb{R}^3$, take $\phi_1(x,y,z)=x$ and $\phi_2(x,y,z)=xyz$.

Set $V$ to be the subbundle of $TM$ whose fiber at $p$ is the span of $\nabla\phi_1(p)$ and $\nabla\phi_2(p)$. Your equation on $u$ is equivalent to requiring that $du$ vanishes on $V$. Frobenius integrability theorem tells you it has local solutions if and only if $V$ is integrable (i.e. it is stable under Lie bracket). It is not too hard to find functions $\phi_1$ and $\phi_2$ such that $[\nabla\phi_1(p),\nabla\phi_2(p)]\notin V$. For instance on $\mathbb{R}^3$, take $\phi_1(x,y,z)=x$ and $\phi_2(x,y,z)=xyz$.

Set $V$ to be the subbundle of $TM$ whose fiber at $p$ is the span of $\nabla\phi_1(p)$ and $\nabla\phi_2(p)$. Your equation on $u$ is equivalent to requiring that $du$ vanishes on $V$. Frobenius integrability theorem tells you it has (EDIT: non constant) local solutions if and only if $V$ is integrable (i.e. it is stable under Lie bracket). It is not too hard to find functions $\phi_1$ and $\phi_2$ such that $[\nabla\phi_1(p),\nabla\phi_2(p)]\notin V$. For instance on $\mathbb{R}^3$, take $\phi_1(x,y,z)=x$ and $\phi_2(x,y,z)=xyz$.

Source Link
Thomas Richard
Thomas Richard
  • 4.1k
  • 1
  • 24
  • 39

Set $V$ to be the subbundle of $TM$ whose fiber at $p$ is the span of $\nabla\phi_1(p)$ and $\nabla\phi_2(p)$. Your equation on $u$ is equivalent to requiring that $du$ vanishes on $V$. Frobenius integrability theorem tells you it has local solutions if and only if $V$ is integrable (i.e. it is stable under Lie bracket). It is not too hard to find functions $\phi_1$ and $\phi_2$ such that $[\nabla\phi_1(p),\nabla\phi_2(p)]\notin V$. For instance on $\mathbb{R}^3$, take $\phi_1(x,y,z)=x$ and $\phi_2(x,y,z)=xyz$.