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Consider $$xU_x +y U_y = 0$$

$$U(x,y) = x, \ \ \ \ on \ \ \ \ x^2 + y^2 = 1$$

has

  1. a solution for all x,y $\in \mathbb R$

  2. an unique solution in $\{ (x,y) \in \mathbb R^2 : (x,y) \neq 0 \}$

  3. a bounded solution in $\{ (x,y) \in \mathbb R^2 : (x,y) \neq 0 \}$

  4. an unique solution in $\{ (x,y) \in \mathbb R^2 : (x,y) \neq 0 \}$, but the solution is unbounded.

We can solve this by Lagranges method, we obtain $U(x,y = f(x-y)$ for some function $f$

We have given that $U(x,y) = x, \ \ \ \ on \ \ \ \ x^2 + y^2 = 1$, so $U(cos(\theta), sin(\theta)) = cos(\theta)$ for all $\theta$.

Please help me check which option is true.

Thank you

Struggler
  • 2,554

1 Answers1

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Our starting point is \begin{equation*} (x\partial _{x}+y\partial _{y})U(x,y)=0 \end{equation*} But \begin{equation*} x\partial _{x}+y\partial _{y}=r\partial _{r},\;r=\sqrt{x^{2}+y^{2}} \end{equation*} so \begin{eqnarray*} r\partial _{r}U(r\cos \theta ,r\sin \theta ) &=&0 \\ \partial _{r}U(r\cos \theta ,r\sin \theta ) &=&0,\;r\neq 0 \end{eqnarray*} and, since \begin{equation*} U(\cos \theta ,\sin \theta )=\cos \theta \end{equation*} we obtain \begin{equation*} U(r\cos \theta ,r\sin \theta )=U(\cos \theta ,\sin \theta )=\cos \theta \end{equation*}

Urgje
  • 1,941