Find coordinates that straightens the vector field (Cauchy's problem












1














Find coordinates that straightens the vector field $X(x_1,x_2)=x_1frac{partial}{partial x_1}+x_2frac{partial}{partial x_2}$.



Suppose that $(y_1, y_2)$ are the coordinates we seek. We have that $$frac{partial}{partial x_1}=frac{partial y_1}{partial x_1} frac{partial}{partial y_1} + frac{partial y_2}{partial x_1} frac{partial}{partial y_2} $$ and $$frac{partial}{partial x_1}=frac{partial y_1}{partial x_2} frac{partial}{partial y_1} + frac{partial y_2}{partial x_2} frac{partial}{partial y_2} $$



I substitute those to $X$ and after some implications we got



$(1); x_1u_{x_1} + x_2u_{x_2}=1$



$(2); x_1u_{x_1} + x_2u_{x_2}=0$



Now I got to find a Cauchy problem for those equations, to find solutions not in the implicit form, but I don't know how to do so. Please help me with the above.






differential-equations pde







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  • Hint: think geometrically. The vector field points in the radial direction. Try polar coordinates.
    – Matthew Kvalheim
    Nov 25 '18 at 9:32










  • For the first one $(1)$ my guess was like: $u(s,0)=h(s)$, but I'm completely not sure.
    – MacAbra
    Nov 25 '18 at 17:02
















1














Find coordinates that straightens the vector field $X(x_1,x_2)=x_1frac{partial}{partial x_1}+x_2frac{partial}{partial x_2}$.



Suppose that $(y_1, y_2)$ are the coordinates we seek. We have that $$frac{partial}{partial x_1}=frac{partial y_1}{partial x_1} frac{partial}{partial y_1} + frac{partial y_2}{partial x_1} frac{partial}{partial y_2} $$ and $$frac{partial}{partial x_1}=frac{partial y_1}{partial x_2} frac{partial}{partial y_1} + frac{partial y_2}{partial x_2} frac{partial}{partial y_2} $$



I substitute those to $X$ and after some implications we got



$(1); x_1u_{x_1} + x_2u_{x_2}=1$



$(2); x_1u_{x_1} + x_2u_{x_2}=0$



Now I got to find a Cauchy problem for those equations, to find solutions not in the implicit form, but I don't know how to do so. Please help me with the above.






differential-equations pde







share|cite|improve this question






















  • Hint: think geometrically. The vector field points in the radial direction. Try polar coordinates.
    – Matthew Kvalheim
    Nov 25 '18 at 9:32










  • For the first one $(1)$ my guess was like: $u(s,0)=h(s)$, but I'm completely not sure.
    – MacAbra
    Nov 25 '18 at 17:02














1












1








1







Find coordinates that straightens the vector field $X(x_1,x_2)=x_1frac{partial}{partial x_1}+x_2frac{partial}{partial x_2}$.



Suppose that $(y_1, y_2)$ are the coordinates we seek. We have that $$frac{partial}{partial x_1}=frac{partial y_1}{partial x_1} frac{partial}{partial y_1} + frac{partial y_2}{partial x_1} frac{partial}{partial y_2} $$ and $$frac{partial}{partial x_1}=frac{partial y_1}{partial x_2} frac{partial}{partial y_1} + frac{partial y_2}{partial x_2} frac{partial}{partial y_2} $$



I substitute those to $X$ and after some implications we got



$(1); x_1u_{x_1} + x_2u_{x_2}=1$



$(2); x_1u_{x_1} + x_2u_{x_2}=0$



Now I got to find a Cauchy problem for those equations, to find solutions not in the implicit form, but I don't know how to do so. Please help me with the above.






differential-equations pde







share|cite|improve this question













Find coordinates that straightens the vector field $X(x_1,x_2)=x_1frac{partial}{partial x_1}+x_2frac{partial}{partial x_2}$.



Suppose that $(y_1, y_2)$ are the coordinates we seek. We have that $$frac{partial}{partial x_1}=frac{partial y_1}{partial x_1} frac{partial}{partial y_1} + frac{partial y_2}{partial x_1} frac{partial}{partial y_2} $$ and $$frac{partial}{partial x_1}=frac{partial y_1}{partial x_2} frac{partial}{partial y_1} + frac{partial y_2}{partial x_2} frac{partial}{partial y_2} $$



I substitute those to $X$ and after some implications we got



$(1); x_1u_{x_1} + x_2u_{x_2}=1$



$(2); x_1u_{x_1} + x_2u_{x_2}=0$



Now I got to find a Cauchy problem for those equations, to find solutions not in the implicit form, but I don't know how to do so. Please help me with the above.






differential-equations pde




differential-equations pde






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asked Nov 25 '18 at 9:20









MacAbra

17719




17719












  • Hint: think geometrically. The vector field points in the radial direction. Try polar coordinates.
    – Matthew Kvalheim
    Nov 25 '18 at 9:32










  • For the first one $(1)$ my guess was like: $u(s,0)=h(s)$, but I'm completely not sure.
    – MacAbra
    Nov 25 '18 at 17:02


















  • Hint: think geometrically. The vector field points in the radial direction. Try polar coordinates.
    – Matthew Kvalheim
    Nov 25 '18 at 9:32










  • For the first one $(1)$ my guess was like: $u(s,0)=h(s)$, but I'm completely not sure.
    – MacAbra
    Nov 25 '18 at 17:02
















Hint: think geometrically. The vector field points in the radial direction. Try polar coordinates.
– Matthew Kvalheim
Nov 25 '18 at 9:32




Hint: think geometrically. The vector field points in the radial direction. Try polar coordinates.
– Matthew Kvalheim
Nov 25 '18 at 9:32












For the first one $(1)$ my guess was like: $u(s,0)=h(s)$, but I'm completely not sure.
– MacAbra
Nov 25 '18 at 17:02




For the first one $(1)$ my guess was like: $u(s,0)=h(s)$, but I'm completely not sure.
– MacAbra
Nov 25 '18 at 17:02










1 Answer
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Using polar coordinates $(r,theta)$ on $mathbb{R}^2setminus {0}$ defined by $x_1 = rcos(theta)$, $x_2 = rsin(theta)$, we have



$X = r frac{partial}{partial r}$



on $Bbb{R}^2 setminus {0}$. (This is obvious from inspection. But you should work through the definitions to convince yourself that this is true.)
Hence the vector field is "straightened."



If you would like coordinates which additionally have the property that the vector field's components become $(1,0,ldots,0)$, define the variable $scolon Bbb{R}^2 to Bbb{R}$ via



$s := ln(r).$



In the coordinates $(s,theta)$, the vector field then takes the form



$X = 1 frac{partial}{partial s}$



as claimed. (You could also discover this additional change of coordinates by inspection, but I figured this out by thinking about the proof of the "flowbox"/"flow straightening"/"rectification" theorem. In fact, you could have discovered this final change of coordinates using the latter method directly from the beginning, which is -- sort of tautologically -- the only general method for doing this.)






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    Using polar coordinates $(r,theta)$ on $mathbb{R}^2setminus {0}$ defined by $x_1 = rcos(theta)$, $x_2 = rsin(theta)$, we have



    $X = r frac{partial}{partial r}$



    on $Bbb{R}^2 setminus {0}$. (This is obvious from inspection. But you should work through the definitions to convince yourself that this is true.)
    Hence the vector field is "straightened."



    If you would like coordinates which additionally have the property that the vector field's components become $(1,0,ldots,0)$, define the variable $scolon Bbb{R}^2 to Bbb{R}$ via



    $s := ln(r).$



    In the coordinates $(s,theta)$, the vector field then takes the form



    $X = 1 frac{partial}{partial s}$



    as claimed. (You could also discover this additional change of coordinates by inspection, but I figured this out by thinking about the proof of the "flowbox"/"flow straightening"/"rectification" theorem. In fact, you could have discovered this final change of coordinates using the latter method directly from the beginning, which is -- sort of tautologically -- the only general method for doing this.)






    share|cite|improve this answer


























      1














      Using polar coordinates $(r,theta)$ on $mathbb{R}^2setminus {0}$ defined by $x_1 = rcos(theta)$, $x_2 = rsin(theta)$, we have



      $X = r frac{partial}{partial r}$



      on $Bbb{R}^2 setminus {0}$. (This is obvious from inspection. But you should work through the definitions to convince yourself that this is true.)
      Hence the vector field is "straightened."



      If you would like coordinates which additionally have the property that the vector field's components become $(1,0,ldots,0)$, define the variable $scolon Bbb{R}^2 to Bbb{R}$ via



      $s := ln(r).$



      In the coordinates $(s,theta)$, the vector field then takes the form



      $X = 1 frac{partial}{partial s}$



      as claimed. (You could also discover this additional change of coordinates by inspection, but I figured this out by thinking about the proof of the "flowbox"/"flow straightening"/"rectification" theorem. In fact, you could have discovered this final change of coordinates using the latter method directly from the beginning, which is -- sort of tautologically -- the only general method for doing this.)






      share|cite|improve this answer
























        1












        1








        1






        Using polar coordinates $(r,theta)$ on $mathbb{R}^2setminus {0}$ defined by $x_1 = rcos(theta)$, $x_2 = rsin(theta)$, we have



        $X = r frac{partial}{partial r}$



        on $Bbb{R}^2 setminus {0}$. (This is obvious from inspection. But you should work through the definitions to convince yourself that this is true.)
        Hence the vector field is "straightened."



        If you would like coordinates which additionally have the property that the vector field's components become $(1,0,ldots,0)$, define the variable $scolon Bbb{R}^2 to Bbb{R}$ via



        $s := ln(r).$



        In the coordinates $(s,theta)$, the vector field then takes the form



        $X = 1 frac{partial}{partial s}$



        as claimed. (You could also discover this additional change of coordinates by inspection, but I figured this out by thinking about the proof of the "flowbox"/"flow straightening"/"rectification" theorem. In fact, you could have discovered this final change of coordinates using the latter method directly from the beginning, which is -- sort of tautologically -- the only general method for doing this.)






        share|cite|improve this answer












        Using polar coordinates $(r,theta)$ on $mathbb{R}^2setminus {0}$ defined by $x_1 = rcos(theta)$, $x_2 = rsin(theta)$, we have



        $X = r frac{partial}{partial r}$



        on $Bbb{R}^2 setminus {0}$. (This is obvious from inspection. But you should work through the definitions to convince yourself that this is true.)
        Hence the vector field is "straightened."



        If you would like coordinates which additionally have the property that the vector field's components become $(1,0,ldots,0)$, define the variable $scolon Bbb{R}^2 to Bbb{R}$ via



        $s := ln(r).$



        In the coordinates $(s,theta)$, the vector field then takes the form



        $X = 1 frac{partial}{partial s}$



        as claimed. (You could also discover this additional change of coordinates by inspection, but I figured this out by thinking about the proof of the "flowbox"/"flow straightening"/"rectification" theorem. In fact, you could have discovered this final change of coordinates using the latter method directly from the beginning, which is -- sort of tautologically -- the only general method for doing this.)







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 25 '18 at 21:08









        Matthew Kvalheim

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