envelope of rays that propagate into the ellipse for $F=u_x^2 + u_y^2 - 1$












0












$begingroup$


I have the equation $F=u_x^2 + u_y^2 - 1$ and that $u=0$ on the boundary of the ellipse $frac{x^2}{a^2} +frac{y^2}{b^2}=1$



At $u=0$ the ellipse can be parametrised as $x=acos(s), y=bsin(s)$



Charpits eqs give me:



$dx/dt= 2p$



$dy/dt = 2q$



$dp/dt = dq/dt = 0$



$du/dt = 2p^2 + 2q^2$



where $p = frac{partial u}{partial x}$ and $q = frac{partial u}{partial y}$



solving charpits eqs I get:



$p = p_0(s)$



$q = q_0(s)$



$x = 2p_0(s)t + acos(s)$



$y = 2q_0(s)t + bsin(s)$



Since $F(p,q) = p^2 + q^2 - 1$, this implies that $p_0^2 + q_0^2 = 1$



To find the envelope i need to find when the jacobian =0 which is when $2bp_0cos(s)+ 2aq_0sin(s) = 0$



I am confused as to how to draw the envelope of rays that propagate into the ellipse and also how to determine the ridge of discontinuity






ordinary-differential-equations pde







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I have the equation $F=u_x^2 + u_y^2 - 1$ and that $u=0$ on the boundary of the ellipse $frac{x^2}{a^2} +frac{y^2}{b^2}=1$



    At $u=0$ the ellipse can be parametrised as $x=acos(s), y=bsin(s)$



    Charpits eqs give me:



    $dx/dt= 2p$



    $dy/dt = 2q$



    $dp/dt = dq/dt = 0$



    $du/dt = 2p^2 + 2q^2$



    where $p = frac{partial u}{partial x}$ and $q = frac{partial u}{partial y}$



    solving charpits eqs I get:



    $p = p_0(s)$



    $q = q_0(s)$



    $x = 2p_0(s)t + acos(s)$



    $y = 2q_0(s)t + bsin(s)$



    Since $F(p,q) = p^2 + q^2 - 1$, this implies that $p_0^2 + q_0^2 = 1$



    To find the envelope i need to find when the jacobian =0 which is when $2bp_0cos(s)+ 2aq_0sin(s) = 0$



    I am confused as to how to draw the envelope of rays that propagate into the ellipse and also how to determine the ridge of discontinuity






    ordinary-differential-equations pde







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I have the equation $F=u_x^2 + u_y^2 - 1$ and that $u=0$ on the boundary of the ellipse $frac{x^2}{a^2} +frac{y^2}{b^2}=1$



      At $u=0$ the ellipse can be parametrised as $x=acos(s), y=bsin(s)$



      Charpits eqs give me:



      $dx/dt= 2p$



      $dy/dt = 2q$



      $dp/dt = dq/dt = 0$



      $du/dt = 2p^2 + 2q^2$



      where $p = frac{partial u}{partial x}$ and $q = frac{partial u}{partial y}$



      solving charpits eqs I get:



      $p = p_0(s)$



      $q = q_0(s)$



      $x = 2p_0(s)t + acos(s)$



      $y = 2q_0(s)t + bsin(s)$



      Since $F(p,q) = p^2 + q^2 - 1$, this implies that $p_0^2 + q_0^2 = 1$



      To find the envelope i need to find when the jacobian =0 which is when $2bp_0cos(s)+ 2aq_0sin(s) = 0$



      I am confused as to how to draw the envelope of rays that propagate into the ellipse and also how to determine the ridge of discontinuity






      ordinary-differential-equations pde







      share|cite|improve this question











      $endgroup$




      I have the equation $F=u_x^2 + u_y^2 - 1$ and that $u=0$ on the boundary of the ellipse $frac{x^2}{a^2} +frac{y^2}{b^2}=1$



      At $u=0$ the ellipse can be parametrised as $x=acos(s), y=bsin(s)$



      Charpits eqs give me:



      $dx/dt= 2p$



      $dy/dt = 2q$



      $dp/dt = dq/dt = 0$



      $du/dt = 2p^2 + 2q^2$



      where $p = frac{partial u}{partial x}$ and $q = frac{partial u}{partial y}$



      solving charpits eqs I get:



      $p = p_0(s)$



      $q = q_0(s)$



      $x = 2p_0(s)t + acos(s)$



      $y = 2q_0(s)t + bsin(s)$



      Since $F(p,q) = p^2 + q^2 - 1$, this implies that $p_0^2 + q_0^2 = 1$



      To find the envelope i need to find when the jacobian =0 which is when $2bp_0cos(s)+ 2aq_0sin(s) = 0$



      I am confused as to how to draw the envelope of rays that propagate into the ellipse and also how to determine the ridge of discontinuity






      ordinary-differential-equations pde




      ordinary-differential-equations pde






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 10 '18 at 16:58









      Harry49

      6,18331132




      6,18331132










      asked Dec 10 '18 at 12:11









      pablo_mathscobarpablo_mathscobar

      996




      996






















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          I think you might have missed a sign in $dot x_0(s)=-asin(s)$. Then the equation for $p_0,q_0$ reads as
          $$
          0=dot z_0(s)=p_0(s)dot x_0(s)+q_0(s)dot y_0(s)=-ap_0sin(s)+bq_0cos(s).
          $$
          This tells us that the unit vector $(p_0,q_0)$ has to be orthogonal to $(-asin(s), bcos(s))$, thus $$(p_0,q_0)=pmfrac{(bcos(s),asin(s))}{sqrt{b^2cos^2(s)+a^2sin^2(s)}}.$$
          As this should point inwards, chose the negative sign.



          To find the characteristic that some point $(x,y)$ lies on, one has to solve
          $$
          left.begin{aligned}
          x&=x_0+2p_0t=acos(s) - r b cos(s)\
          y&=y_0+2q_0t=bsin(s) - r a sin(s)
          end{aligned}right}
          implies frac{x^2}{(a-rb)^2}+frac{y^2}{(b-ar)^2}=1
          $$
          where $r$ is some positive multiple of $t$.
          The last is equivalent to a 4th degree polynomial equation, by the intermediate value theorem there is at least one solution between $r=0$ and $min(frac ba,frac ab)$, there is another solution for $r>max(frac ba,frac ab)$, but that might be inadmissible.




          Two close-by rays associated to the angles $s$ and $s+ds$ intersect at
          $$
          left.begin{aligned}
          (a - r b) cos(s) = (a - (r+dr) b) cos(s+ds)\
          (b - r a) sin(s) = (b - (r+dr) a) sin(s+ds)
          end{aligned}right}
          implies
          left{begin{aligned}
          0=-b,dr,cos(s)-(a-br)sin(s)ds\
          0=-a,dr,sin(s)+(b-ar)cos(s)ds
          end{aligned}right.
          $$
          so that necessarily $$a(a-br)sin^2(s)+b(b-ar)cos^2(s)=0implies r = frac{a^2sin^2s+b^2cos^2s}{ab}$$



          Plotting that curve against the ellipse gives the picture
          enter image description here






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I thought to find the envelope we need to find where the jacobian = 0 so we can find an expression for t? I realise my expression for the jacobian was wrong but the method of solving for t then putting $p_0,q_0,t$ back into the characteristics to find the envelope is still correct , is it not?
            $endgroup$
            – pablo_mathscobar
            Dec 12 '18 at 11:15












          • $begingroup$
            You use $dz=p,dx+q,dy$ along the initial condition curve together with all other previously computed relations to determine $(p_0,q_0)$. The zeros of the determinant of the Jacobian of $$(s,t)mapsto(,x_0(s)+2tp_0(s),, y_0(s)+2tq_0(s),)=(,(a+2tf(s),b)cos(s),,(b+2tf(s),a)sin(s),)$$ tell you where to find the envelope. This is the same as me computing the intersection of neighboring rays with $r=2tf(s)$. The points at $s=kfracpi2$ that are inside the ellipse are the borders of the ridge.
            $endgroup$
            – LutzL
            Dec 12 '18 at 11:47












          • $begingroup$
            I see computing the $frac{partial}{partial s} acos(s) + 2tbcos(s)f(s)$ is quite a messy derivative because of the f(s) term so i guess your way is better? also how did you find the intersection between the 2 rays? what did you do remove the cos(ds) terms
            $endgroup$
            – pablo_mathscobar
            Dec 12 '18 at 12:06












          • $begingroup$
            It does not need to be messy if you leave the derivative at $f'(s)$. But yes, by the chain rule you get $$frac{∂(x,y)}{∂(s,t)}=frac{∂(x,y)}{∂(s,r)}frac{∂(s,r)}{∂(s,t)}text{ with }frac{∂(s,r)}{∂(s,t)}=pmatrix{1&0\2tf'(s)&2f(s)},$$ so that the zeros of the determinants give the same points.
            $endgroup$
            – LutzL
            Dec 12 '18 at 12:10











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          1 Answer
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          1 Answer
          1






          active

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          active

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          0












          $begingroup$

          I think you might have missed a sign in $dot x_0(s)=-asin(s)$. Then the equation for $p_0,q_0$ reads as
          $$
          0=dot z_0(s)=p_0(s)dot x_0(s)+q_0(s)dot y_0(s)=-ap_0sin(s)+bq_0cos(s).
          $$
          This tells us that the unit vector $(p_0,q_0)$ has to be orthogonal to $(-asin(s), bcos(s))$, thus $$(p_0,q_0)=pmfrac{(bcos(s),asin(s))}{sqrt{b^2cos^2(s)+a^2sin^2(s)}}.$$
          As this should point inwards, chose the negative sign.



          To find the characteristic that some point $(x,y)$ lies on, one has to solve
          $$
          left.begin{aligned}
          x&=x_0+2p_0t=acos(s) - r b cos(s)\
          y&=y_0+2q_0t=bsin(s) - r a sin(s)
          end{aligned}right}
          implies frac{x^2}{(a-rb)^2}+frac{y^2}{(b-ar)^2}=1
          $$
          where $r$ is some positive multiple of $t$.
          The last is equivalent to a 4th degree polynomial equation, by the intermediate value theorem there is at least one solution between $r=0$ and $min(frac ba,frac ab)$, there is another solution for $r>max(frac ba,frac ab)$, but that might be inadmissible.




          Two close-by rays associated to the angles $s$ and $s+ds$ intersect at
          $$
          left.begin{aligned}
          (a - r b) cos(s) = (a - (r+dr) b) cos(s+ds)\
          (b - r a) sin(s) = (b - (r+dr) a) sin(s+ds)
          end{aligned}right}
          implies
          left{begin{aligned}
          0=-b,dr,cos(s)-(a-br)sin(s)ds\
          0=-a,dr,sin(s)+(b-ar)cos(s)ds
          end{aligned}right.
          $$
          so that necessarily $$a(a-br)sin^2(s)+b(b-ar)cos^2(s)=0implies r = frac{a^2sin^2s+b^2cos^2s}{ab}$$



          Plotting that curve against the ellipse gives the picture
          enter image description here






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I thought to find the envelope we need to find where the jacobian = 0 so we can find an expression for t? I realise my expression for the jacobian was wrong but the method of solving for t then putting $p_0,q_0,t$ back into the characteristics to find the envelope is still correct , is it not?
            $endgroup$
            – pablo_mathscobar
            Dec 12 '18 at 11:15












          • $begingroup$
            You use $dz=p,dx+q,dy$ along the initial condition curve together with all other previously computed relations to determine $(p_0,q_0)$. The zeros of the determinant of the Jacobian of $$(s,t)mapsto(,x_0(s)+2tp_0(s),, y_0(s)+2tq_0(s),)=(,(a+2tf(s),b)cos(s),,(b+2tf(s),a)sin(s),)$$ tell you where to find the envelope. This is the same as me computing the intersection of neighboring rays with $r=2tf(s)$. The points at $s=kfracpi2$ that are inside the ellipse are the borders of the ridge.
            $endgroup$
            – LutzL
            Dec 12 '18 at 11:47












          • $begingroup$
            I see computing the $frac{partial}{partial s} acos(s) + 2tbcos(s)f(s)$ is quite a messy derivative because of the f(s) term so i guess your way is better? also how did you find the intersection between the 2 rays? what did you do remove the cos(ds) terms
            $endgroup$
            – pablo_mathscobar
            Dec 12 '18 at 12:06












          • $begingroup$
            It does not need to be messy if you leave the derivative at $f'(s)$. But yes, by the chain rule you get $$frac{∂(x,y)}{∂(s,t)}=frac{∂(x,y)}{∂(s,r)}frac{∂(s,r)}{∂(s,t)}text{ with }frac{∂(s,r)}{∂(s,t)}=pmatrix{1&0\2tf'(s)&2f(s)},$$ so that the zeros of the determinants give the same points.
            $endgroup$
            – LutzL
            Dec 12 '18 at 12:10
















          0












          $begingroup$

          I think you might have missed a sign in $dot x_0(s)=-asin(s)$. Then the equation for $p_0,q_0$ reads as
          $$
          0=dot z_0(s)=p_0(s)dot x_0(s)+q_0(s)dot y_0(s)=-ap_0sin(s)+bq_0cos(s).
          $$
          This tells us that the unit vector $(p_0,q_0)$ has to be orthogonal to $(-asin(s), bcos(s))$, thus $$(p_0,q_0)=pmfrac{(bcos(s),asin(s))}{sqrt{b^2cos^2(s)+a^2sin^2(s)}}.$$
          As this should point inwards, chose the negative sign.



          To find the characteristic that some point $(x,y)$ lies on, one has to solve
          $$
          left.begin{aligned}
          x&=x_0+2p_0t=acos(s) - r b cos(s)\
          y&=y_0+2q_0t=bsin(s) - r a sin(s)
          end{aligned}right}
          implies frac{x^2}{(a-rb)^2}+frac{y^2}{(b-ar)^2}=1
          $$
          where $r$ is some positive multiple of $t$.
          The last is equivalent to a 4th degree polynomial equation, by the intermediate value theorem there is at least one solution between $r=0$ and $min(frac ba,frac ab)$, there is another solution for $r>max(frac ba,frac ab)$, but that might be inadmissible.




          Two close-by rays associated to the angles $s$ and $s+ds$ intersect at
          $$
          left.begin{aligned}
          (a - r b) cos(s) = (a - (r+dr) b) cos(s+ds)\
          (b - r a) sin(s) = (b - (r+dr) a) sin(s+ds)
          end{aligned}right}
          implies
          left{begin{aligned}
          0=-b,dr,cos(s)-(a-br)sin(s)ds\
          0=-a,dr,sin(s)+(b-ar)cos(s)ds
          end{aligned}right.
          $$
          so that necessarily $$a(a-br)sin^2(s)+b(b-ar)cos^2(s)=0implies r = frac{a^2sin^2s+b^2cos^2s}{ab}$$



          Plotting that curve against the ellipse gives the picture
          enter image description here






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I thought to find the envelope we need to find where the jacobian = 0 so we can find an expression for t? I realise my expression for the jacobian was wrong but the method of solving for t then putting $p_0,q_0,t$ back into the characteristics to find the envelope is still correct , is it not?
            $endgroup$
            – pablo_mathscobar
            Dec 12 '18 at 11:15












          • $begingroup$
            You use $dz=p,dx+q,dy$ along the initial condition curve together with all other previously computed relations to determine $(p_0,q_0)$. The zeros of the determinant of the Jacobian of $$(s,t)mapsto(,x_0(s)+2tp_0(s),, y_0(s)+2tq_0(s),)=(,(a+2tf(s),b)cos(s),,(b+2tf(s),a)sin(s),)$$ tell you where to find the envelope. This is the same as me computing the intersection of neighboring rays with $r=2tf(s)$. The points at $s=kfracpi2$ that are inside the ellipse are the borders of the ridge.
            $endgroup$
            – LutzL
            Dec 12 '18 at 11:47












          • $begingroup$
            I see computing the $frac{partial}{partial s} acos(s) + 2tbcos(s)f(s)$ is quite a messy derivative because of the f(s) term so i guess your way is better? also how did you find the intersection between the 2 rays? what did you do remove the cos(ds) terms
            $endgroup$
            – pablo_mathscobar
            Dec 12 '18 at 12:06












          • $begingroup$
            It does not need to be messy if you leave the derivative at $f'(s)$. But yes, by the chain rule you get $$frac{∂(x,y)}{∂(s,t)}=frac{∂(x,y)}{∂(s,r)}frac{∂(s,r)}{∂(s,t)}text{ with }frac{∂(s,r)}{∂(s,t)}=pmatrix{1&0\2tf'(s)&2f(s)},$$ so that the zeros of the determinants give the same points.
            $endgroup$
            – LutzL
            Dec 12 '18 at 12:10














          0












          0








          0





          $begingroup$

          I think you might have missed a sign in $dot x_0(s)=-asin(s)$. Then the equation for $p_0,q_0$ reads as
          $$
          0=dot z_0(s)=p_0(s)dot x_0(s)+q_0(s)dot y_0(s)=-ap_0sin(s)+bq_0cos(s).
          $$
          This tells us that the unit vector $(p_0,q_0)$ has to be orthogonal to $(-asin(s), bcos(s))$, thus $$(p_0,q_0)=pmfrac{(bcos(s),asin(s))}{sqrt{b^2cos^2(s)+a^2sin^2(s)}}.$$
          As this should point inwards, chose the negative sign.



          To find the characteristic that some point $(x,y)$ lies on, one has to solve
          $$
          left.begin{aligned}
          x&=x_0+2p_0t=acos(s) - r b cos(s)\
          y&=y_0+2q_0t=bsin(s) - r a sin(s)
          end{aligned}right}
          implies frac{x^2}{(a-rb)^2}+frac{y^2}{(b-ar)^2}=1
          $$
          where $r$ is some positive multiple of $t$.
          The last is equivalent to a 4th degree polynomial equation, by the intermediate value theorem there is at least one solution between $r=0$ and $min(frac ba,frac ab)$, there is another solution for $r>max(frac ba,frac ab)$, but that might be inadmissible.




          Two close-by rays associated to the angles $s$ and $s+ds$ intersect at
          $$
          left.begin{aligned}
          (a - r b) cos(s) = (a - (r+dr) b) cos(s+ds)\
          (b - r a) sin(s) = (b - (r+dr) a) sin(s+ds)
          end{aligned}right}
          implies
          left{begin{aligned}
          0=-b,dr,cos(s)-(a-br)sin(s)ds\
          0=-a,dr,sin(s)+(b-ar)cos(s)ds
          end{aligned}right.
          $$
          so that necessarily $$a(a-br)sin^2(s)+b(b-ar)cos^2(s)=0implies r = frac{a^2sin^2s+b^2cos^2s}{ab}$$



          Plotting that curve against the ellipse gives the picture
          enter image description here






          share|cite|improve this answer









          $endgroup$



          I think you might have missed a sign in $dot x_0(s)=-asin(s)$. Then the equation for $p_0,q_0$ reads as
          $$
          0=dot z_0(s)=p_0(s)dot x_0(s)+q_0(s)dot y_0(s)=-ap_0sin(s)+bq_0cos(s).
          $$
          This tells us that the unit vector $(p_0,q_0)$ has to be orthogonal to $(-asin(s), bcos(s))$, thus $$(p_0,q_0)=pmfrac{(bcos(s),asin(s))}{sqrt{b^2cos^2(s)+a^2sin^2(s)}}.$$
          As this should point inwards, chose the negative sign.



          To find the characteristic that some point $(x,y)$ lies on, one has to solve
          $$
          left.begin{aligned}
          x&=x_0+2p_0t=acos(s) - r b cos(s)\
          y&=y_0+2q_0t=bsin(s) - r a sin(s)
          end{aligned}right}
          implies frac{x^2}{(a-rb)^2}+frac{y^2}{(b-ar)^2}=1
          $$
          where $r$ is some positive multiple of $t$.
          The last is equivalent to a 4th degree polynomial equation, by the intermediate value theorem there is at least one solution between $r=0$ and $min(frac ba,frac ab)$, there is another solution for $r>max(frac ba,frac ab)$, but that might be inadmissible.




          Two close-by rays associated to the angles $s$ and $s+ds$ intersect at
          $$
          left.begin{aligned}
          (a - r b) cos(s) = (a - (r+dr) b) cos(s+ds)\
          (b - r a) sin(s) = (b - (r+dr) a) sin(s+ds)
          end{aligned}right}
          implies
          left{begin{aligned}
          0=-b,dr,cos(s)-(a-br)sin(s)ds\
          0=-a,dr,sin(s)+(b-ar)cos(s)ds
          end{aligned}right.
          $$
          so that necessarily $$a(a-br)sin^2(s)+b(b-ar)cos^2(s)=0implies r = frac{a^2sin^2s+b^2cos^2s}{ab}$$



          Plotting that curve against the ellipse gives the picture
          enter image description here







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 10 '18 at 15:55









          LutzLLutzL

          57.7k42054




          57.7k42054












          • $begingroup$
            I thought to find the envelope we need to find where the jacobian = 0 so we can find an expression for t? I realise my expression for the jacobian was wrong but the method of solving for t then putting $p_0,q_0,t$ back into the characteristics to find the envelope is still correct , is it not?
            $endgroup$
            – pablo_mathscobar
            Dec 12 '18 at 11:15












          • $begingroup$
            You use $dz=p,dx+q,dy$ along the initial condition curve together with all other previously computed relations to determine $(p_0,q_0)$. The zeros of the determinant of the Jacobian of $$(s,t)mapsto(,x_0(s)+2tp_0(s),, y_0(s)+2tq_0(s),)=(,(a+2tf(s),b)cos(s),,(b+2tf(s),a)sin(s),)$$ tell you where to find the envelope. This is the same as me computing the intersection of neighboring rays with $r=2tf(s)$. The points at $s=kfracpi2$ that are inside the ellipse are the borders of the ridge.
            $endgroup$
            – LutzL
            Dec 12 '18 at 11:47












          • $begingroup$
            I see computing the $frac{partial}{partial s} acos(s) + 2tbcos(s)f(s)$ is quite a messy derivative because of the f(s) term so i guess your way is better? also how did you find the intersection between the 2 rays? what did you do remove the cos(ds) terms
            $endgroup$
            – pablo_mathscobar
            Dec 12 '18 at 12:06












          • $begingroup$
            It does not need to be messy if you leave the derivative at $f'(s)$. But yes, by the chain rule you get $$frac{∂(x,y)}{∂(s,t)}=frac{∂(x,y)}{∂(s,r)}frac{∂(s,r)}{∂(s,t)}text{ with }frac{∂(s,r)}{∂(s,t)}=pmatrix{1&0\2tf'(s)&2f(s)},$$ so that the zeros of the determinants give the same points.
            $endgroup$
            – LutzL
            Dec 12 '18 at 12:10


















          • $begingroup$
            I thought to find the envelope we need to find where the jacobian = 0 so we can find an expression for t? I realise my expression for the jacobian was wrong but the method of solving for t then putting $p_0,q_0,t$ back into the characteristics to find the envelope is still correct , is it not?
            $endgroup$
            – pablo_mathscobar
            Dec 12 '18 at 11:15












          • $begingroup$
            You use $dz=p,dx+q,dy$ along the initial condition curve together with all other previously computed relations to determine $(p_0,q_0)$. The zeros of the determinant of the Jacobian of $$(s,t)mapsto(,x_0(s)+2tp_0(s),, y_0(s)+2tq_0(s),)=(,(a+2tf(s),b)cos(s),,(b+2tf(s),a)sin(s),)$$ tell you where to find the envelope. This is the same as me computing the intersection of neighboring rays with $r=2tf(s)$. The points at $s=kfracpi2$ that are inside the ellipse are the borders of the ridge.
            $endgroup$
            – LutzL
            Dec 12 '18 at 11:47












          • $begingroup$
            I see computing the $frac{partial}{partial s} acos(s) + 2tbcos(s)f(s)$ is quite a messy derivative because of the f(s) term so i guess your way is better? also how did you find the intersection between the 2 rays? what did you do remove the cos(ds) terms
            $endgroup$
            – pablo_mathscobar
            Dec 12 '18 at 12:06












          • $begingroup$
            It does not need to be messy if you leave the derivative at $f'(s)$. But yes, by the chain rule you get $$frac{∂(x,y)}{∂(s,t)}=frac{∂(x,y)}{∂(s,r)}frac{∂(s,r)}{∂(s,t)}text{ with }frac{∂(s,r)}{∂(s,t)}=pmatrix{1&0\2tf'(s)&2f(s)},$$ so that the zeros of the determinants give the same points.
            $endgroup$
            – LutzL
            Dec 12 '18 at 12:10
















          $begingroup$
          I thought to find the envelope we need to find where the jacobian = 0 so we can find an expression for t? I realise my expression for the jacobian was wrong but the method of solving for t then putting $p_0,q_0,t$ back into the characteristics to find the envelope is still correct , is it not?
          $endgroup$
          – pablo_mathscobar
          Dec 12 '18 at 11:15






          $begingroup$
          I thought to find the envelope we need to find where the jacobian = 0 so we can find an expression for t? I realise my expression for the jacobian was wrong but the method of solving for t then putting $p_0,q_0,t$ back into the characteristics to find the envelope is still correct , is it not?
          $endgroup$
          – pablo_mathscobar
          Dec 12 '18 at 11:15














          $begingroup$
          You use $dz=p,dx+q,dy$ along the initial condition curve together with all other previously computed relations to determine $(p_0,q_0)$. The zeros of the determinant of the Jacobian of $$(s,t)mapsto(,x_0(s)+2tp_0(s),, y_0(s)+2tq_0(s),)=(,(a+2tf(s),b)cos(s),,(b+2tf(s),a)sin(s),)$$ tell you where to find the envelope. This is the same as me computing the intersection of neighboring rays with $r=2tf(s)$. The points at $s=kfracpi2$ that are inside the ellipse are the borders of the ridge.
          $endgroup$
          – LutzL
          Dec 12 '18 at 11:47






          $begingroup$
          You use $dz=p,dx+q,dy$ along the initial condition curve together with all other previously computed relations to determine $(p_0,q_0)$. The zeros of the determinant of the Jacobian of $$(s,t)mapsto(,x_0(s)+2tp_0(s),, y_0(s)+2tq_0(s),)=(,(a+2tf(s),b)cos(s),,(b+2tf(s),a)sin(s),)$$ tell you where to find the envelope. This is the same as me computing the intersection of neighboring rays with $r=2tf(s)$. The points at $s=kfracpi2$ that are inside the ellipse are the borders of the ridge.
          $endgroup$
          – LutzL
          Dec 12 '18 at 11:47














          $begingroup$
          I see computing the $frac{partial}{partial s} acos(s) + 2tbcos(s)f(s)$ is quite a messy derivative because of the f(s) term so i guess your way is better? also how did you find the intersection between the 2 rays? what did you do remove the cos(ds) terms
          $endgroup$
          – pablo_mathscobar
          Dec 12 '18 at 12:06






          $begingroup$
          I see computing the $frac{partial}{partial s} acos(s) + 2tbcos(s)f(s)$ is quite a messy derivative because of the f(s) term so i guess your way is better? also how did you find the intersection between the 2 rays? what did you do remove the cos(ds) terms
          $endgroup$
          – pablo_mathscobar
          Dec 12 '18 at 12:06














          $begingroup$
          It does not need to be messy if you leave the derivative at $f'(s)$. But yes, by the chain rule you get $$frac{∂(x,y)}{∂(s,t)}=frac{∂(x,y)}{∂(s,r)}frac{∂(s,r)}{∂(s,t)}text{ with }frac{∂(s,r)}{∂(s,t)}=pmatrix{1&0\2tf'(s)&2f(s)},$$ so that the zeros of the determinants give the same points.
          $endgroup$
          – LutzL
          Dec 12 '18 at 12:10




          $begingroup$
          It does not need to be messy if you leave the derivative at $f'(s)$. But yes, by the chain rule you get $$frac{∂(x,y)}{∂(s,t)}=frac{∂(x,y)}{∂(s,r)}frac{∂(s,r)}{∂(s,t)}text{ with }frac{∂(s,r)}{∂(s,t)}=pmatrix{1&0\2tf'(s)&2f(s)},$$ so that the zeros of the determinants give the same points.
          $endgroup$
          – LutzL
          Dec 12 '18 at 12:10


















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