Behavior of the solution of the eikonal equation












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Consider the nonlinear first-order initial-value problem: $$(u_t )^2 + (u_x )^2 = 1$$ with initial condition $u(x, 0) = {−sqrt{1+x^2}}$. Find its solution for all $t>0$ using the method of characteristics.




Let $x(s)=langle x(s),t(s) rangle, p= langle p_1(s),p_2(s) rangle = langle u_x,u_t rangle , z = u(x,t)$



$$F(p,z,x)=p_1^2+p_2^2-1, frac{dF}{dp}=langle 2p_1,2p_2 rangle, frac{dF}{dz}=0, frac{dF}{dx}=langle 0,0 rangle $$



Characteristics:



begin{align}
p_s &= frac{dF}{dx}-frac{dF}{dz}p=-langle 0,0 rangle -0langle p_1,p_2 rangle=langle 0,0 rangle \
z_s &= frac{dF}{dp}p= langle 2p_1,2p_2rangle cdot langle p_1,p_2 rangle =2p_1^2+2p_2^2=2\
x_s &= frac{dF}{dp}= langle 2p_1,2p_2 rangle
end{align}



IVP:



$$ x_0 =r quad t_0 =0, quad u_0=z_0={−sqrt{1+r^2}} $$



$$ u_x(x,0) = -frac{x}{{sqrt{1+x^2}}}implies p_{1_0}= -frac{r}{{sqrt{1+r^2}}}=p_1 $$



$$ p_{1_0}^2+p_{2_0}^2-1 = 0 implies p_{2_0}=pm sqrt{1-frac{r^2}{1+r^2}}=pm frac{1}{{sqrt{1+r^2}}}=p_2$$



Suppose $p_{2_0}=frac{1}{{sqrt{1+r^2}}}$, then $t_s=2p_2=frac{2}{{sqrt{1+r^2}}}$ with $t_0 =0 implies t = frac{2}{{sqrt{1+r^2}}}s implies s=frac{t{sqrt{1+r^2}}}{2}$



$$x_s = 2p_1 = -frac{2r}{{sqrt{1+r^2}}}$$ with $x_0 =r$ implies $$x=-frac{2r}{{sqrt{1+r^2}}}s+r=-tr+r=r(1-t) implies r =frac{x}{1-t}.$$



$z_s=2$ with $z_0={−sqrt{1+r^2}} $ implies $$z=u(x,t)=2s{−sqrt{1+r^2}}=tsqrt{1+r^2}-sqrt{1+r^2}=sqrt{1+r^2}(t-1)=sqrt{1+frac{x^2}{(1-t)^2}}(t-1)=sqrt{(1-t)^2+x^2},$$ but this solution doesn't satisfy the IVP. This seems bizarre to me, since I parametrize the variables to fit the initial conditions. How can this be?






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    $begingroup$



    Consider the nonlinear first-order initial-value problem: $$(u_t )^2 + (u_x )^2 = 1$$ with initial condition $u(x, 0) = {−sqrt{1+x^2}}$. Find its solution for all $t>0$ using the method of characteristics.




    Let $x(s)=langle x(s),t(s) rangle, p= langle p_1(s),p_2(s) rangle = langle u_x,u_t rangle , z = u(x,t)$



    $$F(p,z,x)=p_1^2+p_2^2-1, frac{dF}{dp}=langle 2p_1,2p_2 rangle, frac{dF}{dz}=0, frac{dF}{dx}=langle 0,0 rangle $$



    Characteristics:



    begin{align}
    p_s &= frac{dF}{dx}-frac{dF}{dz}p=-langle 0,0 rangle -0langle p_1,p_2 rangle=langle 0,0 rangle \
    z_s &= frac{dF}{dp}p= langle 2p_1,2p_2rangle cdot langle p_1,p_2 rangle =2p_1^2+2p_2^2=2\
    x_s &= frac{dF}{dp}= langle 2p_1,2p_2 rangle
    end{align}



    IVP:



    $$ x_0 =r quad t_0 =0, quad u_0=z_0={−sqrt{1+r^2}} $$



    $$ u_x(x,0) = -frac{x}{{sqrt{1+x^2}}}implies p_{1_0}= -frac{r}{{sqrt{1+r^2}}}=p_1 $$



    $$ p_{1_0}^2+p_{2_0}^2-1 = 0 implies p_{2_0}=pm sqrt{1-frac{r^2}{1+r^2}}=pm frac{1}{{sqrt{1+r^2}}}=p_2$$



    Suppose $p_{2_0}=frac{1}{{sqrt{1+r^2}}}$, then $t_s=2p_2=frac{2}{{sqrt{1+r^2}}}$ with $t_0 =0 implies t = frac{2}{{sqrt{1+r^2}}}s implies s=frac{t{sqrt{1+r^2}}}{2}$



    $$x_s = 2p_1 = -frac{2r}{{sqrt{1+r^2}}}$$ with $x_0 =r$ implies $$x=-frac{2r}{{sqrt{1+r^2}}}s+r=-tr+r=r(1-t) implies r =frac{x}{1-t}.$$



    $z_s=2$ with $z_0={−sqrt{1+r^2}} $ implies $$z=u(x,t)=2s{−sqrt{1+r^2}}=tsqrt{1+r^2}-sqrt{1+r^2}=sqrt{1+r^2}(t-1)=sqrt{1+frac{x^2}{(1-t)^2}}(t-1)=sqrt{(1-t)^2+x^2},$$ but this solution doesn't satisfy the IVP. This seems bizarre to me, since I parametrize the variables to fit the initial conditions. How can this be?






    pde characteristics







    share|cite|improve this question









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      $begingroup$



      Consider the nonlinear first-order initial-value problem: $$(u_t )^2 + (u_x )^2 = 1$$ with initial condition $u(x, 0) = {−sqrt{1+x^2}}$. Find its solution for all $t>0$ using the method of characteristics.




      Let $x(s)=langle x(s),t(s) rangle, p= langle p_1(s),p_2(s) rangle = langle u_x,u_t rangle , z = u(x,t)$



      $$F(p,z,x)=p_1^2+p_2^2-1, frac{dF}{dp}=langle 2p_1,2p_2 rangle, frac{dF}{dz}=0, frac{dF}{dx}=langle 0,0 rangle $$



      Characteristics:



      begin{align}
      p_s &= frac{dF}{dx}-frac{dF}{dz}p=-langle 0,0 rangle -0langle p_1,p_2 rangle=langle 0,0 rangle \
      z_s &= frac{dF}{dp}p= langle 2p_1,2p_2rangle cdot langle p_1,p_2 rangle =2p_1^2+2p_2^2=2\
      x_s &= frac{dF}{dp}= langle 2p_1,2p_2 rangle
      end{align}



      IVP:



      $$ x_0 =r quad t_0 =0, quad u_0=z_0={−sqrt{1+r^2}} $$



      $$ u_x(x,0) = -frac{x}{{sqrt{1+x^2}}}implies p_{1_0}= -frac{r}{{sqrt{1+r^2}}}=p_1 $$



      $$ p_{1_0}^2+p_{2_0}^2-1 = 0 implies p_{2_0}=pm sqrt{1-frac{r^2}{1+r^2}}=pm frac{1}{{sqrt{1+r^2}}}=p_2$$



      Suppose $p_{2_0}=frac{1}{{sqrt{1+r^2}}}$, then $t_s=2p_2=frac{2}{{sqrt{1+r^2}}}$ with $t_0 =0 implies t = frac{2}{{sqrt{1+r^2}}}s implies s=frac{t{sqrt{1+r^2}}}{2}$



      $$x_s = 2p_1 = -frac{2r}{{sqrt{1+r^2}}}$$ with $x_0 =r$ implies $$x=-frac{2r}{{sqrt{1+r^2}}}s+r=-tr+r=r(1-t) implies r =frac{x}{1-t}.$$



      $z_s=2$ with $z_0={−sqrt{1+r^2}} $ implies $$z=u(x,t)=2s{−sqrt{1+r^2}}=tsqrt{1+r^2}-sqrt{1+r^2}=sqrt{1+r^2}(t-1)=sqrt{1+frac{x^2}{(1-t)^2}}(t-1)=sqrt{(1-t)^2+x^2},$$ but this solution doesn't satisfy the IVP. This seems bizarre to me, since I parametrize the variables to fit the initial conditions. How can this be?






      pde characteristics







      share|cite|improve this question









      $endgroup$





      Consider the nonlinear first-order initial-value problem: $$(u_t )^2 + (u_x )^2 = 1$$ with initial condition $u(x, 0) = {−sqrt{1+x^2}}$. Find its solution for all $t>0$ using the method of characteristics.




      Let $x(s)=langle x(s),t(s) rangle, p= langle p_1(s),p_2(s) rangle = langle u_x,u_t rangle , z = u(x,t)$



      $$F(p,z,x)=p_1^2+p_2^2-1, frac{dF}{dp}=langle 2p_1,2p_2 rangle, frac{dF}{dz}=0, frac{dF}{dx}=langle 0,0 rangle $$



      Characteristics:



      begin{align}
      p_s &= frac{dF}{dx}-frac{dF}{dz}p=-langle 0,0 rangle -0langle p_1,p_2 rangle=langle 0,0 rangle \
      z_s &= frac{dF}{dp}p= langle 2p_1,2p_2rangle cdot langle p_1,p_2 rangle =2p_1^2+2p_2^2=2\
      x_s &= frac{dF}{dp}= langle 2p_1,2p_2 rangle
      end{align}



      IVP:



      $$ x_0 =r quad t_0 =0, quad u_0=z_0={−sqrt{1+r^2}} $$



      $$ u_x(x,0) = -frac{x}{{sqrt{1+x^2}}}implies p_{1_0}= -frac{r}{{sqrt{1+r^2}}}=p_1 $$



      $$ p_{1_0}^2+p_{2_0}^2-1 = 0 implies p_{2_0}=pm sqrt{1-frac{r^2}{1+r^2}}=pm frac{1}{{sqrt{1+r^2}}}=p_2$$



      Suppose $p_{2_0}=frac{1}{{sqrt{1+r^2}}}$, then $t_s=2p_2=frac{2}{{sqrt{1+r^2}}}$ with $t_0 =0 implies t = frac{2}{{sqrt{1+r^2}}}s implies s=frac{t{sqrt{1+r^2}}}{2}$



      $$x_s = 2p_1 = -frac{2r}{{sqrt{1+r^2}}}$$ with $x_0 =r$ implies $$x=-frac{2r}{{sqrt{1+r^2}}}s+r=-tr+r=r(1-t) implies r =frac{x}{1-t}.$$



      $z_s=2$ with $z_0={−sqrt{1+r^2}} $ implies $$z=u(x,t)=2s{−sqrt{1+r^2}}=tsqrt{1+r^2}-sqrt{1+r^2}=sqrt{1+r^2}(t-1)=sqrt{1+frac{x^2}{(1-t)^2}}(t-1)=sqrt{(1-t)^2+x^2},$$ but this solution doesn't satisfy the IVP. This seems bizarre to me, since I parametrize the variables to fit the initial conditions. How can this be?






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      asked Dec 11 '18 at 1:02









      dxdydzdxdydz

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          $begingroup$

          You found $p_{2_0}=pm frac{1}{{sqrt{1+r^2}}}$



          Why do you suppose $p_{2_0}=frac{1}{{sqrt{1+r^2}}}$ instead of $p_{2_0}=-frac{1}{{sqrt{1+r^2}}}$ ?



          One have to examine the both cases and chose which one agrees to the boundary condition.



          The supposition with sign $+$ leads to $z=sqrt{(1-t)^2+x^2}$ which doesn't agrees with $z_0=-sqrt{1+x^2}$.



          The supposition with sign $-$ leads to $z=-sqrt{(1-t)^2+x^2}$ which agrees with $z_0=-sqrt{1+x^2}$.



          Thus the solution is :
          $$u(x,t)=-sqrt{(1-t)^2+x^2}$$
          $u_x=-frac{x}{sqrt{(1-t)^2+x^2}}$



          $u_t=-frac{-(1-t)}{sqrt{(1-t)^2+x^2}}$



          $(u_x)^2+(u_t)^2=frac{x^2}{(1-t)^2+x^2}+frac{(1-t)^2}{(1-t)^2+x^2}=frac{x^2+(1-t)^2}{(1-t)^2+x^2}$
          $$(u_x)^2+(u_t)^2=1$$






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            1 Answer
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            active

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            $begingroup$

            You found $p_{2_0}=pm frac{1}{{sqrt{1+r^2}}}$



            Why do you suppose $p_{2_0}=frac{1}{{sqrt{1+r^2}}}$ instead of $p_{2_0}=-frac{1}{{sqrt{1+r^2}}}$ ?



            One have to examine the both cases and chose which one agrees to the boundary condition.



            The supposition with sign $+$ leads to $z=sqrt{(1-t)^2+x^2}$ which doesn't agrees with $z_0=-sqrt{1+x^2}$.



            The supposition with sign $-$ leads to $z=-sqrt{(1-t)^2+x^2}$ which agrees with $z_0=-sqrt{1+x^2}$.



            Thus the solution is :
            $$u(x,t)=-sqrt{(1-t)^2+x^2}$$
            $u_x=-frac{x}{sqrt{(1-t)^2+x^2}}$



            $u_t=-frac{-(1-t)}{sqrt{(1-t)^2+x^2}}$



            $(u_x)^2+(u_t)^2=frac{x^2}{(1-t)^2+x^2}+frac{(1-t)^2}{(1-t)^2+x^2}=frac{x^2+(1-t)^2}{(1-t)^2+x^2}$
            $$(u_x)^2+(u_t)^2=1$$






            share|cite|improve this answer











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              0












              $begingroup$

              You found $p_{2_0}=pm frac{1}{{sqrt{1+r^2}}}$



              Why do you suppose $p_{2_0}=frac{1}{{sqrt{1+r^2}}}$ instead of $p_{2_0}=-frac{1}{{sqrt{1+r^2}}}$ ?



              One have to examine the both cases and chose which one agrees to the boundary condition.



              The supposition with sign $+$ leads to $z=sqrt{(1-t)^2+x^2}$ which doesn't agrees with $z_0=-sqrt{1+x^2}$.



              The supposition with sign $-$ leads to $z=-sqrt{(1-t)^2+x^2}$ which agrees with $z_0=-sqrt{1+x^2}$.



              Thus the solution is :
              $$u(x,t)=-sqrt{(1-t)^2+x^2}$$
              $u_x=-frac{x}{sqrt{(1-t)^2+x^2}}$



              $u_t=-frac{-(1-t)}{sqrt{(1-t)^2+x^2}}$



              $(u_x)^2+(u_t)^2=frac{x^2}{(1-t)^2+x^2}+frac{(1-t)^2}{(1-t)^2+x^2}=frac{x^2+(1-t)^2}{(1-t)^2+x^2}$
              $$(u_x)^2+(u_t)^2=1$$






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                You found $p_{2_0}=pm frac{1}{{sqrt{1+r^2}}}$



                Why do you suppose $p_{2_0}=frac{1}{{sqrt{1+r^2}}}$ instead of $p_{2_0}=-frac{1}{{sqrt{1+r^2}}}$ ?



                One have to examine the both cases and chose which one agrees to the boundary condition.



                The supposition with sign $+$ leads to $z=sqrt{(1-t)^2+x^2}$ which doesn't agrees with $z_0=-sqrt{1+x^2}$.



                The supposition with sign $-$ leads to $z=-sqrt{(1-t)^2+x^2}$ which agrees with $z_0=-sqrt{1+x^2}$.



                Thus the solution is :
                $$u(x,t)=-sqrt{(1-t)^2+x^2}$$
                $u_x=-frac{x}{sqrt{(1-t)^2+x^2}}$



                $u_t=-frac{-(1-t)}{sqrt{(1-t)^2+x^2}}$



                $(u_x)^2+(u_t)^2=frac{x^2}{(1-t)^2+x^2}+frac{(1-t)^2}{(1-t)^2+x^2}=frac{x^2+(1-t)^2}{(1-t)^2+x^2}$
                $$(u_x)^2+(u_t)^2=1$$






                share|cite|improve this answer











                $endgroup$



                You found $p_{2_0}=pm frac{1}{{sqrt{1+r^2}}}$



                Why do you suppose $p_{2_0}=frac{1}{{sqrt{1+r^2}}}$ instead of $p_{2_0}=-frac{1}{{sqrt{1+r^2}}}$ ?



                One have to examine the both cases and chose which one agrees to the boundary condition.



                The supposition with sign $+$ leads to $z=sqrt{(1-t)^2+x^2}$ which doesn't agrees with $z_0=-sqrt{1+x^2}$.



                The supposition with sign $-$ leads to $z=-sqrt{(1-t)^2+x^2}$ which agrees with $z_0=-sqrt{1+x^2}$.



                Thus the solution is :
                $$u(x,t)=-sqrt{(1-t)^2+x^2}$$
                $u_x=-frac{x}{sqrt{(1-t)^2+x^2}}$



                $u_t=-frac{-(1-t)}{sqrt{(1-t)^2+x^2}}$



                $(u_x)^2+(u_t)^2=frac{x^2}{(1-t)^2+x^2}+frac{(1-t)^2}{(1-t)^2+x^2}=frac{x^2+(1-t)^2}{(1-t)^2+x^2}$
                $$(u_x)^2+(u_t)^2=1$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 11 '18 at 6:59

























                answered Dec 11 '18 at 6:54









                JJacquelinJJacquelin

                44.1k21853




                44.1k21853






























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