Solve IVP of $(u_t )^2 + (u_x )^2 − u^2 = 0$ using method of characteristics












1












$begingroup$



Consider the nonlinear first-order initial-value problem:
$$(u_t )^2 + (u_x )^2 − u^2 = 0$$
with initial condition $u(x, 0) = Ae^{−sqrt{1+x^2}}$.



(a) Find its solution for all $t > 0$ using the method of
characteristics.



(b) Describe its behavior for all $t > 0$. Does it remain a continuous
function of $x$ for all $t$?



(c) Find $lim u(x, t)$.




My attempt:



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



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



Characteristics:



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



IVP:



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



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



$$ p_{1_0}^2+p_{2_0}^2-u_0^2 = 0 implies p_{2_0}^2=A^2e^{-2sqrt{1+r^2}}-frac{A^2e^{−2sqrt{1+r^2}}r^2}{1+r^2}=frac{Ae^{−sqrt{1+r^2}}}{{sqrt{1+r^2}}}
$$



What's the next step? Solving $p_s,z_s,x_s$ seems a little bit painful...






pde characteristics







share|cite|improve this question











$endgroup$

















    1












    $begingroup$



    Consider the nonlinear first-order initial-value problem:
    $$(u_t )^2 + (u_x )^2 − u^2 = 0$$
    with initial condition $u(x, 0) = Ae^{−sqrt{1+x^2}}$.



    (a) Find its solution for all $t > 0$ using the method of
    characteristics.



    (b) Describe its behavior for all $t > 0$. Does it remain a continuous
    function of $x$ for all $t$?



    (c) Find $lim u(x, t)$.




    My attempt:



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



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



    Characteristics:



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



    IVP:



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



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



    $$ p_{1_0}^2+p_{2_0}^2-u_0^2 = 0 implies p_{2_0}^2=A^2e^{-2sqrt{1+r^2}}-frac{A^2e^{−2sqrt{1+r^2}}r^2}{1+r^2}=frac{Ae^{−sqrt{1+r^2}}}{{sqrt{1+r^2}}}
    $$



    What's the next step? Solving $p_s,z_s,x_s$ seems a little bit painful...






    pde characteristics







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$



      Consider the nonlinear first-order initial-value problem:
      $$(u_t )^2 + (u_x )^2 − u^2 = 0$$
      with initial condition $u(x, 0) = Ae^{−sqrt{1+x^2}}$.



      (a) Find its solution for all $t > 0$ using the method of
      characteristics.



      (b) Describe its behavior for all $t > 0$. Does it remain a continuous
      function of $x$ for all $t$?



      (c) Find $lim u(x, t)$.




      My attempt:



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



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



      Characteristics:



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



      IVP:



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



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



      $$ p_{1_0}^2+p_{2_0}^2-u_0^2 = 0 implies p_{2_0}^2=A^2e^{-2sqrt{1+r^2}}-frac{A^2e^{−2sqrt{1+r^2}}r^2}{1+r^2}=frac{Ae^{−sqrt{1+r^2}}}{{sqrt{1+r^2}}}
      $$



      What's the next step? Solving $p_s,z_s,x_s$ seems a little bit painful...






      pde characteristics







      share|cite|improve this question











      $endgroup$





      Consider the nonlinear first-order initial-value problem:
      $$(u_t )^2 + (u_x )^2 − u^2 = 0$$
      with initial condition $u(x, 0) = Ae^{−sqrt{1+x^2}}$.



      (a) Find its solution for all $t > 0$ using the method of
      characteristics.



      (b) Describe its behavior for all $t > 0$. Does it remain a continuous
      function of $x$ for all $t$?



      (c) Find $lim u(x, t)$.




      My attempt:



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



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



      Characteristics:



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



      IVP:



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



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



      $$ p_{1_0}^2+p_{2_0}^2-u_0^2 = 0 implies p_{2_0}^2=A^2e^{-2sqrt{1+r^2}}-frac{A^2e^{−2sqrt{1+r^2}}r^2}{1+r^2}=frac{Ae^{−sqrt{1+r^2}}}{{sqrt{1+r^2}}}
      $$



      What's the next step? Solving $p_s,z_s,x_s$ seems a little bit painful...






      pde characteristics




      pde characteristics






      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 17:06









      Harry49

      6,18331132




      6,18331132










      asked Dec 9 '18 at 19:16









      dxdydzdxdydz

      3009




      3009






















          1 Answer
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          1












          $begingroup$

          $$(u_t )^2 + (u_x )^2 = u^2 $$
          $$u(x, 0) = Ae^{−sqrt{1+x^2}}$$
          HINT:



          $u(x,t)=Ae^{v(x,t)}quad;quad u_x=uv_xquad;quad u_t=uv_t$
          $$(v_t )^2 + (v_x )^2 =1$$
          $$v(x,0)=-sqrt{1+x^2}$$
          This kind of PDE is well known (Eikonal PDE) :



          https://www.math.ualberta.ca/~xinweiyu/436.A1.12f/PDE_Meth_Characteristics.pdf , Eq.(2.208).



          https://web.stanford.edu/class/math220a/handouts/firstorder.pdf , example 11.



          https://people.cam.cornell.edu/~zc227/extras/math6200_pres.pdf , §.1.



          Of course, the boundary conditions are diferent, but with the same method you will find the solution :
          $$v(x,t)=-sqrt{(t+1)^2+x^2}$$
          $$u(x,t)=A:e^{-sqrt{(t+1)^2+x^2}}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So u is a continuous function of x for all t, can we say anything else about the behavior of u?
            $endgroup$
            – dxdydz
            Dec 11 '18 at 0:34










          • $begingroup$
            $u(x,t)$ decreases when $x$ and $t>0$ increase. One can give the maximum and the minimum.
            $endgroup$
            – JJacquelin
            Dec 11 '18 at 13:54











          Your Answer





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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

          votes









          1












          $begingroup$

          $$(u_t )^2 + (u_x )^2 = u^2 $$
          $$u(x, 0) = Ae^{−sqrt{1+x^2}}$$
          HINT:



          $u(x,t)=Ae^{v(x,t)}quad;quad u_x=uv_xquad;quad u_t=uv_t$
          $$(v_t )^2 + (v_x )^2 =1$$
          $$v(x,0)=-sqrt{1+x^2}$$
          This kind of PDE is well known (Eikonal PDE) :



          https://www.math.ualberta.ca/~xinweiyu/436.A1.12f/PDE_Meth_Characteristics.pdf , Eq.(2.208).



          https://web.stanford.edu/class/math220a/handouts/firstorder.pdf , example 11.



          https://people.cam.cornell.edu/~zc227/extras/math6200_pres.pdf , §.1.



          Of course, the boundary conditions are diferent, but with the same method you will find the solution :
          $$v(x,t)=-sqrt{(t+1)^2+x^2}$$
          $$u(x,t)=A:e^{-sqrt{(t+1)^2+x^2}}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So u is a continuous function of x for all t, can we say anything else about the behavior of u?
            $endgroup$
            – dxdydz
            Dec 11 '18 at 0:34










          • $begingroup$
            $u(x,t)$ decreases when $x$ and $t>0$ increase. One can give the maximum and the minimum.
            $endgroup$
            – JJacquelin
            Dec 11 '18 at 13:54
















          1












          $begingroup$

          $$(u_t )^2 + (u_x )^2 = u^2 $$
          $$u(x, 0) = Ae^{−sqrt{1+x^2}}$$
          HINT:



          $u(x,t)=Ae^{v(x,t)}quad;quad u_x=uv_xquad;quad u_t=uv_t$
          $$(v_t )^2 + (v_x )^2 =1$$
          $$v(x,0)=-sqrt{1+x^2}$$
          This kind of PDE is well known (Eikonal PDE) :



          https://www.math.ualberta.ca/~xinweiyu/436.A1.12f/PDE_Meth_Characteristics.pdf , Eq.(2.208).



          https://web.stanford.edu/class/math220a/handouts/firstorder.pdf , example 11.



          https://people.cam.cornell.edu/~zc227/extras/math6200_pres.pdf , §.1.



          Of course, the boundary conditions are diferent, but with the same method you will find the solution :
          $$v(x,t)=-sqrt{(t+1)^2+x^2}$$
          $$u(x,t)=A:e^{-sqrt{(t+1)^2+x^2}}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So u is a continuous function of x for all t, can we say anything else about the behavior of u?
            $endgroup$
            – dxdydz
            Dec 11 '18 at 0:34










          • $begingroup$
            $u(x,t)$ decreases when $x$ and $t>0$ increase. One can give the maximum and the minimum.
            $endgroup$
            – JJacquelin
            Dec 11 '18 at 13:54














          1












          1








          1





          $begingroup$

          $$(u_t )^2 + (u_x )^2 = u^2 $$
          $$u(x, 0) = Ae^{−sqrt{1+x^2}}$$
          HINT:



          $u(x,t)=Ae^{v(x,t)}quad;quad u_x=uv_xquad;quad u_t=uv_t$
          $$(v_t )^2 + (v_x )^2 =1$$
          $$v(x,0)=-sqrt{1+x^2}$$
          This kind of PDE is well known (Eikonal PDE) :



          https://www.math.ualberta.ca/~xinweiyu/436.A1.12f/PDE_Meth_Characteristics.pdf , Eq.(2.208).



          https://web.stanford.edu/class/math220a/handouts/firstorder.pdf , example 11.



          https://people.cam.cornell.edu/~zc227/extras/math6200_pres.pdf , §.1.



          Of course, the boundary conditions are diferent, but with the same method you will find the solution :
          $$v(x,t)=-sqrt{(t+1)^2+x^2}$$
          $$u(x,t)=A:e^{-sqrt{(t+1)^2+x^2}}$$






          share|cite|improve this answer











          $endgroup$



          $$(u_t )^2 + (u_x )^2 = u^2 $$
          $$u(x, 0) = Ae^{−sqrt{1+x^2}}$$
          HINT:



          $u(x,t)=Ae^{v(x,t)}quad;quad u_x=uv_xquad;quad u_t=uv_t$
          $$(v_t )^2 + (v_x )^2 =1$$
          $$v(x,0)=-sqrt{1+x^2}$$
          This kind of PDE is well known (Eikonal PDE) :



          https://www.math.ualberta.ca/~xinweiyu/436.A1.12f/PDE_Meth_Characteristics.pdf , Eq.(2.208).



          https://web.stanford.edu/class/math220a/handouts/firstorder.pdf , example 11.



          https://people.cam.cornell.edu/~zc227/extras/math6200_pres.pdf , §.1.



          Of course, the boundary conditions are diferent, but with the same method you will find the solution :
          $$v(x,t)=-sqrt{(t+1)^2+x^2}$$
          $$u(x,t)=A:e^{-sqrt{(t+1)^2+x^2}}$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 10 '18 at 12:58

























          answered Dec 10 '18 at 12:47









          JJacquelinJJacquelin

          43.4k21853




          43.4k21853












          • $begingroup$
            So u is a continuous function of x for all t, can we say anything else about the behavior of u?
            $endgroup$
            – dxdydz
            Dec 11 '18 at 0:34










          • $begingroup$
            $u(x,t)$ decreases when $x$ and $t>0$ increase. One can give the maximum and the minimum.
            $endgroup$
            – JJacquelin
            Dec 11 '18 at 13:54


















          • $begingroup$
            So u is a continuous function of x for all t, can we say anything else about the behavior of u?
            $endgroup$
            – dxdydz
            Dec 11 '18 at 0:34










          • $begingroup$
            $u(x,t)$ decreases when $x$ and $t>0$ increase. One can give the maximum and the minimum.
            $endgroup$
            – JJacquelin
            Dec 11 '18 at 13:54
















          $begingroup$
          So u is a continuous function of x for all t, can we say anything else about the behavior of u?
          $endgroup$
          – dxdydz
          Dec 11 '18 at 0:34




          $begingroup$
          So u is a continuous function of x for all t, can we say anything else about the behavior of u?
          $endgroup$
          – dxdydz
          Dec 11 '18 at 0:34












          $begingroup$
          $u(x,t)$ decreases when $x$ and $t>0$ increase. One can give the maximum and the minimum.
          $endgroup$
          – JJacquelin
          Dec 11 '18 at 13:54




          $begingroup$
          $u(x,t)$ decreases when $x$ and $t>0$ increase. One can give the maximum and the minimum.
          $endgroup$
          – JJacquelin
          Dec 11 '18 at 13:54


















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