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In my PDE class, we are covering the method of characteristics. I have encountered the following problem

equations of the form $u_t + G(u_x,u,x,t) = 0$ can be solved by the method of characteristics, where $G(p,z,x,t)$ is a scalar function of 4 variables. The ODEs for $x(t)$,$p(t)=u_x(x(t),t)$, $q(t)=u_t(x(t),t)$, and $z(t)=u(x(t),t)$ are \begin{equation} \begin{aligned} \dot x & = G_p(p,z,x,t), \\ \dot z &= p\, G_p + q, \end{aligned} \qquad \begin{aligned} \dot p = -G_x - p\, G_z, \\ \dot q = -G_t - q\, G_z. \end{aligned}\end{equation}

We are asked to solve the equation $u_t + u/u_x = 0$ with initial conditions $u(x,0)=x^2/2$ using the method of characteristics. (Hint: for this problem, the ODE's can be solved one at a time, first for $p$, then $q$, then $z$, and finally $x$. For example, the solution for $p(t)$ is $x_0-t$, where $x_0$ is the initial location of the characteristiccurve. A formula for $u(x,t)$ is then easily derived from $x(t)$ and$z(t)$).

I am a novice in differential equations and I do not really know how to solve this type of equation using the method of characteristics. I cannot imagine how to extract the solution from the ODEs above. I am not quite certain how to proceed. May I please ask someone to help me solve this? I thank all helpers.

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  • $\begingroup$ The first part is just identifying $G$, computing its partial derivatives and inserting them into the equations. How far did you get with that? $\endgroup$ Apr 2 '21 at 6:19
  • $\begingroup$ @LutzLehmann thank you I am a bit unsure here I think G is probably $u/u_x$ but I could not come up with the 4 ODEs $\endgroup$
    – kroner
    Apr 2 '21 at 6:21
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    $\begingroup$ Write this in the curve component names, $G=z/p$. Then $G_x=G_t=G_q=0$, $G_z=1/p$, $G_p=-z/p^2$. $\endgroup$ Apr 2 '21 at 6:25
  • $\begingroup$ @LutzLehmann thanks I think that makes sense to me. But I am not quite certain how to solve the 4 ODEs, I have actually tried with your suggested G but without much luck. And even if I could solve each equation, I still would not be able to extract the solution of the PDE u nor use the initial condition $\endgroup$
    – kroner
    Apr 2 '21 at 6:32
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So you have $$ \frac{dx}{-z/p^2}=\frac{dt}{1}=\frac{dz}{q-z/p}=-\frac{dp}{1}=-\frac{dq}{q/p} $$ leading directly to $$p+t=p_0,~ x+q=x_0+q_0,~ q/p=q_0/p_0,~ z/p^2=z_0/p_0^2,$$ and then in combination $x+z_0/p_0^2t=x_0$.

The PDE at $t=0$ gives $q_0+z_0/p_0=0$. The initial condition evaluates to $z_0=x_0^2/2$, $p_0=u_x(x_0,0)=x_0$, $q_0=-z_0/p_0=-x_0/2$. This simplifies the equations for the characteristic so far to $$ p+t=x_0,~q+x=\tfrac12x_0,~ q/p=-\tfrac12,~z/p^2=\tfrac12,~x+\tfrac12t=x_0 $$ The solution tangent plane equation gives $$ dz = p\,dx+q\,dt = -\tfrac12p\,dt-\tfrac12p\,dt=-(x_0-t)\,dt \\~\\ z=z_0-x_0t+\tfrac12t^2=\tfrac12(x_0-t)^2=\tfrac12(x-\tfrac12t)^2 $$


Often one can condense down such exercise solutions to a much narrower set of identities.

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  • $\begingroup$ thank you so much I think I get this but what is meant by tangent plane? How can we obtain this? $\endgroup$
    – kroner
    Apr 2 '21 at 6:56
  • $\begingroup$ thank you very much, I just don't understand the last step you did with the tangent plane and how you used that to reach the final answer $\endgroup$
    – kroner
    Apr 2 '21 at 7:01
  • $\begingroup$ also you say that the equations lead directly to your characterizations, how do you reach this? Thank you. $\endgroup$
    – kroner
    Apr 2 '21 at 7:08
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    $\begingroup$ Enumerate the quotients in the chain. The first identity is from 2=4, then the second from 1=5 and using the PDE, then 4=5, and 3=4 using again the PDE to eliminate $q$. $\endgroup$ Apr 2 '21 at 7:21
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    $\begingroup$ Which part? $u(x,0)=x^2/2$ is given, the $x$ derivative of that gives $u_x(x,0)=x$. $\endgroup$ Apr 4 '21 at 4:33

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