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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, 2021 at 6:19
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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, 2021 at 6:25

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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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  • $\begingroup$ You have $z=u(x,t)$, thus $dz=u_x\,dx+u_t\,dt=p\,dx+q\,dt$. While the Lagrange-Charpit equations are only valid along the characteristic curves, this is valid in every direction. So apply it to the initial condition, and then also along the tangential direction of the characteristic curves. In the latter, use the known identities to eliminate $p,q$ and then reduce it to integrable terms. There are several paths to do that, all equally valid. $\endgroup$ Apr 2, 2021 at 7:10
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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, 2021 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, 2021 at 4:33

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