Using the method of characteristics to solve a PDE

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.

• The first part is just identifying $G$, computing its partial derivatives and inserting them into the equations. How far did you get with that? Apr 2 '21 at 6:19
• @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 Apr 2 '21 at 6:21
• Write this in the curve component names, $G=z/p$. Then $G_x=G_t=G_q=0$, $G_z=1/p$, $G_p=-z/p^2$. Apr 2 '21 at 6:25
• @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 Apr 2 '21 at 6:32

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 $$dz = p\,dx+q\,dt = -\tfrac12p\,dt-\tfrac12p\,dt=-(x_0-t)\,dt \\~\\ z=z_0-x_0t+\tfrac12t^2=\tfrac12(x_0-t)^2=\tfrac12(x-\tfrac12t)^2$$
• 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 $q$. Apr 2 '21 at 7:21
• Which part? $u(x,0)=x^2/2$ is given, the $x$ derivative of that gives $u_x(x,0)=x$. Apr 4 '21 at 4:33