I know there are some similar problem posted on here, but I can't find the solution to this particular problem:

$$u+\frac{1}{2}(u_x)^2+u_y=0\qquad\text{in }\mathbb{R}\times(-\infty,0)$$ $$u(x,0)=-x^2\qquad x\in\mathbb{R}$$

Here's what I've tried: $$F((p,q),z,(x,y))=z+\frac{1}{2}p^2+q,$$ and $$F_p=p,F_q=1,F_z=1,F_x=F_y=0$$ gives $$\begin{cases} \dot{p}=-p\newline \dot{q}=-q\newline \dot{z}=p^2+q\newline \dot{x}=p\newline \dot{y}=1.\end{cases}$$ Now I don't see how to continue. I've also tried to do this using Lagrange-Charpit equations, but also using this I couldn't get it to work.

I also tried to differentiate the entire expression with respect to $x$ and substitute $v=u_x$, which gave $$vv_x+v_y=-v,$$ which is a quasilinear PDE looking like an inviscid Burger's equation with an extra $-v$ term, but also here I couldn't quite continue after a while.

If anyone could kindly give an answer to this problem, or a hint which method to use, I would very much appreciate it.

(I've been doing characteristics for about 3 days now and every time I think I get it, there seems to be something I'm missing in my particular problem.

  • 2
    $\begingroup$ "Now I don't see how to continue..." - did you try to solve the system of ODEs? You get $p = c_1\exp t$, $q = t+c_2$, $y=t+c_3$, and from that $z$ and $x$. $\endgroup$
    – user8268
    Jan 6 at 14:41
  • $\begingroup$ @user8268 Well the part where I got stuck was finding the $z$ and $x$, but now seeing this that makes sense, but I just didn't think about it. I will try to see if I can continue using this. $\endgroup$ Jan 6 at 14:57
  • $\begingroup$ @user8268 I feel like I'm getting really close, but just can't get it completely, I will give this a go another time again, thanks for your hint, that at least got me almost there! $\endgroup$ Jan 6 at 15:27
  • 1
    $\begingroup$ The first three equations for the characteristic curves are incorrect. The correct ones are $\dot{p}=-p$, $\dot{q}=-1$, and $\dot{z}=p^2+q$. $\endgroup$
    – Gonçalo
    Jan 6 at 22:30
  • 1
    $\begingroup$ There was a mistake in my correction! The second equation should be $\dot{q}=-q$. (Source: en.wikipedia.org/wiki/Method_of_characteristics.) $\endgroup$
    – Gonçalo
    Jan 7 at 7:33

2 Answers 2


Your calculus is correct. $$v\,v_x+v_y=-v\quad\text{is OK.}$$ The Charpit-Lagrange characteristic ODEs are : $$\frac{dx}{v}=\frac{dy}{1}=\frac{dv}{-v}$$ A first characteristic equation comes from solving $\frac{dy}{1}=\frac{dv}{-v}$ : $$v\,e^y=c_1 $$ A second characteristic equation comes from solving $\frac{dx}{v}=\frac{dv}{-v}$ : $$v+x=c_2$$ The general solution of the PDE on the form of implicit equation $c_1=f(c_2)$ is: $$v\,e^y=f(v+x)$$ $f$ is an arbitrary function. $$\boxed{v(x,y)=e^{-y}f\big(v(x,y)+x\big)}$$ $f$ has to be determined according to the boundary condition. $$u(x,0)=-x^2\quad\implies\quad v(x,0)=u_x(x,0)=-2x$$ $$-2x=e^0f(-2x+x)=f(-x)\quad\implies\quad f(X)=-2(-X)=2X$$ The function $f(X)$ is determined. We put it into the above general solution where $X=v+x$ : $$v=e^{-y}\:2(v+x)$$ $$v=\frac{2x}{e^y-2}$$ $$u=\int \frac{2x}{e^y-2} dx=\frac{x^2}{e^y-2}+C$$ $u(x,0)=-x^2\quad\implies\quad C=0$ $$\boxed{u(x,y)=\frac{x^2}{e^y-2}}$$


Hint: Given the boundary condition $u(x,0)=-x^2$, the PDE can be reduced to an ODE with the ansatz $u(x,y)=x^2f(y)$, with $f(0)=-1$. Indeed, plugging it into the PDE, and dividing the result by $x^2$, one obtains $$ f'(y)+2f(y)^2+f(y)=0. $$

  • 1
    $\begingroup$ While this does give a solution route, I am curious about the solution via characteristics. (Mostly because I can't seem to make it work either.) $\endgroup$ Jan 7 at 4:49
  • $\begingroup$ @Semiclassical Same here, perhaps that with @Gonçalo’s correction it does work? $\endgroup$ Jan 7 at 7:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .