# Fully nonlinear 1st order PDE solved with method of characteristics

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.

• "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 $x$. Jan 6 at 14:41
• @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. Jan 6 at 14:57
• @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! Jan 6 at 15:27
• The first three equations for the characteristic curves are incorrect. The correct ones are $\dot{p}=-p$, $\dot{q}=-1$, and $\dot{z}=p^2+q$. Jan 6 at 22:30
• There was a mistake in my correction! The second equation should be $\dot{q}=-q$. (Source: en.wikipedia.org/wiki/Method_of_characteristics.) Jan 7 at 7:33

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 $$f'(y)+2f(y)^2+f(y)=0.$$

• 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.) Jan 7 at 4:49
• @Semiclassical Same here, perhaps that with @Gonçalo’s correction it does work? Jan 7 at 7:15