# PDE $u_t+(1+u)u_x=0$ and method of characteristics

Just started a course on PDE, and I'm trying to understand a specific (perhaps trivial) point in using method of characteristics in solving equations of the form $$u_t+h(u)u_x=0$$ where $$h(u)$$ is some function of $$u$$. Specifically, I was given the problem

$$u_t+(1+u)u_x=0$$

$$u(x,0)=f(x)$$

where

$$f(x)=\begin{cases} 1 & |x|>1\\ 2-|x| & |x|\leq1 \end{cases}$$

I followed an example shown to us in class, to reach the (perhaps wrong) conclusion that $$u(x,t)=f(x-(1+u)t)$$. From here, I'm stuck. The question given to me was to describe and analyze the cahracteristics curves and the solution, but I can't understand how we can disgard the recursive quality of the solution. Trying to figure this, I saw two previously answered problems on this site, here and here, but in both, there isn't a detailed explenation on my specific problems, only final or patial. Will appreciate a detailed explentaion / a method to approaching this step of the problem.

• See also this post and use the piecewise expression of $f$ to solve the equation $u = f(x-(1+u)t)$. Apr 22, 2020 at 16:59

$$u_t+(1+u)u_x=0$$ Characteristic system of ODEs (Charpit-Lagrange) :

https://en.wikipedia.org/wiki/Method_of_characteristics $$\frac{dt}{1}=\frac{dx}{1+u}=\frac{du}{0}$$ First characteristic equation from $$du=0$$ $$u=c_1$$ Second characteristic equation from $$\frac{dt}{1}=\frac{dx}{1+u}=\frac{dx}{1+c_1}$$

$$\frac{x}{1+c_1}=t+c_2$$ $$\frac{x}{1+u}-t=c_2$$ General solution expressed on the form of implicit equation $$c_2=F(c_1)$$ : $$\boxed{\frac{x}{1+u}-t=F(u)}$$ $$F$$ is an arbitrary function to be determined in order to satisfy the condition $$u(x,0)=f(x)$$ .

Condition : $$\frac{x}{1+f(x)}-0=F(f(x))$$

Let $$\quad X=f(x)\quad$$ and the inverse function $$\quad x=f^{-1}(X)$$ $$\frac{f^{-1}(X)}{1+X}=F(X)$$ So the function $$F(X)$$ is determined. We put it into the above general solution where $$X=u$$

$$F(u)=\frac{f^{-1}(u)}{1+u}$$

$$\frac{x}{1+u}-t=\frac{f^{-1}(u)}{1+u}$$ $$f^{-1}(u)=x-(1+u)t$$ The inverse function is : $$u=f\big(x-(1+u)t\big)$$

A good approach is to focus on concepts rather than steps: the concept here is that $$u$$ is constant along characteristics. So, start with a point on the $$x$$ axis, say $$(x_0,0)$$ and ask yourself what the value of $$u$$ is there. Then $$u$$ will have the same value along the characteristic starting at $$(x_0,0)$$. What is that characteristic? We know it has slope $$(1+u)$$, where $$u$$ is the value you just found, and we know one point that it goes through. Therefore you can describe that characteristic. Therefore you can describe all characteristics.