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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.

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

2 Answers 2

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$$u_t+(1+u)u_x=0$$ 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 : $$u=f\big(x-(1+u)t\big)$$

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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.

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