I'm trying to solve the initial value problem with characteristis.: $$ u_t+u^2\cdot u_x=0\quad,\quad u(0,x)=f(x) $$ Where $u$ is a neat function with suitable requirements on its domain and its behaviour as my book says. I am unbelievably akward when it comes to solving with characteristics, so I hope that you can help me solving this.
What I did
First of all, a characteristic line is a line $(X(t),t)$ such that $$ \frac{d}{dt}X(t) \ = \ u(X(t))^2 $$
I read that $u$ does not change on characteristic lines, so $u^2$ shouldn't change either, which means that: $$ X(t) \ = \ u(t,X(t))^2t +k \ = \ u(0,X(0))^2t +k \ = \ f(X(0))^2\cdot t+k $$ Where $k \in \mathbb{R}$ is a value vanishing when taking the derivative. I thougt that I could find $X(0)$ just by filling in: $ \ X(0) \ = \ u(0,X(0))^2 \cdot 0 +k \ = \ k $, which gives us the following: $$ X(t) \ = \ f(k)^2\cdot t +k $$ If we want to find $u(t_0,x_0)$ for an arbitrary point $(t_0,x_0)$ in the domain, we have to follow the characteristic line back to the place where $t=0$, where we know what the function looks like. So: $$ u(t_0,x_0) \ = \ u(0,x_0 - t_0f(k)^2) \ = \ f(x_0 - t_0 f(k)^2) $$ Where a suitable $k$ still has to be found. But that means that the intersection point we are looking for is $k$, and the value we are looking for is $f(k)$.
I guess this might be solvable when $f$ is given, but that's not the case here. Can you help me to solve this?
Recently, I answered to the question : Quasi-linear PDE with Cauchy conditions:
where the PDE was : $$u_t+uu_x=0 \text{ with condition }u(x,0)=\phi(x) $$ This was solved thanks to a change of variable, or alternatively with the method of characteristics, both leading to the solution on implicit form: $$u=\phi (x-ut)$$
Refering to my previous answer, the similar calculus for the new PDE : $$u_t+u^2u_x=0 \text{ with condition }u(x,0)=f(x) $$ will lead to the solution on implicit form : $$u=f(x-u^2t)$$