Show that the solution of the initial value problem for $u_t+u_x=\cos ^2 u$ is given by $u(x,t)=\tan^{-1} \{ \tan [u_o(x-t)]+t\}$, where $u_0(x)$ is the initial condition.
My attempts at a solution: I first tried directly taking the partial derivatives of $u(x,t)$ to plug them in and verify, but I got stuck on how I would do that with the $u_o(x-t)$ part of $u(x,t)$.
I then tried separation of variables, but I couldn't successfully separate the variables $x$ and $t$.
My most recent attempt involved Laplace transform. I got to this: $U_x(x,s)+sU(x,s)=\frac{2}{s(s^2+4)}+u_0(x)$ but didn't know how to proceed (I was trying to teach myself).
Which method would you use? Also, could you please show the first few steps of your method, especially if it's one that I tried and got stuck on?
Thanks!
1 Answer
Here is a solution by the method of characteristics. So, we assume that the equation is $$ u_t+u_x=\cos^2 u,\quad u(x,0)=u_0(x). $$ Consider the characteristics defined by $$ \frac{dx}{ds}=1,\quad x(0)=\tau,\\ \frac{dt}{ds}=1,\quad t(0)=0,\\ \frac{du}{ds}=\cos^2 u,\quad u(0)=u_0(\tau). $$ Obviously, from first two equations $$ x=s+\tau,\quad t=s\implies \tau=x-t. $$ From the third equation $$ \tan u=s+\tan u_0(\tau), $$ or, finally, $$ u=\tan^{-1}(\tan u_0(x-t)+t) $$ as required.
