Consider the first order Partial differential equation (Which is actually Invischid Burger's equation)
$$u_{t}+u(x,t)u_x{}=0$$
With the initial condition ,
$$u(x,0)=\frac{1}{1+x^2}.$$ Then,
$(a)$ Show that there exists $t>0$, such that the problem has unique solution in the strip
$$\mathbb{R} \times (0,t),$$
$(b)$ Show that there may not exist solutions for the strip
$\mathbb{R} \times (0,\infty)$,
$(c)$ Also find a $t'$ such that for $\epsilon \gt0$, there exists solution in $\mathbb{R}\times (0,t')$ but not in
$\mathbb{R} \times (0,t'+\epsilon).$
Best Answer
By Lagrange's Theorem,
$$\frac{dt}{1}=\frac {dx}{u}=\frac{du}{0}$$
So, from the first and last equation, $$\frac{dt}{1}=\frac{du}{0}$$ or,$$u(x,t)=c$$ Similarly, from the first and second equation, $$ \frac{dt}{1}=\frac{dx}{u},$$ or, $$\frac{dt}{1}=\frac{dx}{c},$$ Integrating, we get $$x=c\times {t}+d $$ Hence $$x-u\times{t}=d$$
so,
$$u(x,t)=f(x-ut)$$
initially,
$$u(x,0)=\frac{1}{1+x^{2}}$$
so, $$u(x,t)= \frac{1}{1+(x-ut)^{2}}$$