In the $(x,t)$– plane, the characteristic of the initial value problem $$u_t+uu_x=0$$ with $$u(x,0)=x,0\leq x\leq 1$$ are
$1$. parallel straight lines .
$2.$ straight lines which intersects at $(0,-1)$.
$3.$ non- intersecting parabolas.
$4.$ concentric circles with center at origin.
I am learning partial differential equation so don’t have good knowledge of it . According to me characteristic equations are
$$\frac{dt}{1}=\frac{dx}{u}=\frac{du}{0}$$ Now $u=c$ by last fraction. So by first two fractions I have $x-ct=k$, where $c$ and $k$ are constants. Now I don’t known how to use initial condition of $u(x,0)=x$ and what is final answer? I see that $x-ct-k=0$ are straight lines in $(x,t)$-plane. Please help me to reach at final option . Thank you.
Best Answer
The resolution of the initial value problem is discussed in this post. So we end up with the set of curves $$ u=C_1, \qquad x-ut=C_2 $$ where $C_1$, $C_2$ are constants. These curves are straight lines in the $x$-$t$ plane, thus options 3. and 4. are eliminated. Now we implement the boundary condition $u(x,0)=x$ at $t=0$: $$ x=C_1, \qquad x-x\cdot 0 = C_2 , $$ i.e. $C_1=C_2=c$. To see if the curves $x-ct=c$ are parallel, we look at the slope in $x$-$t$ coordinates, whose value equals $$c=u = \frac{x}{1+t}.$$ Are these curves parallel? Lastly you could check for the intersection of two characteristics by solving the system $$ x-a t = a, \qquad x-b t = b $$ with respect to $(x,t)$ for $0\leq a\neq b \leq 1$.