- #1

evinda

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I want to solve the equation $u_t+u^2u_x=0$ with $u(x,0)=2+x$.

I have tried the following:

The characteristic curves for $u_t+u^2 u_x=0$ are the solutions of the ode $\frac{dx}{dt}=u^2$.

We have that $\frac{d}{dt}u(x(t),t)=0$, implying that $u(x(t),t)=c$.

The characteristic line that passes through $(x,t)$ and $(z,0)$ has slope

$\frac{x-z}{t-0}=\frac{dx}{dt}=u^2(x,t)=u^2(z,0)=(2+z)^2$.

Thus, $x-z=t(2+z)^2 \Rightarrow x-z=t(4+4z+z^2)=4t+4tz+tz^2 \Rightarrow z^2t+(4t+1)z+4t-x=0$.

The discriminant is $\Delta=(4t+1)^2-4t(4t-x)=4tx+8t+1$.

$z_{1,2}=\frac{-(4t+1) \pm \sqrt{4tx+8t+1}}{2t}$ for $t \neq 0$ and $4tx+8t+1 \geq 0$.

Thus $u^2(x,t)=\left( 2+\frac{[-(4t+1) \pm \sqrt{4tx+8t+1}]}{2t}\right)^2$.

We want $(2+x)^2=u^2(x,0)=\lim_{t \to 0} \left( 2+\frac{[-(4t+1) \pm \sqrt{4tx+8t+1}]}{2t}\right)^2$.

When having the + sign, the limit is $+\infty$, so $u$ could not be a solution.

Also, we have that $\lim_{t \to 0} \left( 2+\frac{[-(4t+1) - \sqrt{4tx+8t+1}]}{2t}\right)=-2-x$.

So, $u^2(x,t)=\left( 2+\frac{[-(4t+1)- \sqrt{4tx+8t+1}]}{2t}\right)^2 $.

So $u(x,t)= \pm \left| 2- \frac{(4t+1)+ \sqrt{4tx+8t+1}}{2t}\right|$.

Is everything right or have I done something wrong? (Thinking)