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$$ \frac{\partial u }{\partial t} + u^2 \frac{\partial u}{\partial x}=0 $$ $$ 0 < x < \infty , t>0$$ initial value $u(x,0) = \sqrt{x}, 0< x < \infty$

I've tried to solve this problem but get stuck

this is what I've done so far:

characteristic equations $$\frac{dx}{dt} =u^2$$ $$\frac{du}{dt} = 0$$

since $\frac{du}{dt} = 0$ then $u = A $ , with A is constant from the initial value $$u(x(0),0)=\sqrt{x_0} = B $$ so $u=\sqrt{x_0}$

$\frac{dx}{dt}=u^2 = \left( \sqrt{x_0} \right) ^2 = x_0^2$ then $x = x_0^2 t + x_0$

and what should I do next to find $u(x,t)$ ?

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This question has been solved perfectly. Hope that the asker has been diving enough and accept the answer at an early date. – doraemonpaul Apr 21 '13 at 1:27

Follow the method in

$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$

$\dfrac{du}{ds}=0$ , letting $u(0)=u_0$ , we have $u=u_0$

$\dfrac{dx}{ds}=u^2=u_0^2$ , letting $x(0)=f(u_0)$ , we have $x=u_0^2s+f(u_0)=u^2t+f(u)$ , i.e. $u=F(x-u^2t)$

$u(x,0)=\sqrt{x}$ :


$\therefore u=\sqrt{x-u^2t}$






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what is $F(x-u^2t)$ ? inverse function? – chihiroasleaf Feb 20 '13 at 14:08
@chihiroasleaf: arbitrary function of $x-u^2t$ – doraemonpaul Feb 21 '13 at 0:02

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