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I need help this: Solve the following initial value problem (quasilinear problem)

$u_x + u_y= u^2$, $u(x,0) =h(x)$

Here what I did: The initial curve $\Gamma:<x=s, y=0, z=h(s)>$ and the characteristic equations: $dx/dt = 1$, $dy/dt = 1$ and $dz/dt =u^2 =z^2$. So I get $x = t + s$, $y = t$ and $z=u^2t + h(s)$. Then solution is $u=u^2y + h(x-y)$ I am not sure about this answer. Please advise. Thanks

Vui Tinh
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  • The point is that along the characteristics, $\frac{du}{dt}=u^2$. Explicitly, $\frac{d}{dt} u(x_0+t,t)=u(x_0+t,t)^2$ for any $x_0$. But the solution to $\frac{du}{dt}=u^2$ does not look like what you just said. – Ian Sep 26 '17 at 17:33

2 Answers2

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The characteristic equations are given by:

$$dx/1=dy/1=du/u^2.$$

From the first equality, we obtain $x-y=c_1$. From the second equality, we obtain $-c_2+y=-1/u \implies u= \frac{1}{c_2-y}=\frac{1}{F(c_1)-y}=\frac{1}{F(x-y)-y}.$

From $u(x,0)=h(x)$ we obtain $F(x)=1/h(x)$, hence

$$u = \frac{h(x-y)}{1-yh(x-y)}$$

MrYouMath
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You have solved the ODE $z'(t)=z^2(t)$ incorrectly. One can solve this ODE e.g. via separation of variables to get

$$z(t)=\frac{1}{\frac{1}{h(s)}-t}.$$

Now we have $x=t+s=y+s$ hence $s=x-y$. Thus we have the solution

$$u(x,y)=\frac{1}{\frac{1}{h(x-y)}-y}.$$

Cahn
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