Solve: $u_t+(u_x)^ 2=0$ , $u(x, 0)=-x^2$ and show that the solution breaks down when $t=\dfrac{1}{4}$ .

Question

Since the differential equation is not linear, I like to know how to solve such equation. If you are in details It would be great!

Takeing the $x$-derivative and defining $w=2u_x$ we get $w_t + w w_x = 0$ which is a Burgers-equation. This can also be transformed into the linear Heat equation $\phi_t = \phi_xx + g(t)\phi$ if wanted by taking $w = -2\frac{\phi_x}{\phi}$ (the Cole–Hopf transformation). To solve your problem you can try to apply the method of characteristics on the Burgers equation. If the characteristics intersect at some $t$ then the solution sieze to exist. — Winther, Feb 14 '16 at 22:34
See this question for an analysis of the very similar PDE $u_t + (u_x)^2 = t$. — Winther, Feb 14 '16 at 22:37

doraemonpaul · Answer 1 · 2016-02-20T19:23:50.787

$u_t+(u_x)^2=0$

$u_{tx}+2u_xu_{xx}=0$

Let $v=u_x$ ,

Then $v_t+2vv_x=0$ with $v(x,0)=-2x$

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

$\dfrac{dv}{ds}=0$ , letting $v(0)=v_0$ , we have $v=v_0$

$\dfrac{dx}{ds}=2v=2v_0$ , letting $x(0)=f(v_0)$ , we have $x=2v_0s+f(v_0)=2vt+f(v)$ , i.e. $v=F(x-2vt)$

$v(x,0)=-2x$ :

$F(x)=-2x$

$\therefore v=-2(x-2vt)$

$v=-2x+4vt$

$v=\dfrac{2x}{4t-1}$

$u_x=\dfrac{2x}{4t-1}$

$u(x,t)=\dfrac{x^2}{4t-1}+g(t)$

$u_t=-\dfrac{4x^2}{(4t-1)^2}+g_t(t)$

$\therefore-\dfrac{4x^2}{(4t-1)^2}+g_t(t)+\dfrac{4x^2}{(4t-1)^2}=0$

$g_t(t)=0$

$g(t)=C$

$\therefore u(x,t)=\dfrac{x^2}{4t-1}+C$

$u(x,0)=-x^2$ :

$C=0$

$\therefore u(x,t)=\dfrac{x^2}{4t-1}$

Hence the solution breaks down when $t=\dfrac{1}{4}$ .

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