Solving PDE using Hopf-Lax formula How can I solve this pde by using the Hopf-Lax formula?
$$\frac{\partial u}{\partial t}\cdot \frac{\partial u}{\partial x}=1,\; u(x,0)=x$$
Thanks  lot!
 A: I believe that the Hopf Lax formula is for the equation $u_t+(f(u))_x=0$.
But how about $u=x+t$ for yours?
A: $\dfrac{\partial u}{\partial t}\dfrac{\partial u}{\partial x}=1$
$\dfrac{\partial u}{\partial t}-\dfrac{1}{\dfrac{\partial u}{\partial x}}=0$
$\dfrac{\partial^2u}{\partial x\partial t}+\dfrac{\dfrac{\partial^2u}{\partial x^2}}{\left(\dfrac{\partial u}{\partial x}\right)^2}=0$
Let $v=\dfrac{\partial u}{\partial x}$ ,
Then $\dfrac{\partial v}{\partial t}+\dfrac{1}{v^2}\dfrac{\partial v}{\partial x}=0$ with $v(x,0)=1$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$
$\dfrac{dv}{ds}=0$ , letting $v(0)=v_0$ , we have $v=v_0$
$\dfrac{dx}{ds}=\dfrac{1}{v^2}=\dfrac{1}{v_0^2}$ , letting $x(0)=f(v_0)$ , we have $x=\dfrac{s}{v_0^2}+f(v_0)=\dfrac{t}{v^2}+f(v)$ , i.e. $v=F\left(x-\dfrac{t}{v^2}\right)$
$v(x,0)=1$ :
$F(x)=1$
$\therefore v=1$
$\dfrac{\partial u}{\partial x}=1$
$u(x,t)=x+g(t)$
$\dfrac{\partial u}{\partial t}=\dfrac{\partial g(t)}{\partial t}$
$\therefore\dfrac{\partial g(t)}{\partial t}=1$
$g(t)=t+C$
$\therefore u(x,t)=x+t+C$
$u(x,0)=x$ :
$C=0$
$\therefore u(x,t)=x+t$
