Solving PDE using Hopf-Lax formula

Question

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!

Answer 1

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?

Answer 2

$\dfrac{\partial u}{\partial t}\dfrac{\partial u}{\partial x}=1$

$\dfrac{\partial u}{\partial t}-\dfrac{1}{\dfrac{\partial u}{\partial x}}=0$

$\dfrac{\partial^2u}{\partial x\partial t}+\dfrac{\dfrac{\partial^2u}{\partial x^2}}{\left(\dfrac{\partial u}{\partial x}\right)^2}=0$

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. $v=F\left(x-\dfrac{t}{v^2}\right)$

$v(x,0)=1$ :

$F(x)=1$

$\therefore v=1$

$\dfrac{\partial u}{\partial x}=1$

$u(x,t)=x+g(t)$

$\dfrac{\partial u}{\partial t}=\dfrac{\partial g(t)}{\partial t}$

$\therefore\dfrac{\partial g(t)}{\partial t}=1$

$g(t)=t+C$

$\therefore u(x,t)=x+t+C$

$u(x,0)=x$ :

$C=0$

$\therefore u(x,t)=x+t$