non linear first order with ln I am trying to solve $u_{t}+\ln(u)u_{x}=0$ ; $u(x,0)=e^x$
using the method of characteristics similar to solving an equation of the form $au_x+bu_y=0$ as a reference, can I say that $u(x,t)=f(t\ln(u)-x)$ is a solution? Plugging in the initial value we get $u(t,0)=f(0-x)= f(-x)=e^x$ so that $u(x,t)=e^{-(t\ln(u)-x)}$
Is this correct? How is this equation different from the linear case (no $u$ term)
 A: Your result on the form of implicit equation $u=e^{-(t\ln(u)-x)}$ is correct but can be simplified to the explicit solution.
First, checking your result :
$$u_t+\ln(u)u_x=0$$
The Charpit-Lagrange system of characteristic ODEs is :
$$\frac{dt}{1}=\frac{dx}{\ln(u)}=\frac{du}{0}$$
A first characteristic equation comes from necessarily $du=0$
$$u=c_1$$
A second characteristic equation comes from solving $\frac{dt}{1}=\frac{dx}{\ln(c_1)}$
$$t=\frac{x}{\ln(c_1)}+\text{constant}$$
$$t\ln(c_1)-x=c_2$$
The general solution of the PDE expressed on the form of implicit equation $c_1=F(c_2)$ is :
$$\boxed{u=F\left(t\ln(u)-x\right)}$$
$F$ is an arbitrary function (to be determined according to some boundary condition).
Condition : $u(x,0)=e^x$
$$u(x,0)=e^x=F\left(0\ln(e^x)-x\right)=F(-x)$$
Let $X=-x$
$$F(X)=e^{-X}$$
The function $F$ is known. We put it into the above general solution where $X=t\ln(u)-x$ thus $F(X)=F(t\ln(u)-x)=e^{-(t\ln(u)-x)}=e^xu^{-t}$
$$u=e^{-(t\ln(u)-x)}=e^xu^{-t}$$
This is consistent with your own result.
On explicit form :
$u=e^xu^{-t}$ simplified gives $1=e^xu^{-t-1}$
$$\boxed{u(x,t)=\exp\left(\frac{x}{t+1}\right)}$$
