How do we solve the ODE for the initial value problem with Burger's equation?

[docnet]

Homework Statement

please see below

Relevant Equations

please see below

Sorry the problem is a bit long to read. thank you to anyone who comments.

We consider the initial value problem for the Burger's equation with viscosity given by
$$\begin{cases} \partial_t u-\partial^2_xu+u\partial_xu=0 & \text{in}\quad (1,T)\times R\\\quad \quad \quad \quad \quad u(0,\cdot)=g & \text{on}\quad \{t=0\}\times R \end{cases}$$
(1) We know that $u(0,x)=g$ so the substitution of the Burger equation gives $$u(0,x)=\boxed{g=\frac{-2}{\phi(0,x)}\frac{\partial}{\partial x}\phi(0,x)}$$ which is a separable ODE for $\psi:=\phi(0,x)$. \\\\To solve the ODE, we re-write the equation
$$\Rightarrow \frac{\partial}{\partial x}\Big[ln(\phi(0,x))\Big]=-\frac{g}{2}$$
and integrate
$$\Rightarrow \int\frac{\partial}{\partial x}\Big[ln(\phi(0,x))\Big]dx=-\int \frac{g}{2} dx$$
$$\Rightarrow ln(\phi(0,x))=-\int \frac{g}{2} dx$$
then exponentiate
$$\Rightarrow \phi(0,x)=\boxed{exp(-\int\frac{g}{2}dx)}$$

 

[haruspex]

Looks ok. Are you just seeking confirmation?
You did misquote a boundary as (1,T) instead of (0,T).

 

[Orodruin]

Technically you are missing a constant of integration. This will not affect $u(0,\cdot)$ however as it will cancel between denominator and numerator in the definition of $u$ in terms of $\phi$.

Was there a question here?

 

[haruspex]

you are missing a constant of integration

It was left as an indefinite integral.

 

[docnet]

Looks ok. Are you just seeking confirmation?
You did misquote a boundary as (1,T) instead of (0,T).

Technically you are missing a constant of integration. This will not affect $u(0,\cdot)$ however as it will cancel between denominator and numerator in the definition of $u$ in terms of $\phi$.

Was there a question here?

Thanks! No questions, but I was unsure about the following step.

and integrate
$$\Rightarrow \int\frac{\partial}{\partial x}\Big[ln(\phi(0,x))\Big]dx=-\int \frac{g}{2} dx$$
$$\Rightarrow ln(\phi(0,x))=-\int \frac{g}{2} dx$$

 

[haruspex]

Thanks! No questions, but I was unsure about the following step.

Yes, that's fine.