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

  • Thread starter Thread starter docnet
  • Start date Start date
  • Tags Tags
    Ode Pde
docnet
Messages
796
Reaction score
488
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.

Screen Shot 2021-03-25 at 11.41.05 PM.png

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
$$\Rightarrow \phi(0,x)=\boxed{exp(-\int\frac{g}{2}dx)}$$
 

Attachments

  • Screen Shot 2021-03-25 at 11.41.05 PM.png
    Screen Shot 2021-03-25 at 11.41.05 PM.png
    28.4 KB · Views: 180
Physics news on Phys.org
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?
 
Orodruin said:
you are missing a constant of integration
It was left as an indefinite integral.
 
haruspex said:
Looks ok. Are you just seeking confirmation?
You did misquote a boundary as (1,T) instead of (0,T).
Orodruin said:
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! :bow: No questions, but I was unsure about the following step.

docnet said:
and integrate
$$\Rightarrow \int\frac{\partial}{\partial x}\Big[ln(\phi(0,x))\Big]dx=-\int \frac{g}{2} dx$$
$$\Rightarrow ln(\phi(0,x))=-\int \frac{g}{2} dx$$
 
docnet said:
Thanks! :bow: No questions, but I was unsure about the following step.
Yes, that's fine.
 
Back
Top