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

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In summary, the conversation discusses the initial value problem for Burger's equation with viscosity and the steps to solve it. There is also a mention of a missing constant of integration, but it is noted that it will not affect the solution. The individual asking for confirmation on their work and clarification on a specific step. No further questions were asked.
  • #1
docnet
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Homework Statement
please see below
Relevant Equations
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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
$$\Rightarrow \phi(0,x)=\boxed{exp(-\int\frac{g}{2}dx)}$$
 

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  • #2
Looks ok. Are you just seeking confirmation?
You did misquote a boundary as (1,T) instead of (0,T).
 
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  • #3
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?
 
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  • #4
Orodruin said:
you are missing a constant of integration
It was left as an indefinite integral.
 
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  • #5
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
$$\Rightarrow \int\frac{\partial}{\partial x}\Big[ln(\phi(0,x))\Big]dx=-\int \frac{g}{2} dx$$
$$\Rightarrow ln(\phi(0,x))=-\int \frac{g}{2} dx$$
 
  • #6
docnet said:
Thanks! :bow: No questions, but I was unsure about the following step.
Yes, that's fine.
 
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Related to How do we solve the ODE for the initial value problem with Burger's equation?

1. What is the difference between a PDE and an ODE?

A PDE (partial differential equation) involves multiple independent variables and their respective partial derivatives, while an ODE (ordinary differential equation) involves a single independent variable and its derivatives.

2. Why would someone want to convert a PDE to an ODE?

Converting a PDE to an ODE can make the problem simpler and easier to solve. It can also help in modeling physical systems, as many real-world phenomena can be described by ODEs.

3. How do you convert a PDE to an ODE?

The process of converting a PDE to an ODE involves finding a suitable change of variables that reduces the number of independent variables in the equation. This can be done through a variety of techniques, such as separation of variables, substitution, or using special functions.

4. Are there any limitations to converting a PDE to an ODE?

Converting a PDE to an ODE may not always be possible, as some PDEs are inherently more complex and cannot be simplified to an ODE. Additionally, the conversion process may result in loss of information or accuracy in the solution.

5. Can converting a PDE to an ODE be applied to all types of PDEs?

No, not all PDEs can be converted to ODEs. The feasibility of conversion depends on the specific form and properties of the PDE. Some PDEs, such as nonlinear or higher-order PDEs, may not be amenable to conversion.

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