Second order PDE with variable coefficients

In summary: The differential equation ##\partial x/\partial s = -2ax## tells you that the flow is in the negative x-direction and that the pressure is constant.
  • #1
grquanti
17
0
Hello,
I have an equation of the form:

##\partial_t f(x,t)+a\partial_x^2 f(x,t)+g(x)\partial_xf(x,t)=0 ##

(In my particular case ##g(x)=kx## with ##k>0## and ##a=2k=2g'(x)##)

I'd like to know if there is some general technique that i can use to solve my problem (for example: in the first order case I know the method of characteristics is widely used).

Thanks in advance.

P.s. I have searched for the answer in the forum, but I can't find it. If a similar question was already answered,please, forgive me.​
 
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  • #2
I suggest you use the method of characteristics to rewrite the first order part as a single partial derivative. You can then deal with the second order derivative later on. You should get a differential equation reminiscent of the heat equation.
 
  • #3
Orodruin said:
I suggest you use the method of characteristics to rewrite the first order part as a single partial derivative. You can then deal with the second order derivative later on.

Forgive me, I'm not an expert (actually I don't know nothing about the method of characteristics), can you be a little more explicit?

Orodruin said:
You should get a differential equation reminiscent of the heat equation.

If it's what you mean, my starting point is

##\partial_t f(x,t) = k\partial_x^2 xf(x,t)##

with ##k>0## (there is an error in the first post: when putted in the homogeneous form, the ##+## singns should be ##-##.
For instance, it's a Fokker-Planck equation with no drift.
 
  • #4
grquanti said:
Forgive me, I'm not an expert (actually I don't know nothing about the method of characteristics), can you be a little more explicit?
Perfect opportunity to learn then. If given a partial differential equation on the form ##\partial_t f + \vec v(\vec x,t) \cdot \nabla f = \ldots##, where ##\vec v## can be regarded as a velocity field you can introduce new variables ##s## and ##\vec \xi## such that the left-hand side is equal to ##\partial_s f## by finding the characteristics of the velocity field ##\vec v##, i.e., its flow lines. You do this by letting ##s = t## and solving the resulting ordinary differential equation ##\partial{\vec x}/\partial s = \vec v## (note that the resulting ##\vec x## will generally depend on both ##\vec \xi## and ##s##!). In your case, your space is one-dimensional and so you can do away with all the vector arrows. I suggest trying to introduce these new variables and writing down the differential equation in terms of them.
 
  • #5
If I understand, you mean:

from

##\partial_t f(x,t) -2a\partial_x f(x,t) = ax\partial_x^2 f(x,t)##

I take ##s## such that

##\partial_s f(x,t) = \partial_t f(x,t) -2a\partial_x f(x,t)##

This implies

## \frac{d}{ds}t=1##
## \frac{d}{ds}x=-2a##

So that

##t(s)=s+c_1##
##x(s)=-2as+c_2##

and I'm left with

##\partial_s f(x(s),t(s))=ax(s)\partial_x^2f(x(s),t(s))##

But I can't understand how to use the parameter ##ξ## and, more in general, I have seen the method is applied when the right hand side of the equation is independent from the solution: if the right hand side is a function ##g(x,t)## , in the characteristic equations I should also consider

##\frac{d}{ds}z=g(x(s),t(x))##

but in my case the function ##g## is ##g=ax\partial_x^2 f## which depends on ##f##

Where are my mistakes?
 
  • #6
grquanti said:
and I'm left with

##\partial_s f(x(s),t(s))=ax(s)\partial_x^2f(x(s),t(s))##

This is not necessarily true. Note that
$$
\partial_x = \frac{\partial s}{\partial x} \partial_s + \frac{\partial \xi}{\partial x}\partial_\xi.
$$
This is the reason to call both ##s## and ##\xi## something different from ##t## and ##x## - to keep track of which variables are which.

grquanti said:
But I can't understand how to use the parameter ξ
A good choice would be to let ##\xi = x## for ##s = 0##. Also, you can choose to let your integration constant ##c_1## be equal to 0 as this just represents a shift in ##s##. What does the differential equation ##\partial x/\partial s = -2ax## really tell you? (Don't forget the ##x##! As you defined the problem, it is here and not in the second order term.).
 

Related to Second order PDE with variable coefficients

1. What is a second order PDE with variable coefficients?

A second order PDE with variable coefficients is a type of partial differential equation (PDE) that involves a second derivative of a function with respect to two or more independent variables, where the coefficients of the second derivative are not constant but instead vary with the independent variables. This makes the equation more complex and difficult to solve compared to PDEs with constant coefficients.

2. What are some examples of second order PDEs with variable coefficients?

Examples of second order PDEs with variable coefficients include the heat equation, wave equation, and Laplace's equation. These equations are commonly used in physics and engineering to model the behavior of heat, waves, and electric potentials, respectively.

3. How are second order PDEs with variable coefficients solved?

Solving a second order PDE with variable coefficients can be a challenging task and often requires advanced mathematical techniques. One common approach is to use separation of variables, where the solution is expressed as a product of functions of individual variables. Other methods include the method of characteristics, Green's functions, and numerical methods such as finite difference or finite element methods.

4. What are the applications of second order PDEs with variable coefficients?

Second order PDEs with variable coefficients have many applications in various fields, particularly in physics, engineering, and finance. They are used to model and understand complex physical phenomena such as heat transfer, wave propagation, and fluid flow. They also play a crucial role in financial mathematics, where they are used to price derivatives and model the behavior of financial markets.

5. What are the challenges of working with second order PDEs with variable coefficients?

One of the main challenges of working with second order PDEs with variable coefficients is their complexity and difficulty in finding exact solutions. In many cases, numerical methods are needed to approximate the solutions, which can be computationally intensive. Additionally, the behavior of these equations can be highly sensitive to small changes in the coefficients, making it challenging to analyze their long-term behavior.

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