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I Second order PDE with variable coefficients

  1. Jul 18, 2017 #1
    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.​
     
  2. jcsd
  3. Jul 18, 2017 #2

    Orodruin

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    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.
     
  4. Jul 19, 2017 #3
    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?

    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.
     
  5. Jul 19, 2017 #4

    Orodruin

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    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.
     
  6. Jul 21, 2017 #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?
     
  7. Jul 21, 2017 #6

    Orodruin

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    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.

    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.).
     
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