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First order non linear boundary value pb

  1. Mar 30, 2012 #1
    Hi there:

    I am trying to solve a two points boundary value problem.
    Consider a function f:[x1,x2]->[x2,x3]

    x1 and x3 are knowns
    x2 is an unknown parameter

    f'(x) = exp( -a*x + b*f(x) )

    where b>a>0

    Two boundaries conditions:

    f(x1) = x2
    f(x2) = x3

    Does anyone know how to solve it?

    Thanks!!!!
     
  2. jcsd
  3. Mar 30, 2012 #2
    Consider the equation:

    [tex]f'=e^{-ax}e^{bf}[/tex]

    now, just separate variables and integrate.
     
  4. Mar 31, 2012 #3
    I agree, solving the ODE is rather easy. But the difficulty isn't here.
    The boundary conditions raise a very difficult problem:
    f(x1)=x2 and f(x2)=x3 with given x1, x3 and unknown x2
    In fact, x2 cannot be expressed with a combination of a finit number of usual functions of x1 and x3.
    In this case, we need numerical solving in order to compute x2 and then, the unknown constant C which is necessary to fully express f(x).
     

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  5. Apr 1, 2012 #4

    hunt_mat

    User Avatar
    Homework Helper

    The problem is that you have a first order equation and two unknowns and that isn't possible, for a boundary value problem you need a second order equation. One of your points could be an attractor which is possible but you would need to check that to begin with.
     
  6. Apr 2, 2012 #5
    I do not agree. A second order equation is not needed.
    Apparently, there are two boundary conditions f(x1)=x2 and f(x2)=x3. But, in fact, there is only one bondary condition, which is : f(f(x1))=x3 with known x1 and x3. Since x2 is not known, only one relationship f(f(x1))=x3 is remaining.
    The ODE f'(x) = exp( -a*x + b*f(x) ) and the boundary condition f(f(x1))=x3, with given a, b, x1, x3 is a complete system, needing nothing more to be fully defined.
     
  7. Apr 2, 2012 #6

    hunt_mat

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    Homework Helper

    So I misread the question, apologies, you basically have a differential equation of the form:
    [tex]
    e^{-bf(x)}\frac{df}{dx}=e^{-ax}
    [/tex]
    So to begin with, integrate between [itex]x_{1}[/itex] and [itex]x[/itex] to obtain:
    [tex]
    e^{-bf(x)}=e^{-bx_{2}}+\frac{b}{a}e^{-ax}-\frac{b}{a}e^{-ax_{1}}
    [/tex]
    And integrating between [itex]x[/itex] and [itex]x_{2}[/itex] to obtain:
    [tex]
    e^{-bf(x)}=e^{-bx_{3}}+\frac{b}{a}e^{-ax}-\frac{b}{a}e^{-ax_{2}}
    [/tex]
    Now from here I see two possible ways forward, you equate that equations and obtain an equation for [itex]x_{2}[/itex] and use Newtons method to find that or you can solve for the unknown term and obtain a implicit expression for [itex]f(x)[/itex] depending on the property of b/a. It's your choice. Do you have numerical values for [itex]x_{1}[/itex],[itex]x_{3}[/itex], a and b?
     
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