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Bernoulli differential equation, Cauchy problem

  1. Sep 23, 2015 #1
    1. The problem statement, all variables and given/known data
    Observe a Cauchy problem [tex]\begin{cases}y' + p(x)y =q(x)y^n\\ y(x_0) = y_0\end{cases}[/tex]
    Assume ##p(x), q(x)## are continous for some ##(a,b)\subseteq\mathbb{R}##
    Verify the equation has a solution and determine the condition for there to be exactly one solution.

    2. Relevant equations
    Theorem:
    Assume ##y' = f(x,y)## and ##y(x_0) = y_0##, where ##x_0\in (a,b)##. If ##f## is continous in some ##D\subseteq\mathbb{R}^2##, there exists continuous ##f_y := \frac{\partial f}{\partial y}## in ##D## and ##(x_0,y_0)\in D## then the Cauchy problem has exactly one solution.


    3. The attempt at a solution
    Solve the equation. If ##y^n\neq 0## then multiplying both sides by ##y^{-n}##:
    [tex]\frac{y'}{y^n}+p(x)\frac{y}{y^n}=q(x)[/tex]
    Substituting ##w=y^{1-n}## then ##w' = (1-n)y^{-n}\cdot y'## Resulting in [tex]w' + p(x)w(1-n) = q(x)(1-n)[/tex]
    From Lagrange's method we can say ##w = C(x)e^{-\int p(x)dx}##
    Apply the theorem:
    ##w' = f(x,w)## and we have ##f(x,w) = q(x)(1-n) - p(x)w(1-n)##. From which we can say ##\frac{\partial f}{\partial w} = -p(x)(1-n)##. As ##p(x)## is assumed to be continous then ##f_w## is defined and continous. ##(x_0, y_0)## is guaranteed to be in ##D## because ##D## is only bound by ##a## and ##b## on the X-axis. There are no limitations on the Y-axis.

    Problem is how do I justify that ##f(x,w)## is necessarely continous in ##D##. I know that ##f_w## is continous, if somehow ##f_x## would be continous then ##f## itself is continous. From analysis I remember that if ##f## is an elementary function, then it and its differentials are continous, but ##p(x) , q(x)## don't have to be elementary functions, do they?
     
    Last edited: Sep 23, 2015
  2. jcsd
  3. Sep 23, 2015 #2

    verty

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    Homework Helper

    Should it be ##w' + p(x)w(1 - n) = q(x)(1 - n)##?
     
  4. Sep 23, 2015 #3
    Yes, correct you are. Thanks.
     
  5. Sep 24, 2015 #4
    Edit
    NOPE, still not right
     
    Last edited: Sep 24, 2015
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