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Prove that a differential function is bounded by 1/2

  1. Oct 25, 2013 #1
    1. The problem statement, all variables and given/known data
    Suppose ##\phi(x)## is a function with a continuous derivative on ##0\leq x<\infty## such that ##\phi'(x)+2\phi(x)\leq 1## for all such ##x## and ##\phi(0)=0##. Show that ##\phi(x)<\frac{1}{2}## for ##x\geq 0##.


    3. The attempt at a solution
    I tried to solve this like I would any other first order differential equation.
    $$
    \phi'(x)+2\phi(x)\leq 1\Leftrightarrow e^{2x}(\phi'(x)+2\phi(x))\leq e^{2x}\Leftrightarrow e^{2x}\phi(x)\leq e^{2x}
    $$
    so
    $$
    \phi(x)\leq e^{-2x}\int\limits_{x_{0}}^{x}e^{2t}dt + ce^{-2x}=\frac{1}{2}+ce^{-2x}
    $$
    but that was as far as I could get. Any help would be greatly appreciated.
     
  2. jcsd
  3. Oct 25, 2013 #2

    LCKurtz

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    Science Advisor
    Homework Helper
    Gold Member

    You almost have it. After you multiply by your integrating factor you have$$
    (e^{2x}\phi(x))'\le e^{2x}$$Instead of doing an indefinite integral, integrate this from ##0## to ##x##:$$
    \int_0^x(e^{2t}\phi(t))'~dt\le \int_0^x e^{2t}~dt$$and see what happens.
     
  4. Oct 25, 2013 #3
    Ahh! I see it! Thanks!
     
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