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Constant solution

  1. May 4, 2010 #1
    Let [tex] f \in \mathcal{C}(\mathbb{R})[/tex] be a continuous function such that [tex] tf(t) \geq 0[/tex] [tex]\forall t [/tex]. I must prove that
    [tex] y''+e^{-x}f(y)=0 [/tex]
    [tex] y(0)=y'(0)=0 [/tex]

    has [tex] y \equiv 0 [/tex] as unique solution. No idea whatsoever up to this moment, so... thanks in adv.
  2. jcsd
  3. May 4, 2010 #2


    Staff: Mentor

    I haven't worked this through, but here are some thoughts.

    Since tf(t) >= 0 for all t, it must be true that f(t) >= 0 for t >= 0, and f(t) <= 0 for t <= 0. For this reason, and assuming that y is some twice differentiable function of x, g(x), f(y) >= 0 for y >= 0 and f(y) <=0 for y <= 0.

    Looking at the differential equation, e-x > 0 for all x. For any y >= 0, f(y) >= 0, hence y'' must be <= 0 (because y'' plus a positive number has to equal zero).

    In a similar vein, for any y <= 0, f(y) <= 0, so y'' must be >= 0.

    Where I would go next is to assume that y'' > 0 for y (= g(x)) < 0, and y'' < 0 for y > 0, and work toward a contradiction.
  4. May 4, 2010 #3


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    Is this homework?
  5. May 5, 2010 #4
    This is homework of a past course I did not attend, so I don't *have to* do this stuff. I'm doing it 'cause I'm attending some different stuff about ODEs and lack some of the prerequisites. Anyway I have absolutely no problem in posting any future question in the homework section if you tell me to.
  6. May 5, 2010 #5


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    Hi, yes please use the homework section in the future. Our "homework" rules apply to independent study problems from textbooks as well.

    Oh, and ... welcome to Physics Forums! :smile:
  7. May 5, 2010 #6
    Hi Mark. You were very helpful indeed but I can't figure out what makes the whole thing go wrong and forces y to be constant.
  8. May 5, 2010 #7
    Perhaps you might wanna try the following. Suppose y is not zero all around. Then there must exist an interval (x1,x2) where either y > 0 or y < 0.

    Without loss of generality, assume y < 0.

    Look up the Strong Maximum Principle and that will help you out.
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