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Homework Help: Maximum principle - don't understand why RHS is negative.

  1. Nov 12, 2011 #1
    1. The problem statement, all variables and given/known data

    Consider the linear inhomogeneous second order two point BVP;

    -a(x)u''(x) + b(x)u'(x) = f(x) for 0 < x < 1

    for some functions a, f, b where a(x) > 0 for all x

    1) If f(x) < 0 for x = [0,1], show that u(x) attains its maximum value at one of the two end points x = 0, 1. - I've done this, anyone who has done the maximum principle should be able to as well.

    2) Substitute v(x) = u(x) + [itex]\epsilon[/itex] [itex]e^{\lambda x}[/itex], show that if f(x) [itex]\leq[/itex] 0 then u(x) attains its maximum value at one of the end points x = 0,1

    3. The attempt at a solution

    Question two is what I'm stuck on. If we substitute v(x) in to the equation instead we get

    [PLAIN]http://img819.imageshack.us/img819/9646/unledmug.jpg [Broken]

    It says the RHS is strictly negative since a(x) > 0. How can we know this, when we don't know what sign b(x) may come out as for any x? It may turn out b(x) is positive and [itex]\lambda b[/itex] is greater than |[itex]\lambda^{2 } a[/itex]| and therefore what is contained within the brackets is positive, and may be bigger than the absolute value of f(x). Making the whole thing positive.

    Am I missing something?
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 12, 2011 #2


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    Science Advisor

    for any x, b(x)λ is a linear function in λ.

    but a(x)λ2 is quadratic, so for sufficiently large λ, this term will dominate.
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