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

  • #1
130
1

Homework Statement



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


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?
 
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Answers and Replies

  • #2
Deveno
Science Advisor
906
6
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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