A problem from finite element book

  • Thread starter ericm1234
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  • #1
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..though I figure it's sort of an analysis type problem.
∫wvdx=0 (int from 0 to 1) for all v in V. w is continuous on [0,1]. What it means to be in V: v in V satisfies being continuous on [0,1], v(0)=v(1)=0, and derivatives of v are piecewise continuous .
Problem is:
Show that w(x)=0 for x in [0,1].

I have spent hours with this. The book I'm looking at describes a couple problems:
for u in V, u is a solution to -u''=f(x), u is also a solution to (u',v')=(f,v) and also a solution to
1/2(u',u')-(f,u) less than or equal to 1/2(v',v')-(f,v).

I have tried in many combinations to use this information, and this is the only given info in the chapter.

Help.
 

Answers and Replies

  • #2
AlephZero
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Suppose ##w(x) \ne 0## somewhere in [0,1].

Now invent a function ##v(x)## such that ##\int_0^1 w(x)v(x)\,dx \ne 0##.

Hint: think how you could make ## w(x)v(x) \ge 0## everywhere in [0, 1].
 
  • #3
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Maybe you could elaborate a bit as my proof skills are not strong.
Youre saying assume w does not equal 0. But then why would we consider a single arbitrary v, when this condition must hold for all v?
Are you saying that, if we can find a v where the condition doesnt hold, then we have a contradiction?
 
  • #4
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In other words, is finding an arbitrary v that contradicts the given conditions really something that leads to the conclusion w must equal 0?
 
  • #5
AlephZero
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The theorem says that if ##\int_0^1 wv\,dx = 0## for every function ##v##, then ##w## must be 0.

Yes, I'm suggesting you prove this by contradition. Suppose ##w \ne 0## and the integral = 0 for every possible function ##v##. But you can show that for any given continuous nonzero function ##w##, you can find a function ##v## that makes the integral non-zero, so it is impossible that ##w \ne 0##.
 
  • #6
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thank you sir
 
  • #7
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so, wait: Please confirm this is a valid write up:
Suppose w does not equal 0. Then choose v=sin(x^2-x) (a negative function on [0,1] I just randomly came up with which satisfies being 0 at x=0 and x=1 and is continuous, with a continuous derivative, hence all that satisfies v in space V). Then if w<0 on [0,1], the integral is positive. If w>0 on [0,1], the integral is negative. This is a contradiction, so w must be 0.
Would this be a 'proper' proof?
 
  • #8
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Is that a valid approach? I lack analysis skills.
 

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