A problem from finite element book

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Discussion Overview

The discussion revolves around a problem from finite element analysis concerning the properties of a continuous function \( w \) defined on the interval [0,1]. Participants are exploring the implications of the integral condition \( \int_0^1 wv \, dx = 0 \) for all functions \( v \) in a specified function space \( V \), where \( V \) consists of continuous functions that vanish at the endpoints and have piecewise continuous derivatives. The goal is to show that \( w(x) = 0 \) for all \( x \) in [0,1].

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents the integral condition and seeks help in proving that \( w(x) = 0 \) based on this condition.
  • Another participant suggests assuming \( w(x) \neq 0 \) at some point and constructing a function \( v(x) \) such that the integral \( \int_0^1 w(x)v(x) \, dx \neq 0 \), hinting at a contradiction.
  • A participant questions the validity of considering a single arbitrary function \( v \) when the condition must hold for all functions in \( V \).
  • There is a discussion about whether finding a specific \( v \) that contradicts the integral condition can lead to the conclusion that \( w \) must be zero.
  • Another participant reiterates the theorem stating that if the integral equals zero for every function \( v \), then \( w \) must be zero, suggesting a proof by contradiction.
  • A participant proposes a specific function \( v = \sin(x^2 - x) \) as a candidate to demonstrate the contradiction, analyzing the sign of the integral based on the sign of \( w \).
  • There is uncertainty expressed regarding the validity of this approach and the proof's rigor.

Areas of Agreement / Disagreement

Participants express differing views on the approach to proving that \( w(x) = 0 \). Some support the idea of using contradiction with specific functions, while others question the generality of the argument. The discussion remains unresolved regarding the sufficiency of the proposed proof.

Contextual Notes

Participants highlight the need for careful consideration of the conditions under which the integral holds and the implications of choosing specific functions from the space \( V \). There is an acknowledgment of the complexity involved in proving the statement rigorously.

ericm1234
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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.
 
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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].
 
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 doesn't hold, then we have a contradiction?
 
In other words, is finding an arbitrary v that contradicts the given conditions really something that leads to the conclusion w must equal 0?
 
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##.
 
thank you sir
 
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?
 
Is that a valid approach? I lack analysis skills.
 

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