# A problem from finite element book

1. Mar 25, 2013

### ericm1234

..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.

2. Mar 25, 2013

### AlephZero

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. Mar 25, 2013

### ericm1234

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. Mar 25, 2013

### ericm1234

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. Mar 25, 2013

### AlephZero

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. Mar 25, 2013

### ericm1234

thank you sir

7. Mar 25, 2013

### ericm1234

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. Mar 27, 2013

### ericm1234

Is that a valid approach? I lack analysis skills.