Showing that the functional dot product must be positive.

Question

The Question:

Given $Lu=-u''$ with $D(L)=\{u(x)|u \in C^2 [0,1], u(0)=0, u(1)=u'(1)\},$ show that $\langle Lu,u \rangle \geq 0$ for all $u \in D(L)$ where

$$ \langle f,g \rangle =\int_{0}^{1} f(x)g(x) dx. $$

My attempt:

I did integration by parts to get the expression

$$ -u'(1)^2 + \int_{0}^{1} u'(x)^2 dx $$ I then attempted Simpson's Rule on the integral to try and show that the integral must be larger than $u'(1)^2$ but that doesn't seem to work. I think using a cubic spline might work but that seems like more trouble than it's worth. Any hints (please just a hint) on where to proceed would be appreciated.

Hint: a certain inequality regarding inner products might help. — Bruno B, May 02 '23 at 19:01
Do I apply this inequality to $-u'(1)^2 + \int$ or do I start over at the definition? — , May 02 '23 at 20:46
It's really just the same hint as Suspicious Fred's. Either way you use the same inequality. Hint: remember that $1 = \int_0^1 1 \mathrm{d}x$. — Bruno B, May 02 '23 at 20:53
Ah. The cousin of adding zero: multiplying by 1. Thanks, I figured it out. — , May 02 '23 at 21:14

Suspicious Fred · Accepted Answer · 2023-05-02T19:04:19.293

0

Hint : use $u'(1)^2= u(1)^2 = (\int_0^1 u'(x)dx)^2$ and an inequality on inner products.

edited May 02 '23 at 19:04

answered May 02 '23 at 19:02

Suspicious Fred

426

Showing that the functional dot product must be positive.

1 Answers1