Finding a constant such that the following integral inequality holds:Constant:&nbsp;c&gt;0&nbsp;such that for all&nbsp;C1&nbsp;function in&nbsp;(0,&nbsp;1)cu(0)2≤∫01u′2+u2&nbsp;dt&nbsp;

Step 1Using the fundamental theorem of calculus,u(0)=−∫0tu′(s)ds+u(t),thus by Holder inequality,Step 2u2(0)≤2(∫0tu'(s))2ds+2u2(t)&nbsp;≤2(∫0t12ds∫0tu'2(s)ds)ds+2u2(t)&nbsp;≤2∫01u'2(s)ds+2u2(t)Integrating with respect to t from&nbsp;0&nbsp;to&nbsp;1,u2(0)≤2∫01u′2(s)+2∫01u2(t)&nbsp;dt&nbsp;.

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Calculus 2
Differential equations

runddansm88e

Answered question

2022-04-07

Finding a constant such that the following integral inequality holds:
Constant: $c > 0$ such that for all $C^{1}$ function in $(0, 1)$
$c u {(0)}^{2} \leq \int_{0}^{1} u^{' 2} + u^{2} d t$

Answer & Explanation

blessgansgxei

Beginner2022-04-08Added 10 answers

Step 1
Using the fundamental theorem of calculus,
$u (0) = - \int_{0}^{t} u^{'} (s) d s + u (t),$
thus by Holder inequality,
Step 2
$\begin{aligned} u^{2} (0) & \leq 2 (\int_{0}^{t} u^{'} (s))^{2} d s + 2 u^{2} (t) \\ \leq 2 (\int_{0}^{t} 1^{2} d s \int_{0}^{t} u^{' 2} (s) d s) d s + 2 u^{2} (t) \\ \leq 2 \int_{0}^{1} u^{' 2} (s) d s + 2 u^{2} (t) \end{aligned}$
Integrating with respect to t from $0$ to $1$ ,
$u^{2} (0) \leq 2 \int_{0}^{1} u^{' 2} (s) + 2 \int_{0}^{1} u^{2} (t) d t .$

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