How to prove $fequiv 0$ $forall xin [a,b]$?$quad$($f''+pf'+qf=0$ with $f(a)=f(b)=0$)












5












$begingroup$


Define $f in C^{2}[a,b]$ satisfying $f''+pf'+qf=0$ with $f(a)=f(b)=0$, where $pin C^{0}[a,b]$ and $qin C^{0}[a,b]$ are two functions.



If $qleq0$, can we prove $fequiv 0$ $forall xin [a,b]$ ?



My try:

If $fnotequiv 0$, without loss of generality, we assume that the maximum of $f$ on $[a,b]$ is greater than zero, while notating $f(x_0)=displaystylemax_{[a,b]} f$.



Then we have $f(x_0) > 0$, $f'(x_0) = 0$ and $f''(x_0) leq 0$.



I figured out that if we alter the condition $qleq0$ into $q(x)<0$ there evidently exists contradiction.



But how to analyze further with the condition $qleq0$? Can we still find contradiction if $q(x_0)=0$ and $f''(x_0)=0$ ?



Any ideas would be highy appreciated!










share|cite|improve this question











$endgroup$












  • $begingroup$
    Maybe it is meant that $f in C^2[a,b]$ ?
    $endgroup$
    – Rebellos
    Nov 28 '18 at 13:20










  • $begingroup$
    I'm not sure if this approach is useful but I was thinking of taking an inner product with $f$ and see if you run into a contradiction of any kind that way.
    $endgroup$
    – Cameron Williams
    Nov 28 '18 at 13:20










  • $begingroup$
    @Rebellos Thanks for reminding me. I've modified it
    $endgroup$
    – Zero
    Nov 28 '18 at 13:22






  • 3




    $begingroup$
    This is a duplicate of math.stackexchange.com/q/3016693.
    $endgroup$
    – Paul Frost
    Nov 28 '18 at 14:14










  • $begingroup$
    Possible duplicate of $f''+pf'+qf=0$ where $qleq0$ and $f(0)=f(1)=0$ prove $f=0$ ($f$, $p$, $q$ defined on $[0,1]$)
    $endgroup$
    – Paul Frost
    Nov 30 '18 at 11:30
















5












$begingroup$


Define $f in C^{2}[a,b]$ satisfying $f''+pf'+qf=0$ with $f(a)=f(b)=0$, where $pin C^{0}[a,b]$ and $qin C^{0}[a,b]$ are two functions.



If $qleq0$, can we prove $fequiv 0$ $forall xin [a,b]$ ?



My try:

If $fnotequiv 0$, without loss of generality, we assume that the maximum of $f$ on $[a,b]$ is greater than zero, while notating $f(x_0)=displaystylemax_{[a,b]} f$.



Then we have $f(x_0) > 0$, $f'(x_0) = 0$ and $f''(x_0) leq 0$.



I figured out that if we alter the condition $qleq0$ into $q(x)<0$ there evidently exists contradiction.



But how to analyze further with the condition $qleq0$? Can we still find contradiction if $q(x_0)=0$ and $f''(x_0)=0$ ?



Any ideas would be highy appreciated!










share|cite|improve this question











$endgroup$












  • $begingroup$
    Maybe it is meant that $f in C^2[a,b]$ ?
    $endgroup$
    – Rebellos
    Nov 28 '18 at 13:20










  • $begingroup$
    I'm not sure if this approach is useful but I was thinking of taking an inner product with $f$ and see if you run into a contradiction of any kind that way.
    $endgroup$
    – Cameron Williams
    Nov 28 '18 at 13:20










  • $begingroup$
    @Rebellos Thanks for reminding me. I've modified it
    $endgroup$
    – Zero
    Nov 28 '18 at 13:22






  • 3




    $begingroup$
    This is a duplicate of math.stackexchange.com/q/3016693.
    $endgroup$
    – Paul Frost
    Nov 28 '18 at 14:14










  • $begingroup$
    Possible duplicate of $f''+pf'+qf=0$ where $qleq0$ and $f(0)=f(1)=0$ prove $f=0$ ($f$, $p$, $q$ defined on $[0,1]$)
    $endgroup$
    – Paul Frost
    Nov 30 '18 at 11:30














5












5








5


3



$begingroup$


Define $f in C^{2}[a,b]$ satisfying $f''+pf'+qf=0$ with $f(a)=f(b)=0$, where $pin C^{0}[a,b]$ and $qin C^{0}[a,b]$ are two functions.



If $qleq0$, can we prove $fequiv 0$ $forall xin [a,b]$ ?



My try:

If $fnotequiv 0$, without loss of generality, we assume that the maximum of $f$ on $[a,b]$ is greater than zero, while notating $f(x_0)=displaystylemax_{[a,b]} f$.



Then we have $f(x_0) > 0$, $f'(x_0) = 0$ and $f''(x_0) leq 0$.



I figured out that if we alter the condition $qleq0$ into $q(x)<0$ there evidently exists contradiction.



But how to analyze further with the condition $qleq0$? Can we still find contradiction if $q(x_0)=0$ and $f''(x_0)=0$ ?



Any ideas would be highy appreciated!










share|cite|improve this question











$endgroup$




Define $f in C^{2}[a,b]$ satisfying $f''+pf'+qf=0$ with $f(a)=f(b)=0$, where $pin C^{0}[a,b]$ and $qin C^{0}[a,b]$ are two functions.



If $qleq0$, can we prove $fequiv 0$ $forall xin [a,b]$ ?



My try:

If $fnotequiv 0$, without loss of generality, we assume that the maximum of $f$ on $[a,b]$ is greater than zero, while notating $f(x_0)=displaystylemax_{[a,b]} f$.



Then we have $f(x_0) > 0$, $f'(x_0) = 0$ and $f''(x_0) leq 0$.



I figured out that if we alter the condition $qleq0$ into $q(x)<0$ there evidently exists contradiction.



But how to analyze further with the condition $qleq0$? Can we still find contradiction if $q(x_0)=0$ and $f''(x_0)=0$ ?



Any ideas would be highy appreciated!







real-analysis differential-equations continuity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 '18 at 17:57









rafa11111

1,106417




1,106417










asked Nov 28 '18 at 13:12









ZeroZero

33610




33610












  • $begingroup$
    Maybe it is meant that $f in C^2[a,b]$ ?
    $endgroup$
    – Rebellos
    Nov 28 '18 at 13:20










  • $begingroup$
    I'm not sure if this approach is useful but I was thinking of taking an inner product with $f$ and see if you run into a contradiction of any kind that way.
    $endgroup$
    – Cameron Williams
    Nov 28 '18 at 13:20










  • $begingroup$
    @Rebellos Thanks for reminding me. I've modified it
    $endgroup$
    – Zero
    Nov 28 '18 at 13:22






  • 3




    $begingroup$
    This is a duplicate of math.stackexchange.com/q/3016693.
    $endgroup$
    – Paul Frost
    Nov 28 '18 at 14:14










  • $begingroup$
    Possible duplicate of $f''+pf'+qf=0$ where $qleq0$ and $f(0)=f(1)=0$ prove $f=0$ ($f$, $p$, $q$ defined on $[0,1]$)
    $endgroup$
    – Paul Frost
    Nov 30 '18 at 11:30


















  • $begingroup$
    Maybe it is meant that $f in C^2[a,b]$ ?
    $endgroup$
    – Rebellos
    Nov 28 '18 at 13:20










  • $begingroup$
    I'm not sure if this approach is useful but I was thinking of taking an inner product with $f$ and see if you run into a contradiction of any kind that way.
    $endgroup$
    – Cameron Williams
    Nov 28 '18 at 13:20










  • $begingroup$
    @Rebellos Thanks for reminding me. I've modified it
    $endgroup$
    – Zero
    Nov 28 '18 at 13:22






  • 3




    $begingroup$
    This is a duplicate of math.stackexchange.com/q/3016693.
    $endgroup$
    – Paul Frost
    Nov 28 '18 at 14:14










  • $begingroup$
    Possible duplicate of $f''+pf'+qf=0$ where $qleq0$ and $f(0)=f(1)=0$ prove $f=0$ ($f$, $p$, $q$ defined on $[0,1]$)
    $endgroup$
    – Paul Frost
    Nov 30 '18 at 11:30
















$begingroup$
Maybe it is meant that $f in C^2[a,b]$ ?
$endgroup$
– Rebellos
Nov 28 '18 at 13:20




$begingroup$
Maybe it is meant that $f in C^2[a,b]$ ?
$endgroup$
– Rebellos
Nov 28 '18 at 13:20












$begingroup$
I'm not sure if this approach is useful but I was thinking of taking an inner product with $f$ and see if you run into a contradiction of any kind that way.
$endgroup$
– Cameron Williams
Nov 28 '18 at 13:20




$begingroup$
I'm not sure if this approach is useful but I was thinking of taking an inner product with $f$ and see if you run into a contradiction of any kind that way.
$endgroup$
– Cameron Williams
Nov 28 '18 at 13:20












$begingroup$
@Rebellos Thanks for reminding me. I've modified it
$endgroup$
– Zero
Nov 28 '18 at 13:22




$begingroup$
@Rebellos Thanks for reminding me. I've modified it
$endgroup$
– Zero
Nov 28 '18 at 13:22




3




3




$begingroup$
This is a duplicate of math.stackexchange.com/q/3016693.
$endgroup$
– Paul Frost
Nov 28 '18 at 14:14




$begingroup$
This is a duplicate of math.stackexchange.com/q/3016693.
$endgroup$
– Paul Frost
Nov 28 '18 at 14:14












$begingroup$
Possible duplicate of $f''+pf'+qf=0$ where $qleq0$ and $f(0)=f(1)=0$ prove $f=0$ ($f$, $p$, $q$ defined on $[0,1]$)
$endgroup$
– Paul Frost
Nov 30 '18 at 11:30




$begingroup$
Possible duplicate of $f''+pf'+qf=0$ where $qleq0$ and $f(0)=f(1)=0$ prove $f=0$ ($f$, $p$, $q$ defined on $[0,1]$)
$endgroup$
– Paul Frost
Nov 30 '18 at 11:30










1 Answer
1






active

oldest

votes


















2












$begingroup$

The following is based on the classical proof of the maximum principle.



Let $L(f)=f''+p,f'+q,f$. If $L(f)>0$, then your argument shows that $f$ must be identically $0$.



But we have $L(f)=0$, not $>0$. What can we do? Take $M>0$ such that $M^2+M,p(x)+q(x)>0$ for all $xin[a,b]$ and let $epsilon>0$. Then
$$
L(f+epsilon,e^{Mx})=epsilon,e^{Mx}(M^2+M,p(x)+q(x))>0quadforall xin[a,b].
$$

Then
$$
max_{ale xle b}(f+epsilon,e^{Mx})=maxbigl(f(a)+epsilon,e^{Ma},f(b)+epsilon,e^{Mb}bigr)=epsilon,e^{Mb}.
$$

Letting $epsilonto0$ gives the desired result.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Your answer does help! I got lost in finding contradiction by analyzing $f(x_0 +s)$ ($sto 0$), which is not as brilliant as your answer is.
    $endgroup$
    – Zero
    Nov 28 '18 at 16:06












  • $begingroup$
    All I did was an adaptation of the proof of the maximum principle.
    $endgroup$
    – Julián Aguirre
    Nov 28 '18 at 17:33











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3017125%2fhow-to-prove-f-equiv-0-forall-x-in-a-b-quadf%25ef%25bc%258bpf%25ef%25bc%258bqf%25ef%25bc%259d0-with-fa%25ef%25bc%259d%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

The following is based on the classical proof of the maximum principle.



Let $L(f)=f''+p,f'+q,f$. If $L(f)>0$, then your argument shows that $f$ must be identically $0$.



But we have $L(f)=0$, not $>0$. What can we do? Take $M>0$ such that $M^2+M,p(x)+q(x)>0$ for all $xin[a,b]$ and let $epsilon>0$. Then
$$
L(f+epsilon,e^{Mx})=epsilon,e^{Mx}(M^2+M,p(x)+q(x))>0quadforall xin[a,b].
$$

Then
$$
max_{ale xle b}(f+epsilon,e^{Mx})=maxbigl(f(a)+epsilon,e^{Ma},f(b)+epsilon,e^{Mb}bigr)=epsilon,e^{Mb}.
$$

Letting $epsilonto0$ gives the desired result.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Your answer does help! I got lost in finding contradiction by analyzing $f(x_0 +s)$ ($sto 0$), which is not as brilliant as your answer is.
    $endgroup$
    – Zero
    Nov 28 '18 at 16:06












  • $begingroup$
    All I did was an adaptation of the proof of the maximum principle.
    $endgroup$
    – Julián Aguirre
    Nov 28 '18 at 17:33
















2












$begingroup$

The following is based on the classical proof of the maximum principle.



Let $L(f)=f''+p,f'+q,f$. If $L(f)>0$, then your argument shows that $f$ must be identically $0$.



But we have $L(f)=0$, not $>0$. What can we do? Take $M>0$ such that $M^2+M,p(x)+q(x)>0$ for all $xin[a,b]$ and let $epsilon>0$. Then
$$
L(f+epsilon,e^{Mx})=epsilon,e^{Mx}(M^2+M,p(x)+q(x))>0quadforall xin[a,b].
$$

Then
$$
max_{ale xle b}(f+epsilon,e^{Mx})=maxbigl(f(a)+epsilon,e^{Ma},f(b)+epsilon,e^{Mb}bigr)=epsilon,e^{Mb}.
$$

Letting $epsilonto0$ gives the desired result.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Your answer does help! I got lost in finding contradiction by analyzing $f(x_0 +s)$ ($sto 0$), which is not as brilliant as your answer is.
    $endgroup$
    – Zero
    Nov 28 '18 at 16:06












  • $begingroup$
    All I did was an adaptation of the proof of the maximum principle.
    $endgroup$
    – Julián Aguirre
    Nov 28 '18 at 17:33














2












2








2





$begingroup$

The following is based on the classical proof of the maximum principle.



Let $L(f)=f''+p,f'+q,f$. If $L(f)>0$, then your argument shows that $f$ must be identically $0$.



But we have $L(f)=0$, not $>0$. What can we do? Take $M>0$ such that $M^2+M,p(x)+q(x)>0$ for all $xin[a,b]$ and let $epsilon>0$. Then
$$
L(f+epsilon,e^{Mx})=epsilon,e^{Mx}(M^2+M,p(x)+q(x))>0quadforall xin[a,b].
$$

Then
$$
max_{ale xle b}(f+epsilon,e^{Mx})=maxbigl(f(a)+epsilon,e^{Ma},f(b)+epsilon,e^{Mb}bigr)=epsilon,e^{Mb}.
$$

Letting $epsilonto0$ gives the desired result.






share|cite|improve this answer











$endgroup$



The following is based on the classical proof of the maximum principle.



Let $L(f)=f''+p,f'+q,f$. If $L(f)>0$, then your argument shows that $f$ must be identically $0$.



But we have $L(f)=0$, not $>0$. What can we do? Take $M>0$ such that $M^2+M,p(x)+q(x)>0$ for all $xin[a,b]$ and let $epsilon>0$. Then
$$
L(f+epsilon,e^{Mx})=epsilon,e^{Mx}(M^2+M,p(x)+q(x))>0quadforall xin[a,b].
$$

Then
$$
max_{ale xle b}(f+epsilon,e^{Mx})=maxbigl(f(a)+epsilon,e^{Ma},f(b)+epsilon,e^{Mb}bigr)=epsilon,e^{Mb}.
$$

Letting $epsilonto0$ gives the desired result.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 28 '18 at 17:33

























answered Nov 28 '18 at 14:52









Julián AguirreJulián Aguirre

67.9k24094




67.9k24094












  • $begingroup$
    Your answer does help! I got lost in finding contradiction by analyzing $f(x_0 +s)$ ($sto 0$), which is not as brilliant as your answer is.
    $endgroup$
    – Zero
    Nov 28 '18 at 16:06












  • $begingroup$
    All I did was an adaptation of the proof of the maximum principle.
    $endgroup$
    – Julián Aguirre
    Nov 28 '18 at 17:33


















  • $begingroup$
    Your answer does help! I got lost in finding contradiction by analyzing $f(x_0 +s)$ ($sto 0$), which is not as brilliant as your answer is.
    $endgroup$
    – Zero
    Nov 28 '18 at 16:06












  • $begingroup$
    All I did was an adaptation of the proof of the maximum principle.
    $endgroup$
    – Julián Aguirre
    Nov 28 '18 at 17:33
















$begingroup$
Your answer does help! I got lost in finding contradiction by analyzing $f(x_0 +s)$ ($sto 0$), which is not as brilliant as your answer is.
$endgroup$
– Zero
Nov 28 '18 at 16:06






$begingroup$
Your answer does help! I got lost in finding contradiction by analyzing $f(x_0 +s)$ ($sto 0$), which is not as brilliant as your answer is.
$endgroup$
– Zero
Nov 28 '18 at 16:06














$begingroup$
All I did was an adaptation of the proof of the maximum principle.
$endgroup$
– Julián Aguirre
Nov 28 '18 at 17:33




$begingroup$
All I did was an adaptation of the proof of the maximum principle.
$endgroup$
– Julián Aguirre
Nov 28 '18 at 17:33


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3017125%2fhow-to-prove-f-equiv-0-forall-x-in-a-b-quadf%25ef%25bc%258bpf%25ef%25bc%258bqf%25ef%25bc%259d0-with-fa%25ef%25bc%259d%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Plaza Victoria

Puebla de Zaragoza

Musa