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!






real-analysis differential-equations continuity







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!






real-analysis differential-equations continuity







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!






real-analysis differential-equations continuity







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




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











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1 Answer
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active

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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


















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