How to prove $fequiv 0$ $forall xin [a,b]$?$quad$($f''＋pf'＋qf＝0$ with $f(a)＝f(b)＝0$)
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
edited Nov 28 at 17:57
rafa11111
1,124417
asked Nov 28 at 13:12
Zero
1929
add a comment |
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
edited Nov 28 at 17:57
rafa11111
1,124417
asked Nov 28 at 13:12
Zero
1929
Maybe it is meant that $f in C^2[a,b]$ ?
– Rebellos
Nov 28 at 13:20
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.
– Cameron Williams
Nov 28 at 13:20
@Rebellos Thanks for reminding me. I've modified it
– Zero
Nov 28 at 13:22
3
This is a duplicate of math.stackexchange.com/q/3016693.
– Paul Frost
Nov 28 at 14:14
Possible duplicate of $f''+pf'+qf=0$ where $qleq0$ and $f(0)=f(1)=0$ prove $f=0$ ($f$, $p$, $q$ defined on $[0,1]$)
– Paul Frost
Nov 30 at 11:30
add a comment |
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
edited Nov 28 at 17:57
rafa11111
1,124417
asked Nov 28 at 13:12
Zero
1929
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
edited Nov 28 at 17:57
rafa11111
1,124417
asked Nov 28 at 13:12
Zero
1929
edited Nov 28 at 17:57
rafa11111
1,124417
asked Nov 28 at 13:12
Zero
1929
edited Nov 28 at 17:57
rafa11111
1,124417
edited Nov 28 at 17:57
rafa11111
1,124417
edited Nov 28 at 17:57
rafa11111
1,124417
1,124417
asked Nov 28 at 13:12
Zero
1929
asked Nov 28 at 13:12
Zero
1929
asked Nov 28 at 13:12
Zero
1929
1929
Maybe it is meant that $f in C^2[a,b]$ ?
– Rebellos
Nov 28 at 13:20
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.
– Cameron Williams
Nov 28 at 13:20
@Rebellos Thanks for reminding me. I've modified it
– Zero
Nov 28 at 13:22
3
This is a duplicate of math.stackexchange.com/q/3016693.
– Paul Frost
Nov 28 at 14:14
Possible duplicate of $f''+pf'+qf=0$ where $qleq0$ and $f(0)=f(1)=0$ prove $f=0$ ($f$, $p$, $q$ defined on $[0,1]$)
– Paul Frost
Nov 30 at 11:30
add a comment |
Maybe it is meant that $f in C^2[a,b]$ ?
– Rebellos
Nov 28 at 13:20
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.
– Cameron Williams
Nov 28 at 13:20
@Rebellos Thanks for reminding me. I've modified it
– Zero
Nov 28 at 13:22
3
This is a duplicate of math.stackexchange.com/q/3016693.
– Paul Frost
Nov 28 at 14:14
Possible duplicate of $f''+pf'+qf=0$ where $qleq0$ and $f(0)=f(1)=0$ prove $f=0$ ($f$, $p$, $q$ defined on $[0,1]$)
– Paul Frost
Nov 30 at 11:30
Maybe it is meant that $f in C^2[a,b]$ ?
– Rebellos
Nov 28 at 13:20
Maybe it is meant that $f in C^2[a,b]$ ?
– Rebellos
Nov 28 at 13:20
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.
– Cameron Williams
Nov 28 at 13:20
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.
– Cameron Williams
Nov 28 at 13:20
@Rebellos Thanks for reminding me. I've modified it
– Zero
Nov 28 at 13:22
@Rebellos Thanks for reminding me. I've modified it
– Zero
Nov 28 at 13:22
3
3
This is a duplicate of math.stackexchange.com/q/3016693.
– Paul Frost
Nov 28 at 14:14
This is a duplicate of math.stackexchange.com/q/3016693.
– Paul Frost
Nov 28 at 14:14
Possible duplicate of $f''+pf'+qf=0$ where $qleq0$ and $f(0)=f(1)=0$ prove $f=0$ ($f$, $p$, $q$ defined on $[0,1]$)
– Paul Frost
Nov 30 at 11:30
Possible duplicate of $f''+pf'+qf=0$ where $qleq0$ and $f(0)=f(1)=0$ prove $f=0$ ($f$, $p$, $q$ defined on $[0,1]$)
– Paul Frost
Nov 30 at 11:30
add a comment |
1 Answer
1
active
oldest
votes
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.
answered Nov 28 at 14:52
Julián Aguirre
67.4k24094
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.
– Zero
Nov 28 at 16:06
All I did was an adaptation of the proof of the maximum principle.
– Julián Aguirre
Nov 28 at 17:33
add a comment |
Your Answer
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1 Answer
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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.
answered Nov 28 at 14:52
Julián Aguirre
67.4k24094
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.
– Zero
Nov 28 at 16:06
All I did was an adaptation of the proof of the maximum principle.
– Julián Aguirre
Nov 28 at 17:33
add a comment |
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.
answered Nov 28 at 14:52
Julián Aguirre
67.4k24094
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.
– Zero
Nov 28 at 16:06
All I did was an adaptation of the proof of the maximum principle.
– Julián Aguirre
Nov 28 at 17:33
add a comment |
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.
answered Nov 28 at 14:52
Julián Aguirre
67.4k24094
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.
answered Nov 28 at 14:52
Julián Aguirre
67.4k24094
edited Nov 28 at 17:33
answered Nov 28 at 14:52
Julián Aguirre
67.4k24094
answered Nov 28 at 14:52
Julián Aguirre
67.4k24094
answered Nov 28 at 14:52
Julián Aguirre
67.4k24094
67.4k24094
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.
– Zero
Nov 28 at 16:06
All I did was an adaptation of the proof of the maximum principle.
– Julián Aguirre
Nov 28 at 17:33
add a comment |
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.
– Zero
Nov 28 at 16:06
All I did was an adaptation of the proof of the maximum principle.
– Julián Aguirre
Nov 28 at 17:33
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.
– Zero
Nov 28 at 16:06
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.
– Zero
Nov 28 at 16:06
All I did was an adaptation of the proof of the maximum principle.
– Julián Aguirre
Nov 28 at 17:33
All I did was an adaptation of the proof of the maximum principle.
– Julián Aguirre
Nov 28 at 17:33
add a comment |
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Maybe it is meant that $f in C^2[a,b]$ ?
– Rebellos
Nov 28 at 13:20
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.
– Cameron Williams
Nov 28 at 13:20
@Rebellos Thanks for reminding me. I've modified it
– Zero
Nov 28 at 13:22
3
This is a duplicate of math.stackexchange.com/q/3016693.
– Paul Frost
Nov 28 at 14:14
Possible duplicate of $f''+pf'+qf=0$ where $qleq0$ and $f(0)=f(1)=0$ prove $f=0$ ($f$, $p$, $q$ defined on $[0,1]$)
– Paul Frost
Nov 30 at 11:30