$f''+pf'+qf=0$ where $qleq0$ and $f(0)=f(1)=0$ prove $f=0$ ($f$, $p$, $q$ defined on $[0,1]$)
For $f$ defined on $[0,1]$ twice differentiable, and two continuous function defined on $[0,1]$ named $p(x)$, $q(x)$, satisfying
$f''+pf'+qf=0$, where $qleq0$ and $f(0)=f(1)=0$,
prove $f=0$.
I haven't got an idea about the question. Any hint will be appreciated.
Edited: for $q(x_0)<0$ where $f'(x_0)=0$ the problem can easily be solved.
So now I'm mainly concerned about $q(x_0)=0$. It appears on my textbook, so i guess it's probably true. Counter examples are also welcomed.
real-analysis differential-equations
edited Nov 28 '18 at 18:44
LutzL
56.3k42054
asked Nov 28 '18 at 3:57
Oolong milk tea
737
add a comment |
For $f$ defined on $[0,1]$ twice differentiable, and two continuous function defined on $[0,1]$ named $p(x)$, $q(x)$, satisfying
$f''+pf'+qf=0$, where $qleq0$ and $f(0)=f(1)=0$,
prove $f=0$.
I haven't got an idea about the question. Any hint will be appreciated.
Edited: for $q(x_0)<0$ where $f'(x_0)=0$ the problem can easily be solved.
So now I'm mainly concerned about $q(x_0)=0$. It appears on my textbook, so i guess it's probably true. Counter examples are also welcomed.
real-analysis differential-equations
edited Nov 28 '18 at 18:44
LutzL
56.3k42054
asked Nov 28 '18 at 3:57
Oolong milk tea
737
1
Are $p,q$ constants or functions?
– edm
Nov 28 '18 at 5:33
1
For strictly negative $q$ one can use the same argument as in math.stackexchange.com/q/2998648/42969.
– Martin R
Nov 28 '18 at 6:12
@edm functions,sorry for my unclear statement
– Oolong milk tea
Nov 28 '18 at 9:30
add a comment |
For $f$ defined on $[0,1]$ twice differentiable, and two continuous function defined on $[0,1]$ named $p(x)$, $q(x)$, satisfying
$f''+pf'+qf=0$, where $qleq0$ and $f(0)=f(1)=0$,
prove $f=0$.
I haven't got an idea about the question. Any hint will be appreciated.
Edited: for $q(x_0)<0$ where $f'(x_0)=0$ the problem can easily be solved.
So now I'm mainly concerned about $q(x_0)=0$. It appears on my textbook, so i guess it's probably true. Counter examples are also welcomed.
real-analysis differential-equations
edited Nov 28 '18 at 18:44
LutzL
56.3k42054
asked Nov 28 '18 at 3:57
Oolong milk tea
737
For $f$ defined on $[0,1]$ twice differentiable, and two continuous function defined on $[0,1]$ named $p(x)$, $q(x)$, satisfying
$f''+pf'+qf=0$, where $qleq0$ and $f(0)=f(1)=0$,
prove $f=0$.
I haven't got an idea about the question. Any hint will be appreciated.
Edited: for $q(x_0)<0$ where $f'(x_0)=0$ the problem can easily be solved.
So now I'm mainly concerned about $q(x_0)=0$. It appears on my textbook, so i guess it's probably true. Counter examples are also welcomed.
real-analysis differential-equations
real-analysis differential-equations
edited Nov 28 '18 at 18:44
LutzL
56.3k42054
asked Nov 28 '18 at 3:57
Oolong milk tea
737
edited Nov 28 '18 at 18:44
LutzL
56.3k42054
asked Nov 28 '18 at 3:57
Oolong milk tea
737
edited Nov 28 '18 at 18:44
LutzL
56.3k42054
edited Nov 28 '18 at 18:44
LutzL
56.3k42054
edited Nov 28 '18 at 18:44
LutzL
56.3k42054
56.3k42054
asked Nov 28 '18 at 3:57
Oolong milk tea
737
asked Nov 28 '18 at 3:57
Oolong milk tea
737
asked Nov 28 '18 at 3:57
Oolong milk tea
737
737
1
Are $p,q$ constants or functions?
– edm
Nov 28 '18 at 5:33
1
For strictly negative $q$ one can use the same argument as in math.stackexchange.com/q/2998648/42969.
– Martin R
Nov 28 '18 at 6:12
@edm functions,sorry for my unclear statement
– Oolong milk tea
Nov 28 '18 at 9:30
add a comment |
1
Are $p,q$ constants or functions?
– edm
Nov 28 '18 at 5:33
1
For strictly negative $q$ one can use the same argument as in math.stackexchange.com/q/2998648/42969.
– Martin R
Nov 28 '18 at 6:12
@edm functions,sorry for my unclear statement
– Oolong milk tea
Nov 28 '18 at 9:30
1
1
Are $p,q$ constants or functions?
– edm
Nov 28 '18 at 5:33
Are $p,q$ constants or functions?
– edm
Nov 28 '18 at 5:33
1
1
For strictly negative $q$ one can use the same argument as in math.stackexchange.com/q/2998648/42969.
– Martin R
Nov 28 '18 at 6:12
For strictly negative $q$ one can use the same argument as in math.stackexchange.com/q/2998648/42969.
– Martin R
Nov 28 '18 at 6:12
@edm functions,sorry for my unclear statement
– Oolong milk tea
Nov 28 '18 at 9:30
@edm functions,sorry for my unclear statement
– Oolong milk tea
Nov 28 '18 at 9:30
add a comment |
2 Answers
2
active
oldest
votes
As $f'(0)ne 0$ by the uniqueness theorem, we can set $f'(0)=1$, as the linearity allows arbitrary rescaling.
Then let $$a=min{sin(0,1]colon f(s)le0}$$ be the smallest positive root of $f$, such that $f(x)>0$ for $xin (0,a)$.
On this interval one has then,
begin{align}
f''(x)+p(x)f'(x)=(-q(x))f(x)ge 0,
end{align}
and with the anti-derivative $P(x)=int_0^xp(s)ds$,
begin{align}
(e^{P(x)}f'(x))'ge 0.
end{align}
Integrating twice results in
begin{align}
&e^{P(x)}f'(x)ge e^{P(0)}f'(0)=1\[1em]
text{ and }~~ &f(x)ge int_0^xe^{-P(s)} dsge e^{-max_{sin[0,1]} P(s)}x
end{align}
This is a contradiction, as $f(a)=0$ and $f(a)ge e^{-max_{sin[0,1]} P(s)}a>0$ can not be true at the same time.
This result can easily extended to the more general claim that:
Any solution of the given differential equation can have at most one root.
answered Nov 28 '18 at 13:54
LutzL
56.3k42054
A little problem: how do you make the conclusion $f'(0)>0$ by uniqueness theorem? Maybe we have fuction like $f＝e^{-frac{1}{x}}$ around $0$.
– Zero
Nov 29 '18 at 11:41
@Zero : Because then the initial values are $f(0)=0$ and $f'(0)=0$, and $f=0$ is a solution. The ODE is linear, satisfies thus the Lipschitz condition, so that uniqueness follows. Now take the assumed solution and divide by $f'(0)$ to get another solution with $f'(0)=1$.
– LutzL
Nov 29 '18 at 13:02
Thank you, I get what you mean now..
– Zero
Nov 29 '18 at 13:09
add a comment |
Your differential equation is a homogeneous second order ODE (I take it the coefficients are constant, it's not clear). That means that the characteristic polynomial should give us the possible solutions. The polynomial would be:
$$ r^{2} + p r + q = 0 $$
And the roots are, then:
$$ r_{1} = frac{-p + sqrt{p^{2} - 4q}}{2} $$
$$ r_{2} = frac{-p - sqrt{p^{2} - 4q}}{2} $$
Now, there are three different cases for the solutions of homogenous second order ODEs with constant coefficients, either the roots are both real and different, they are both real and the same, or they are both complex. But notice that $ p in mathbb{R} $ implies that $ p^{2} geq 0 $, and $q leq 0 $ of course implies that $-4q geq 0$. Then, the square root expression can only be a non-negative number, and there are your 2 remaining possible cases, so you'll just need to evaluate the general solution for either of them.
On one case, you have that both roots are equal. Think about what conditions are required for $p,q$ for that to be possible. And in the case that the roots are different, you will need to evaluate the values you are given in the general solution, of the form:
$$y(t) = c_{1} e^{r_{1}t} + c_{2} e^{r_{2}t}$$
answered Nov 28 '18 at 5:30
Joaquin C.
154
5
The $p$ and $q$ are functions, not constants.
– RRL
Nov 28 '18 at 5:40
add a comment |
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As $f'(0)ne 0$ by the uniqueness theorem, we can set $f'(0)=1$, as the linearity allows arbitrary rescaling.
Then let $$a=min{sin(0,1]colon f(s)le0}$$ be the smallest positive root of $f$, such that $f(x)>0$ for $xin (0,a)$.
On this interval one has then,
begin{align}
f''(x)+p(x)f'(x)=(-q(x))f(x)ge 0,
end{align}
and with the anti-derivative $P(x)=int_0^xp(s)ds$,
begin{align}
(e^{P(x)}f'(x))'ge 0.
end{align}
Integrating twice results in
begin{align}
&e^{P(x)}f'(x)ge e^{P(0)}f'(0)=1\[1em]
text{ and }~~ &f(x)ge int_0^xe^{-P(s)} dsge e^{-max_{sin[0,1]} P(s)}x
end{align}
This is a contradiction, as $f(a)=0$ and $f(a)ge e^{-max_{sin[0,1]} P(s)}a>0$ can not be true at the same time.
This result can easily extended to the more general claim that:
Any solution of the given differential equation can have at most one root.
answered Nov 28 '18 at 13:54
LutzL
56.3k42054
A little problem: how do you make the conclusion $f'(0)>0$ by uniqueness theorem? Maybe we have fuction like $f＝e^{-frac{1}{x}}$ around $0$.
– Zero
Nov 29 '18 at 11:41
@Zero : Because then the initial values are $f(0)=0$ and $f'(0)=0$, and $f=0$ is a solution. The ODE is linear, satisfies thus the Lipschitz condition, so that uniqueness follows. Now take the assumed solution and divide by $f'(0)$ to get another solution with $f'(0)=1$.
– LutzL
Nov 29 '18 at 13:02
Thank you, I get what you mean now..
– Zero
Nov 29 '18 at 13:09
add a comment |
As $f'(0)ne 0$ by the uniqueness theorem, we can set $f'(0)=1$, as the linearity allows arbitrary rescaling.
Then let $$a=min{sin(0,1]colon f(s)le0}$$ be the smallest positive root of $f$, such that $f(x)>0$ for $xin (0,a)$.
On this interval one has then,
begin{align}
f''(x)+p(x)f'(x)=(-q(x))f(x)ge 0,
end{align}
and with the anti-derivative $P(x)=int_0^xp(s)ds$,
begin{align}
(e^{P(x)}f'(x))'ge 0.
end{align}
Integrating twice results in
begin{align}
&e^{P(x)}f'(x)ge e^{P(0)}f'(0)=1\[1em]
text{ and }~~ &f(x)ge int_0^xe^{-P(s)} dsge e^{-max_{sin[0,1]} P(s)}x
end{align}
This is a contradiction, as $f(a)=0$ and $f(a)ge e^{-max_{sin[0,1]} P(s)}a>0$ can not be true at the same time.
This result can easily extended to the more general claim that:
Any solution of the given differential equation can have at most one root.
answered Nov 28 '18 at 13:54
LutzL
56.3k42054
A little problem: how do you make the conclusion $f'(0)>0$ by uniqueness theorem? Maybe we have fuction like $f＝e^{-frac{1}{x}}$ around $0$.
– Zero
Nov 29 '18 at 11:41
@Zero : Because then the initial values are $f(0)=0$ and $f'(0)=0$, and $f=0$ is a solution. The ODE is linear, satisfies thus the Lipschitz condition, so that uniqueness follows. Now take the assumed solution and divide by $f'(0)$ to get another solution with $f'(0)=1$.
– LutzL
Nov 29 '18 at 13:02
Thank you, I get what you mean now..
– Zero
Nov 29 '18 at 13:09
add a comment |
As $f'(0)ne 0$ by the uniqueness theorem, we can set $f'(0)=1$, as the linearity allows arbitrary rescaling.
Then let $$a=min{sin(0,1]colon f(s)le0}$$ be the smallest positive root of $f$, such that $f(x)>0$ for $xin (0,a)$.
On this interval one has then,
begin{align}
f''(x)+p(x)f'(x)=(-q(x))f(x)ge 0,
end{align}
and with the anti-derivative $P(x)=int_0^xp(s)ds$,
begin{align}
(e^{P(x)}f'(x))'ge 0.
end{align}
Integrating twice results in
begin{align}
&e^{P(x)}f'(x)ge e^{P(0)}f'(0)=1\[1em]
text{ and }~~ &f(x)ge int_0^xe^{-P(s)} dsge e^{-max_{sin[0,1]} P(s)}x
end{align}
This is a contradiction, as $f(a)=0$ and $f(a)ge e^{-max_{sin[0,1]} P(s)}a>0$ can not be true at the same time.
This result can easily extended to the more general claim that:
Any solution of the given differential equation can have at most one root.
answered Nov 28 '18 at 13:54
LutzL
56.3k42054
As $f'(0)ne 0$ by the uniqueness theorem, we can set $f'(0)=1$, as the linearity allows arbitrary rescaling.
Then let $$a=min{sin(0,1]colon f(s)le0}$$ be the smallest positive root of $f$, such that $f(x)>0$ for $xin (0,a)$.
On this interval one has then,
begin{align}
f''(x)+p(x)f'(x)=(-q(x))f(x)ge 0,
end{align}
and with the anti-derivative $P(x)=int_0^xp(s)ds$,
begin{align}
(e^{P(x)}f'(x))'ge 0.
end{align}
Integrating twice results in
begin{align}
&e^{P(x)}f'(x)ge e^{P(0)}f'(0)=1\[1em]
text{ and }~~ &f(x)ge int_0^xe^{-P(s)} dsge e^{-max_{sin[0,1]} P(s)}x
end{align}
This is a contradiction, as $f(a)=0$ and $f(a)ge e^{-max_{sin[0,1]} P(s)}a>0$ can not be true at the same time.
This result can easily extended to the more general claim that:
Any solution of the given differential equation can have at most one root.
answered Nov 28 '18 at 13:54
LutzL
56.3k42054
edited Nov 28 '18 at 19:10
answered Nov 28 '18 at 13:54
LutzL
56.3k42054
answered Nov 28 '18 at 13:54
LutzL
56.3k42054
answered Nov 28 '18 at 13:54
LutzL
56.3k42054
56.3k42054
A little problem: how do you make the conclusion $f'(0)>0$ by uniqueness theorem? Maybe we have fuction like $f＝e^{-frac{1}{x}}$ around $0$.
– Zero
Nov 29 '18 at 11:41
@Zero : Because then the initial values are $f(0)=0$ and $f'(0)=0$, and $f=0$ is a solution. The ODE is linear, satisfies thus the Lipschitz condition, so that uniqueness follows. Now take the assumed solution and divide by $f'(0)$ to get another solution with $f'(0)=1$.
– LutzL
Nov 29 '18 at 13:02
Thank you, I get what you mean now..
– Zero
Nov 29 '18 at 13:09
add a comment |
A little problem: how do you make the conclusion $f'(0)>0$ by uniqueness theorem? Maybe we have fuction like $f＝e^{-frac{1}{x}}$ around $0$.
– Zero
Nov 29 '18 at 11:41
@Zero : Because then the initial values are $f(0)=0$ and $f'(0)=0$, and $f=0$ is a solution. The ODE is linear, satisfies thus the Lipschitz condition, so that uniqueness follows. Now take the assumed solution and divide by $f'(0)$ to get another solution with $f'(0)=1$.
– LutzL
Nov 29 '18 at 13:02
Thank you, I get what you mean now..
– Zero
Nov 29 '18 at 13:09
A little problem: how do you make the conclusion $f'(0)>0$ by uniqueness theorem? Maybe we have fuction like $f＝e^{-frac{1}{x}}$ around $0$.
– Zero
Nov 29 '18 at 11:41
A little problem: how do you make the conclusion $f'(0)>0$ by uniqueness theorem? Maybe we have fuction like $f＝e^{-frac{1}{x}}$ around $0$.
– Zero
Nov 29 '18 at 11:41
@Zero : Because then the initial values are $f(0)=0$ and $f'(0)=0$, and $f=0$ is a solution. The ODE is linear, satisfies thus the Lipschitz condition, so that uniqueness follows. Now take the assumed solution and divide by $f'(0)$ to get another solution with $f'(0)=1$.
– LutzL
Nov 29 '18 at 13:02
@Zero : Because then the initial values are $f(0)=0$ and $f'(0)=0$, and $f=0$ is a solution. The ODE is linear, satisfies thus the Lipschitz condition, so that uniqueness follows. Now take the assumed solution and divide by $f'(0)$ to get another solution with $f'(0)=1$.
– LutzL
Nov 29 '18 at 13:02
Thank you, I get what you mean now..
– Zero
Nov 29 '18 at 13:09
Thank you, I get what you mean now..
– Zero
Nov 29 '18 at 13:09
add a comment |
Your differential equation is a homogeneous second order ODE (I take it the coefficients are constant, it's not clear). That means that the characteristic polynomial should give us the possible solutions. The polynomial would be:
$$ r^{2} + p r + q = 0 $$
And the roots are, then:
$$ r_{1} = frac{-p + sqrt{p^{2} - 4q}}{2} $$
$$ r_{2} = frac{-p - sqrt{p^{2} - 4q}}{2} $$
Now, there are three different cases for the solutions of homogenous second order ODEs with constant coefficients, either the roots are both real and different, they are both real and the same, or they are both complex. But notice that $ p in mathbb{R} $ implies that $ p^{2} geq 0 $, and $q leq 0 $ of course implies that $-4q geq 0$. Then, the square root expression can only be a non-negative number, and there are your 2 remaining possible cases, so you'll just need to evaluate the general solution for either of them.
On one case, you have that both roots are equal. Think about what conditions are required for $p,q$ for that to be possible. And in the case that the roots are different, you will need to evaluate the values you are given in the general solution, of the form:
$$y(t) = c_{1} e^{r_{1}t} + c_{2} e^{r_{2}t}$$
answered Nov 28 '18 at 5:30
Joaquin C.
154
5
The $p$ and $q$ are functions, not constants.
– RRL
Nov 28 '18 at 5:40
add a comment |
Your differential equation is a homogeneous second order ODE (I take it the coefficients are constant, it's not clear). That means that the characteristic polynomial should give us the possible solutions. The polynomial would be:
$$ r^{2} + p r + q = 0 $$
And the roots are, then:
$$ r_{1} = frac{-p + sqrt{p^{2} - 4q}}{2} $$
$$ r_{2} = frac{-p - sqrt{p^{2} - 4q}}{2} $$
Now, there are three different cases for the solutions of homogenous second order ODEs with constant coefficients, either the roots are both real and different, they are both real and the same, or they are both complex. But notice that $ p in mathbb{R} $ implies that $ p^{2} geq 0 $, and $q leq 0 $ of course implies that $-4q geq 0$. Then, the square root expression can only be a non-negative number, and there are your 2 remaining possible cases, so you'll just need to evaluate the general solution for either of them.
On one case, you have that both roots are equal. Think about what conditions are required for $p,q$ for that to be possible. And in the case that the roots are different, you will need to evaluate the values you are given in the general solution, of the form:
$$y(t) = c_{1} e^{r_{1}t} + c_{2} e^{r_{2}t}$$
answered Nov 28 '18 at 5:30
Joaquin C.
154
5
The $p$ and $q$ are functions, not constants.
– RRL
Nov 28 '18 at 5:40
add a comment |
Your differential equation is a homogeneous second order ODE (I take it the coefficients are constant, it's not clear). That means that the characteristic polynomial should give us the possible solutions. The polynomial would be:
$$ r^{2} + p r + q = 0 $$
And the roots are, then:
$$ r_{1} = frac{-p + sqrt{p^{2} - 4q}}{2} $$
$$ r_{2} = frac{-p - sqrt{p^{2} - 4q}}{2} $$
Now, there are three different cases for the solutions of homogenous second order ODEs with constant coefficients, either the roots are both real and different, they are both real and the same, or they are both complex. But notice that $ p in mathbb{R} $ implies that $ p^{2} geq 0 $, and $q leq 0 $ of course implies that $-4q geq 0$. Then, the square root expression can only be a non-negative number, and there are your 2 remaining possible cases, so you'll just need to evaluate the general solution for either of them.
On one case, you have that both roots are equal. Think about what conditions are required for $p,q$ for that to be possible. And in the case that the roots are different, you will need to evaluate the values you are given in the general solution, of the form:
$$y(t) = c_{1} e^{r_{1}t} + c_{2} e^{r_{2}t}$$
answered Nov 28 '18 at 5:30
Joaquin C.
154
Your differential equation is a homogeneous second order ODE (I take it the coefficients are constant, it's not clear). That means that the characteristic polynomial should give us the possible solutions. The polynomial would be:
$$ r^{2} + p r + q = 0 $$
And the roots are, then:
$$ r_{1} = frac{-p + sqrt{p^{2} - 4q}}{2} $$
$$ r_{2} = frac{-p - sqrt{p^{2} - 4q}}{2} $$
Now, there are three different cases for the solutions of homogenous second order ODEs with constant coefficients, either the roots are both real and different, they are both real and the same, or they are both complex. But notice that $ p in mathbb{R} $ implies that $ p^{2} geq 0 $, and $q leq 0 $ of course implies that $-4q geq 0$. Then, the square root expression can only be a non-negative number, and there are your 2 remaining possible cases, so you'll just need to evaluate the general solution for either of them.
On one case, you have that both roots are equal. Think about what conditions are required for $p,q$ for that to be possible. And in the case that the roots are different, you will need to evaluate the values you are given in the general solution, of the form:
$$y(t) = c_{1} e^{r_{1}t} + c_{2} e^{r_{2}t}$$
answered Nov 28 '18 at 5:30
Joaquin C.
154
edited Nov 28 '18 at 5:37
answered Nov 28 '18 at 5:30
Joaquin C.
154
answered Nov 28 '18 at 5:30
Joaquin C.
154
answered Nov 28 '18 at 5:30
Joaquin C.
154
154
5
The $p$ and $q$ are functions, not constants.
– RRL
Nov 28 '18 at 5:40
add a comment |
5
The $p$ and $q$ are functions, not constants.
– RRL
Nov 28 '18 at 5:40
5
5
The $p$ and $q$ are functions, not constants.
– RRL
Nov 28 '18 at 5:40
The $p$ and $q$ are functions, not constants.
– RRL
Nov 28 '18 at 5:40
add a comment |
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1
Are $p,q$ constants or functions?
– edm
Nov 28 '18 at 5:33
1
For strictly negative $q$ one can use the same argument as in math.stackexchange.com/q/2998648/42969.
– Martin R
Nov 28 '18 at 6:12
@edm functions,sorry for my unclear statement
– Oolong milk tea
Nov 28 '18 at 9:30