Mixing
If $f(x)=int_{x-1}^x f(s)ds$, is $f$ constant? Periodic?
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I was thinking of periodic functions, and in particular the following type of condition:
If a function $f:mathbb{R}tomathbb{R}$ always "tends to its average", then it should be periodic.
To make things more formal, by "tending to the average" we could say something like $f(x)=int_{x-1}^x f(s)ds$. This is only the average depending on a previous time interval of length $1$, but it seems an interesting enough property. However, the only type of functions which I could find that satisfies this property are the constant ones!
Question: If $f:mathbb{R}tomathbb{R}$ is continuous (or more generaly measurable) and $f(x)=int_{x-1}^x f(s)ds$ for (almost) every $xinmathbb{R}$, then is $f$ constant (a.e.)? Periodic (a.e.)?
Here is a first try for $C^1$ functions (see edit below!): If $f$ is $C^1$ and $x$ is fixed, we can use Taylor expansion $f(s)=f(x)+O(s-x)$ (and similarly for $x-1$) to obtain
begin{align*}
f(x+t)-f(x)&=int_x^{x+t}f(s)ds-int_{x-1}^{x-1+t}f(s)ds\
&=int_x^{x+t}f(x)+O(s-x)ds-int_{x-1}^{x-1+t}f(x-1)+O(s-x+1)ds\
&=t(f(x)-f(x-1))+O(t^2),
end{align*}
so $f'(x)=f(x)-f(x-1)$. This is obviously true if $f$ is constant, but the converse is not clear to me at the moment.
Edit: From a comment and answer below, the equation $f'(x)=f(x)-f(x-1)$ has non-periodic solutions on $mathbb{R}setminusmathbb{Z}$, so this should not be the way to go for $C^1$ functions. However, even in this case it is not clear that any solution of this equation will satisfy $f(x)=int_{x-1}^x f(s)ds$, which is the question: All I can obtain, in principle, is $f(x)-f(x-1)=int_x^{x-1}f(s)ds-int_{x-2}^{x-1}f(s)ds$.
calculus integration recreational-mathematics average
asked Jan 7 at 8:36
QuestionerQuestioner
549321
$endgroup$
add a comment |
$begingroup$
I was thinking of periodic functions, and in particular the following type of condition:
If a function $f:mathbb{R}tomathbb{R}$ always "tends to its average", then it should be periodic.
To make things more formal, by "tending to the average" we could say something like $f(x)=int_{x-1}^x f(s)ds$. This is only the average depending on a previous time interval of length $1$, but it seems an interesting enough property. However, the only type of functions which I could find that satisfies this property are the constant ones!
Question: If $f:mathbb{R}tomathbb{R}$ is continuous (or more generaly measurable) and $f(x)=int_{x-1}^x f(s)ds$ for (almost) every $xinmathbb{R}$, then is $f$ constant (a.e.)? Periodic (a.e.)?
Here is a first try for $C^1$ functions (see edit below!): If $f$ is $C^1$ and $x$ is fixed, we can use Taylor expansion $f(s)=f(x)+O(s-x)$ (and similarly for $x-1$) to obtain
begin{align*}
f(x+t)-f(x)&=int_x^{x+t}f(s)ds-int_{x-1}^{x-1+t}f(s)ds\
&=int_x^{x+t}f(x)+O(s-x)ds-int_{x-1}^{x-1+t}f(x-1)+O(s-x+1)ds\
&=t(f(x)-f(x-1))+O(t^2),
end{align*}
so $f'(x)=f(x)-f(x-1)$. This is obviously true if $f$ is constant, but the converse is not clear to me at the moment.
Edit: From a comment and answer below, the equation $f'(x)=f(x)-f(x-1)$ has non-periodic solutions on $mathbb{R}setminusmathbb{Z}$, so this should not be the way to go for $C^1$ functions. However, even in this case it is not clear that any solution of this equation will satisfy $f(x)=int_{x-1}^x f(s)ds$, which is the question: All I can obtain, in principle, is $f(x)-f(x-1)=int_x^{x-1}f(s)ds-int_{x-2}^{x-1}f(s)ds$.
calculus integration recreational-mathematics average
asked Jan 7 at 8:36
QuestionerQuestioner
549321
$endgroup$
1
$begingroup$
This is a delay differential equation. Not all solutions are constant or periodic.
$endgroup$
– Dylan
Jan 7 at 9:06
add a comment |
$begingroup$
I was thinking of periodic functions, and in particular the following type of condition:
If a function $f:mathbb{R}tomathbb{R}$ always "tends to its average", then it should be periodic.
To make things more formal, by "tending to the average" we could say something like $f(x)=int_{x-1}^x f(s)ds$. This is only the average depending on a previous time interval of length $1$, but it seems an interesting enough property. However, the only type of functions which I could find that satisfies this property are the constant ones!
Question: If $f:mathbb{R}tomathbb{R}$ is continuous (or more generaly measurable) and $f(x)=int_{x-1}^x f(s)ds$ for (almost) every $xinmathbb{R}$, then is $f$ constant (a.e.)? Periodic (a.e.)?
Here is a first try for $C^1$ functions (see edit below!): If $f$ is $C^1$ and $x$ is fixed, we can use Taylor expansion $f(s)=f(x)+O(s-x)$ (and similarly for $x-1$) to obtain
begin{align*}
f(x+t)-f(x)&=int_x^{x+t}f(s)ds-int_{x-1}^{x-1+t}f(s)ds\
&=int_x^{x+t}f(x)+O(s-x)ds-int_{x-1}^{x-1+t}f(x-1)+O(s-x+1)ds\
&=t(f(x)-f(x-1))+O(t^2),
end{align*}
so $f'(x)=f(x)-f(x-1)$. This is obviously true if $f$ is constant, but the converse is not clear to me at the moment.
Edit: From a comment and answer below, the equation $f'(x)=f(x)-f(x-1)$ has non-periodic solutions on $mathbb{R}setminusmathbb{Z}$, so this should not be the way to go for $C^1$ functions. However, even in this case it is not clear that any solution of this equation will satisfy $f(x)=int_{x-1}^x f(s)ds$, which is the question: All I can obtain, in principle, is $f(x)-f(x-1)=int_x^{x-1}f(s)ds-int_{x-2}^{x-1}f(s)ds$.
calculus integration recreational-mathematics average
asked Jan 7 at 8:36
QuestionerQuestioner
549321
$endgroup$
I was thinking of periodic functions, and in particular the following type of condition:
If a function $f:mathbb{R}tomathbb{R}$ always "tends to its average", then it should be periodic.
To make things more formal, by "tending to the average" we could say something like $f(x)=int_{x-1}^x f(s)ds$. This is only the average depending on a previous time interval of length $1$, but it seems an interesting enough property. However, the only type of functions which I could find that satisfies this property are the constant ones!
Question: If $f:mathbb{R}tomathbb{R}$ is continuous (or more generaly measurable) and $f(x)=int_{x-1}^x f(s)ds$ for (almost) every $xinmathbb{R}$, then is $f$ constant (a.e.)? Periodic (a.e.)?
Here is a first try for $C^1$ functions (see edit below!): If $f$ is $C^1$ and $x$ is fixed, we can use Taylor expansion $f(s)=f(x)+O(s-x)$ (and similarly for $x-1$) to obtain
begin{align*}
f(x+t)-f(x)&=int_x^{x+t}f(s)ds-int_{x-1}^{x-1+t}f(s)ds\
&=int_x^{x+t}f(x)+O(s-x)ds-int_{x-1}^{x-1+t}f(x-1)+O(s-x+1)ds\
&=t(f(x)-f(x-1))+O(t^2),
end{align*}
so $f'(x)=f(x)-f(x-1)$. This is obviously true if $f$ is constant, but the converse is not clear to me at the moment.
Edit: From a comment and answer below, the equation $f'(x)=f(x)-f(x-1)$ has non-periodic solutions on $mathbb{R}setminusmathbb{Z}$, so this should not be the way to go for $C^1$ functions. However, even in this case it is not clear that any solution of this equation will satisfy $f(x)=int_{x-1}^x f(s)ds$, which is the question: All I can obtain, in principle, is $f(x)-f(x-1)=int_x^{x-1}f(s)ds-int_{x-2}^{x-1}f(s)ds$.
calculus integration recreational-mathematics average
calculus integration recreational-mathematics average
asked Jan 7 at 8:36
QuestionerQuestioner
549321
asked Jan 7 at 8:36
QuestionerQuestioner
549321
edited Jan 7 at 9:15
Questioner
asked Jan 7 at 8:36
QuestionerQuestioner
549321
asked Jan 7 at 8:36
QuestionerQuestioner
549321
asked Jan 7 at 8:36
QuestionerQuestioner
549321
549321
1
$begingroup$
This is a delay differential equation. Not all solutions are constant or periodic.
$endgroup$
– Dylan
Jan 7 at 9:06
add a comment |
1
$begingroup$
This is a delay differential equation. Not all solutions are constant or periodic.
$endgroup$
– Dylan
Jan 7 at 9:06
1
1
$begingroup$
This is a delay differential equation. Not all solutions are constant or periodic.
$endgroup$
– Dylan
Jan 7 at 9:06
$begingroup$
This is a delay differential equation. Not all solutions are constant or periodic.
$endgroup$
– Dylan
Jan 7 at 9:06
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Here is a typical argument using characteristic equation:
Let $alpha in mathbb{C}setminus{0}$ solve the equation $alpha = 1-e^{-alpha}$. One can indeed prove that such solution exists. One such solution is numerically given by $-2.08884 + 7.46149 i$. Then
$$ int_{x-1}^{x} e^{alpha t} , mathrm{d}t = frac{1 - e^{-alpha}}{alpha} e^{alpha x} = e^{alpha x}, $$
hence $f(x) = e^{alpha x}$ is one (complex-valued) solution of the equation
$$ f(x) = int_{x-1}^{x} f(t) , mathrm{d}t tag{*}$$
If one is interested in real-valued solutions only, then one can consider both the real part and the imaginary part of $e^{alpha x}$. In particular, this tells that there exists an analytic solution of $text{(*)}$ which is neither constant nor having real-period.
Addendum. We prove the following claim:
Claim. There exists a non-zero solution of $alpha = 1 - e^{-alpha}$ in $mathbb{C}$.
Proof. We first note that $varphi(x) = x(1-log x)$ satisfies $varphi(0^+) = 0$ and $varphi(1) = 1$. Next, let $k$ be a positive integer. Then
There exists $y in (2kpi, (2k+frac{1}{2})pi)$ such that $ varphi(sin(y)/y) = cos (y) $, by the intermediate-value theorem.
Set $x = log(sin(y)/y)$.
We claim that $ alpha = x + iy $ solves the equation. Indeed, it is clear that $ e^{-x}sin y = y $ holds. Moreover,
$$ (1-x)e^x = varphi(sin(y)/y) = cos(y), $$
and so, $1 - x = e^{-x}cos(y)$. Combining altogether,
$$ 1 - alpha = 1 - x - iy = e^{-x}cos(y) - ie^{-x}sin(y) = e^{-x-iy} = e^{-alpha}. $$
Therefore the claim follows. ////
(A careful inespection shows that this construction produces all the solutions of $alpha = 1 - e^{-alpha}$ in the upper half-plane.)
answered Jan 12 at 6:40
Sangchul LeeSangchul Lee
92.6k12167268
$endgroup$
$begingroup$
I think you want $varphi(x)=x(1-log x)$ instead. It satisfies the same properties, and $$(1-x)e^x=(1-log(sin(y)/y))sin(y)/y.$$ Other than that, the real part $f(t)$ of $tmapsto e^{alpha t}$ will be $f(t)=e^{xt}cos(yt)$, a non-periodic solution of the original equation.
$endgroup$
– Questioner
Jan 16 at 11:59
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@Questioner, That's a nice catch! I will fix it :)
$endgroup$
– Sangchul Lee
Jan 19 at 10:22
add a comment |
$begingroup$
This is a partial answer where I provided some properties for $f$ that complies with
$$
f(x)=int_{x-1}^xf(s),{rm d}s
$$
for all $xinmathbb{R}$. I will update this thread as I make any further progress.
1. $fin C^{infty}(mathbb{R})$ if $fin L_{rm loc}^1(mathbb{R})$ ($f$ must be smooth if it is locally integrable).
Proof. $forall,x_0inmathbb{R}$, $forall,xge x_0+1$, we have
$$
f(x)=int_{x_0}^xf(s),{rm d}s-int_{x_0}^{x-1}f(s),{rm d}s.
$$
Since $fin L_{rm loc}^1(mathbb{R})$, it follows that
$$
int_{x_0}^xf(s),{rm d}squadtext{and}quadint_{x_0}^{x-1}f(s),{rm d}s
$$
are absolutely continuous. Thus $f$ is absolutely continuous on $left(x_0+1,inftyright)$. The arbitrariness of $x_0$ implies that $f$ is absolutely continuous on $mathbb{R}$. Thus necessarily, $fin C(mathbb{R})$.
Likewise, since $fin C(mathbb{R})$, it follows that
$$
int_{x_0}^xf(s),{rm d}squadtext{and}quadint_{x_0}^{x-1}f(s),{rm d}s
$$
are continuously differentiable, which leads to $fin C^1(mathbb{R})$.
Repeat the above reasoning inductively, and we eventually obtain $fin C^{infty}(mathbb{R})$.$#$
This conclusion suggests that, at least for a most general case, we shall only consider those $f$'s that are smooth on $mathbb{R}$.
2. $f=0$ if $fin L^1(mathbb{R})$ ($f$ must be zero if it is integrable on $mathbb{R}$).
Proof. Since $fin L^1(mathbb{R})$, it is obvious that
$$
int_{x-1}^xf(s),{rm d}s=int_{mathbb{R}}1_{left[0,1right]}(x-s)f(s),{rm d}s=left(1_{left[0,1right]}*fright)(x).
$$
Hence, the original relation is equivalent to the following convolution equation on $mathbb{R}$:
$$
f=1_{left[0,1right]}*f.
$$
Note that $fin L^1(mathbb{R})$, and its Fourier transform $hat{f}$ is well-defined. By the convolution theorem,
$$
hat{f}=widehat{1_{left[0,1right]}},hat{f}Longrightarrowleft(1-widehat{1_{left[0,1right]}}right)hat{f}=0.
$$
This implies that $hat{f}=hat{f}(xi)=0$ for all $xine 0$. Besides, the continuity of $hat{f}$ yields $hat{f}(0)=0$. Consequently, we have
$$
hat{f}=0iff f=0.#
$$
This conclusion suggests that any non-trivial solution to the original equation must be non-integrable on $mathbb{R}$, e.g., non-zero constants. Nevertheless, given that it is reasonable to assume $f$ to be locally integrable, the original equation can always be formulated in the convolution form $f=1_{left[0,1right]}*f$. Just note that $hat{f}$ is not defined and the convolution theorem no longer applies if $f$ is locally integrable but not integrable on $mathbb{R}$.
3. $fequivtext{const}$ if $fin L_{rm loc}^1(mathbb{R})cap C_T(mathbb{R})$ ($f$ must be constant if it is locally integrable and $T$-periodic).
Proof. Since $fin L_{rm loc}^1(mathbb{R})$, we have $fin C^{infty}(left[0,Tright])$. Thanks to the periodicity, $f$ observes its Fourier series on $left[0,Tright]$
$$
f(x)simsum_{ninmathbb{Z}}a_n,e^{frac{2ipi nx}{T}}.
$$
Since $fin C^{infty}(left[0,Tright])$, the Fourier series of $f$ converges absolutely and uniformly to $f$. Therefore, we have for one thing,
$$
f(x)=sum_{ninmathbb{Z}}a_n,e^{frac{2ipi nx}{T}}.
$$
For another,
begin{align}
int_{x-1}^xf(s),{rm d}s&=int_{x-1}^xsum_{ninmathbb{Z}}a_n,e^{frac{2ipi ns}{T}}{rm d}s\
&=sum_{ninmathbb{Z}}a_nint_{x-1}^xe^{frac{2ipi ns}{T}}{rm d}s\
&=sum_{ninmathbb{Z}}a_nfrac{1-e^{-theta_n}}{theta_n}e^{frac{2ipi nx}{T}},
end{align}
where $theta_n=2ipi n/T$, and $left(1-e^{-theta_0}right)/theta_0=1$ (this is defined so that the form of the series preserves; otherwise, one may carry out the integration separately for $n=0$ and $nne 0$, and may find the results identical).
Thanks to this result, the original equation requires
$$
a_n=a_nfrac{1-e^{-theta_n}}{theta_n}iffleft(1-frac{1-e^{-theta_n}}{theta_n}right)a_n=0.
$$
Note that
$$
1-frac{1-e^{-z}}{z}=0
$$
yields only one solution on $imathbb{R}$ (the imaginary axis), i.e., $z=0$. Thus since $theta_nne 0$ for all $nne 0$, it is a must that $a_n=0$ for all $nne 0$. This leads to $f(x)=a_0$, i.e., $f$ is constant on $mathbb{R}$.$#$
This conclusion suggests that any continuous periodic function that satisfies the original equation must be constant.
[TBC...]
[Following @Empy2's answer, I believe the existence of some non-periodic solution to the original equation. Yet as per the above properties, this solution has to be smooth and most likely unbounded. Trying the polynomial series $f(x)=sum_{n=0}^{infty}a_nx^n$ could be promising, but it leads to an infinite dimensional linear system, and its convergence also remains unknown, which challenges the commutativity of summation and integration...]
answered Jan 12 at 4:17
hypernovahypernova
4,534313
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add a comment |
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Define $f(x)=3x^2-4x+1$ for $xin(0,1)$ so the differential equation is true for $x=1$.
For $xin(1,2)$, solve the differential equation
$$frac{df}{dx}=f(x)-(3(x-1)^2-4(x-1)+1)$$
Iterate the procedure, for $xin (2,3)$ and so on.
answered Jan 7 at 9:02
Empy2Empy2
33.5k12261
$endgroup$
$begingroup$
That's nice. However, how do we define $f(x)$ for $x<0$? Also, I have not proved yet that the equation $f'(x)=f(x)-f(x-1)$ implies that $f$ satisfies the property described above.
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– Questioner
Jan 7 at 9:09
$begingroup$
Nevermind the first part of the comment above: For $xin[-1,0]$ we have $f'(x+1)=f(x+1)-f(x)$, so this defines $f$ on $[-1,1]$ and we iterate...
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– Questioner
Jan 7 at 9:13
$begingroup$
@Questioner Note that$$f(1)=int_0^1f(t),mathrm dt=0,$$thus integrating $f'(t)=f(t)-f(t-1)$ for $tin[1,x]$ yields$$f(x)=int_1^xf(t),mathrm dt-int_0^{x-1}f(t),mathrm dt=int_{x-1}^xf(t),mathrm dt$$for $xgeqslant1$.
$endgroup$
– Saad
Jan 12 at 2:09
add a comment |
$begingroup$
Let
$$f(x) = dfrac{a_0}2+sumlimits_{n=1}^inftyleft(a_ncos2pi nx + a_nsin2pi nxright),$$
then
$$intlimits_{x-1}^x f(x),mathrm dx = dfrac {a_0}2,$$
so $f(x)$ is a constant.
answered Jan 12 at 0:10
Yuri NegometyanovYuri Negometyanov
11.1k1728
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add a comment |
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4 Answers
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4 Answers
4
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$begingroup$
Here is a typical argument using characteristic equation:
Let $alpha in mathbb{C}setminus{0}$ solve the equation $alpha = 1-e^{-alpha}$. One can indeed prove that such solution exists. One such solution is numerically given by $-2.08884 + 7.46149 i$. Then
$$ int_{x-1}^{x} e^{alpha t} , mathrm{d}t = frac{1 - e^{-alpha}}{alpha} e^{alpha x} = e^{alpha x}, $$
hence $f(x) = e^{alpha x}$ is one (complex-valued) solution of the equation
$$ f(x) = int_{x-1}^{x} f(t) , mathrm{d}t tag{*}$$
If one is interested in real-valued solutions only, then one can consider both the real part and the imaginary part of $e^{alpha x}$. In particular, this tells that there exists an analytic solution of $text{(*)}$ which is neither constant nor having real-period.
Addendum. We prove the following claim:
Claim. There exists a non-zero solution of $alpha = 1 - e^{-alpha}$ in $mathbb{C}$.
Proof. We first note that $varphi(x) = x(1-log x)$ satisfies $varphi(0^+) = 0$ and $varphi(1) = 1$. Next, let $k$ be a positive integer. Then
There exists $y in (2kpi, (2k+frac{1}{2})pi)$ such that $ varphi(sin(y)/y) = cos (y) $, by the intermediate-value theorem.
Set $x = log(sin(y)/y)$.
We claim that $ alpha = x + iy $ solves the equation. Indeed, it is clear that $ e^{-x}sin y = y $ holds. Moreover,
$$ (1-x)e^x = varphi(sin(y)/y) = cos(y), $$
and so, $1 - x = e^{-x}cos(y)$. Combining altogether,
$$ 1 - alpha = 1 - x - iy = e^{-x}cos(y) - ie^{-x}sin(y) = e^{-x-iy} = e^{-alpha}. $$
Therefore the claim follows. ////
(A careful inespection shows that this construction produces all the solutions of $alpha = 1 - e^{-alpha}$ in the upper half-plane.)
answered Jan 12 at 6:40
Sangchul LeeSangchul Lee
92.6k12167268
$endgroup$
$begingroup$
I think you want $varphi(x)=x(1-log x)$ instead. It satisfies the same properties, and $$(1-x)e^x=(1-log(sin(y)/y))sin(y)/y.$$ Other than that, the real part $f(t)$ of $tmapsto e^{alpha t}$ will be $f(t)=e^{xt}cos(yt)$, a non-periodic solution of the original equation.
$endgroup$
– Questioner
Jan 16 at 11:59
$begingroup$
@Questioner, That's a nice catch! I will fix it :)
$endgroup$
– Sangchul Lee
Jan 19 at 10:22
add a comment |
$begingroup$
Here is a typical argument using characteristic equation:
Let $alpha in mathbb{C}setminus{0}$ solve the equation $alpha = 1-e^{-alpha}$. One can indeed prove that such solution exists. One such solution is numerically given by $-2.08884 + 7.46149 i$. Then
$$ int_{x-1}^{x} e^{alpha t} , mathrm{d}t = frac{1 - e^{-alpha}}{alpha} e^{alpha x} = e^{alpha x}, $$
hence $f(x) = e^{alpha x}$ is one (complex-valued) solution of the equation
$$ f(x) = int_{x-1}^{x} f(t) , mathrm{d}t tag{*}$$
If one is interested in real-valued solutions only, then one can consider both the real part and the imaginary part of $e^{alpha x}$. In particular, this tells that there exists an analytic solution of $text{(*)}$ which is neither constant nor having real-period.
Addendum. We prove the following claim:
Claim. There exists a non-zero solution of $alpha = 1 - e^{-alpha}$ in $mathbb{C}$.
Proof. We first note that $varphi(x) = x(1-log x)$ satisfies $varphi(0^+) = 0$ and $varphi(1) = 1$. Next, let $k$ be a positive integer. Then
There exists $y in (2kpi, (2k+frac{1}{2})pi)$ such that $ varphi(sin(y)/y) = cos (y) $, by the intermediate-value theorem.
Set $x = log(sin(y)/y)$.
We claim that $ alpha = x + iy $ solves the equation. Indeed, it is clear that $ e^{-x}sin y = y $ holds. Moreover,
$$ (1-x)e^x = varphi(sin(y)/y) = cos(y), $$
and so, $1 - x = e^{-x}cos(y)$. Combining altogether,
$$ 1 - alpha = 1 - x - iy = e^{-x}cos(y) - ie^{-x}sin(y) = e^{-x-iy} = e^{-alpha}. $$
Therefore the claim follows. ////
(A careful inespection shows that this construction produces all the solutions of $alpha = 1 - e^{-alpha}$ in the upper half-plane.)
answered Jan 12 at 6:40
Sangchul LeeSangchul Lee
92.6k12167268
$endgroup$
$begingroup$
I think you want $varphi(x)=x(1-log x)$ instead. It satisfies the same properties, and $$(1-x)e^x=(1-log(sin(y)/y))sin(y)/y.$$ Other than that, the real part $f(t)$ of $tmapsto e^{alpha t}$ will be $f(t)=e^{xt}cos(yt)$, a non-periodic solution of the original equation.
$endgroup$
– Questioner
Jan 16 at 11:59
$begingroup$
@Questioner, That's a nice catch! I will fix it :)
$endgroup$
– Sangchul Lee
Jan 19 at 10:22
add a comment |
$begingroup$
Here is a typical argument using characteristic equation:
Let $alpha in mathbb{C}setminus{0}$ solve the equation $alpha = 1-e^{-alpha}$. One can indeed prove that such solution exists. One such solution is numerically given by $-2.08884 + 7.46149 i$. Then
$$ int_{x-1}^{x} e^{alpha t} , mathrm{d}t = frac{1 - e^{-alpha}}{alpha} e^{alpha x} = e^{alpha x}, $$
hence $f(x) = e^{alpha x}$ is one (complex-valued) solution of the equation
$$ f(x) = int_{x-1}^{x} f(t) , mathrm{d}t tag{*}$$
If one is interested in real-valued solutions only, then one can consider both the real part and the imaginary part of $e^{alpha x}$. In particular, this tells that there exists an analytic solution of $text{(*)}$ which is neither constant nor having real-period.
Addendum. We prove the following claim:
Claim. There exists a non-zero solution of $alpha = 1 - e^{-alpha}$ in $mathbb{C}$.
Proof. We first note that $varphi(x) = x(1-log x)$ satisfies $varphi(0^+) = 0$ and $varphi(1) = 1$. Next, let $k$ be a positive integer. Then
There exists $y in (2kpi, (2k+frac{1}{2})pi)$ such that $ varphi(sin(y)/y) = cos (y) $, by the intermediate-value theorem.
Set $x = log(sin(y)/y)$.
We claim that $ alpha = x + iy $ solves the equation. Indeed, it is clear that $ e^{-x}sin y = y $ holds. Moreover,
$$ (1-x)e^x = varphi(sin(y)/y) = cos(y), $$
and so, $1 - x = e^{-x}cos(y)$. Combining altogether,
$$ 1 - alpha = 1 - x - iy = e^{-x}cos(y) - ie^{-x}sin(y) = e^{-x-iy} = e^{-alpha}. $$
Therefore the claim follows. ////
(A careful inespection shows that this construction produces all the solutions of $alpha = 1 - e^{-alpha}$ in the upper half-plane.)
answered Jan 12 at 6:40
Sangchul LeeSangchul Lee
92.6k12167268
$endgroup$
Here is a typical argument using characteristic equation:
Let $alpha in mathbb{C}setminus{0}$ solve the equation $alpha = 1-e^{-alpha}$. One can indeed prove that such solution exists. One such solution is numerically given by $-2.08884 + 7.46149 i$. Then
$$ int_{x-1}^{x} e^{alpha t} , mathrm{d}t = frac{1 - e^{-alpha}}{alpha} e^{alpha x} = e^{alpha x}, $$
hence $f(x) = e^{alpha x}$ is one (complex-valued) solution of the equation
$$ f(x) = int_{x-1}^{x} f(t) , mathrm{d}t tag{*}$$
If one is interested in real-valued solutions only, then one can consider both the real part and the imaginary part of $e^{alpha x}$. In particular, this tells that there exists an analytic solution of $text{(*)}$ which is neither constant nor having real-period.
Addendum. We prove the following claim:
Claim. There exists a non-zero solution of $alpha = 1 - e^{-alpha}$ in $mathbb{C}$.
Proof. We first note that $varphi(x) = x(1-log x)$ satisfies $varphi(0^+) = 0$ and $varphi(1) = 1$. Next, let $k$ be a positive integer. Then
There exists $y in (2kpi, (2k+frac{1}{2})pi)$ such that $ varphi(sin(y)/y) = cos (y) $, by the intermediate-value theorem.
Set $x = log(sin(y)/y)$.
We claim that $ alpha = x + iy $ solves the equation. Indeed, it is clear that $ e^{-x}sin y = y $ holds. Moreover,
$$ (1-x)e^x = varphi(sin(y)/y) = cos(y), $$
and so, $1 - x = e^{-x}cos(y)$. Combining altogether,
$$ 1 - alpha = 1 - x - iy = e^{-x}cos(y) - ie^{-x}sin(y) = e^{-x-iy} = e^{-alpha}. $$
Therefore the claim follows. ////
(A careful inespection shows that this construction produces all the solutions of $alpha = 1 - e^{-alpha}$ in the upper half-plane.)
answered Jan 12 at 6:40
Sangchul LeeSangchul Lee
92.6k12167268
edited Jan 19 at 10:23
answered Jan 12 at 6:40
Sangchul LeeSangchul Lee
92.6k12167268
answered Jan 12 at 6:40
Sangchul LeeSangchul Lee
92.6k12167268
answered Jan 12 at 6:40
Sangchul LeeSangchul Lee
92.6k12167268
92.6k12167268
$begingroup$
I think you want $varphi(x)=x(1-log x)$ instead. It satisfies the same properties, and $$(1-x)e^x=(1-log(sin(y)/y))sin(y)/y.$$ Other than that, the real part $f(t)$ of $tmapsto e^{alpha t}$ will be $f(t)=e^{xt}cos(yt)$, a non-periodic solution of the original equation.
$endgroup$
– Questioner
Jan 16 at 11:59
$begingroup$
@Questioner, That's a nice catch! I will fix it :)
$endgroup$
– Sangchul Lee
Jan 19 at 10:22
add a comment |
$begingroup$
I think you want $varphi(x)=x(1-log x)$ instead. It satisfies the same properties, and $$(1-x)e^x=(1-log(sin(y)/y))sin(y)/y.$$ Other than that, the real part $f(t)$ of $tmapsto e^{alpha t}$ will be $f(t)=e^{xt}cos(yt)$, a non-periodic solution of the original equation.
$endgroup$
– Questioner
Jan 16 at 11:59
$begingroup$
@Questioner, That's a nice catch! I will fix it :)
$endgroup$
– Sangchul Lee
Jan 19 at 10:22
$begingroup$
I think you want $varphi(x)=x(1-log x)$ instead. It satisfies the same properties, and $$(1-x)e^x=(1-log(sin(y)/y))sin(y)/y.$$ Other than that, the real part $f(t)$ of $tmapsto e^{alpha t}$ will be $f(t)=e^{xt}cos(yt)$, a non-periodic solution of the original equation.
$endgroup$
– Questioner
Jan 16 at 11:59
$begingroup$
I think you want $varphi(x)=x(1-log x)$ instead. It satisfies the same properties, and $$(1-x)e^x=(1-log(sin(y)/y))sin(y)/y.$$ Other than that, the real part $f(t)$ of $tmapsto e^{alpha t}$ will be $f(t)=e^{xt}cos(yt)$, a non-periodic solution of the original equation.
$endgroup$
– Questioner
Jan 16 at 11:59
$begingroup$
@Questioner, That's a nice catch! I will fix it :)
$endgroup$
– Sangchul Lee
Jan 19 at 10:22
$begingroup$
@Questioner, That's a nice catch! I will fix it :)
$endgroup$
– Sangchul Lee
Jan 19 at 10:22
add a comment |
$begingroup$
This is a partial answer where I provided some properties for $f$ that complies with
$$
f(x)=int_{x-1}^xf(s),{rm d}s
$$
for all $xinmathbb{R}$. I will update this thread as I make any further progress.
1. $fin C^{infty}(mathbb{R})$ if $fin L_{rm loc}^1(mathbb{R})$ ($f$ must be smooth if it is locally integrable).
Proof. $forall,x_0inmathbb{R}$, $forall,xge x_0+1$, we have
$$
f(x)=int_{x_0}^xf(s),{rm d}s-int_{x_0}^{x-1}f(s),{rm d}s.
$$
Since $fin L_{rm loc}^1(mathbb{R})$, it follows that
$$
int_{x_0}^xf(s),{rm d}squadtext{and}quadint_{x_0}^{x-1}f(s),{rm d}s
$$
are absolutely continuous. Thus $f$ is absolutely continuous on $left(x_0+1,inftyright)$. The arbitrariness of $x_0$ implies that $f$ is absolutely continuous on $mathbb{R}$. Thus necessarily, $fin C(mathbb{R})$.
Likewise, since $fin C(mathbb{R})$, it follows that
$$
int_{x_0}^xf(s),{rm d}squadtext{and}quadint_{x_0}^{x-1}f(s),{rm d}s
$$
are continuously differentiable, which leads to $fin C^1(mathbb{R})$.
Repeat the above reasoning inductively, and we eventually obtain $fin C^{infty}(mathbb{R})$.$#$
This conclusion suggests that, at least for a most general case, we shall only consider those $f$'s that are smooth on $mathbb{R}$.
2. $f=0$ if $fin L^1(mathbb{R})$ ($f$ must be zero if it is integrable on $mathbb{R}$).
Proof. Since $fin L^1(mathbb{R})$, it is obvious that
$$
int_{x-1}^xf(s),{rm d}s=int_{mathbb{R}}1_{left[0,1right]}(x-s)f(s),{rm d}s=left(1_{left[0,1right]}*fright)(x).
$$
Hence, the original relation is equivalent to the following convolution equation on $mathbb{R}$:
$$
f=1_{left[0,1right]}*f.
$$
Note that $fin L^1(mathbb{R})$, and its Fourier transform $hat{f}$ is well-defined. By the convolution theorem,
$$
hat{f}=widehat{1_{left[0,1right]}},hat{f}Longrightarrowleft(1-widehat{1_{left[0,1right]}}right)hat{f}=0.
$$
This implies that $hat{f}=hat{f}(xi)=0$ for all $xine 0$. Besides, the continuity of $hat{f}$ yields $hat{f}(0)=0$. Consequently, we have
$$
hat{f}=0iff f=0.#
$$
This conclusion suggests that any non-trivial solution to the original equation must be non-integrable on $mathbb{R}$, e.g., non-zero constants. Nevertheless, given that it is reasonable to assume $f$ to be locally integrable, the original equation can always be formulated in the convolution form $f=1_{left[0,1right]}*f$. Just note that $hat{f}$ is not defined and the convolution theorem no longer applies if $f$ is locally integrable but not integrable on $mathbb{R}$.
3. $fequivtext{const}$ if $fin L_{rm loc}^1(mathbb{R})cap C_T(mathbb{R})$ ($f$ must be constant if it is locally integrable and $T$-periodic).
Proof. Since $fin L_{rm loc}^1(mathbb{R})$, we have $fin C^{infty}(left[0,Tright])$. Thanks to the periodicity, $f$ observes its Fourier series on $left[0,Tright]$
$$
f(x)simsum_{ninmathbb{Z}}a_n,e^{frac{2ipi nx}{T}}.
$$
Since $fin C^{infty}(left[0,Tright])$, the Fourier series of $f$ converges absolutely and uniformly to $f$. Therefore, we have for one thing,
$$
f(x)=sum_{ninmathbb{Z}}a_n,e^{frac{2ipi nx}{T}}.
$$
For another,
begin{align}
int_{x-1}^xf(s),{rm d}s&=int_{x-1}^xsum_{ninmathbb{Z}}a_n,e^{frac{2ipi ns}{T}}{rm d}s\
&=sum_{ninmathbb{Z}}a_nint_{x-1}^xe^{frac{2ipi ns}{T}}{rm d}s\
&=sum_{ninmathbb{Z}}a_nfrac{1-e^{-theta_n}}{theta_n}e^{frac{2ipi nx}{T}},
end{align}
where $theta_n=2ipi n/T$, and $left(1-e^{-theta_0}right)/theta_0=1$ (this is defined so that the form of the series preserves; otherwise, one may carry out the integration separately for $n=0$ and $nne 0$, and may find the results identical).
Thanks to this result, the original equation requires
$$
a_n=a_nfrac{1-e^{-theta_n}}{theta_n}iffleft(1-frac{1-e^{-theta_n}}{theta_n}right)a_n=0.
$$
Note that
$$
1-frac{1-e^{-z}}{z}=0
$$
yields only one solution on $imathbb{R}$ (the imaginary axis), i.e., $z=0$. Thus since $theta_nne 0$ for all $nne 0$, it is a must that $a_n=0$ for all $nne 0$. This leads to $f(x)=a_0$, i.e., $f$ is constant on $mathbb{R}$.$#$
This conclusion suggests that any continuous periodic function that satisfies the original equation must be constant.
[TBC...]
[Following @Empy2's answer, I believe the existence of some non-periodic solution to the original equation. Yet as per the above properties, this solution has to be smooth and most likely unbounded. Trying the polynomial series $f(x)=sum_{n=0}^{infty}a_nx^n$ could be promising, but it leads to an infinite dimensional linear system, and its convergence also remains unknown, which challenges the commutativity of summation and integration...]
answered Jan 12 at 4:17
hypernovahypernova
4,534313
$endgroup$
add a comment |
$begingroup$
This is a partial answer where I provided some properties for $f$ that complies with
$$
f(x)=int_{x-1}^xf(s),{rm d}s
$$
for all $xinmathbb{R}$. I will update this thread as I make any further progress.
1. $fin C^{infty}(mathbb{R})$ if $fin L_{rm loc}^1(mathbb{R})$ ($f$ must be smooth if it is locally integrable).
Proof. $forall,x_0inmathbb{R}$, $forall,xge x_0+1$, we have
$$
f(x)=int_{x_0}^xf(s),{rm d}s-int_{x_0}^{x-1}f(s),{rm d}s.
$$
Since $fin L_{rm loc}^1(mathbb{R})$, it follows that
$$
int_{x_0}^xf(s),{rm d}squadtext{and}quadint_{x_0}^{x-1}f(s),{rm d}s
$$
are absolutely continuous. Thus $f$ is absolutely continuous on $left(x_0+1,inftyright)$. The arbitrariness of $x_0$ implies that $f$ is absolutely continuous on $mathbb{R}$. Thus necessarily, $fin C(mathbb{R})$.
Likewise, since $fin C(mathbb{R})$, it follows that
$$
int_{x_0}^xf(s),{rm d}squadtext{and}quadint_{x_0}^{x-1}f(s),{rm d}s
$$
are continuously differentiable, which leads to $fin C^1(mathbb{R})$.
Repeat the above reasoning inductively, and we eventually obtain $fin C^{infty}(mathbb{R})$.$#$
This conclusion suggests that, at least for a most general case, we shall only consider those $f$'s that are smooth on $mathbb{R}$.
2. $f=0$ if $fin L^1(mathbb{R})$ ($f$ must be zero if it is integrable on $mathbb{R}$).
Proof. Since $fin L^1(mathbb{R})$, it is obvious that
$$
int_{x-1}^xf(s),{rm d}s=int_{mathbb{R}}1_{left[0,1right]}(x-s)f(s),{rm d}s=left(1_{left[0,1right]}*fright)(x).
$$
Hence, the original relation is equivalent to the following convolution equation on $mathbb{R}$:
$$
f=1_{left[0,1right]}*f.
$$
Note that $fin L^1(mathbb{R})$, and its Fourier transform $hat{f}$ is well-defined. By the convolution theorem,
$$
hat{f}=widehat{1_{left[0,1right]}},hat{f}Longrightarrowleft(1-widehat{1_{left[0,1right]}}right)hat{f}=0.
$$
This implies that $hat{f}=hat{f}(xi)=0$ for all $xine 0$. Besides, the continuity of $hat{f}$ yields $hat{f}(0)=0$. Consequently, we have
$$
hat{f}=0iff f=0.#
$$
This conclusion suggests that any non-trivial solution to the original equation must be non-integrable on $mathbb{R}$, e.g., non-zero constants. Nevertheless, given that it is reasonable to assume $f$ to be locally integrable, the original equation can always be formulated in the convolution form $f=1_{left[0,1right]}*f$. Just note that $hat{f}$ is not defined and the convolution theorem no longer applies if $f$ is locally integrable but not integrable on $mathbb{R}$.
3. $fequivtext{const}$ if $fin L_{rm loc}^1(mathbb{R})cap C_T(mathbb{R})$ ($f$ must be constant if it is locally integrable and $T$-periodic).
Proof. Since $fin L_{rm loc}^1(mathbb{R})$, we have $fin C^{infty}(left[0,Tright])$. Thanks to the periodicity, $f$ observes its Fourier series on $left[0,Tright]$
$$
f(x)simsum_{ninmathbb{Z}}a_n,e^{frac{2ipi nx}{T}}.
$$
Since $fin C^{infty}(left[0,Tright])$, the Fourier series of $f$ converges absolutely and uniformly to $f$. Therefore, we have for one thing,
$$
f(x)=sum_{ninmathbb{Z}}a_n,e^{frac{2ipi nx}{T}}.
$$
For another,
begin{align}
int_{x-1}^xf(s),{rm d}s&=int_{x-1}^xsum_{ninmathbb{Z}}a_n,e^{frac{2ipi ns}{T}}{rm d}s\
&=sum_{ninmathbb{Z}}a_nint_{x-1}^xe^{frac{2ipi ns}{T}}{rm d}s\
&=sum_{ninmathbb{Z}}a_nfrac{1-e^{-theta_n}}{theta_n}e^{frac{2ipi nx}{T}},
end{align}
where $theta_n=2ipi n/T$, and $left(1-e^{-theta_0}right)/theta_0=1$ (this is defined so that the form of the series preserves; otherwise, one may carry out the integration separately for $n=0$ and $nne 0$, and may find the results identical).
Thanks to this result, the original equation requires
$$
a_n=a_nfrac{1-e^{-theta_n}}{theta_n}iffleft(1-frac{1-e^{-theta_n}}{theta_n}right)a_n=0.
$$
Note that
$$
1-frac{1-e^{-z}}{z}=0
$$
yields only one solution on $imathbb{R}$ (the imaginary axis), i.e., $z=0$. Thus since $theta_nne 0$ for all $nne 0$, it is a must that $a_n=0$ for all $nne 0$. This leads to $f(x)=a_0$, i.e., $f$ is constant on $mathbb{R}$.$#$
This conclusion suggests that any continuous periodic function that satisfies the original equation must be constant.
[TBC...]
[Following @Empy2's answer, I believe the existence of some non-periodic solution to the original equation. Yet as per the above properties, this solution has to be smooth and most likely unbounded. Trying the polynomial series $f(x)=sum_{n=0}^{infty}a_nx^n$ could be promising, but it leads to an infinite dimensional linear system, and its convergence also remains unknown, which challenges the commutativity of summation and integration...]
answered Jan 12 at 4:17
hypernovahypernova
4,534313
$endgroup$
add a comment |
$begingroup$
This is a partial answer where I provided some properties for $f$ that complies with
$$
f(x)=int_{x-1}^xf(s),{rm d}s
$$
for all $xinmathbb{R}$. I will update this thread as I make any further progress.
1. $fin C^{infty}(mathbb{R})$ if $fin L_{rm loc}^1(mathbb{R})$ ($f$ must be smooth if it is locally integrable).
Proof. $forall,x_0inmathbb{R}$, $forall,xge x_0+1$, we have
$$
f(x)=int_{x_0}^xf(s),{rm d}s-int_{x_0}^{x-1}f(s),{rm d}s.
$$
Since $fin L_{rm loc}^1(mathbb{R})$, it follows that
$$
int_{x_0}^xf(s),{rm d}squadtext{and}quadint_{x_0}^{x-1}f(s),{rm d}s
$$
are absolutely continuous. Thus $f$ is absolutely continuous on $left(x_0+1,inftyright)$. The arbitrariness of $x_0$ implies that $f$ is absolutely continuous on $mathbb{R}$. Thus necessarily, $fin C(mathbb{R})$.
Likewise, since $fin C(mathbb{R})$, it follows that
$$
int_{x_0}^xf(s),{rm d}squadtext{and}quadint_{x_0}^{x-1}f(s),{rm d}s
$$
are continuously differentiable, which leads to $fin C^1(mathbb{R})$.
Repeat the above reasoning inductively, and we eventually obtain $fin C^{infty}(mathbb{R})$.$#$
This conclusion suggests that, at least for a most general case, we shall only consider those $f$'s that are smooth on $mathbb{R}$.
2. $f=0$ if $fin L^1(mathbb{R})$ ($f$ must be zero if it is integrable on $mathbb{R}$).
Proof. Since $fin L^1(mathbb{R})$, it is obvious that
$$
int_{x-1}^xf(s),{rm d}s=int_{mathbb{R}}1_{left[0,1right]}(x-s)f(s),{rm d}s=left(1_{left[0,1right]}*fright)(x).
$$
Hence, the original relation is equivalent to the following convolution equation on $mathbb{R}$:
$$
f=1_{left[0,1right]}*f.
$$
Note that $fin L^1(mathbb{R})$, and its Fourier transform $hat{f}$ is well-defined. By the convolution theorem,
$$
hat{f}=widehat{1_{left[0,1right]}},hat{f}Longrightarrowleft(1-widehat{1_{left[0,1right]}}right)hat{f}=0.
$$
This implies that $hat{f}=hat{f}(xi)=0$ for all $xine 0$. Besides, the continuity of $hat{f}$ yields $hat{f}(0)=0$. Consequently, we have
$$
hat{f}=0iff f=0.#
$$
This conclusion suggests that any non-trivial solution to the original equation must be non-integrable on $mathbb{R}$, e.g., non-zero constants. Nevertheless, given that it is reasonable to assume $f$ to be locally integrable, the original equation can always be formulated in the convolution form $f=1_{left[0,1right]}*f$. Just note that $hat{f}$ is not defined and the convolution theorem no longer applies if $f$ is locally integrable but not integrable on $mathbb{R}$.
3. $fequivtext{const}$ if $fin L_{rm loc}^1(mathbb{R})cap C_T(mathbb{R})$ ($f$ must be constant if it is locally integrable and $T$-periodic).
Proof. Since $fin L_{rm loc}^1(mathbb{R})$, we have $fin C^{infty}(left[0,Tright])$. Thanks to the periodicity, $f$ observes its Fourier series on $left[0,Tright]$
$$
f(x)simsum_{ninmathbb{Z}}a_n,e^{frac{2ipi nx}{T}}.
$$
Since $fin C^{infty}(left[0,Tright])$, the Fourier series of $f$ converges absolutely and uniformly to $f$. Therefore, we have for one thing,
$$
f(x)=sum_{ninmathbb{Z}}a_n,e^{frac{2ipi nx}{T}}.
$$
For another,
begin{align}
int_{x-1}^xf(s),{rm d}s&=int_{x-1}^xsum_{ninmathbb{Z}}a_n,e^{frac{2ipi ns}{T}}{rm d}s\
&=sum_{ninmathbb{Z}}a_nint_{x-1}^xe^{frac{2ipi ns}{T}}{rm d}s\
&=sum_{ninmathbb{Z}}a_nfrac{1-e^{-theta_n}}{theta_n}e^{frac{2ipi nx}{T}},
end{align}
where $theta_n=2ipi n/T$, and $left(1-e^{-theta_0}right)/theta_0=1$ (this is defined so that the form of the series preserves; otherwise, one may carry out the integration separately for $n=0$ and $nne 0$, and may find the results identical).
Thanks to this result, the original equation requires
$$
a_n=a_nfrac{1-e^{-theta_n}}{theta_n}iffleft(1-frac{1-e^{-theta_n}}{theta_n}right)a_n=0.
$$
Note that
$$
1-frac{1-e^{-z}}{z}=0
$$
yields only one solution on $imathbb{R}$ (the imaginary axis), i.e., $z=0$. Thus since $theta_nne 0$ for all $nne 0$, it is a must that $a_n=0$ for all $nne 0$. This leads to $f(x)=a_0$, i.e., $f$ is constant on $mathbb{R}$.$#$
This conclusion suggests that any continuous periodic function that satisfies the original equation must be constant.
[TBC...]
[Following @Empy2's answer, I believe the existence of some non-periodic solution to the original equation. Yet as per the above properties, this solution has to be smooth and most likely unbounded. Trying the polynomial series $f(x)=sum_{n=0}^{infty}a_nx^n$ could be promising, but it leads to an infinite dimensional linear system, and its convergence also remains unknown, which challenges the commutativity of summation and integration...]
answered Jan 12 at 4:17
hypernovahypernova
4,534313
$endgroup$
This is a partial answer where I provided some properties for $f$ that complies with
$$
f(x)=int_{x-1}^xf(s),{rm d}s
$$
for all $xinmathbb{R}$. I will update this thread as I make any further progress.
1. $fin C^{infty}(mathbb{R})$ if $fin L_{rm loc}^1(mathbb{R})$ ($f$ must be smooth if it is locally integrable).
Proof. $forall,x_0inmathbb{R}$, $forall,xge x_0+1$, we have
$$
f(x)=int_{x_0}^xf(s),{rm d}s-int_{x_0}^{x-1}f(s),{rm d}s.
$$
Since $fin L_{rm loc}^1(mathbb{R})$, it follows that
$$
int_{x_0}^xf(s),{rm d}squadtext{and}quadint_{x_0}^{x-1}f(s),{rm d}s
$$
are absolutely continuous. Thus $f$ is absolutely continuous on $left(x_0+1,inftyright)$. The arbitrariness of $x_0$ implies that $f$ is absolutely continuous on $mathbb{R}$. Thus necessarily, $fin C(mathbb{R})$.
Likewise, since $fin C(mathbb{R})$, it follows that
$$
int_{x_0}^xf(s),{rm d}squadtext{and}quadint_{x_0}^{x-1}f(s),{rm d}s
$$
are continuously differentiable, which leads to $fin C^1(mathbb{R})$.
Repeat the above reasoning inductively, and we eventually obtain $fin C^{infty}(mathbb{R})$.$#$
This conclusion suggests that, at least for a most general case, we shall only consider those $f$'s that are smooth on $mathbb{R}$.
2. $f=0$ if $fin L^1(mathbb{R})$ ($f$ must be zero if it is integrable on $mathbb{R}$).
Proof. Since $fin L^1(mathbb{R})$, it is obvious that
$$
int_{x-1}^xf(s),{rm d}s=int_{mathbb{R}}1_{left[0,1right]}(x-s)f(s),{rm d}s=left(1_{left[0,1right]}*fright)(x).
$$
Hence, the original relation is equivalent to the following convolution equation on $mathbb{R}$:
$$
f=1_{left[0,1right]}*f.
$$
Note that $fin L^1(mathbb{R})$, and its Fourier transform $hat{f}$ is well-defined. By the convolution theorem,
$$
hat{f}=widehat{1_{left[0,1right]}},hat{f}Longrightarrowleft(1-widehat{1_{left[0,1right]}}right)hat{f}=0.
$$
This implies that $hat{f}=hat{f}(xi)=0$ for all $xine 0$. Besides, the continuity of $hat{f}$ yields $hat{f}(0)=0$. Consequently, we have
$$
hat{f}=0iff f=0.#
$$
This conclusion suggests that any non-trivial solution to the original equation must be non-integrable on $mathbb{R}$, e.g., non-zero constants. Nevertheless, given that it is reasonable to assume $f$ to be locally integrable, the original equation can always be formulated in the convolution form $f=1_{left[0,1right]}*f$. Just note that $hat{f}$ is not defined and the convolution theorem no longer applies if $f$ is locally integrable but not integrable on $mathbb{R}$.
3. $fequivtext{const}$ if $fin L_{rm loc}^1(mathbb{R})cap C_T(mathbb{R})$ ($f$ must be constant if it is locally integrable and $T$-periodic).
Proof. Since $fin L_{rm loc}^1(mathbb{R})$, we have $fin C^{infty}(left[0,Tright])$. Thanks to the periodicity, $f$ observes its Fourier series on $left[0,Tright]$
$$
f(x)simsum_{ninmathbb{Z}}a_n,e^{frac{2ipi nx}{T}}.
$$
Since $fin C^{infty}(left[0,Tright])$, the Fourier series of $f$ converges absolutely and uniformly to $f$. Therefore, we have for one thing,
$$
f(x)=sum_{ninmathbb{Z}}a_n,e^{frac{2ipi nx}{T}}.
$$
For another,
begin{align}
int_{x-1}^xf(s),{rm d}s&=int_{x-1}^xsum_{ninmathbb{Z}}a_n,e^{frac{2ipi ns}{T}}{rm d}s\
&=sum_{ninmathbb{Z}}a_nint_{x-1}^xe^{frac{2ipi ns}{T}}{rm d}s\
&=sum_{ninmathbb{Z}}a_nfrac{1-e^{-theta_n}}{theta_n}e^{frac{2ipi nx}{T}},
end{align}
where $theta_n=2ipi n/T$, and $left(1-e^{-theta_0}right)/theta_0=1$ (this is defined so that the form of the series preserves; otherwise, one may carry out the integration separately for $n=0$ and $nne 0$, and may find the results identical).
Thanks to this result, the original equation requires
$$
a_n=a_nfrac{1-e^{-theta_n}}{theta_n}iffleft(1-frac{1-e^{-theta_n}}{theta_n}right)a_n=0.
$$
Note that
$$
1-frac{1-e^{-z}}{z}=0
$$
yields only one solution on $imathbb{R}$ (the imaginary axis), i.e., $z=0$. Thus since $theta_nne 0$ for all $nne 0$, it is a must that $a_n=0$ for all $nne 0$. This leads to $f(x)=a_0$, i.e., $f$ is constant on $mathbb{R}$.$#$
This conclusion suggests that any continuous periodic function that satisfies the original equation must be constant.
[TBC...]
[Following @Empy2's answer, I believe the existence of some non-periodic solution to the original equation. Yet as per the above properties, this solution has to be smooth and most likely unbounded. Trying the polynomial series $f(x)=sum_{n=0}^{infty}a_nx^n$ could be promising, but it leads to an infinite dimensional linear system, and its convergence also remains unknown, which challenges the commutativity of summation and integration...]
answered Jan 12 at 4:17
hypernovahypernova
4,534313
edited Jan 13 at 0:31
answered Jan 12 at 4:17
hypernovahypernova
4,534313
answered Jan 12 at 4:17
hypernovahypernova
4,534313
answered Jan 12 at 4:17
hypernovahypernova
4,534313
4,534313
add a comment |
add a comment |
$begingroup$
Define $f(x)=3x^2-4x+1$ for $xin(0,1)$ so the differential equation is true for $x=1$.
For $xin(1,2)$, solve the differential equation
$$frac{df}{dx}=f(x)-(3(x-1)^2-4(x-1)+1)$$
Iterate the procedure, for $xin (2,3)$ and so on.
answered Jan 7 at 9:02
Empy2Empy2
33.5k12261
$endgroup$
$begingroup$
That's nice. However, how do we define $f(x)$ for $x<0$? Also, I have not proved yet that the equation $f'(x)=f(x)-f(x-1)$ implies that $f$ satisfies the property described above.
$endgroup$
– Questioner
Jan 7 at 9:09
$begingroup$
Nevermind the first part of the comment above: For $xin[-1,0]$ we have $f'(x+1)=f(x+1)-f(x)$, so this defines $f$ on $[-1,1]$ and we iterate...
$endgroup$
– Questioner
Jan 7 at 9:13
$begingroup$
@Questioner Note that$$f(1)=int_0^1f(t),mathrm dt=0,$$thus integrating $f'(t)=f(t)-f(t-1)$ for $tin[1,x]$ yields$$f(x)=int_1^xf(t),mathrm dt-int_0^{x-1}f(t),mathrm dt=int_{x-1}^xf(t),mathrm dt$$for $xgeqslant1$.
$endgroup$
– Saad
Jan 12 at 2:09
add a comment |
$begingroup$
Define $f(x)=3x^2-4x+1$ for $xin(0,1)$ so the differential equation is true for $x=1$.
For $xin(1,2)$, solve the differential equation
$$frac{df}{dx}=f(x)-(3(x-1)^2-4(x-1)+1)$$
Iterate the procedure, for $xin (2,3)$ and so on.
answered Jan 7 at 9:02
Empy2Empy2
33.5k12261
$endgroup$
$begingroup$
That's nice. However, how do we define $f(x)$ for $x<0$? Also, I have not proved yet that the equation $f'(x)=f(x)-f(x-1)$ implies that $f$ satisfies the property described above.
$endgroup$
– Questioner
Jan 7 at 9:09
$begingroup$
Nevermind the first part of the comment above: For $xin[-1,0]$ we have $f'(x+1)=f(x+1)-f(x)$, so this defines $f$ on $[-1,1]$ and we iterate...
$endgroup$
– Questioner
Jan 7 at 9:13
$begingroup$
@Questioner Note that$$f(1)=int_0^1f(t),mathrm dt=0,$$thus integrating $f'(t)=f(t)-f(t-1)$ for $tin[1,x]$ yields$$f(x)=int_1^xf(t),mathrm dt-int_0^{x-1}f(t),mathrm dt=int_{x-1}^xf(t),mathrm dt$$for $xgeqslant1$.
$endgroup$
– Saad
Jan 12 at 2:09
add a comment |
$begingroup$
Define $f(x)=3x^2-4x+1$ for $xin(0,1)$ so the differential equation is true for $x=1$.
For $xin(1,2)$, solve the differential equation
$$frac{df}{dx}=f(x)-(3(x-1)^2-4(x-1)+1)$$
Iterate the procedure, for $xin (2,3)$ and so on.
answered Jan 7 at 9:02
Empy2Empy2
33.5k12261
$endgroup$
Define $f(x)=3x^2-4x+1$ for $xin(0,1)$ so the differential equation is true for $x=1$.
For $xin(1,2)$, solve the differential equation
$$frac{df}{dx}=f(x)-(3(x-1)^2-4(x-1)+1)$$
Iterate the procedure, for $xin (2,3)$ and so on.
answered Jan 7 at 9:02
Empy2Empy2
33.5k12261
answered Jan 7 at 9:02
Empy2Empy2
33.5k12261
answered Jan 7 at 9:02
Empy2Empy2
33.5k12261
answered Jan 7 at 9:02
Empy2Empy2
33.5k12261
33.5k12261
$begingroup$
That's nice. However, how do we define $f(x)$ for $x<0$? Also, I have not proved yet that the equation $f'(x)=f(x)-f(x-1)$ implies that $f$ satisfies the property described above.
$endgroup$
– Questioner
Jan 7 at 9:09
$begingroup$
Nevermind the first part of the comment above: For $xin[-1,0]$ we have $f'(x+1)=f(x+1)-f(x)$, so this defines $f$ on $[-1,1]$ and we iterate...
$endgroup$
– Questioner
Jan 7 at 9:13
$begingroup$
@Questioner Note that$$f(1)=int_0^1f(t),mathrm dt=0,$$thus integrating $f'(t)=f(t)-f(t-1)$ for $tin[1,x]$ yields$$f(x)=int_1^xf(t),mathrm dt-int_0^{x-1}f(t),mathrm dt=int_{x-1}^xf(t),mathrm dt$$for $xgeqslant1$.
$endgroup$
– Saad
Jan 12 at 2:09
add a comment |
$begingroup$
That's nice. However, how do we define $f(x)$ for $x<0$? Also, I have not proved yet that the equation $f'(x)=f(x)-f(x-1)$ implies that $f$ satisfies the property described above.
$endgroup$
– Questioner
Jan 7 at 9:09
$begingroup$
Nevermind the first part of the comment above: For $xin[-1,0]$ we have $f'(x+1)=f(x+1)-f(x)$, so this defines $f$ on $[-1,1]$ and we iterate...
$endgroup$
– Questioner
Jan 7 at 9:13
$begingroup$
@Questioner Note that$$f(1)=int_0^1f(t),mathrm dt=0,$$thus integrating $f'(t)=f(t)-f(t-1)$ for $tin[1,x]$ yields$$f(x)=int_1^xf(t),mathrm dt-int_0^{x-1}f(t),mathrm dt=int_{x-1}^xf(t),mathrm dt$$for $xgeqslant1$.
$endgroup$
– Saad
Jan 12 at 2:09
$begingroup$
That's nice. However, how do we define $f(x)$ for $x<0$? Also, I have not proved yet that the equation $f'(x)=f(x)-f(x-1)$ implies that $f$ satisfies the property described above.
$endgroup$
– Questioner
Jan 7 at 9:09
$begingroup$
That's nice. However, how do we define $f(x)$ for $x<0$? Also, I have not proved yet that the equation $f'(x)=f(x)-f(x-1)$ implies that $f$ satisfies the property described above.
$endgroup$
– Questioner
Jan 7 at 9:09
$begingroup$
Nevermind the first part of the comment above: For $xin[-1,0]$ we have $f'(x+1)=f(x+1)-f(x)$, so this defines $f$ on $[-1,1]$ and we iterate...
$endgroup$
– Questioner
Jan 7 at 9:13
$begingroup$
Nevermind the first part of the comment above: For $xin[-1,0]$ we have $f'(x+1)=f(x+1)-f(x)$, so this defines $f$ on $[-1,1]$ and we iterate...
$endgroup$
– Questioner
Jan 7 at 9:13
$begingroup$
@Questioner Note that$$f(1)=int_0^1f(t),mathrm dt=0,$$thus integrating $f'(t)=f(t)-f(t-1)$ for $tin[1,x]$ yields$$f(x)=int_1^xf(t),mathrm dt-int_0^{x-1}f(t),mathrm dt=int_{x-1}^xf(t),mathrm dt$$for $xgeqslant1$.
$endgroup$
– Saad
Jan 12 at 2:09
$begingroup$
@Questioner Note that$$f(1)=int_0^1f(t),mathrm dt=0,$$thus integrating $f'(t)=f(t)-f(t-1)$ for $tin[1,x]$ yields$$f(x)=int_1^xf(t),mathrm dt-int_0^{x-1}f(t),mathrm dt=int_{x-1}^xf(t),mathrm dt$$for $xgeqslant1$.
$endgroup$
– Saad
Jan 12 at 2:09
add a comment |
$begingroup$
Let
$$f(x) = dfrac{a_0}2+sumlimits_{n=1}^inftyleft(a_ncos2pi nx + a_nsin2pi nxright),$$
then
$$intlimits_{x-1}^x f(x),mathrm dx = dfrac {a_0}2,$$
so $f(x)$ is a constant.
answered Jan 12 at 0:10
Yuri NegometyanovYuri Negometyanov
11.1k1728
$endgroup$
add a comment |
$begingroup$
Let
$$f(x) = dfrac{a_0}2+sumlimits_{n=1}^inftyleft(a_ncos2pi nx + a_nsin2pi nxright),$$
then
$$intlimits_{x-1}^x f(x),mathrm dx = dfrac {a_0}2,$$
so $f(x)$ is a constant.
answered Jan 12 at 0:10
Yuri NegometyanovYuri Negometyanov
11.1k1728
$endgroup$
add a comment |
$begingroup$
Let
$$f(x) = dfrac{a_0}2+sumlimits_{n=1}^inftyleft(a_ncos2pi nx + a_nsin2pi nxright),$$
then
$$intlimits_{x-1}^x f(x),mathrm dx = dfrac {a_0}2,$$
so $f(x)$ is a constant.
answered Jan 12 at 0:10
Yuri NegometyanovYuri Negometyanov
11.1k1728
$endgroup$
Let
$$f(x) = dfrac{a_0}2+sumlimits_{n=1}^inftyleft(a_ncos2pi nx + a_nsin2pi nxright),$$
then
$$intlimits_{x-1}^x f(x),mathrm dx = dfrac {a_0}2,$$
so $f(x)$ is a constant.
answered Jan 12 at 0:10
Yuri NegometyanovYuri Negometyanov
11.1k1728
answered Jan 12 at 0:10
Yuri NegometyanovYuri Negometyanov
11.1k1728
answered Jan 12 at 0:10
Yuri NegometyanovYuri Negometyanov
11.1k1728
answered Jan 12 at 0:10
Yuri NegometyanovYuri Negometyanov
11.1k1728
11.1k1728
add a comment |
add a comment |
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$begingroup$
This is a delay differential equation. Not all solutions are constant or periodic.
$endgroup$
– Dylan
Jan 7 at 9:06