Finding an exact solution for $u(x)+int_0^{2π}cos(x+t)u(t),mathrm dt=(π+1)cos x$












3












$begingroup$



I need to find an exact solution for$$u(x)+int_0^{2π}cos(x+t)u(t),mathrm dt=(π+1)cos x.$$




So far, I have tried to integrate by parts but it does not seem to be helpful because I got this:
$$u(t)sin(x+t)bigg|_0^{2pi} - int_{0}^{2pi}sin(x+t)u'(t),mathrm{d}t = (pi + 1)cos x - u(x).$$
I think it is a wrong way to solve this equation.



Any suggestion for how to solve it would be appreciated.






trigonometry definite-integrals







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$endgroup$








  • 1




    $begingroup$
    What about the sum of angle formula for the $cos$ term?
    $endgroup$
    – baharampuri
    Dec 3 '18 at 12:18










  • $begingroup$
    After your advice we got: $$cos xint_{0}^{2pi} cos t, u(t),dt - sin xint_{0}^{2pi} sin(t), u(t), dt = (pi + 1)cos x - u(x)$$ And I still have no idea of how to solve this (following integration by parts didn't help either). I suppose we cannot solve it until we set $u(x)$ and $u(t)$ manually
    $endgroup$
    – George Zorikov
    Dec 3 '18 at 12:38


















3












$begingroup$



I need to find an exact solution for$$u(x)+int_0^{2π}cos(x+t)u(t),mathrm dt=(π+1)cos x.$$




So far, I have tried to integrate by parts but it does not seem to be helpful because I got this:
$$u(t)sin(x+t)bigg|_0^{2pi} - int_{0}^{2pi}sin(x+t)u'(t),mathrm{d}t = (pi + 1)cos x - u(x).$$
I think it is a wrong way to solve this equation.



Any suggestion for how to solve it would be appreciated.






trigonometry definite-integrals







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What about the sum of angle formula for the $cos$ term?
    $endgroup$
    – baharampuri
    Dec 3 '18 at 12:18










  • $begingroup$
    After your advice we got: $$cos xint_{0}^{2pi} cos t, u(t),dt - sin xint_{0}^{2pi} sin(t), u(t), dt = (pi + 1)cos x - u(x)$$ And I still have no idea of how to solve this (following integration by parts didn't help either). I suppose we cannot solve it until we set $u(x)$ and $u(t)$ manually
    $endgroup$
    – George Zorikov
    Dec 3 '18 at 12:38
















3












3








3


1



$begingroup$



I need to find an exact solution for$$u(x)+int_0^{2π}cos(x+t)u(t),mathrm dt=(π+1)cos x.$$




So far, I have tried to integrate by parts but it does not seem to be helpful because I got this:
$$u(t)sin(x+t)bigg|_0^{2pi} - int_{0}^{2pi}sin(x+t)u'(t),mathrm{d}t = (pi + 1)cos x - u(x).$$
I think it is a wrong way to solve this equation.



Any suggestion for how to solve it would be appreciated.






trigonometry definite-integrals







share|cite|improve this question











$endgroup$





I need to find an exact solution for$$u(x)+int_0^{2π}cos(x+t)u(t),mathrm dt=(π+1)cos x.$$




So far, I have tried to integrate by parts but it does not seem to be helpful because I got this:
$$u(t)sin(x+t)bigg|_0^{2pi} - int_{0}^{2pi}sin(x+t)u'(t),mathrm{d}t = (pi + 1)cos x - u(x).$$
I think it is a wrong way to solve this equation.



Any suggestion for how to solve it would be appreciated.






trigonometry definite-integrals




trigonometry definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 '18 at 12:48









Saad

19.7k92352




19.7k92352










asked Dec 3 '18 at 12:09









George ZorikovGeorge Zorikov

273




273








  • 1




    $begingroup$
    What about the sum of angle formula for the $cos$ term?
    $endgroup$
    – baharampuri
    Dec 3 '18 at 12:18










  • $begingroup$
    After your advice we got: $$cos xint_{0}^{2pi} cos t, u(t),dt - sin xint_{0}^{2pi} sin(t), u(t), dt = (pi + 1)cos x - u(x)$$ And I still have no idea of how to solve this (following integration by parts didn't help either). I suppose we cannot solve it until we set $u(x)$ and $u(t)$ manually
    $endgroup$
    – George Zorikov
    Dec 3 '18 at 12:38
















  • 1




    $begingroup$
    What about the sum of angle formula for the $cos$ term?
    $endgroup$
    – baharampuri
    Dec 3 '18 at 12:18










  • $begingroup$
    After your advice we got: $$cos xint_{0}^{2pi} cos t, u(t),dt - sin xint_{0}^{2pi} sin(t), u(t), dt = (pi + 1)cos x - u(x)$$ And I still have no idea of how to solve this (following integration by parts didn't help either). I suppose we cannot solve it until we set $u(x)$ and $u(t)$ manually
    $endgroup$
    – George Zorikov
    Dec 3 '18 at 12:38










1




1




$begingroup$
What about the sum of angle formula for the $cos$ term?
$endgroup$
– baharampuri
Dec 3 '18 at 12:18




$begingroup$
What about the sum of angle formula for the $cos$ term?
$endgroup$
– baharampuri
Dec 3 '18 at 12:18












$begingroup$
After your advice we got: $$cos xint_{0}^{2pi} cos t, u(t),dt - sin xint_{0}^{2pi} sin(t), u(t), dt = (pi + 1)cos x - u(x)$$ And I still have no idea of how to solve this (following integration by parts didn't help either). I suppose we cannot solve it until we set $u(x)$ and $u(t)$ manually
$endgroup$
– George Zorikov
Dec 3 '18 at 12:38






$begingroup$
After your advice we got: $$cos xint_{0}^{2pi} cos t, u(t),dt - sin xint_{0}^{2pi} sin(t), u(t), dt = (pi + 1)cos x - u(x)$$ And I still have no idea of how to solve this (following integration by parts didn't help either). I suppose we cannot solve it until we set $u(x)$ and $u(t)$ manually
$endgroup$
– George Zorikov
Dec 3 '18 at 12:38












2 Answers
2






active

oldest

votes


















3












$begingroup$

From
$$cos(a+b)=cos(a)cos(b)-sin(a)sin(b)$$
We have that
$$u(x)+cos(x)int_{0}^{2pi} cos(t)u(t)mathrm{d}t-sin(x)int_{0}^{2pi} sin(t)u(t)mathrm{d}t=(pi+1)cos(t)$$
But the integrals are independent of $x$, so they are constant! So the form of $u$ will be
$$u(x)=(pi+1)cos(x)-c_1cos(x)+c_2sin(x)$$
And you can calculate the value of the constants easily:
$$c_1=int_{0}^{2pi} cos(t) u(t) mathrm{d}t$$
$$c_1=int_{0}^{2pi} cos(t)[(pi+1)cos(x)-c_1cos(x)+c_2sin(x)] mathrm{d}t$$
$$c_1=(pi+1)int_{0}^{2pi} cos^2(x) mathrm{d}x-c_1int_{0}^{2pi} cos^2(x) mathrm{d}x +c_2 int_{0}^{2pi} sin(x)cos(x) mathrm{d}x$$
You can easily calculate these integrals with the double-angle formulas:
$$c_1=(pi+1)pi-c_1 pi+c_2 0$$
$$c_1(1+pi)=(pi+1)pi$$
$$c_1=pi$$
And you can do the same to get $c_2$:
$$c_2=int_{0}^{2pi} sin(t) u(t) mathrm{d}t$$
So we have that $c_1=pi$ and$c_2=0$, which implies that
$$u(x)=cos(x)$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you! But may I ask you one more question? When I try to find $c_1$ I use integration by parts. And I cannot solve it (especially $c_1 = pi$) because I have $u(t)$ undefined. Maybe I should calculate it using another technique?
    $endgroup$
    – George Zorikov
    Dec 3 '18 at 13:44












  • $begingroup$
    @GeorgeZorikov Is it better now?
    $endgroup$
    – Botond
    Dec 3 '18 at 13:50










  • $begingroup$
    now I get it! Thank you again
    $endgroup$
    – George Zorikov
    Dec 3 '18 at 13:57





















2












$begingroup$

$defR{mathbb{R}}defd{mathrm{d}}defpeq{mathrel{phantom{=}}{}}$Assume that $u$ is integrable on $[0, 2π]$. For $x in R$ and $Δx ≠ 0$, becausebegin{align*}
frac{1}{Δx} (u(x + Δx) - u(x)) &= -int_0^{2π} frac{1}{Δx} (cos(x + t + Δx) - cos(x + t)) u(t) ,d t\
&peq + (π + 1) frac{1}{Δx} (cos(x + Δx) - cos x),
end{align*}
making $Δx → 0$ with the dominated convergence theorem yields$$
u'(x) = int_0^{2π} sin(x + t) u(t) ,d t - (π + 1) sin x.
$$
Analogously,$$
u''(x) = int_0^{2π} cos(x + t) u(t) ,d t - (π + 1) cos x.
$$
Thus $u'' + u = 0$, which implies $u(x) = c_1 cos x + c_2 sin x$ for some constant $c_1$ and $c_2$.



Now, sincebegin{align*}
int_0^{2π} cos(x + t) u(t) ,d t &= c_1 int_0^{2π} cos(x + t) cos t ,d t + c_2 int_0^{2π} cos(x + t) sin t ,d t\
&= π(c_1 cos x - c_2 sin x),
end{align*}
then$$
u(x) + int_0^{2π} cos(x + t) u(t) ,d t = c_1 (π + 1) cos x - c_2 π sin x,
$$
which implies $c_1 = 1$, $c_2 = 0$. Therefore, $u(x) = cos x$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much!
    $endgroup$
    – George Zorikov
    Dec 3 '18 at 13:16











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

From
$$cos(a+b)=cos(a)cos(b)-sin(a)sin(b)$$
We have that
$$u(x)+cos(x)int_{0}^{2pi} cos(t)u(t)mathrm{d}t-sin(x)int_{0}^{2pi} sin(t)u(t)mathrm{d}t=(pi+1)cos(t)$$
But the integrals are independent of $x$, so they are constant! So the form of $u$ will be
$$u(x)=(pi+1)cos(x)-c_1cos(x)+c_2sin(x)$$
And you can calculate the value of the constants easily:
$$c_1=int_{0}^{2pi} cos(t) u(t) mathrm{d}t$$
$$c_1=int_{0}^{2pi} cos(t)[(pi+1)cos(x)-c_1cos(x)+c_2sin(x)] mathrm{d}t$$
$$c_1=(pi+1)int_{0}^{2pi} cos^2(x) mathrm{d}x-c_1int_{0}^{2pi} cos^2(x) mathrm{d}x +c_2 int_{0}^{2pi} sin(x)cos(x) mathrm{d}x$$
You can easily calculate these integrals with the double-angle formulas:
$$c_1=(pi+1)pi-c_1 pi+c_2 0$$
$$c_1(1+pi)=(pi+1)pi$$
$$c_1=pi$$
And you can do the same to get $c_2$:
$$c_2=int_{0}^{2pi} sin(t) u(t) mathrm{d}t$$
So we have that $c_1=pi$ and$c_2=0$, which implies that
$$u(x)=cos(x)$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you! But may I ask you one more question? When I try to find $c_1$ I use integration by parts. And I cannot solve it (especially $c_1 = pi$) because I have $u(t)$ undefined. Maybe I should calculate it using another technique?
    $endgroup$
    – George Zorikov
    Dec 3 '18 at 13:44












  • $begingroup$
    @GeorgeZorikov Is it better now?
    $endgroup$
    – Botond
    Dec 3 '18 at 13:50










  • $begingroup$
    now I get it! Thank you again
    $endgroup$
    – George Zorikov
    Dec 3 '18 at 13:57


















3












$begingroup$

From
$$cos(a+b)=cos(a)cos(b)-sin(a)sin(b)$$
We have that
$$u(x)+cos(x)int_{0}^{2pi} cos(t)u(t)mathrm{d}t-sin(x)int_{0}^{2pi} sin(t)u(t)mathrm{d}t=(pi+1)cos(t)$$
But the integrals are independent of $x$, so they are constant! So the form of $u$ will be
$$u(x)=(pi+1)cos(x)-c_1cos(x)+c_2sin(x)$$
And you can calculate the value of the constants easily:
$$c_1=int_{0}^{2pi} cos(t) u(t) mathrm{d}t$$
$$c_1=int_{0}^{2pi} cos(t)[(pi+1)cos(x)-c_1cos(x)+c_2sin(x)] mathrm{d}t$$
$$c_1=(pi+1)int_{0}^{2pi} cos^2(x) mathrm{d}x-c_1int_{0}^{2pi} cos^2(x) mathrm{d}x +c_2 int_{0}^{2pi} sin(x)cos(x) mathrm{d}x$$
You can easily calculate these integrals with the double-angle formulas:
$$c_1=(pi+1)pi-c_1 pi+c_2 0$$
$$c_1(1+pi)=(pi+1)pi$$
$$c_1=pi$$
And you can do the same to get $c_2$:
$$c_2=int_{0}^{2pi} sin(t) u(t) mathrm{d}t$$
So we have that $c_1=pi$ and$c_2=0$, which implies that
$$u(x)=cos(x)$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you! But may I ask you one more question? When I try to find $c_1$ I use integration by parts. And I cannot solve it (especially $c_1 = pi$) because I have $u(t)$ undefined. Maybe I should calculate it using another technique?
    $endgroup$
    – George Zorikov
    Dec 3 '18 at 13:44












  • $begingroup$
    @GeorgeZorikov Is it better now?
    $endgroup$
    – Botond
    Dec 3 '18 at 13:50










  • $begingroup$
    now I get it! Thank you again
    $endgroup$
    – George Zorikov
    Dec 3 '18 at 13:57
















3












3








3





$begingroup$

From
$$cos(a+b)=cos(a)cos(b)-sin(a)sin(b)$$
We have that
$$u(x)+cos(x)int_{0}^{2pi} cos(t)u(t)mathrm{d}t-sin(x)int_{0}^{2pi} sin(t)u(t)mathrm{d}t=(pi+1)cos(t)$$
But the integrals are independent of $x$, so they are constant! So the form of $u$ will be
$$u(x)=(pi+1)cos(x)-c_1cos(x)+c_2sin(x)$$
And you can calculate the value of the constants easily:
$$c_1=int_{0}^{2pi} cos(t) u(t) mathrm{d}t$$
$$c_1=int_{0}^{2pi} cos(t)[(pi+1)cos(x)-c_1cos(x)+c_2sin(x)] mathrm{d}t$$
$$c_1=(pi+1)int_{0}^{2pi} cos^2(x) mathrm{d}x-c_1int_{0}^{2pi} cos^2(x) mathrm{d}x +c_2 int_{0}^{2pi} sin(x)cos(x) mathrm{d}x$$
You can easily calculate these integrals with the double-angle formulas:
$$c_1=(pi+1)pi-c_1 pi+c_2 0$$
$$c_1(1+pi)=(pi+1)pi$$
$$c_1=pi$$
And you can do the same to get $c_2$:
$$c_2=int_{0}^{2pi} sin(t) u(t) mathrm{d}t$$
So we have that $c_1=pi$ and$c_2=0$, which implies that
$$u(x)=cos(x)$$






share|cite|improve this answer











$endgroup$



From
$$cos(a+b)=cos(a)cos(b)-sin(a)sin(b)$$
We have that
$$u(x)+cos(x)int_{0}^{2pi} cos(t)u(t)mathrm{d}t-sin(x)int_{0}^{2pi} sin(t)u(t)mathrm{d}t=(pi+1)cos(t)$$
But the integrals are independent of $x$, so they are constant! So the form of $u$ will be
$$u(x)=(pi+1)cos(x)-c_1cos(x)+c_2sin(x)$$
And you can calculate the value of the constants easily:
$$c_1=int_{0}^{2pi} cos(t) u(t) mathrm{d}t$$
$$c_1=int_{0}^{2pi} cos(t)[(pi+1)cos(x)-c_1cos(x)+c_2sin(x)] mathrm{d}t$$
$$c_1=(pi+1)int_{0}^{2pi} cos^2(x) mathrm{d}x-c_1int_{0}^{2pi} cos^2(x) mathrm{d}x +c_2 int_{0}^{2pi} sin(x)cos(x) mathrm{d}x$$
You can easily calculate these integrals with the double-angle formulas:
$$c_1=(pi+1)pi-c_1 pi+c_2 0$$
$$c_1(1+pi)=(pi+1)pi$$
$$c_1=pi$$
And you can do the same to get $c_2$:
$$c_2=int_{0}^{2pi} sin(t) u(t) mathrm{d}t$$
So we have that $c_1=pi$ and$c_2=0$, which implies that
$$u(x)=cos(x)$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 3 '18 at 14:04









A.Γ.

22.7k32656




22.7k32656










answered Dec 3 '18 at 13:06









BotondBotond

5,6632732




5,6632732












  • $begingroup$
    Thank you! But may I ask you one more question? When I try to find $c_1$ I use integration by parts. And I cannot solve it (especially $c_1 = pi$) because I have $u(t)$ undefined. Maybe I should calculate it using another technique?
    $endgroup$
    – George Zorikov
    Dec 3 '18 at 13:44












  • $begingroup$
    @GeorgeZorikov Is it better now?
    $endgroup$
    – Botond
    Dec 3 '18 at 13:50










  • $begingroup$
    now I get it! Thank you again
    $endgroup$
    – George Zorikov
    Dec 3 '18 at 13:57




















  • $begingroup$
    Thank you! But may I ask you one more question? When I try to find $c_1$ I use integration by parts. And I cannot solve it (especially $c_1 = pi$) because I have $u(t)$ undefined. Maybe I should calculate it using another technique?
    $endgroup$
    – George Zorikov
    Dec 3 '18 at 13:44












  • $begingroup$
    @GeorgeZorikov Is it better now?
    $endgroup$
    – Botond
    Dec 3 '18 at 13:50










  • $begingroup$
    now I get it! Thank you again
    $endgroup$
    – George Zorikov
    Dec 3 '18 at 13:57


















$begingroup$
Thank you! But may I ask you one more question? When I try to find $c_1$ I use integration by parts. And I cannot solve it (especially $c_1 = pi$) because I have $u(t)$ undefined. Maybe I should calculate it using another technique?
$endgroup$
– George Zorikov
Dec 3 '18 at 13:44






$begingroup$
Thank you! But may I ask you one more question? When I try to find $c_1$ I use integration by parts. And I cannot solve it (especially $c_1 = pi$) because I have $u(t)$ undefined. Maybe I should calculate it using another technique?
$endgroup$
– George Zorikov
Dec 3 '18 at 13:44














$begingroup$
@GeorgeZorikov Is it better now?
$endgroup$
– Botond
Dec 3 '18 at 13:50




$begingroup$
@GeorgeZorikov Is it better now?
$endgroup$
– Botond
Dec 3 '18 at 13:50












$begingroup$
now I get it! Thank you again
$endgroup$
– George Zorikov
Dec 3 '18 at 13:57






$begingroup$
now I get it! Thank you again
$endgroup$
– George Zorikov
Dec 3 '18 at 13:57













2












$begingroup$

$defR{mathbb{R}}defd{mathrm{d}}defpeq{mathrel{phantom{=}}{}}$Assume that $u$ is integrable on $[0, 2π]$. For $x in R$ and $Δx ≠ 0$, becausebegin{align*}
frac{1}{Δx} (u(x + Δx) - u(x)) &= -int_0^{2π} frac{1}{Δx} (cos(x + t + Δx) - cos(x + t)) u(t) ,d t\
&peq + (π + 1) frac{1}{Δx} (cos(x + Δx) - cos x),
end{align*}
making $Δx → 0$ with the dominated convergence theorem yields$$
u'(x) = int_0^{2π} sin(x + t) u(t) ,d t - (π + 1) sin x.
$$
Analogously,$$
u''(x) = int_0^{2π} cos(x + t) u(t) ,d t - (π + 1) cos x.
$$
Thus $u'' + u = 0$, which implies $u(x) = c_1 cos x + c_2 sin x$ for some constant $c_1$ and $c_2$.



Now, sincebegin{align*}
int_0^{2π} cos(x + t) u(t) ,d t &= c_1 int_0^{2π} cos(x + t) cos t ,d t + c_2 int_0^{2π} cos(x + t) sin t ,d t\
&= π(c_1 cos x - c_2 sin x),
end{align*}
then$$
u(x) + int_0^{2π} cos(x + t) u(t) ,d t = c_1 (π + 1) cos x - c_2 π sin x,
$$
which implies $c_1 = 1$, $c_2 = 0$. Therefore, $u(x) = cos x$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much!
    $endgroup$
    – George Zorikov
    Dec 3 '18 at 13:16
















2












$begingroup$

$defR{mathbb{R}}defd{mathrm{d}}defpeq{mathrel{phantom{=}}{}}$Assume that $u$ is integrable on $[0, 2π]$. For $x in R$ and $Δx ≠ 0$, becausebegin{align*}
frac{1}{Δx} (u(x + Δx) - u(x)) &= -int_0^{2π} frac{1}{Δx} (cos(x + t + Δx) - cos(x + t)) u(t) ,d t\
&peq + (π + 1) frac{1}{Δx} (cos(x + Δx) - cos x),
end{align*}
making $Δx → 0$ with the dominated convergence theorem yields$$
u'(x) = int_0^{2π} sin(x + t) u(t) ,d t - (π + 1) sin x.
$$
Analogously,$$
u''(x) = int_0^{2π} cos(x + t) u(t) ,d t - (π + 1) cos x.
$$
Thus $u'' + u = 0$, which implies $u(x) = c_1 cos x + c_2 sin x$ for some constant $c_1$ and $c_2$.



Now, sincebegin{align*}
int_0^{2π} cos(x + t) u(t) ,d t &= c_1 int_0^{2π} cos(x + t) cos t ,d t + c_2 int_0^{2π} cos(x + t) sin t ,d t\
&= π(c_1 cos x - c_2 sin x),
end{align*}
then$$
u(x) + int_0^{2π} cos(x + t) u(t) ,d t = c_1 (π + 1) cos x - c_2 π sin x,
$$
which implies $c_1 = 1$, $c_2 = 0$. Therefore, $u(x) = cos x$.






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2












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$begingroup$

$defR{mathbb{R}}defd{mathrm{d}}defpeq{mathrel{phantom{=}}{}}$Assume that $u$ is integrable on $[0, 2π]$. For $x in R$ and $Δx ≠ 0$, becausebegin{align*}
frac{1}{Δx} (u(x + Δx) - u(x)) &= -int_0^{2π} frac{1}{Δx} (cos(x + t + Δx) - cos(x + t)) u(t) ,d t\
&peq + (π + 1) frac{1}{Δx} (cos(x + Δx) - cos x),
end{align*}
making $Δx → 0$ with the dominated convergence theorem yields$$
u'(x) = int_0^{2π} sin(x + t) u(t) ,d t - (π + 1) sin x.
$$
Analogously,$$
u''(x) = int_0^{2π} cos(x + t) u(t) ,d t - (π + 1) cos x.
$$
Thus $u'' + u = 0$, which implies $u(x) = c_1 cos x + c_2 sin x$ for some constant $c_1$ and $c_2$.



Now, sincebegin{align*}
int_0^{2π} cos(x + t) u(t) ,d t &= c_1 int_0^{2π} cos(x + t) cos t ,d t + c_2 int_0^{2π} cos(x + t) sin t ,d t\
&= π(c_1 cos x - c_2 sin x),
end{align*}
then$$
u(x) + int_0^{2π} cos(x + t) u(t) ,d t = c_1 (π + 1) cos x - c_2 π sin x,
$$
which implies $c_1 = 1$, $c_2 = 0$. Therefore, $u(x) = cos x$.






share|cite|improve this answer









$endgroup$



$defR{mathbb{R}}defd{mathrm{d}}defpeq{mathrel{phantom{=}}{}}$Assume that $u$ is integrable on $[0, 2π]$. For $x in R$ and $Δx ≠ 0$, becausebegin{align*}
frac{1}{Δx} (u(x + Δx) - u(x)) &= -int_0^{2π} frac{1}{Δx} (cos(x + t + Δx) - cos(x + t)) u(t) ,d t\
&peq + (π + 1) frac{1}{Δx} (cos(x + Δx) - cos x),
end{align*}
making $Δx → 0$ with the dominated convergence theorem yields$$
u'(x) = int_0^{2π} sin(x + t) u(t) ,d t - (π + 1) sin x.
$$
Analogously,$$
u''(x) = int_0^{2π} cos(x + t) u(t) ,d t - (π + 1) cos x.
$$
Thus $u'' + u = 0$, which implies $u(x) = c_1 cos x + c_2 sin x$ for some constant $c_1$ and $c_2$.



Now, sincebegin{align*}
int_0^{2π} cos(x + t) u(t) ,d t &= c_1 int_0^{2π} cos(x + t) cos t ,d t + c_2 int_0^{2π} cos(x + t) sin t ,d t\
&= π(c_1 cos x - c_2 sin x),
end{align*}
then$$
u(x) + int_0^{2π} cos(x + t) u(t) ,d t = c_1 (π + 1) cos x - c_2 π sin x,
$$
which implies $c_1 = 1$, $c_2 = 0$. Therefore, $u(x) = cos x$.







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answered Dec 3 '18 at 12:44









SaadSaad

19.7k92352




19.7k92352












  • $begingroup$
    Thank you very much!
    $endgroup$
    – George Zorikov
    Dec 3 '18 at 13:16


















  • $begingroup$
    Thank you very much!
    $endgroup$
    – George Zorikov
    Dec 3 '18 at 13:16
















$begingroup$
Thank you very much!
$endgroup$
– George Zorikov
Dec 3 '18 at 13:16




$begingroup$
Thank you very much!
$endgroup$
– George Zorikov
Dec 3 '18 at 13:16


















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