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

Question

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: $$u(t)\sin(x+t)\bigg|_0^{2\pi} - \int_{0}^{2\pi}\sin(x+t)u'(t)\,\mathrm{d}t = (\pi + 1)\cos x - u(x).$$ I think it is a wrong way to solve this equation.

Any suggestion for how to solve it would be appreciated.

Answer 1

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

Answer 2

$\def\R{\mathbb{R}}\def\d{\mathrm{d}}\def\peq{\mathrel{\phantom{=}}{}}$Assume that $u$ is integrable on $[0, 2π]$. For $x \in \R$ and $Δx ≠ 0$, because\begin{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, since\begin{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, $u(x) = \cos x$.