I am working on exercise in Fourier analysis, but it really confused me since it involves some differential equation.
Define $B_{1}:=\{x\in\mathbb{R}^{2}:x_{1}^{2}+x_{2}^{2}<1\}$. Note that $\partial B_{1}=\mathbb{S}^{1}$. Let $f\in C^{\infty}(\mathbb{S}^{1})$ and $u$ be the harmonic extension of $f$ to $B_{1}$. Let $\nu$ be the unit outer normal direction of $\mathbb{S}^{1}$, we define the Dirichlet to Neumann operator $\mathcal{A}$ by $\mathcal{A}f:=\dfrac{\partial u}{\partial \nu}.$
(a) Let $\alpha>0$, $\alpha\notin\mathbb{N}$. If $f\in C^{\infty}(\mathbb{S}^{1})$ is a solution to $\mathcal{A}f+\alpha=e^{f},$ then show that $f$ must be a constant, i.e. $f=\log\alpha$.
(b) What happens if $\alpha\in \mathbb{N}$?
For the first one, I tired to use Fourier expansion of $f$ to compute the Fourier coefficients but I failed...
For the second one, what is the difference between $\alpha\in\mathbb{N}$ and $\alpha\notin\mathbb{N}$?
I am sorry for not giving enough details since I really don't have idea about this exercise..
Thank you!
Edit 1: (Partial Solution)
Okay, I figured out a proof for $\alpha\notin\mathbb{N}$. I also worked out some part of $\alpha\in\mathbb{N}$ but could not finish, so if anyone has a refined proof, please let me know.
I believe if $\alpha\notin\mathbb{N}$ does not make the solution different, but the case of $\alpha\in\mathbb{N}$ is much much more complicated, and I could not prove it completely.
I have answered my own post.
Let $f\in C^{\infty}(\mathbb{S}^{1})$ satisfy $\mathcal{A}f+\alpha=e^{f}$. Then, $f$ has the Fourier expansion $f(\theta)=\sum_{k=-\infty}^{\infty}\widehat{f}(k)e_{k}(\theta),$ where $e_{k}(\theta)=e^{ik\theta}.$ Then, we know that the harmonic extension $u$ of it has the form $$u(re^{i\theta})=\sum_{k=-\infty}^{\infty}\widehat{f}(k)r^{|k|}e_{k}(\theta)=(P_{r}*f)(\theta),\ \text{where}\ P_{r}\ \text{is the Poisson kernel.}$$
Now, note that for any point $(x_{1}, x_{2})\in\partial B_{1}=\mathbb{S}^{1}$, we always have $\nu=(x_{1}, x_{2})$, so that $$\dfrac{\partial }{\partial \nu}=x_{1}\dfrac{\partial }{\partial x_{1}}+x_{2}\dfrac{\partial }{\partial x_{2}},$$ but the parametrization of $x_{1}=r\cos(\theta)$ and $x_{2}=r\sin(\theta)$ yields us $$x_{1}\dfrac{\partial }{\partial x_{1}}+x_{2}\dfrac{\partial}{\partial x_{2}}=r\dfrac{\partial }{\partial r}.$$ Hence, we have \begin{align*} \mathcal{A}f=\dfrac{\partial u}{\partial \nu}=r\dfrac{\partial u}{\partial r}&=\Big(r\sum_{k=-\infty}^{\infty}\widehat{f}(k)|k|r^{|k|-1}e_{k}(\theta)\Big)_{r=1}\\ &=\Big(\sum_{k=-\infty}^{\infty}\widehat{f}(k)|k|r^{|k|}e_{k}(\theta)\Big)_{r=1}\\ &=\sum_{k=-\infty}^{\infty}\widehat{f}(k)|k|e_{k}(\theta). \end{align*}
Now, take derivative with respect to $\theta$ on the both side of $\mathcal{A}f+\alpha=e^{f}$, which gives us $$\dfrac{d\mathcal{A}f}{d\theta}=\dfrac{df}{d\theta}e^{f}=\dfrac{df}{d\theta}(\mathcal{A}f+\alpha).$$ Then, using the above Fours expansion, the LHS can be computed as $$\dfrac{d\mathcal{A}f}{d\theta}=\sum_{k=-\infty}^{\infty}\widehat{f}(k)|k|\dfrac{de_{k}(\theta)}{d\theta}=\sum_{k=-\infty}^{\infty}\widehat{f}(k)|k|ike_{k}(\theta),$$ while the RHS is $$\dfrac{df}{d\theta}(\mathcal{A}f+\alpha)=\sum_{k=-\infty}^{\infty}\widehat{f}(k)ike_{k}(\theta)\Big(\sum_{k=-\infty}^{\infty}\widehat{f}(k)|k|e_{k}(\theta)+\alpha\Big).$$
Hence, we have $$\sum_{k=-\infty}^{\infty}\widehat{f}(k)|k|ike_{k}(\theta)=\sum_{k=-\infty}^{\infty}\widehat{f}(k)ike_{k}(\theta)\Big(\sum_{k=-\infty}^{\infty}\widehat{f}(k)|k|e_{k}(\theta)+\alpha\Big),$$ which gives us $$\sum_{k=-\infty}^{\infty}\widehat{f}(k)(|k|-\alpha)ke_{k}=\sum_{k=-\infty}^{\infty}\Big(\sum_{m=-\infty}^{\infty}\widehat{f}(k-m)\cdot (k-m)\cdot \widehat{f}(m)\cdot |m|\Big)e_{k}(\theta),$$ which implies that the below identity $$\widehat{f}(k)(|k|-\alpha)k=\sum_{m=-\infty}^{\infty}\widehat{f}(k-m)\widehat{f}(m)(k-m)|m|\ \ \ \ \ \ (*)$$ holds for all $k$.
If $k>0$, we can write the RHS as $$RHS=\sum_{m=-\infty}^{-1}\widehat{f}(k-m)\widehat{f}(m)(k-m)(-m)+\sum_{m=0}^{k}\widehat{f}(k-m)\widehat{f}(m)(k-m)m+\sum_{m=k+1}^{\infty}\widehat{f}(k-m)\widehat{f}(m)(k-m)m,$$ and thus if we do the change of variable $n:=k-m$ in the third term, we then have $$RHS=\sum_{m=-\infty}^{-1}\widehat{f}(k-m)\widehat{f}(m)(k-m)(-m)+\sum_{m=0}^{k}\widehat{f}(k-m)\widehat{f}(m)(k-m)m+\sum_{n=-\infty}^{-1}\widehat{f}(k-n)\widehat{f}(n)(k-n)n,$$ so that the first term and the third term cancel, which leaves us $$RHS=\sum_{m=0}^{k}\widehat{f}(k-m)\widehat{f}(m)(k-m)m.$$
Hence, for all $k>0$, $(*)$ could be rewritten as $$\widehat{f}(k)(|k|-\alpha)k=\sum_{m=0}^{k}\widehat{f}(k-m)\widehat{f}(m)(k-m)m.$$
To continue, let us firstly suppose $\alpha>0, \alpha\notin\mathbb{N}$. Then, I claim that with the above new identity, the Fourier coefficient $\widehat{f}(k)=0$ for all $k\geq 1$.
We will prove it by (strong) induction on $k$.
Firstly, consider this new identity with $k=1$, we see that $LHS=\widehat{f}(1)(1-\alpha)$ while $RHS=\widehat{f}(1)\widehat{f}(0)(1-0)\cdot 0+\widehat{f}(0)\widehat{f}(1)(1-1)\cdot 1=0+0=0,$ and thus we have $$\widehat{f}(1)(1-\alpha)=0,$$ but $\alpha\notin\mathbb{N}$, so we must have $\widehat{f}(1)=0$.
For a fixed $n\geq 2$, suppose $\widehat{f}(k)=0$ for all $2\leq k\leq n$, then consider the case of $k=n+1$. Then, $LHS=\widehat{f}(n+1)(n+1-\alpha)(n+1)$ while $RHS=\sum_{m=0}^{n+1}\widehat{f}(n+1-n )\widehat{f}(m)(n+1-m)m=0$ by induction hypothesis, and thus we have $$\widehat{f}(n+1)(n+1-\alpha)(n+1)=0,$$ but $\alpha\notin\mathbb{N}$, so $\widehat{f}(n+1)=0.$
Hence, it follows by induction that $\widehat{f}(k)=0$ for all $k\geq 1$.
On the other hand, for all $k<0$, using the similar computation, we see that $(*)$ can be rewritten as $$\widehat{f}(k)(|k|-\alpha)k=\sum_{m=k}^{0}\widehat{f}(k-m)\widehat{f}(m)(k-m)m.$$ Then assuming $\alpha>0, \alpha\notin\mathbb{N}$, and using the similar computation, we see that when $k=-1$, we have $$-\widehat{f}(-1)(1-\alpha)=0$$ but $\alpha\notin\mathbb{N}$, so we must have $\widehat{f}(-1)=0$.
Then a similar inductive argument enables us to conclude $\widehat{f}(k)=0$ for all $k\leq -1$. Hence, for all $k\in\mathbb{Z}\setminus\{0\}$, if $\alpha\notin\mathbb{N}$, we must have $\widehat{f}(k)=0$.
Hence, if $\alpha\notin\mathbb{N}$, $f=\widehat{f}(0)$ and thus must be a constant function. Further, if $\widehat{f}(k)=0$ for all $k\neq 0$, then $\mathcal{A}f=\widehat{f}(0)\cdot |0|\cdot e_{0}(\theta)=0$, but $f$ satisfies the differential equation, so we have $0+\alpha=e^{f}$ and thus $f=\log\alpha$.
In conclusion, in the case of $\alpha>0,\alpha\notin\mathbb{N}$, $\widehat{f}(k)=0$ for all $k\neq 0$, and consequently $f=\log\alpha$, a constant.
Now, we discuss the the case where $\alpha\in\mathbb{N}_{>0}$. Since we did not change anything else, we still have the below identity for all $k>0$: $$\widehat{f}(k)(|k|-\alpha)k=\sum_{m=0}^{k}\widehat{f}(k-m)\widehat{f}(m)(k-m)m.$$
Then for all $0<k<\alpha$, \textit{(this case does not exist if $\alpha=1$)}, by the spirit of our proof in $\alpha\notin\mathbb{N}$, we see that $\widehat{f}(k)=0$. Indeed, see in the case of $k=1$, we have $\widehat{f}(1)(1-\alpha)=0$ but $\alpha\neq 1$, so $\widehat{f}(1)=0$. Then if $k=2$, then the sum in RHS is again $0+0+0$, and since $\alpha\neq 2$, we see $\widehat{f}(2)=0$. Iterating, we need to stop at $k=\alpha-1$, and one could conclude that $\widehat{f}(k)=0$ for all $1\leq k\leq \alpha-1$.
For the case of $k=\alpha$, we cannot say anything about $\widehat{f}(\alpha)$. Since $LHS=\widehat{f}(\alpha)\cdot(\alpha-\alpha)\cdot \alpha$ and $RHS=0+0+0+\cdots+0=0$, so the above identity holds for any value of $\widehat{f}(\alpha)$.
Then, for all $k\geq \alpha+1$ that is not a multiple of $\alpha$, \textit{(this case does not exist if $\alpha=1$, since everything is a multiple of it)}, we see that $\widehat{f}(k)=0$ again. One could see this by observing that the first term and last term of the summation in $RHS$ is always $0$, and for all other terms, you always have a multiplication of $\widehat{f}(\ell)=0$ for some $1\leq\ell\leq \alpha-1$.
However, for $k>\alpha+1$ and $k$ is a multiple of $\alpha$, that is $k=m\alpha$ for some $m>1$, then by some computation, one could see that, the sum on the RHS has the first and last term being $0$ and other terms having a multiplication of $1\leq\ell\leq \alpha-1$, except one term where you have a multiplication of $m$ numbers of $\widehat{f}(\alpha)$ with coefficients as $1/m$ multiple of the coefficients in the LHS. Thus, we have in this case, $$\widehat{f}(k)=\widehat{f}(m\alpha)=\dfrac{1}{m}[\widehat{f}(\alpha)]^{m}.$$
Thus, overall for all $k>0$, $k\neq m\alpha$ for $m\geq 1$, $\widehat{f}(k)=0$, and for all $k>0$, $k=m\alpha$ for $m\geq 1$, we have $\widehat{f}(k)=\frac{1}{m}[\widehat{f}(\alpha)]^{m}$.
Similarly, for all $k<0$, the below identity still holds: $$\widehat{f}(k)(|k|-\alpha)k=\sum_{m=k}^{0}\widehat{f}(k-m)\widehat{f}(m)(k-m)m.$$ By the similar argument, we see that for all $k<0$, $k\neq -m\alpha$ for all $m\geq 1$, $\widehat{f}(k)=0$, and for all $k<0$, $k=-m\alpha$ for all $m\geq 1$, we have $\widehat{f}(k)=-\frac{1}{m}[\widehat{f}(-\alpha)]^{m}.$
Thus, we can write the Fourier expansion of $f$ as: \begin{align*} f&=\widehat{f}(0)+\sum_{m=1}^{\infty}\dfrac{1}{m}[\widehat{f}(\alpha)]^{m}e_{\alpha m}(\theta)-\sum_{m=1}^{\infty}\dfrac{1}{m}[\widehat{f}(-\alpha)]^{m}e_{-\alpha m}(\theta) \end{align*} and \begin{align*} \mathcal{A}f&=\sum_{m=1}^{\infty}\dfrac{1}{m}[\widehat{f}(\alpha)]^{m}|\alpha m|e_{\alpha m}(\theta)-\sum_{m=1}^{\infty}\dfrac{1}{m}[\widehat{f}(-\alpha)]^{m}|-\alpha m|e_{-\alpha m}(\theta)\\ &=\sum_{m=1}^{\infty}\alpha [\widehat{f}(\alpha)]^{m}e_{\alpha m}(\theta)-\sum_{m=1}^{\infty}\alpha [\widehat{f}(-\alpha)]^{m}e_{-\alpha m}(\theta)\\ &=\sum_{m=1}^{\infty}\alpha[\widehat{f}(\alpha)e_{\alpha}(\theta)]^{m}-\sum_{m=1}^{\infty}\alpha[\widehat{f}(-\alpha)e_{-\alpha}(\theta)]^{m} \end{align*}
Now, we compute $\frac{df}{d\theta}$, which is \begin{align*} \dfrac{df}{d\theta}&=\sum_{m=1}^{\infty}\dfrac{1}{m}[\widehat{f}(\alpha)]^{m}i\alpha me_{\alpha m}(\theta)-\sum_{m=1}^{\infty}\dfrac{1}{m}[\widehat{f}(-\alpha)]^{m}(-i\alpha m)e_{-\alpha m}(\theta)\\ &=\sum_{m=1}^{\infty}i\alpha[\widehat{f}(\alpha)e_{\alpha}(\theta)]^{m}+\sum_{m=1}^{\infty}i\alpha[\widehat{f}(-\alpha)e_{-\alpha}(\theta)]^{m}. \end{align*}
Note that by Riemann Lebesgue lemma, we see that $\frac{1}{m}[\widehat{f}(\alpha)]^{m}\longrightarrow 0$ and thus $|\widehat{f}(\alpha)|\leq 1$. Similarly, we see that $|\widehat{f}(-\alpha)|\leq 1.$
If $|\widehat{f}(\alpha)|< 1$ and $|\widehat{f}(-\alpha)|<1,$ then the above derivative converges as a geometric series: $$\dfrac{df}{d\theta}=i\alpha\dfrac{\widehat{f}(\alpha)e_{\alpha}(\theta)}{1-\widehat{f}(\alpha)e_{\alpha}(\theta)}+i\alpha\dfrac{\widehat{f}(-\alpha)e_{-\alpha}(\theta)}{1-\widehat{f}(-\alpha)e_{-\alpha}(\theta)},$$ so that an integral with respect to $\theta$ gives back the formula of $f(\theta)$: $$f(\theta)=-\alpha\log[1-\widehat{f}(\alpha)e_{\alpha}(\theta)]+\alpha\log[1-\widehat{f}(-\alpha)e_{-\alpha}(\theta)]+C_{0},$$ and thus $$e^{f}=C_{1}\dfrac{[1-\widehat{f}(-\alpha)e_{-\alpha}(\theta)]^{\alpha}}{[1-\widehat{f}(\alpha)e_{\alpha}(\theta)]^{\alpha}}.$$
Now, note that we can write \begin{align*} \dfrac{[1-\widehat{f}(-\alpha)e_{-\alpha}(\theta)]}{[1-\widehat{f}(\alpha)e_{\alpha}(\theta)]}&=[1-\widehat{f}(-\alpha)e_{-\alpha}(\theta)]\sum_{k=0}^{\infty}[\widehat{f}(\alpha)e_{\alpha}(\theta)]^{k}\\ &=\sum_{k=0}^{\infty}\Big([\widehat{f}(\alpha)e_{\alpha}(\theta)]^{k}-[\widehat{f}(\alpha)]^{k}\widehat{f}(-\alpha)[e_{\alpha}(\theta)]^{k-1}\Big)\\ &=(\text{something})+(\text{something})e_{-\alpha}(\theta)+(\text{something})e_{\alpha}(\theta)+(\text{something})e_{2\alpha}(\theta)+\cdots, \end{align*} here the coefficients are not important, we only need the ``most negative term'' is finite.
Using this, we can further write \begin{align*} e^{f}=(\text{something})e_{-\alpha^{2}}(\theta)&+(\text{something})e_{-\alpha^{2}+\alpha}(\theta)+\cdots\\ &+(\text{something})e_{-\alpha}(\theta)+(\text{something})\cdot 1+(\text{something})e_{\alpha}(\theta)+\cdots\\ &=\mathcal{A}f+\alpha. \end{align*} But $e^{f}$ depends only up to $e_{-k}$ for $k\leq \alpha^{2}$, and thus $\widehat{f}(k)=0$ for all $k\leq -\alpha^{2}$.
This implies that we can rewrite $$f=\widehat{f}(0)+\sum_{k=1}^{\infty}\dfrac{1}{k}[\widehat{f}(\alpha)e_{\alpha}(\theta)]^{k}\ \text{and}\ \mathcal{A}f=\sum_{k=1}^{\infty}[\widehat{f}(\alpha)e_{\alpha}(\theta)]^{k},$$ then note that $$\dfrac{df}{d\theta}=\sum_{k=1}^{\infty}i\alpha[\widehat{f}(\alpha)e_{\alpha}(\theta)]^{k}, $$ and thus $$\dfrac{\frac{df}{d\theta}}{i\alpha}=\mathcal{A}f,$$ and hence $$\dfrac{1}{i\alpha}\dfrac{df}{d\theta}+\alpha=e^{f}.$$
Set $F:=e^{f}$, then we have $$\dfrac{1}{\alpha}\dfrac{\frac{dF}{d\theta}}{F}+\alpha=F$$ $$\implies -\dfrac{1}{i\alpha}\dfrac{d}{d\theta}\Big(\dfrac{1}{F}\Big)+\dfrac{\alpha}{F}=1,$$ so if we set $G:=\dfrac{1}{F}$, we have $$\dfrac{i}{\alpha}\dfrac{dG}{d\theta}+\alpha G=1$$ $$\implies \dfrac{d}{d\theta}\Big(G\dfrac{i}{\alpha}e^{\frac{\alpha^{2}}{i}x}\Big)=e^{\frac{\alpha^{2}}{i}x}$$ $$\implies G\dfrac{i}{\alpha}e^{\frac{\alpha^{2}}{i}x}=\dfrac{i}{\alpha^{2}}e^{\frac{\alpha^{2}x}{i}}+C$$ $$\implies G=\dfrac{1}{\alpha}+C\dfrac{i}{\alpha}e^{-\frac{\alpha^{2}}{i}x}=\dfrac{1}{\alpha}\Big(1+Ce^{-\frac{\alpha^{2}}{i}x}\Big)$$ $$\implies F=\dfrac{1}{G}=\dfrac{\alpha}{1+Ce^{\frac{-\alpha^{2}}{i}x}}$$ $$\implies f=\log F=\log\Big(\dfrac{\alpha}{1+Ce^{\frac{-\alpha^{2}}{i}x}}\Big).$$
This is all I can get, and I do not know how to deal with the case of $|\widehat{f}(\alpha)|=1$ and $|\widehat{f}(-\alpha)|=1.$