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The normal trick is to set it up as an eigenvalue problem, namely to look instead at $$\Delta u = 0, ~~\frac{\partial u}{\partial n} = \lambda b(x) u(x)~~ \forall x \in \partial D. (1)$$ You first establish that this corresponds to an eigenvalue problem on $L^2(\partial \Omega)$ for a compact operator, thanks to the compact embedding of $H^{1/2}(\partial D)$ into $L^2((\partial D)$.

Then the usual machinery rolls out. There is a discrete spectrum, and there are eigenvalues. And if it so happens that $\lambda=1$ is one of those, then your problem has non trivial solutions.

The advantage of this point of view is that the problem is well posed, there is a clear functional analytic setup, and you don't need much on $b$, some integrability (if it is bounded as you liked to consider, it is fine).

It is done in the following way. Given $f \in H^{-1/2}(\partial D)$ such that $\int_{\partial\Omega} f d\sigma= 0$ consider the so called Neumann-to-Dirichlet map \begin{eqnarray*} NtD : H^{-1/2}(\partial D)/\mathbb R &\to& H^{-1/2}(\partial D)/\mathbb R \\ f &\to& u|_{\partial D} : \begin{cases} -\Delta u =0& \text{ in }D \\ \partial_n u =f &\text{ on }\partial D \end{cases} \end{eqnarray*} the quotient over $\mathbb R$ means that we impose $\int_{\partial\Omega} f d\sigma= \int_{\partial\Omega} u d\sigma= 0.$

The operator $NtD$ is compact : we are going to use this operator on a restricted domain, where the compactness is very easy to see.

Restrict this operator to $L^2(\partial D)/ \mathbb R$, and you can write it explicitely in terms of Fourier coefficients. Write $$ f=\sum_{n=1}^\infty a_n \cos n \theta + b_n \sin n \theta, \sum |a_n|^2 + |b_n|^2 <\infty $$ then $$ NtD(f)= \sum_{n=1}^\infty \frac{a_n}{n} \cos n \theta + \frac{b_n}{n} \sin n \theta. $$ It is easy to see that it is a compact operator from $L^{2}(\partial D)/\mathbb R$ to $L^{2}(\partial D)/\mathbb R$ (the $n^{-1}$ helps) .

Supppose $\| b \|_{L^\infty(\partial D)} <\infty$. Then \begin{eqnarray*} B : L^{2}(\partial D)/\mathbb R &\to& L^{2}(\partial D)/\mathbb R \\ f &\to& b f - \frac{1}{\left|\partial D\right|} \int_{\partial D}bf \end{eqnarray*} is a continuous, bounded operator from $ L^{2}(\partial D)/\mathbb R$ into itself.

So the map \begin{eqnarray*} T : L^{2}(\partial D)/\mathbb R &\to& L^{2}(\partial D)/\mathbb R \\ f &\to& B\left(NtD \left(f\right)\right) \end{eqnarray*} is a compact, linear operator. Its spectrum is discrete, it has a maximal eigenvalue etc. and (1) is $$ f=\lambda T f. $$

Regarding regularity of the solutions. Naturally, they are analytic inside $D(0,\rho)$ for any $\rho<1$, this is just interior regularity of harmonic functions. At the boundary, the regularity is dictated by $b$. Indeed, suppose $b$ is piecewise constant. We have $Du= \partial_r u e_r + \frac{1}{r} \partial_\theta u e_\theta= \lambda b u e_r + \frac{1}{r} \partial_\theta u e_\theta $. If $u$ is regular, this formula can be "extended" a little inside. Then you see immediately that you cannot differentiate $u$ a second time: all terms are differentiable, except one, $b$.

The normal trick is to set it up as an eigenvalue problem, namely to look instead at $$\Delta u = 0, ~~\frac{\partial u}{\partial n} = \lambda b(x) u(x)~~ \forall x \in \partial D. (1)$$ You first establish that this corresponds to an eigenvalue problem on $L^2(\partial \Omega)$ for a compact operator, thanks to the compact embedding of $H^{1/2}(\partial D)$ into $L^2((\partial D)$.

Then the usual machinery rolls out. There is a discrete spectrum, and there are eigenvalues. And if it so happens that $\lambda=1$ is one of those, then your problem has non trivial solutions.

The advantage of this point of view is that the problem is well posed, there is a clear functional analytic setup, and you don't need much on $b$, some integrability (if it is bounded as you liked to consider, it is fine).

It is done in the following way. Given $f \in H^{-1/2}(\partial D)$ such that $\int_{\partial\Omega} f d\sigma= 0$ consider the so called Neumann-to-Dirichlet map \begin{eqnarray*} NtD : H^{-1/2}(\partial D)/\mathbb R &\to& H^{-1/2}(\partial D)/\mathbb R \\ f &\to& u|_{\partial D} : \begin{cases} -\Delta u =0& \text{ in }D \\ \partial_n u =f &\text{ on }\partial D \end{cases} \end{eqnarray*} the quotient over $\mathbb R$ means that we impose $\int_{\partial\Omega} f d\sigma= \int_{\partial\Omega} u d\sigma= 0.$

The operator $NtD$ is compact : we are going to use this operator on a restricted domain, where the compactness is very easy to see.

Restrict this operator to $L^2(\partial D)/ \mathbb R$, and you can write it explicitely in terms of Fourier coefficients. Write $$ f=\sum_{n=1}^\infty a_n \cos n \theta + b_n \sin n \theta, \sum |a_n|^2 + |b_n|^2 <\infty $$ then $$ NtD(f)= \sum_{n=1}^\infty \frac{a_n}{n} \cos n \theta + \frac{b_n}{n} \sin n \theta. $$ It is easy to see that it is a compact operator from $L^{2}(\partial D)/\mathbb R$ to $L^{2}(\partial D)/\mathbb R$ (the $n^{-1}$ helps) .

Supppose $\| b \|_{L^\infty(\partial D)} <\infty$. Then \begin{eqnarray*} B : L^{2}(\partial D)/\mathbb R &\to& L^{2}(\partial D)/\mathbb R \\ f &\to& b f - \frac{1}{\left|\partial D\right|} \int_{\partial D}bf \end{eqnarray*} is a continuous, bounded operator from $ L^{2}(\partial D)/\mathbb R$ into itself.

So the map \begin{eqnarray*} T : L^{2}(\partial D)/\mathbb R &\to& L^{2}(\partial D)/\mathbb R \\ f &\to& B\left(NtD \left(f\right)\right) \end{eqnarray*} is a compact, linear operator. Its spectrum is discrete, it has a maximal eigenvalue etc. and (1) is $$ f=\lambda T f. $$

Regarding regularity of the solutions. Naturally, they are analytic inside $D(0,\rho)$ for any $\rho<1$, this is just interior regularity of harmonic functions. At the boundary, the regularity is dictated by $b$. Indeed, suppose $b$ is piecewise constant. We have $Du= \partial_r u e_r + \frac{1}{r} \partial_\theta u e_\theta= \lambda b u e_r + \frac{1}{r} \partial_\theta u e_\theta $. If $u$ is regular, this formula can be "extended" a little inside. Then you see immediately that you cannot differentiate $u$ a second time: all terms are differentiable, except one, $b$. However, nothing forbid $Du$ to be very integrable, so it is: $u\in W^{1,p}(D)$ for every $p<\infty$ (and maybe $\infty$ as well). So $u\in W^{1-1/p,p}(\partial D)$.

The normal trick is to set it up as an eigenvalue problem, namely to look instead at $$\Delta u = 0, ~~\frac{\partial u}{\partial n} = \lambda b(x) u(x)~~ \forall x \in \partial D. (1)$$ You first establish that this corresponds to an eigenvalue problem on $L^2(\partial \Omega)$ for a compact operator, thanks to the compact embedding of $H^{1/2}(\partial D)$ into $L^2((\partial D)$.

Then the usual machinery rolls out. There is a discrete spectrum, and there are eigenvalues. And if it so happens that $\lambda=1$ is one of those, then your problem has non trivial solutions.

The advantage of this point of view is that the problem is well posed, there is a clear functional analytic setup, and you don't need much on $b$, some integrability (if it is bounded as you liked to consider, it is fine).

It is done in the following way. Given $f \in H^{-1/2}(\partial D)$ such that $\int_{\partial\Omega} f d\sigma= 0$ consider the so called Neumann-to-Dirichlet map \begin{eqnarray*} NtD : H^{-1/2}(\partial D)/\mathbb R &\to& H^{-1/2}(\partial D)/\mathbb R \\ f &\to& u|_{\partial D} : \begin{cases} -\Delta u =0& \text{ in }D \\ \partial_n u =f &\text{ on }\partial D \end{cases} \end{eqnarray*} the quotient over $\mathbb R$ means that we impose $\int_{\partial\Omega} f d\sigma= \int_{\partial\Omega} u d\sigma= 0.$

The operator $NtD$ is compact, but we are going to use only that it is well defined and continuous.

Now restrict this operator to $L^2(\partial D)/ \mathbb R$, and you can even write it explicitely in terms of Fourier coefficients. $$ f=\sum_{n=1}^\infty a_n \cos n \theta + b_n \sin n \theta, \sum |a_n|^2 + |b_n|^2 <\infty $$ then $$ NtD(f)= \sum_{n=1}^\infty \frac{a_n}{n} \cos n \theta + \frac{b_n}{n} \sin n \theta. $$ It is easy to see that it is a compact operator (the $n^{-1}$ helps).

Supppose $\| b \|_{L^\infty(\partial D)} <\infty$. Then \begin{eqnarray*} B : L^{2}(\partial D)/\mathbb R &\to& L^{2}(\partial D)/\mathbb R \\ f &\to& b f - \frac{1}{\left|\partial D\right|} \int_{\partial D}bf \end{eqnarray*} is a continuous, bounded operator from $ L^{2}(\partial D)/\mathbb R$ into itself.

So the map \begin{eqnarray*} T : L^{2}(\partial D)/\mathbb R &\to& L^{2}(\partial D)/\mathbb R \\ f &\to& B\left(NtD \left(f\right)\right) \end{eqnarray*} is a compact, linear operator. Its spectrum is discrete, it has a maximal eigenvalue etc. and (1) is $$ f=\lambda T f. $$

The normal trick is to set it up as an eigenvalue problem, namely to look instead at $$\Delta u = 0, ~~\frac{\partial u}{\partial n} = \lambda b(x) u(x)~~ \forall x \in \partial D. (1)$$ You first establish that this corresponds to an eigenvalue problem on $L^2(\partial \Omega)$ for a compact operator, thanks to the compact embedding of $H^{1/2}(\partial D)$ into $L^2((\partial D)$.

Then the usual machinery rolls out. There is a discrete spectrum, and there are eigenvalues. And if it so happens that $\lambda=1$ is one of those, then your problem has non trivial solutions.

The advantage of this point of view is that the problem is well posed, there is a clear functional analytic setup, and you don't need much on $b$, some integrability (if it is bounded as you liked to consider, it is fine).

It is done in the following way. Given $f \in H^{-1/2}(\partial D)$ such that $\int_{\partial\Omega} f d\sigma= 0$ consider the so called Neumann-to-Dirichlet map \begin{eqnarray*} NtD : H^{-1/2}(\partial D)/\mathbb R &\to& H^{-1/2}(\partial D)/\mathbb R \\ f &\to& u|_{\partial D} : \begin{cases} -\Delta u =0& \text{ in }D \\ \partial_n u =f &\text{ on }\partial D \end{cases} \end{eqnarray*} the quotient over $\mathbb R$ means that we impose $\int_{\partial\Omega} f d\sigma= \int_{\partial\Omega} u d\sigma= 0.$

The operator $NtD$ is compact : we are going to use this operator on a restricted domain, where the compactness is very easy to see.

Restrict this operator to $L^2(\partial D)/ \mathbb R$, and you can write it explicitely in terms of Fourier coefficients. Write $$ f=\sum_{n=1}^\infty a_n \cos n \theta + b_n \sin n \theta, \sum |a_n|^2 + |b_n|^2 <\infty $$ then $$ NtD(f)= \sum_{n=1}^\infty \frac{a_n}{n} \cos n \theta + \frac{b_n}{n} \sin n \theta. $$ It is easy to see that it is a compact operator from $L^{2}(\partial D)/\mathbb R$ to $L^{2}(\partial D)/\mathbb R$ (the $n^{-1}$ helps)  .

Supppose $\| b \|_{L^\infty(\partial D)} <\infty$. Then \begin{eqnarray*} B : L^{2}(\partial D)/\mathbb R &\to& L^{2}(\partial D)/\mathbb R \\ f &\to& b f - \frac{1}{\left|\partial D\right|} \int_{\partial D}bf \end{eqnarray*} is a continuous, bounded operator from $ L^{2}(\partial D)/\mathbb R$ into itself.

So the map \begin{eqnarray*} T : L^{2}(\partial D)/\mathbb R &\to& L^{2}(\partial D)/\mathbb R \\ f &\to& B\left(NtD \left(f\right)\right) \end{eqnarray*} is a compact, linear operator. Its spectrum is discrete, it has a maximal eigenvalue etc. and (1) is $$ f=\lambda T f. $$

Regarding regularity of the solutions. Naturally, they are analytic inside $D(0,\rho)$ for any $\rho<1$, this is just interior regularity of harmonic functions. At the boundary, the regularity is dictated by $b$. Indeed, suppose $b$ is piecewise constant. We have $Du= \partial_r u e_r + \frac{1}{r} \partial_\theta u e_\theta= \lambda b u e_r + \frac{1}{r} \partial_\theta u e_\theta $. If $u$ is regular, this formula can be "extended" a little inside. Then you see immediately that you cannot differentiate $u$ a second time: all terms are differentiable, except one, $b$.

The normal trick is to set it up as an eigenvalue problem, namely to look instead at $$\Delta u = 0, ~~\frac{\partial u}{\partial n} = \lambda b(x) u(x)~~ \forall x \in \partial D.$$ You first establish that this corresponds to an eigenvalue problem on $L^2(\partial \Omega)$ for a compact operator, thanks to the compact embedding of $H^{1/2}(\partial \Omega)$ into $L^2((\partial \Omega)$.

Then the usual machinery rolls out. There is a discrete spectrum, and there are eigenvalues. And if it so happens that $\lambda=1$ is one of those, then your problem has non trivial solutions.

The advantage of this point of view is that the problem is well posed, there is a clear functional analytic setup, and you don't need much on $b$, some integrability (if it is bounded as you liked to consider, it is fine).

The normal trick is to set it up as an eigenvalue problem, namely to look instead at $$\Delta u = 0, ~~\frac{\partial u}{\partial n} = \lambda b(x) u(x)~~ \forall x \in \partial D. (1)$$ You first establish that this corresponds to an eigenvalue problem on $L^2(\partial \Omega)$ for a compact operator, thanks to the compact embedding of $H^{1/2}(\partial D)$ into $L^2((\partial D)$.

Then the usual machinery rolls out. There is a discrete spectrum, and there are eigenvalues. And if it so happens that $\lambda=1$ is one of those, then your problem has non trivial solutions.

The advantage of this point of view is that the problem is well posed, there is a clear functional analytic setup, and you don't need much on $b$, some integrability (if it is bounded as you liked to consider, it is fine).

It is done in the following way. Given $f \in H^{-1/2}(\partial D)$ such that $\int_{\partial\Omega} f d\sigma= 0$ consider the so called Neumann-to-Dirichlet map \begin{eqnarray*} NtD : H^{-1/2}(\partial D)/\mathbb R &\to& H^{-1/2}(\partial D)/\mathbb R \\ f &\to& u|_{\partial D} : \begin{cases} -\Delta u =0& \text{ in }D \\ \partial_n u =f &\text{ on }\partial D \end{cases} \end{eqnarray*} the quotient over $\mathbb R$ means that we impose $\int_{\partial\Omega} f d\sigma= \int_{\partial\Omega} u d\sigma= 0.$

The operator $NtD$ is compact, but we are going to use only that it is well defined and continuous.

Now restrict this operator to $L^2(\partial D)/ \mathbb R$, and you can even write it explicitely in terms of Fourier coefficients. $$ f=\sum_{n=1}^\infty a_n \cos n \theta + b_n \sin n \theta, \sum |a_n|^2 + |b_n|^2 <\infty $$ then $$ NtD(f)= \sum_{n=1}^\infty \frac{a_n}{n} \cos n \theta + \frac{b_n}{n} \sin n \theta. $$ It is easy to see that it is a compact operator (the $n^{-1}$ helps).

Supppose $\| b \|_{L^\infty(\partial D)} <\infty$. Then \begin{eqnarray*} B : L^{2}(\partial D)/\mathbb R &\to& L^{2}(\partial D)/\mathbb R \\ f &\to& b f - \frac{1}{\left|\partial D\right|} \int_{\partial D}bf \end{eqnarray*} is a continuous, bounded operator from $ L^{2}(\partial D)/\mathbb R$ into itself.

So the map \begin{eqnarray*} T : L^{2}(\partial D)/\mathbb R &\to& L^{2}(\partial D)/\mathbb R \\ f &\to& B\left(NtD \left(f\right)\right) \end{eqnarray*} is a compact, linear operator. Its spectrum is discrete, it has a maximal eigenvalue etc. and (1) is $$ f=\lambda T f. $$