Brezis' exercise 8.30.8: how to find the eigenvalues of this self-adjoint compact operator?

Question

Let $I$ be the open interval $(0, 1)$. Let $k \in \mathbb R \setminus \{1\}$. We consider the space $$ V := \{v \in H^1 (I) : v(0) = kv(1)\}, $$ and the symmetric bilinear form $a$ defined on $V$ by $$ a(u, v) = \int_I [ u'v' + uv ] - \left ( \int_I u \right) \left ( \int_I v \right). $$

I am trying to solve a problem in Brezis' Functional Analysis

Exercise 8.30

1. Check that $V$ is a closed subspace of $H^1 (I)$.

In what follows, $V$ is equipped with the Hilbert structure induced by the $H^1$ inner product.

2. Prove that $a$ is a continuous and coercive bilinear form on $V$.
3. Deduce that for every $f \in L^2 (I)$ there exists a unique solution of the problem $$ (1) \quad u \in V \quad \text{and} \quad a(u, v)=\int_I f v \quad \forall v \in V. $$
4. Show that the solution $u$ of $(1)$ belongs to $H^2 (I)$ and satisfies $$ (2) \quad \begin{cases} -u'' + u-\int_I u = f \quad \text {on} \quad I, \\ u(0)=k u(1) \text { and } u'(1)=k u'(0). \end{cases} $$
5. Conversely, prove that any function $u \in H^2(I)$ satisfying $(2)$ is a solution of $(1)$.
6. Let $k_n \in \mathbb{R} \setminus \{1\}$ for all $n$. Assume $k_n \xrightarrow{n \to \infty} k \neq 1$. Set $$ V_n = \{v \in H^1(I) : v(0)=k_n v(1)\} . $$ Given $f \in L^2 (I)$, let $u_n$ be the solution of $$ (1_n) \quad u_n \in V_n \quad \text { and } \quad a(u_n, v) = \int_I f v \quad \forall v \in V_n . $$ Prove that $u_n \to u$ in $H^1 (I)$, where $u$ is the solution of $(1)$. Deduce that $u_n \to u$ in $H^2 (I)$.
7. What happens to the sequence $\left(u_n\right)$ if $k_n$ converges to $1$?
8. Consider the operator $T: L^2 (I) \rightarrow L^2 (I)$ defined by $T f=u$, where $u$ is the solution of $(1)$. Show that $T$ is self-adjoint and compact. Study the set $E V(T)$ of eigenvalues of $T$.

I am trying to solve (8.) In below attempt, I come up with a system of $4$ linear equations to solve for $(A, B, C, \beta)$. Unfortunately, I don't know how to tackle this system. Could you elaborate on how to solve (8.) efficiently?

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Clearly, $T$ is injective, so $0 \notin EV (T)$. Let $\frac{1}{\lambda}$ be an eigenvalue and $u$ the corresponding eigenfunction. Then $$ (3) \quad \begin{cases} -u'' - \alpha u = \beta \quad \text {on} \quad I, \\ u(0)=k u(1) \\ u'(1)=k u'(0), \\ \alpha = \lambda-1, \\ \beta = \int_I u. \end{cases} $$

We consider three cases.

• $\alpha < 0$. Then $u$ has the form $$ u(x) = A e^{\sqrt{-\alpha} x} + B e^{-\sqrt{-\alpha} x} +C, $$ for some $A,B, C \in \mathbb R$ to be chosen. We have $$ \begin{align*} u'(x) &= A \sqrt{-\alpha} e^{\sqrt{-\alpha} x} - B \sqrt{-\alpha} e^{-\sqrt{-\alpha} x}, \\ u''(x) &= -A \alpha e^{\sqrt{-\alpha} x} - B \alpha e^{-\sqrt{-\alpha} x}. \end{align*} $$ Then $$ \begin{align*} u (0) &= A+B+C, \\ u (1) &= A e^{\sqrt{-\alpha}} + B e^{-\sqrt{-\alpha}} +C, \\ u' (0) &= A \sqrt{-\alpha} - B \sqrt{-\alpha} , \\ u' (1) &= A \sqrt{-\alpha} e^{\sqrt{-\alpha}} - B \sqrt{-\alpha} e^{-\sqrt{-\alpha}}. \end{align*} $$ We have $-u'' - \alpha u = \beta$ implies $-C\alpha = \beta$. We have $u'(1)=k u'(0)$ implies $A e^{\sqrt{-\alpha}} - B e^{-\sqrt{-\alpha}} = k(A-B)$. We have $u(0)=k u(1)$ implies $A+B+C = k (A e^{\sqrt{-\alpha}} + B e^{-\sqrt{-\alpha}} +C)$. We have $\beta = \int_I u$ implies $\frac{A e^{\sqrt{-\alpha}} - B e^{-\sqrt{-\alpha}}}{\sqrt{-\alpha}} + \frac{B-A}{\sqrt{-\alpha}} = \beta -C$. So we have the system $$ (4) \quad \begin{cases} \alpha C + \beta &=0 , \\ ( e^{\sqrt{-\alpha}} - k ) A + ( k - e^{-\sqrt{-\alpha}} ) B &=0, \\ ( e^{\sqrt{-\alpha}} - 1 ) A + ( 1 - e^{-\sqrt{-\alpha}} ) B + \sqrt{-\alpha} C - \sqrt{-\alpha} \beta &=0, \\ ( ke^{\sqrt{-\alpha}} - 1 ) A + ( ke^{-\sqrt{-\alpha}} -1 ) B + (k-1) C &=0, \end{cases} $$ to solve for $(A, B, C, \beta)$.

Answer 1

If $\beta=0$, the solution of (3) is $u=0$. Now one can assume $\beta\neq 0$. Let $v=\frac u\beta$ and then (3) is equivalent to $$ (4) \quad \begin{cases} -v'' - (\lambda-1) v = 1 \quad \text {on} \quad I, \\ v(0)=k v(1) \\ v'(1)=k v'(0), \end{cases} $$ and $\int_Iv=1$.

If $\lambda=1$ and $k=1$, it is easy to see that (4) has no solution.

If $\lambda=1$ and $k\not=1$, (4) has the solution $$ u=\frac{k^2+k}{2 (k-1)^2}-\frac{t}{k-1}-\frac{t^2}{2}. $$ Applying $\int_Iv=1$ gives $$ \frac{\left(k^2+k+1\right)}{3 (k-1)^2}=1.$$ So (4) has a unique solution if $$ k=\frac14 (7 \pm \sqrt{33}). $$

Now one can assume $\lambda<1$. Let $\mu=\sqrt{1-\lambda}$. Clearly the corresponding homogeneous equation of the DE of (4) has the general solution $$ v_c=c_1e^{\mu t}+c_2e^{-\mu t} $$ and the DE of (4) has a particular solution $v_p=\frac{1}{\mu^2}$. Thus the general solution of the DE of (3) is $$ v=v_c+v_p=c_1e^{\mu t}+c_2e^{-\mu t}+\frac{1}{\mu^2}. $$ Now one can apply the boundary conditions $v(0) = k v(1), v'(1) = k v'(0)$ to get $c_1,c_2$.

Case 1: $k=1$. it is not hard to get $c_1=c_2=0$ and hence $v=v_p$. Now applying $\int_Iv=1$ gives $$ \frac{1}{\lambda-1} =1$$ which implies $\lambda=2$ which is against $\lambda<1$. So if $\lambda<1,k=1$, (4) has no solution.

Case 2: $k\not=1$. it is not hard to get $c_1,c_2$ which are the following $$ c_1= \frac{(k-1) \left(k-e^{-\mu}\right)}{2\mu^2 ((k^2+1)\cosh(\mu)-2 k)},\\ c_2= \frac{(k-1) \left(k-e^{\mu}\right)}{2\mu^2 ((k^2+1)\cosh(\mu)-2k)}. $$ Finally, using $\int_Iv=1$, one can obtain the following condition $$ \frac1{\mu^3}\bigg(\mu-\frac{(k-1)^2\sinh(\mu)}{-2k+(k^2+1)\cosh(mu)}\bigg)=1. $$ or $$ \mu -\frac{(k-1)^2 \sinh (\mu )}{\left(k^2+1\right) \cosh (\mu )-2 k}=\mu ^3. \tag{*}$$ Thus if $\lambda\neq1,k\neq 1$ satisfy (*), (4) has a unique solution.

If $\lambda>1$, one can perform the same trick and I omit the detail.