The results of the previous section will be the key to prove the differentiability of eigenvalues. In this section, our aim is to illustrate that the isolated eigenvalues are differentiable with respect to all the data. For this purpose, we will make use of the definition of Frechet derivatives and we recall Lemma 3.2 of [5] here.
Definition 4.1
Let X and Y be Banach spaces. A map \(\mathbf{T}:\mathbf{X}\to\mathbf{Y}\) is a linear map, \(x\in\mathbf{X}\) is a given point, if it is satisfied by a bounded linear operator \(d\mathbf{T}_{x}:\mathbf{X}\rightarrow\mathbf{Y} \), for \(h\in\mathbf{X}\) and as \(h\rightarrow0\), and
$$ \bigl\vert \mathbf{T}(x+h)-\mathbf{T}(x)-d\mathbf{T}_{x} \bigr\vert =o(h), $$
we call the map T differentiable at x.
Lemma 4.1
We suppose that
u
and
v
are two solutions of (2.1) with different
\(\lambda=\mu\)
and
\(\lambda=\nu\). Then
$$\begin{aligned}{} [u, v] _{a}^{b} :={}& \Biggl[(-1)^{m}\sum _{i=0}^{2m-1}(-1)^{2m+1-i}u^{[i]}\bar {v}^{[2m-i-1]} \Biggr]_{a}^{b} \\ ={}&(\mu-\nu) \int_{a}^{b}u\bar{v}w\,ds. \end{aligned}$$
(4.1)
Proof
Since u and v are solutions of (2.1) with \(\lambda=\mu\), \(\lambda=\nu\), respectively, we have
$$\begin{aligned} (\mu-\nu) \int_{a}^{b}u\bar{v}w\,ds &= \bigl(w^{-1}lu, v \bigr)- \bigl(u, w^{-1}lv \bigr) \\ &= \int_{a}^{b} \Biggl[\sum _{i=0}^{m}(-1)^{i} \bigl(p_{i}(x)u^{(i)} \bigr)^{(i)}\overline {v}-u\sum_{i=0}^{m}(-1)^{i} \bigl(p_{i}(x){\overline{v}}^{(i)} \bigr)^{(i)} \Biggr]\,ds. \end{aligned}$$
The integration by parts directly leads to the desired result. □
Lemma 4.2
[5], Lemma 3.2
We suppose that
\(f\in\mathbf {L}_{\mathrm{loc}}(a^{\prime}, b^{\prime})\)
is a real valued function. We deduce that
$$ \lim_{h\rightarrow0 } \frac{1}{h} \int _{x}^{x+h}f(s)\,ds=f(x)\quad \textit{a.e. } \bigl(a^{\prime}, b^{\prime} \bigr). $$
In the sequel of this paper, we will investigate the differentiability of the eigenvalue as one of the components of
ω, as the other components of
ω
are considered to be the same.
Theorem 4.1
Eigenvalue-eigenfunction differential equation for special case of separated BVPs
Assume (2.2) and
\({\boldsymbol {\omega}= (\mathbf{A}, \mathbf{B}, a, b,1/p_{m}, p_{m-1}, \ldots, p_{0}, w)}\in \mathbf{W}\)
hold. We suppose further that either (i) the multiplicity of
\(\lambda(\omega)\)
is 1 in some neighborhood
\(M\subset\mathbf{W}\)
of
ω, or (ii) \(\lambda(\omega)\)
is an eigenvalue of multiplicity
\(l\ (l=2, 3, \ldots, 2m)\)
for each
\(\omega\in M\), \(M\subset\mathbf{W}\).
-
(1)
Consider the spectral problems (2.1), (2.9) and (2.10) with
\(0\leq \alpha<\pi\)
and
\(\beta=\pi\). Let
\(\lambda=\lambda(b)\)
be a function of endpoint
b, and
\(u=u(\cdot, b)\)
the corresponding eigenfunction. We see that
λ
is differentiable a.e. and it satisfies:
-
if
m
is even, then
$$\begin{aligned} &\lambda^{\prime}(b)=\sum_{i=1} ^{[\frac{m}{2}]} \bigl(2u^{[2i-1]}(b)u^{[2(n-i)+1]}(b) +p_{2i-1}(b) \bigl\vert u^{[2i-1]}(b) \bigr\vert ^{2} \bigr) \\ &\quad\textit{a.e. in } \bigl(a, b^{\prime} \bigr); \end{aligned}$$
(4.2)
-
if
m
is odd, then
$$\begin{aligned} & \lambda^{\prime}(b)=-\sum_{i=1} ^{[\frac{m}{2}]} \bigl(2u^{[2i-1]}(b)u^{[2(n-i)+1]}(b) -p_{2i-1}(b) \bigl\vert u^{[2i-1]}(b) \bigr\vert ^{2} \bigr)-\frac { \vert u^{[m]}(b) \vert ^{2}}{p_{m}(b)} \\ & \quad \textit{a.e. in } \bigl(a, b^{\prime} \bigr). \end{aligned}$$
(4.3)
-
(2)
Consider the spectral problems (2.1), (2.11) and (2.12) with
\(0\leq \varphi<\pi\)
and
\(\psi=\pi\). Let
\(\lambda=\lambda(b)\)
be a function of endpoint
b, and
\(u=u(\cdot, b)\)
the corresponding eigenfunction. Then
λ
is differentiable a.e. and it satisfies
$$ \bigl(p_{m}\lambda^{\prime} \bigr) (b)=- \bigl\vert u^{[m]}(b) \bigr\vert ^{2} \quad\textit{a.e. in } \bigl(a, b^{\prime} \bigr). $$
(4.4)
-
(3)
Consider the spectral problems (2.1), (2.13) and (2.14) with
\(0\leq \eta<\pi\)
and
\(\tau=\pi\). Let
\(\lambda=\lambda(b)\)
be a function of endpoint
b, and
\(u=u(\cdot, b)\)
the corresponding eigenfunction. We see that
λ
is differentiable a.e. and it satisfies
$$ \bigl(p_{m}\lambda^{\prime} \bigr) (b)=- \bigl\vert u^{[m]}(b) \bigr\vert ^{2} \quad\textit{a.e. in } \bigl(a, b^{\prime} \bigr). $$
(4.5)
Proof
(1) For sufficiently small h in (4.1), we denote by \(\mu =\lambda(b), u=u(\cdot, b)\) and \(\nu=\lambda(b+h), v=u(\cdot, b+h)\) the corresponding eigenvalues and eigenfunctions, respectively. Since \([u, v](a)=0, u(b, b)=0\) and \(u^{[2j]}(b, b)=0\ (j=0, 1, \ldots, m-1)\), we infer
$$\begin{aligned} &\bigl(\lambda(b)-\lambda(b+h) \bigr) \int_{a}^{b}u(s, b)u(s, b+h)w(s)\,ds \\ &\quad=(-1)^{m}\sum_{i=1} ^{m}u^{[2i-1]}(b, b)u^{[2(n-i)]}(b, b+h). \end{aligned}$$
(4.6)
By normalizing the eigenfunction and applying Theorem 3.2, we deduce that
$$ \int_{a}^{b}u(s, b)u(s, b+h)w(s)\,ds \rightarrow \int_{a}^{b}u^{2}(s, b)w(s)\,ds=1, \quad h \rightarrow0. $$
(4.7)
When m is even, by (4.6), we obtain
$$\begin{aligned} & \bigl(\lambda(b)-\lambda(b+h) \bigr) \int_{a}^{b}u(s, b)u(s, b+h)w(s)\,ds \\ &\quad =(-1)^{m}\sum_{i=1} ^{\frac{m}{2}}u^{[2i-1]}(b, b)u^{[2(m-i)]}(b, b+h) \\ &\qquad{}+(-1)^{m}\sum_{\frac{m}{2}+1} ^{m}u^{[2i-1]}(b, b)u^{[2(m-i)]}(b, b+h). \end{aligned}$$
(4.8)
For \(i=1, 2, \ldots, m/2, u^{[2(m-i)]}(b, b)=0\), we obtain
$$\begin{aligned} u^{[2(m-i)]}(b, b+h)&=u^{[2(m-i)]}(b, b+h)-u^{[2(m-i)]}(b+h, b+h) \\ &=- \int_{b}^{b+h} \bigl(u^{[2(m-i)]} \bigr)^{\prime}(s, b+h)\,ds. \end{aligned}$$
(4.9)
It follows from Theorem 3.2 that as \(h\rightarrow0\), \(u_{m}^{[j]}(\cdot, b+h)\longrightarrow u_{m}^{[j]}(\cdot, b)\ (j=0, 1, 2, \ldots, 2m-1)\) uniformly hold on any compact subinterval of \((a^{\prime}, b^{\prime})\), by (4.9) and Lemma 4.2, we get
$$ \lim_{h\rightarrow0}\frac{(-1)^{m}u^{[2(m-i)]}(b, b+h)}{h}=(-1)^{m+1} \bigl(u^{[2(m-i)]} \bigr)^{\prime}(b, b), \quad\text{a.e. in } \bigl(a, b^{\prime} \bigr). $$
Since \((u^{[2(m-i)]})^{\prime}=u^{[2(m-i)+1]}-(-1)^{m-2i+1}p_{2i-1}y^{(2i-1)}\), we conclude that
$$ \lim_{h\rightarrow0}\frac{(-1)^{m}u^{[2(m-i)]}(b, b+h)}{h} =(-1)^{m+1}u^{[2(m-i)+1]}(b, b)-p_{2i-1}(b)u^{[2i-1]}(b, b). $$
(4.10)
In a similar way, for \(i=n/2+1, n/2+2, \ldots, n, u^{[2(m-i)]}(b, b)=0\), we have
$$\begin{aligned} u^{[2(m-i)]}(b, b+h)&=u^{(2m-2i)}(b, b+h)-u^{(2m-2i)}(b+h, b+h) \\ &=- \int_{b}^{b+h}u^{(2m-2i+1)}(s, b+h)\,ds. \end{aligned}$$
(4.11)
Now (4.11) and Lemma 4.2 imply that
$$ \lim_{h\rightarrow0}\frac{u^{[2(m-i)]}(b, b+h)}{h}=-u^{[2(m-i)+1]}(b, b), \quad \text{a.e. in } \bigl(a, b^{\prime} \bigr). $$
(4.12)
We divide (4.8) by h, combine (4.7), (4.10) and (4.12), and as \(h\rightarrow0\), we get
$$\begin{aligned} -\lambda^{\prime}(b) ={}&\sum_{i=1} ^{\frac{m}{2}}u^{[2i-1]}(b, b) \bigl(-u^{[2(m-i)+1]}(b, b)-p_{2i-1}(b)u^{2i-1}(b, b) \bigr)\\ &{}- \sum _{\frac{m}{2}+1} ^{m}u^{[2i-1]}(b, b)u^{[2(m-i)+1]}(b, b) \\ ={}&{-}\sum_{i=1} ^{\frac{m}{2}} \bigl(2u^{[2i-1]}(b, b)u^{[2(m-i)+1]}(b, b) +p_{2i-1}(b) \bigl(u^{[2i-1]} \bigr)^{2}(b, b) \bigr). \end{aligned}$$
So, we have (4.2);
if m is odd, by (4.6) we infer
$$\begin{aligned} & \bigl(\lambda(b)-\lambda(b+h) \bigr) \int_{a}^{b}u(s, b)u(s, b+h)w(s)\,ds \\ &\quad = (-1)^{m}\sum_{i=1} ^{\frac{m-1}{2}}u^{[2i-1]}(b, b)u^{[2(m-i)]}(b, b+h) +(-1)^{m}u^{[m]}(b, b)u^{[m-1]}(b, b+h) \\ &\qquad{}+(-1)^{m}\sum_{\frac{m+3}{2}} ^{m}u^{[2i-1]}(b, b)u^{[2(m-i)]}(b, b+h), \end{aligned}$$
(4.13)
for \(i=m-1, u^{(m-1)}(b, b)=0\); we get
$$\begin{aligned} u^{[m-1]}(b, b+h)&=u^{(m-1)}(b, b+h)-u^{(m-1)}(b+h, b+h) \\ & =- \int_{b}^{b+h}u^{(n)}(s, b+h)\,ds \\ &=- \int_{b}^{b+h}\frac{p_{m}u^{(n)}(s, b+h)}{p_{m}(s)}\,ds =- \int_{b}^{b+h}\frac{u^{[m]}(s, b+h)}{p_{m}(s)}\,ds. \end{aligned}$$
(4.14)
Applying Lemma 4.1, Lemma 4.2 and (4.14), we get
$$ \lim_{h\rightarrow0}\frac{u^{[m-1]}(b, b+h)}{h}=-\frac {u^{[m]}(b, b)}{p_{m}(b)},\quad \text{a.e. in } \bigl(a, b^{\prime} \bigr). $$
(4.15)
We divide (4.13) by h, as \(h\rightarrow0\) and use (4.7), (4.10), (4.12) and (4.13) to get
$$\begin{aligned} -\lambda^{\prime}(b) ={}&\sum_{i=1} ^{\frac{m-1}{2}}u^{[2i-1]}(b, b) \bigl(u^{[2(m-i)+1]}(b, b)-p_{2i-1}(b)u^{[2i-1]}(b, b) \bigr) +\frac{(u^{[m]})^{2}(b, b)}{p_{m}(b)} \\ &{}+\sum_{\frac{m+3}{2}} ^{m}u^{[2i-1]}(b, b)u^{[2(m-i)+1]}(b, b+h) \\ ={}&\sum_{i=1} ^{\frac{m-1}{2}} \bigl(2u^{[2i-1]}(b, b)u^{[2(m-i)+1]}(b, b) -p_{2i-1}(b) \bigl(u^{[2i-1]} \bigr)^{2}(b, b) \bigr)\\ &{}+\frac{(u^{[m]})^{2}(b, b)}{p_{m}(b)}. \end{aligned}$$
So, we obtain (4.3).
(2) For sufficiently small h in (4.1), we denote by \(\mu=\lambda(b), u=u(\cdot, b)\) and \(\nu=\lambda(b+h), v=u(\cdot, b+h)\) the corresponding eigenvalues and eigenfunctions, respectively. Since \([u, v](a)=0, u(b, b)=0\) and \(u^{[2j]}(b, b)=0\ (j=0, 1, \ldots, m-1)\), we infer
$$\begin{aligned} & \bigl(\lambda(b)-\lambda(b+h) \bigr) \int_{a}^{b}u(s, b)u(s, b+h)w(s)\,ds \\ &\quad=(-1)^{m}\sum_{i=n} ^{2m-1}(-1)^{2m+1-i}u^{[i]}(b, b)u^{[2m-i-1]}(b, b+h) \\ &\quad=-u^{[m]}(b, b)u^{[m-1]}(b, b+h) \\ &\qquad{}+(-1)^{m}\sum _{i=n+1} ^{2m-1}(-1)^{2m+1-i}u^{[i]}(b, b)u^{[2m-i-1]}(b, b+h), \end{aligned}$$
(4.16)
for \(i=m+1, m+2, \ldots, 2m-1, u^{[2m-i-1]}(b, b)=0\). Now proceeding with the proof of (4.12), we have
$$ \lim_{h\rightarrow0}\frac{u^{[2m-i-1]}(b, b+h)}{h}=-u^{[2m-i]}(b, b), \quad \text{a.e. in } \bigl(a, b^{\prime} \bigr). $$
(4.17)
Dividing (4.16) by h, using (4.7) (4.15) and (4.17), as \(h\rightarrow0\) we get
$$ -\lambda^{\prime}(b)=\frac{(u^{[m]})^{2}(b, b)}{p_{m}(b)} +\sum_{i=n+1} ^{2m-1}(-1)^{m-i}u^{[i]}(b, b)u^{[2m-i]}(b, b). $$
By \(u^{[j]}(b, b)=0\ (j=0, 1, \ldots, m-1)\), we obtain \(-\lambda ^{\prime}(b)=\frac{ \vert u^{[m]}(b) \vert ^{2}}{p_{m}(b)}\) and (4.4) holds. The proof of (4.5) is similar to the proof of (4.4), so we omit it. □
Corollary 4.1
Suppose (2.2) holds and we consider the BVPs (2.1) and (2.11)-(2.14) with
\(0\leq\varphi<\pi, \psi=\pi\), or
\(0\leq\eta<\pi, \tau=\pi\). Let
\(\lambda=\lambda(b)\)
be a function of endpoint
b, and
\(u=u(\cdot, b)\)
the corresponding eigenfunction. If
\(p_{m}\geq0\), a.e., then
\(\lambda(b)\)
is strictly decreasing on
\((a, b^{\prime})\).
Theorem 4.2
Eigenvalue-eigenfunction differential equation for special case of separated BVPs
Assume (2.2) and
\({\boldsymbol {\omega}=(\mathbf{A}, \mathbf{B}, a, b, 1/p_{m}, p_{m-1}, \ldots, p_{0}, w)}\in\mathbf{W}\)
hold. We suppose further that either (i) the multiplicity of
\(\lambda(\omega)\)
is 1 in some neighborhood
\(M\subset \mathbf{W}\)
of
ω, or (ii) \(\lambda(\omega)\)
is an eigenvalue of multiplicity
\(l\ (l=2, 3, \ldots, 2m)\)
for each
\(\omega\in M\), \(M\subset \mathbf{W}\).
-
(1)
Consider the spectral problems (2.1), (2.9) and (2.10) with
\(0\leq \alpha<\pi\)
and
\(\beta=\frac{\pi}{2}\). Let
\(\lambda=\lambda(b)\)
be a function of endpoint
b, and
\(u=u(\cdot, b)\)
the corresponding eigenfunction. We see that
λ
is differentiable a.e. and it satisfies:
-
if
m
is even, then
$$\begin{aligned} \lambda^{\prime}(b) = {}&\bigl\vert u(b) \bigr\vert ^{2} \bigl(p_{0}(b)-\lambda(b)w(b) \bigr)- \frac{ \vert u^{[m]}(b) \vert ^{2}}{p_{m}(b)} \\ &{} -\sum_{i=1} ^{\frac{m-2}{2}} \bigl(2u^{[2i]}(b)u^{[2m-2i]}(b) -p_{2i}(b) \bigl\vert u^{[2i]}(b) \bigr\vert ^{2} \bigr)\quad \textit{a.e. in } \bigl(a, b^{\prime} \bigr); \end{aligned}$$
(4.18)
-
if
m
is odd, then
$$\begin{aligned} \lambda^{\prime}(b)={}& \bigl\vert u(b) \bigr\vert ^{2} \bigl(p_{0}(b)-\lambda(b)w(b) \bigr) \\ &{} +\sum _{i=1} ^{\frac{m-1}{2}} \bigl(2u^{[2i]}(b)u^{[2(m-i)]}(b)+p_{2i}(b) \bigl\vert u^{[2i]}(b) \bigr\vert ^{2} \bigr) \quad \textit{a.e. in } \bigl(a, b^{\prime} \bigr). \end{aligned}$$
(4.19)
-
(2)
Consider the spectral problems (2.1), (2.11) and (2.12) with
\(0\leq \varphi<\pi\)
and
\(\psi=\pi/2\). Let
\(\lambda=\lambda(b)\)
be a function of endpoint
b, and
\(u=u(\cdot, b)\)
the corresponding eigenfunction. We see that
λ
is differentiable a.e. and it satisfies
$$ \lambda^{\prime}(b)= \bigl\vert u(b) \bigr\vert ^{2} \bigl(p_{0}(b)-\lambda(b)w(b) \bigr) +\sum _{i=1} ^{m-1}p_{i}(b) \bigl\vert u^{[i]}(b) \bigr\vert ^{2}\quad \textit{a.e. in } \bigl(a, b^{\prime} \bigr); $$
(4.20)
-
(3)
Consider the spectral problems (2.1), (2.13) and (2.14) with
\(0\leq \eta<\pi\)
and
\(\tau=\pi/2\). Let
\(\lambda=\lambda(b)\)
be a function of endpoint
b, and
\(u=u(\cdot, b)\)
the corresponding eigenfunction. Then
λ
is differentiable a.e. and it satisfies
$$ \lambda^{\prime}(b)= \bigl\vert u(b) \bigr\vert ^{2} \bigl(p_{0}(b)-\lambda(b)w(b) \bigr) +\sum _{i=1} ^{m-1}p_{i}(b) \bigl\vert u^{[i]}(b) \bigr\vert ^{2} \quad\textit{a.e. in } \bigl(a, b^{\prime} \bigr). $$
(4.21)
Proof
(1) For sufficiently small h in (4.1), we denote by \(\mu =\lambda(b), u=u(\cdot, b)\), \(\nu=\lambda(b+h), v=u(\cdot, b+h)\) the corresponding eigenvalues and eigenfunctions, respectively. Since \([u, v](a)=0\) and \(u^{[2j+1]}(b, b)=0\ (j=0, 1, \ldots, m-1)\), we have
$$\begin{aligned} & \int_{a}^{b}u(s, b)u(s, b+h)w(s)\,ds \\ &\quad=(-1)^{m}\sum_{i=0} ^{m-1}(-1)^{2m+1-2i}u^{[2i]}(b, b)u^{[2(m-i)-1]}(b, b+h) \\ &\quad=(-1)^{m+1}\sum_{i=0} ^{m-1}u^{[2i]}(b, b)u^{[2(m-i)-1]}(b, b+h). \end{aligned}$$
(4.22)
If m is even, we get
$$\begin{aligned} & \bigl(\lambda(b)-\lambda(b+h) \bigr) \int_{a}^{b}u(s, b)u(s, b+h)w(s)\,ds \\ &\quad =-(-1)^{m}u(b, b)u^{[2m-1]}(b, b+h) -(-1)^{m} \sum_{i=1} ^{\frac{m-2}{2}}u^{[2i]}(b, b)u^{[2(m-i)-1]}(b, b+h) \\ &\qquad{}-u^{[m]}(b, b)u^{[m-1]}(b, b+h) -\sum _{\frac{m+2}{2}} ^{m-1}u^{[2i]}(b, b)u^{[2(m-i)-1]}(b, b+h), \end{aligned}$$
(4.23)
for \(i=2m-1, u^{[2m-1]}(b, b)=0\); we infer
$$\begin{aligned} (-1)^{m}u^{[2m-1]}(b, b+h)&=(-1)^{m}u^{[2m-1]}(b, b+h)-(-1)^{m}u^{[2m-1]}(b+h, b+h) \\ &=- \int_{b}^{b+h}(-1)^{m} \bigl(u^{[2m-1]} \bigr)^{\prime}(s, b+h)\,ds. \end{aligned}$$
By \(u^{[2m]}=(u^{[2m-1]})^{\prime}+(-1)^{m}p_{0}y\) and (2.4), we obtain
$$\begin{aligned} &(-1)^{m}u^{[2m-1]}(b, b+h) \\ &\quad=- \int_{b}^{b+h}(-1)^{m} \bigl(u^{[2m]}(s, b+h)-(-1)^{m}p_{0}(s)u(s, b+h) \bigr)\,ds \\ &\quad= \int_{b}^{b+h} \bigl[p_{0}(s)u(s, b+h)- \lambda(b+h)u(s, b+h)w(s) \bigr]\,ds \\ &\quad = \int_{b}^{b+h}p_{0}(s)u(s, b)\,ds+ \int_{b}^{b+h} p_{0} (s) \bigl[u(s, b+h)-u(s, b) \bigr]\,ds-\lambda(b+h) \\ &\qquad{}\times \int_{b}^{b+h}u(s, b)w(s)\,ds+\lambda(b+h) \int_{b}^{b+h} \bigl[u(s, b)-u(s, b+h) \bigr]w(s) \,ds. \end{aligned}$$
(4.24)
Applying Theorem 3.2, as \(h\rightarrow0\), \(u_{m}^{[j]}(\cdot , b+h)\longrightarrow u_{m}^{[j]}(\cdot, b)\ (j=0, 1, 2, \ldots, 2m-1)\) uniformly hold on the compact subinterval of \((a^{\prime}, b^{\prime})\). We get by (4.24) and Lemma 4.2
$$ \lim_{h\rightarrow0}\frac{(-1)^{m}u^{[2m-1]}(b, b+h)}{h}=u(b, b) \bigl[p_{0}(b)- \lambda(b)w(b) \bigr] \quad\text{a.e. in } \bigl(a, b^{\prime} \bigr), $$
(4.25)
for \(i=1, 2, \ldots, (m-2)/2, u^{[2(m-i)-1]}(b, b)=0\). We obtain
$$\begin{aligned} u^{[2(m-i)-1]}(b, b+h)&=u^{[2(m-i)-1]}(b, b+h)-u^{[2(m-i)-1]}(b+h, b+h) \\ &=- \int_{b}^{b+h} \bigl(u^{[2(m-i)-1]} \bigr)^{\prime}(s, b+h)\,ds. \end{aligned}$$
(4.26)
Applying Lemma 4.2 and (4.26), we get
$$ \lim_{h\rightarrow0}\frac{(-1)^{m}u^{[2(m-i)-1]}(b, b+h)}{h}=(-1)^{m+1} \bigl(u^{[2(m-i)-1]} \bigr)^{\prime}(b, b) \quad\text{a.e. in } \bigl(a, b^{\prime} \bigr). $$
By \((u^{[2(m-i)-1]})^{\prime}=u^{[2(m-i)]}-(-1)^{m-2i}p_{2i}y^{(2i)}\), we conclude that
$$\begin{aligned} \lim_{h\rightarrow0}\frac{(-1)^{m}u^{[2(m-i)-1]}(b, b+h)}{h} &=(-1)^{m+1}u^{[2(m-i)]}(b, b)+p_{2i}(b)u^{[2i]}(b, b) \\ &=-u^{[2(m-i)]}(b, b)+p_{2i}(b)u^{[2i]}(b, b). \end{aligned}$$
(4.27)
Similarly, for \(i=(m+2)/2, (m+4)/2, \ldots, m-1\), then \(u^{[2(m-i)-1]}(b, b)=0\), hence
$$ \lim_{h\rightarrow0}\frac{u^{[2(m-i)-1]}(b, b+h)}{h}=-u^{[2(m-i)]}(b, b). $$
(4.28)
Dividing (4.23) by h, as \(h\rightarrow0\) and using (4.5), (4.10), (4.25), (4.27) and (4.28), we get
$$\begin{aligned} -\lambda^{\prime}(b) ={}&{-}u^{2}(b, b) \bigl[p_{0}(b)- \lambda(b)w(b) \bigr]+\frac{(u^{[m]})^{2}(b, b)}{p_{m}(b)} \\ &{}+\sum_{i=1} ^{\frac{m-2}{2}} \bigl(2u^{[2i]}(b, b)u^{[2(m-i)]}(b, b) -p_{2i}(b) \bigl(u^{[2i]} \bigr)^{2}(b, b) \bigr). \end{aligned}$$
Then we obtain (4.17);
if m is odd, we get
$$\begin{aligned} &\bigl[\lambda(b)-\lambda(b+h) \bigr] \int_{a}^{b}u(s, b)u(s, b+h)w(s)\,ds \\ &\quad =-(-1)^{m} \Biggl[u(b, b)u^{[2m-1]}(b, b+h) +\sum _{i=1} ^{\frac{m-1}{2}}u^{[2i]}(b, b)u^{[2(m-i)-1]}(b, b+h) \Biggr] \\ &\qquad{}-(-1)^{m}\sum_{i=\frac{m+1}{2}} ^{m-1}u^{[2i]}(b, b)u^{[2(m-i)-1]}(b, b+h). \end{aligned}$$
(4.29)
Dividing (4.29) by h, as \(h\rightarrow0\) and using (4.7) (4.25), (4.27) and (4.28), we conclude that
$$\begin{aligned} -\lambda^{\prime}(b)={}&{-}u^{2}(b, b) \bigl[p_{0}(b)- \lambda(b)w(b) \bigr] \\ &{} -\sum_{i=1} ^{\frac{m-1}{2}} \bigl(2u^{[2i]}(b, b)u^{[2(m-i)]}(b, b)+p_{2i}(b) \bigl(u^{[2i]} \bigr)^{2}(b, b) \bigr). \end{aligned}$$
Then we obtain (4.19).
For sufficiently small h in (4.1), we denote by \(\mu=\lambda(b), u=u(\cdot, b)\) and \(\nu=\lambda(b+h), v=u(\cdot, b+h)\) the corresponding eigenvalues and eigenfunctions, respectively. Since \([u, v](a)=0, u(b, b)=0\) and \(u^{[j]}(b, b)=0\ (j=m, m+1, \ldots, 2m-1)\), we have
$$\begin{aligned} & \bigl(\lambda(b)-\lambda(b+h) \bigr) \int_{a}^{b}u(s, b)u(s, b+h)w(s)\,ds \\ &\quad=(-1)^{m}\sum_{i=0} ^{m-1}(-1)^{2m+1-i}u^{[i]}(b, b)u^{[2m-i-1]}(b, b+h) \\ &\quad =-(-1)^{m}u(b, b)u^{[2m-1]}(b, b+h) \\ &\qquad{}+(-1)^{m} \sum_{i=1} ^{m-1}(-1)^{2m+1-i}u^{[i]}(b, b)u^{[2m-i-1]}(b, b+h), \end{aligned}$$
(4.30)
for \(i=1, 2, \ldots, m-1\), \(u^{[2m-i-1]}(b, b)=0\); hence
$$\begin{aligned} u^{[2m-i-1]}(b, b+h)&=u^{[2m-i-1]}(b, b+h)-u^{[2m-i-1]}(b+h, b+h) \\ &=- \int_{b}^{b+h} \bigl(u^{[2m-i-1]} \bigr)^{\prime}(s, b+h)\,ds. \end{aligned}$$
(4.31)
By Lemma 4.2 and (4.31), we get
$$ \lim_{h\rightarrow0}\frac{(-1)^{m}u^{[2m-i-1]}(b, b+h)}{h}=(-1)^{m+1} \bigl(u^{[2m-i-1]} \bigr)^{\prime}(b, b) \quad\text{a.e. in } \bigl(a, b^{\prime} \bigr). $$
Combining \((u^{[2m-i-1]})^{\prime}=u^{[2m-i]}-(-1)^{m-i}p_{i}y^{(i)}\), we get
$$\begin{aligned} &\lim_{h\rightarrow0}\frac{(-1)^{m}u^{[2m-i-1]}(b, b+h)}{h} \\ &\quad=(-1)^{m+1}u^{[2m-i]}(b, b)-(-1)^{2m-i+1}p_{i}(b)u^{[i]}(b, b) \\ &\quad=-(-1)^{2m-i+1}p_{i}(b)u^{[i]}(b, b). \end{aligned}$$
(4.32)
We divide (4.30) by h, as \(h\rightarrow0\), and use (4.7) (4.25) and (4.32) to get
$$ -\lambda^{\prime}(b)=-u^{2}(b, b) \bigl[p_{0}(b)- \lambda(b)w(b) \bigr] -\sum_{i=1} ^{m-1}p_{i}(b) \bigl(u^{[i]} \bigr)^{2}(b, b). $$
We obtain (4.20). The proof of (4.21) is similar to that of (4.20), so we omit it. □
Theorem 4.3
Eigenvalue-eigenfunction differential equation for separated BVPs
Assume (2.2) and
\({\boldsymbol {\omega}=(\mathbf {A}, \mathbf{B}, a, b, 1/p_{m}, p_{m-1}, \ldots, p_{0}, w)}\in\mathbf {W}\)
hold. Consider the BVPs (2.1), (2.9) and (2.10) with
\(0\leq\alpha <\pi\)
and
\(0<\beta\leq\pi\), or BVP (2.1), (2.11) and (2.12) with
\(0\leq \varphi<\pi\)
and
\(0<\psi\leq\pi\), or BVP (2.1), (2.13) and (2.14) with
\(0\leq\eta<\pi\)
and
\(0<\tau\leq\pi\). We suppose further that either (i) the multiplicity of
\(\lambda(\omega)\)
is 1 in some neighborhood
\(M\subset\mathbf{W}\)
of
ω, or (ii) \(\lambda(\omega)\)
is an eigenvalue of multiplicity
\(l\ (l=2, 3, \ldots, 2m)\)
for each
\(\omega \in M\), \(M\subset\mathbf{W}\).
(1) Let
\(\lambda=\lambda(a)\)
be a function of endpoint
a, and
\(u=u(\cdot, a)\)
the corresponding eigenfunction. We see that
λ
is differentiable a.e. and it satisfies
$$\begin{aligned} \lambda^{\prime}(a)={}& \bigl\vert u(a) \bigr\vert ^{2} \bigl[p_{0}(b)-\lambda(b)w(b) \bigr]-\frac { \vert u^{[m]}(a) \vert ^{2}}{p_{m}(a)} \\ &{}-\sum_{i=1} ^{m-1} \bigl(p_{i}(a) \bigl\vert u^{[i]}(a) \bigr\vert ^{2}-(-1)^{m-i}2u^{[i]}(a)u^{[2m-i]}(a) \bigr) \quad\textit{a.e. in } \bigl(a^{\prime}, b \bigr), \end{aligned}$$
(4.33)
particularly, if
\(p_{m}, p_{m-1}, \ldots, p_{0}\)
and
w
are continuous at
a
and
\(p_{m}(a)\neq0\); then (4.33) holds at
a.
(2) Let
\(\lambda=\lambda_{m}\)
and
\(u=u_{m}\). We see that
λ
is differentiable a.e. and it satisfies
$$\begin{aligned} \lambda^{\prime}(b)={}& \bigl\vert u(b) \bigr\vert ^{2} \bigl[p_{0}(b)-\lambda(b)w(b) \bigr]-\frac { \vert u^{[m]}(b) \vert ^{2}}{p_{m}(b)} \\ &{}-\sum_{i=1} ^{m-1} \bigl(p_{i}(b) \bigl\vert u^{[i]}(b) \bigr\vert ^{2}-(-1)^{m-i}2u^{[i]}(b)u^{[2m-i]}(b) \bigr) \quad\textit{a.e. in } \bigl(a, b^{\prime} \bigr), \end{aligned}$$
(4.34)
particularly, if
\(p_{m}, p_{m-1}, \ldots, p_{0}\)
and
w
are continuous at
b
and
\(p_{m}(b)\neq0\); then (4.34) holds at
b.
Proof
The proofs of (4.33) and (4.34) are similar, so we will prove (4.34) only. For sufficiently small h in (4.1), we denote by \(\mu=\lambda(b), u=u(\cdot, b)\) and \(\nu=\lambda(b+h), v=u(\cdot, b+h)\) the corresponding eigenvalues and eigenfunctions, respectively. Since \([u, v](a)=0\), we have
$$\begin{aligned} & \bigl(\lambda(b)-\lambda(b+h) \bigr) \int_{a}^{b}u(s, b)u(s, b+h)w(s)\,ds \\ &\quad=(-1)^{m}\sum_{i=0} ^{2m-1}(-1)^{2m+1-i}u^{[i]}(b, b)u^{[2m-i-1]}(b, b+h) \\ &\quad=-(-1)^{m}u(b, b)u^{[2m-1]}(b, b+h) \\ &\qquad{}+(-1)^{m} \sum_{i=1} ^{m-1}(-1)^{2m+1-i}u^{[i]}(b, b)u^{[2m-i-1]}(b, b+h) \\ &\qquad{}-u^{[m]}(b, b)u^{[m-1]}(b, b+h) \\ &\qquad{}+(-1)^{m}\sum _{i=n+1} ^{2m-1}(-1)^{2m+1-i}u^{[i]}(b, b)u^{[2m-i-1]}(b, b+h). \end{aligned}$$
(4.35)
We divide (4.35) by h and take the limit \(h\rightarrow0\), use (4.5) (4.17) (4.25) and (4.32), and apply the continuity of λ at b in Theorem 3.2 and Lemma 4.2, so we get (4.34). □
Theorem 4.4
Eigenvalue-eigenfunction differential equation for coupled BVPs
Assume that
\({\boldsymbol {\omega}=(\mathbf{A}, \mathbf{B}, a, b, 1/p_{m}, p_{m-1}, \ldots, p_{0}, w)}\in\mathbf{W}\)
and (2.2) hold. We suppose further that either (i) the multiplicity of
\(\lambda(\omega)\)
is 1 in some neighborhood
\(M\subset\mathbf{W}\)
of
ω, or (ii) \(\lambda(\omega)\)
is an eigenvalue of multiplicity
\(l\ (l=2, 3, \ldots, 2m)\)
for each
\(\omega\in M\), \(M\subset\mathbf{W}\). Consider the spectral problems (2.1), (2.16) and (2.17) with
\(-\pi< \theta\leq\pi\).
(1) Let
\(\lambda=\lambda(a)\)
be a function of endpoint
a, and
\(u=u(\cdot, a)\)
the corresponding eigenfunction. We see that
λ
is differentiable a.e. and it satisfies:
if
m
is even, then
$$\begin{aligned} \lambda^{\prime}(a)={}&{-} \bigl\vert u(a) \bigr\vert ^{2} \bigl(p_{0}(a)-\lambda(a)w(a) \bigr) +2 \operatorname{Re} \sum _{i=1} ^{m-1}(-1)^{i}u^{[i]}(a) \overline{u}^{[2m-i]}(a) \\ &{}+\frac {u^{[m]}(a)}{p_{m}(a)} -\sum _{i=1} ^{m-1}p_{i}(a) \bigl\vert \overline{u}^{[i]}(a) \bigr\vert ^{2}; \end{aligned}$$
(4.36)
if
m
is odd, then
$$\begin{aligned} \lambda^{\prime}(a)={}&{-} \bigl\vert u(a) \bigr\vert ^{2} \bigl(p_{0}(a)-\lambda(a)w(a) \bigr) -2 \operatorname{Re} \sum _{i=1} ^{m-1}(-1)^{i}u^{[i]}(a) \overline{u}^{[2m-i]}(a) \\ &{}+\frac {u^{[m]}(a)}{p_{m}(a)} -\sum _{i=1} ^{m-1}p_{i}(a) \bigl\vert \overline{u}^{[i]}(a) \bigr\vert ^{2}, \end{aligned}$$
(4.37)
particularly, if
\(p_{m}, p_{m-1}, \ldots, p_{0}\)
and
w
are continuous at
\(a\in(a^{\prime}, b]\)
and
\(p_{m}(a)\neq0\), then equations (4.36) and (4.37) hold at
a.
(2) Let
\(\lambda=\lambda(b)\)
be a function of endpoint
b, and
\(u=u(\cdot, b)\)
the corresponding eigenfunction. We see that
λ
is differentiable a.e. and it satisfies:
if
m
is even, then
$$\begin{aligned} \lambda^{\prime}(b)={}& \bigl\vert u(b) \bigr\vert ^{2} \bigl(p_{0}(b)-\lambda(b)w(b) \bigr) -2\operatorname{Re} \sum _{i=1} ^{m-1}(-1)^{i}u^{[i]}(b) \overline{u}^{[2m-i]}(b) \\ &{}-\frac {u^{[m]}(b)}{p_{m}(b)} +\sum _{i=1} ^{m-1}p_{i}(b) \bigl\vert \overline{u}^{[i]}(b) \bigr\vert ^{2}; \end{aligned}$$
(4.38)
if
m
is odd, then
$$\begin{aligned} \lambda^{\prime}(b)={}& \bigl\vert u(b) \bigr\vert ^{2} \bigl(p_{0}(b)-\lambda(b)w(b) \bigr) +2\operatorname{Re} \sum _{i=1} ^{m-1}(-1)^{i}u^{[i]}(b) \overline{u}^{[2m-i]}(b) \\ &{}-\frac {u^{[m]}(b)}{p_{m}(b)} +\sum _{i=1} ^{m-1}p_{i}(b) \bigl\vert \overline{u}^{[i]}(b) \bigr\vert ^{2}, \end{aligned}$$
(4.39)
particularly, if
\(p_{m}, p_{m-1}, \ldots, p_{0}\)
and
w
are continuous at
\(b\in[a, b^{\prime})\)
and
\(p_{m}(b)\neq0\), then equations (4.38) and (4.39) hold at
b.
Proof
The proofs of (4.36) and (4.37) are similar to (4.38) and (4.39), respectively, so we prove (4.38) and (4.39) only. For sufficiently small h in (4.1), we denote by \(\mu=\lambda(b), \nu =\lambda(b+h)\) and \(u=u(\cdot, b), v=u(\cdot, b+h)\) the eigenvalues and eigenfunctions.
$$ \bigl[\lambda(b)-\lambda(b+h) \bigr] \int_{a}^{b} u\overline{v}w\,ds = \Biggl[(-1)^{m}\sum_{i=0}^{2m-1}(-1)^{2m+1-i}u^{[i]} \bar {v}^{[2m-i-1]} \Biggr]_{a}^{b}. $$
If m is even,
$$\begin{aligned} & \bigl(\lambda(b)-\lambda(b+h) \bigr) \int_{a}^{b} u\overline{v}w\,ds \\ &\quad =- \begin{pmatrix} \overline{v}^{[2m-1]}, & -\overline{v}^{[2m-2]}, &\cdots, & \overline {v}^{[1]}, & -\overline{v} \end{pmatrix} (b) \begin{pmatrix}{c} u \\ u^{[1]} \\ \vdots\\ u^{[2m-2]} \\ u^{[2m-1]} \end{pmatrix} (b) \\ &\qquad{} + \begin{pmatrix} \overline{v}^{[2m-1]}, & -\overline{v}^{[2m-2]}, &\cdots, & \overline {v}^{[1]}, & -\overline{v} \end{pmatrix} (a) \begin{pmatrix}{c} u \\ u^{[1]} \\ \vdots\\ u^{[2m-2]} \\ u^{[2m-1]} \end{pmatrix} (a), \\ & \begin{pmatrix} u^{[2m-1]}, & -u^{[2m-2]}, & u^{[2m-3]}, & \cdots, & -u^{[2]}, & u^{[1]} & -u \end{pmatrix} (b) \\ &\quad= \begin{pmatrix} u, & u^{[1]}, & u^{[2]}, & \cdots, & u^{[2m-3]}, & u^{[2m-2]}, & u^{[2m-1]} \end{pmatrix} (b)\\ &\qquad{}\times \begin{pmatrix} 0 & 0 & 0 & \cdots& 0 & 0 & -1 \\ 0 & 0 & 0 & \cdots& 0 & 1 & 0\\ 0 & 0& 0 & \cdots& -1 & 0 & 0\\ \vdots& \vdots& \vdots& \vdots& \vdots& \vdots& \vdots\\ 0 & 0 & 1 & \cdots& 0 & 0& 0 \\ 0 & -1 & 0 & \cdots& 0 & 0& 0 \\ 1 & 0& 0 & \cdots& 0& 0 & 0 \end{pmatrix} \\ &\quad=e^{i\theta} \begin{pmatrix} u, & u^{[1]}, & u^{[2]}, & \cdots, & u^{[2m-3]}, & u^{[2m-2]}, & u^{[2m-1]} \end{pmatrix} (a) \mathbf{K} ^{T}\\ &\qquad{}\times \begin{pmatrix} 0 & 0 & 0 & \cdots& 0 & 0 & -1 \\ 0 & 0 & 0 & \cdots& 0 & 1 & 0\\ 0 & 0& 0 & \cdots& -1 & 0 & 0\\ \vdots& \vdots& \vdots& \vdots& \vdots& \vdots& \vdots\\ 0 & 0 & 1 & \cdots& 0 & 0& 0 \\ 0 & -1 & 0 & \cdots& 0 & 0& 0 \\ 1 & 0& 0 & \cdots& 0& 0 & 0 \end{pmatrix} \\ &\quad=e^{i\theta} \begin{pmatrix} u^{[2m-1]}, & -u^{[2m-2]}, & u^{[2m-3]}, & \cdots, & - u^{[2]}, & u^{[1]}, & - u \end{pmatrix}(a) \\ &\qquad{}\times \begin{pmatrix} 0 & 0 & 0 & \cdots& 0 & 0 & 1 \\ 0 & 0 & 0 & \cdots& 0 & -1 & 0 \\ 0 & 0 & 0 & \cdots& 1 & 0 & 0 \\ \vdots& \vdots& \vdots& \vdots& \vdots& \vdots& \vdots\\ 0 & 0 & -1 & \cdots& 0 & 0 & 0 \\ 0 & 1 & 0 & \cdots& 0 & 0 & 0 \\ -1 & 0 & 0 & \cdots& 0 & 0 & 0 \end{pmatrix} \mathbf{K}^{T} \begin{pmatrix} 0 & 0 & 0 & \cdots& 0 & 0 & -1 \\ 0 & 0 & 0 & \cdots& 0 & 1 & 0\\ 0 & 0& 0 & \cdots& -1 & 0 & 0\\ \vdots& \vdots& \vdots& \vdots& \vdots& \vdots& \vdots\\ 0 & 0 & 1 & \cdots& 0 & 0& 0 \\ 0 & -1 & 0 & \cdots& 0 & 0& 0 \\ 1 & 0& 0 & \cdots& 0& 0 & 0 \end{pmatrix} \\ &\quad=e^{i\theta} \begin{pmatrix} u^{[2m-1]}, & -u^{[2m-2]}, & u^{[2m-3]}, & \cdots, & - u^{[2]}, & u^{[1]}, & - u \end{pmatrix} (a) \mathbf{K}^{-1}, \end{aligned}$$
hence
$$\begin{aligned} & \begin{pmatrix} u^{[2m-1]}, & -u^{[2m-2]}, & \cdots, & u^{[1]}, & - u \end{pmatrix} (a) \\ &\quad= e^{-i\theta} \begin{pmatrix} u^{[2m-1]}, & -u^{[2m-2]}, & \cdots, & u^{[1]}, & - u \end{pmatrix} (b) \mathbf{K}, \end{aligned}$$
or
$$\begin{aligned} & \begin{pmatrix} \overline{u}^{[2m-1]}, & -\overline{u}^{[2m-2]}, & \cdots, & \overline { u}^{[1]}, & - \overline{u} \end{pmatrix} (a) \\ &\quad= e^{i\theta} \begin{pmatrix} \overline{u}^{[2m-1]}, & -\overline{u}^{[2m-2]}, & \cdots, & \overline {u}^{[1]}, & - \overline{u} \end{pmatrix} (b) \mathbf{K}. \end{aligned}$$
(4.40)
Then
$$\begin{aligned} & \bigl(\lambda(b)-\lambda(b+h) \bigr) \int_{a}^{b} u\overline{v}w\,ds \\ &\quad=- \begin{pmatrix} \overline{v}^{[2m-1]}, & -\overline{v}^{[2m-2]}, &\cdots, & \overline {v}^{[1]}, & -\overline{v} \end{pmatrix} (b)\begin{pmatrix} u \\ u^{[1]} \\ \vdots\\ u^{[2m-2]} \\ u^{[2m-1]} \end{pmatrix} (b) \\ &\qquad{} + \begin{pmatrix} \overline{v}^{[2m-1]}, & -\overline{v}^{[2m-2]}, &\cdots, & \overline {v}^{[1]}, & -\overline{v} \end{pmatrix} (b+h)\begin{pmatrix} u \\ u^{[1]} \\ \vdots\\ u^{[2m-2]} \\ u^{[2m-1]} \end{pmatrix} \\ &\quad = \left[ \begin{pmatrix} \overline{v}^{[2m-1]}, & -\overline{v}^{[2m-2]}, &\cdots, & \overline {v}^{[1]}, & -\overline{v} \end{pmatrix} (b+h)\right. \\ &\qquad{}\left.- \begin{pmatrix} \overline{v}^{[2m-1]}, & -\overline{v}^{[2m-2]}, &\cdots, & \overline {v}^{[1]}, & -\overline{v} \end{pmatrix}(b) \right] \\ &\qquad{} \times\begin{pmatrix} u \\ u^{[1]} \\ \vdots\\ u^{[2m-2]} \\ u^{[2m-1]} \end{pmatrix} (b). \end{aligned}$$
(4.41)
Now proceeding as in Theorem 4.1 and Theorem 4.2, we get
$$\begin{aligned} &\lim_{h\rightarrow0} \frac{\overline {v}^{[2m-1]}(b+h)-\overline{v}^{[2m-1]}(b)}{h} = \bigl(\lambda(b)w(b)-p_{0}(b) \bigr)\overline{u}(b), \\ &\lim_{h\rightarrow0} \frac{\overline {v}^{[2m-i-1]}(b+h)-\overline{v}^{[2m-i-1]}(b)}{h} =\overline{u}^{[2m-i]}(b)-(-1)^{m-i}p_{i}(b)u^{[i]}(b), \quad i=1, 2, \ldots, m-1, \\ &\lim_{h\rightarrow0} \frac{\overline {v}^{[m-1]}(b+h)-\overline{v}^{[m-1]}(b)}{h} =\frac{\overline{u}^{[m]}(b)}{p_{m}(b)}, \\ & \lim_{h\rightarrow0} \frac{\overline {v}^{[i]}(b+h)-\overline{v}^{[i]}(b)}{h} = \overline{u}^{[i+1]}(b), \quad i=1, 2, \ldots, m-1. \end{aligned}$$
Dividing (4.41) by h, as \(h\rightarrow0\), we obtain
and (4.38) holds.
If m is odd, we infer
$$\begin{aligned} & \bigl(\lambda(b)-\lambda(b+h) \bigr) \int_{a}^{b} u\overline{v}w\,ds \\ &\quad =- \left[ \begin{pmatrix} \overline{v}^{[2m-1]}, & -\overline{v}^{[2m-2]}, &\cdots, & \overline {v}^{[1]}, & -\overline{v} \end{pmatrix} (b+h)\right. \\ &\qquad{}\left.- \begin{pmatrix} \overline{v}^{[2m-1]}, & -\overline{v}^{[2m-2]}, &\cdots, & \overline {v}^{[1]}, & -\overline{v} \end{pmatrix} (b) \right] \\ &\qquad{} \times\begin{pmatrix} u \\ u^{[1]} \\ \vdots\\ u^{[2m-2]} \\ u^{[2m-1]} \end{pmatrix}(b) \end{aligned}$$
(4.42)
and
$$\begin{aligned} &\lim_{h\rightarrow0}\frac{\overline {v}^{[2m-1]}(b+h)-\overline{v}^{[2m-1]}(b)}{h} = \bigl(p_{0}(b)- \lambda(b)w(b) \bigr)\overline{u}(b), \\ &\lim_{h\rightarrow0}\frac{\overline {v}^{[2m-i-1]}(b+h)-\overline{v}^{[2m-i-1]}(b)}{h} =\overline{u}^{[2m-i]}(b)-(-1)^{m-i}p_{i}(b)u^{[i]}(b), \quad i=1, 2, \ldots, m-1, \\ &\lim_{h\rightarrow0}\frac{\overline {v}^{[m-1]}(b+h)-\overline{v}^{[m-1]}(b)}{h} =\frac{\overline{u}^{[m]}(b)}{p_{m}(b)}, \\ & \lim_{h\rightarrow0}\frac{\overline {v}^{[i]}(b+h)-\overline{v}^{[i]}(b)}{h} = \overline{u}^{[i+1]}(b), \quad i=1, 2, \ldots, m-1. \end{aligned}$$
We divide (4.42) by h, and take the limit as \(h\rightarrow0\) to obtain
and (4.39) holds. □
Theorem 4.5
For the BVPs (2.1) and (2.9)-(2.17), let
\({\boldsymbol {\omega}=(\mathbf{A}, \mathbf{B},a, b, 1/p_{m}, p_{m-1}, \ldots , p_{0}, w)}\in\mathbf{W}\), and
\(\lambda=\lambda(\omega)\)
and
\(u=u(\cdot , \omega)\)
be the eigenvalue and normalized eigenfunction. We suppose further that either (i) the multiplicity of
\(\lambda(\omega)\)
is 1 in some neighborhood
\(M\subset\mathbf{W}\)
of
ω, or (ii) \(\lambda (\omega)\)
is an eigenvalue of multiplicity
\(l\ (l=2, 3, \ldots, 2m)\)
for each
\(\omega\in M\), \(M\subset\mathbf{W}\). Then
λ
is continuously differentiable with respect to each variable of
ω
in the appropriate sense. Their derivatives are given as follows.
-
1.
Let
\(\lambda=\lambda(\alpha)\)
be a function of
α, and
\(u=u(\cdot, \alpha)\)
the corresponding eigenfunction. We see that
λ
is differentiable and it satisfies:
if
m
is even, then
$$ \lambda^{\prime}(\alpha)=2\operatorname{Re}\sum _{i=1} ^{\frac{m-1}{2}} \bigl(u^{[2i]}(a) \overline{u}^{[2m-2i]}(a) +u^{[2i+1]}(a)\overline{u}^{[2m-2i-1]}(a) \bigr); $$
(4.43)
if
m
is odd, then
$$\begin{aligned} \lambda^{\prime}(\alpha)={}&{-}2\operatorname{Re}\sum _{i=0} ^{\frac{m-3}{2}} \bigl(u^{[2i]}(a) \overline{u}^{[2m-2i-2]}(a) +u^{[2i+1]}(a)\overline{u}^{[2m-2i-1]}(a) \bigr) \\ &{}- \bigl\vert u^{[m-1]}(a) \bigr\vert ^{2}- \bigl\vert u^{[m]}(a) \bigr\vert ^{2}. \end{aligned}$$
(4.44)
-
2.
Let
\(\lambda=\lambda(\beta)\)
be a function of
β, and
\(u=u(\cdot, \beta)\)
the corresponding eigenfunction. We see that
λ
is differentiable and it satisfies: if
m
is even, then
$$ \lambda^{\prime}(\beta)=-2\operatorname{Re}\sum _{i=1} ^{\frac{m-1}{2}} \bigl(u^{[2i]}(b) \overline{u}^{[2m-2i]}(b) +u^{[2i+1]}(b)\overline{u}^{[2m-2i-1]}(b) \bigr); $$
(4.45)
if
m
is odd, then
$$\begin{aligned} \lambda^{\prime}(\beta)={}&2\operatorname{Re}\sum _{i=0} ^{\frac{m-3}{2}} \bigl(u^{[2i]}(b) \overline{u}^{[2m-2i-2]}(b)+u^{[2i+1]}(b) \overline{u}^{[2m-2i-1]}(b) \bigr) \\ &{}+ \bigl\vert u^{[m-1]}(b) \bigr\vert ^{2}+ \bigl\vert u^{[m]}(b) \bigr\vert ^{2}. \end{aligned}$$
(4.46)
-
3.
Let
\(\lambda=\lambda(\varphi)\)
be a function of
φ, and
\(u=u(\cdot, \varphi)\)
the corresponding eigenfunction. We see that
λ
is differentiable and it satisfies:
if
m
is even, then
$$ \lambda^{\prime}(\varphi)=-2\operatorname{Re}\sum _{i=0} ^{\frac{m-2}{2}} \bigl(u^{[i]}(a) \overline{u}^{[m-i-1]}(a) -u^{[m+i]}(a)\overline{u}^{[2m-i-1]}(a) \bigr); $$
(4.47)
if
m
is odd, then
$$\begin{aligned} \lambda^{\prime}(\varphi)={}&{-}2\operatorname{Re}\sum _{i=0} ^{\frac{m-3}{2}} \bigl(u^{[i]}(a) \overline{u}^{[m-i-1]}(a) +u^{[m+i]}(a)\overline{u}^{[2m-i-1]}(a) \bigr) \\ &{}- \bigl\vert u^{[\frac{m-1}{2}]}(a) \bigr\vert ^{2}- \bigl\vert u^{[\frac{3m-1}{2}]}(a) \bigr\vert ^{2}. \end{aligned}$$
(4.48)
-
4.
Let
\(\lambda=\lambda(\psi)\)
be a function of
ψ, and
\(u=u(\cdot, \psi)\)
the corresponding eigenfunction. We see that
λ
is differentiable and it satisfies:
if
m
is even, then
$$ \lambda^{\prime}(\psi)=2\operatorname{Re}\sum_{i=0} ^{\frac{m-2}{2}} \bigl(u^{[i]}(b)\overline{u}^{[m-i-1]}(b) -u^{[m+i]}(b)\overline{u}^{[2m-i-1]}(b) \bigr); $$
(4.49)
if
m
is odd, then
$$\begin{aligned} \lambda^{\prime}(\psi)={}&2\operatorname{Re}\sum_{i=0} ^{\frac{m-3}{2}} \bigl(u^{[i]}(b)\overline{u}^{[m-i-1]}(b) +u^{[m+i]}(b)\overline{u}^{[2m-i-1]}(b) \bigr) \\ &{}+ \bigl\vert u^{[\frac{m-1}{2}]}(b) \bigr\vert ^{2}+ \bigl\vert u^{[\frac{3m-1}{2}]}(b) \bigr\vert ^{2}. \end{aligned}$$
(4.50)
-
5.
Let
\(\lambda=\lambda(\eta)\)
be a function of
η, and
\(u=u(\cdot, \eta)\)
the corresponding eigenfunction. We see that
λ
is differentiable and it satisfies:
$$ \lambda^{\prime}(\eta)=(-1)^{(n)}\sum _{i=0} ^{m-1} \bigl( \bigl\vert u^{[i]}(a) \bigr\vert ^{2}+ \bigl\vert u^{[m+i]}(a) \bigr\vert ^{2} \bigr). $$
(4.51)
-
6.
Let
\(\lambda=\lambda(\tau)\)
be a function of
τ, and
\(u=u(\cdot, \tau)\)
the corresponding eigenfunction. We see that
λ
is differentiable and it satisfies
$$ \lambda^{\prime}(\tau)=(-1)^{(n+1)}\sum _{i=0} ^{m-1} \bigl( \bigl\vert u^{[i]}(b) \bigr\vert ^{2}+ \bigl\vert u^{[m+i]}(b) \bigr\vert ^{2} \bigr). $$
(4.52)
-
7.
Let
\(\lambda=\lambda(\theta)\)
be a function of
θ, and
\(u=u(\cdot, \theta)\)
the corresponding eigenfunction. We see that
λ
is differentiable and it satisfies
$$ \lambda^{\prime}(\theta)=(-1)^{m} 2\operatorname{Im}\sum _{i=0} ^{m-1}(-1)^{i}u^{[i]}(b) \overline{u}^{[2m-i-1]}(b). $$
(4.53)
-
8.
Let
\(\lambda=\lambda(\mathbf{K})\)
be a function of matrix
K, and
\(u=u(\cdot, \mathbf{K})\)
the corresponding eigenfunction. We suppose that
K
fulfills (2.16). Then we see that
λ
is differentiable, and its Frechet derivative is
$$\begin{aligned} d\lambda_{K}(\mathbf{H})={}&(-1)^{m+1} \begin{pmatrix} \overline{u}^{[2m-1]}, & -\overline{u}^{[2m-2]}, &\cdots, & \overline {u}^{[1]}, & -\overline{u} \end{pmatrix} (b) \\ &{}\times\mathbf{HK^{-1}}\begin{pmatrix} u \\ u^{[1]} \\ \vdots\\ u^{[2m-2]} \\ u^{[2m-1]} \end{pmatrix} (b). \end{aligned}$$
(4.54)
-
9.
Consider
λ
as a function of
\(p_{0}\in\mathbf {L}^{1}(a, b)\). We see that
λ
is differentiable and its Frechet derivative is
$$ d\lambda_{p_{0}}(h)= \int_{a}^{b} \vert u \vert ^{2}h. $$
(4.55)
-
10.
Consider
λ
as a function
\(w\in\mathbf{L}^{1}(a, b)\). We see that
λ
is differentiable and its Frechet derivative is
$$ d\lambda_{w}(h)=-\lambda \int_{a}^{b} \vert u \vert ^{2}h. $$
(4.56)
Proof
Since the proofs of (4.43)-(4.50) are similar we prove (4.45) and (4.46) only. Assume \(\beta\neq\pi/ 2\). When \(h\in\mathbb{R}\) is small enough, we denote by \(u=u(\cdot, \beta)\) and \(v=u(\cdot, \beta+h)\) the normalized real valued eigenfunctions of \(\mu =\lambda(\beta)\) and \(\nu=\lambda(\beta+h)\). By (2.1) we obtain
$$ \bigl[\lambda(\beta)-\lambda(\beta+h) \bigr] \int_{a}^{b} uvw\,ds = \Biggl[(-1)^{m} \sum_{i=0}^{2m-1}(-1)^{2m+1-i}u^{[i]} \bar{v}^{[2m-i-1]} \Biggr]_{a}^{b}. $$
If m is even,
$$\begin{aligned} & \bigl(\lambda(\beta)-\lambda(\beta+h) \bigr) \int_{a}^{b}uvw\,ds \\ &\quad= \bigl[-u\overline{v}^{[2m-1]}+u^{[1]} \overline{v}^{[2m-2]}-u^{[2]} \overline {v}^{[2m-3]}+\cdots +u^{[2m-3]}\overline{v}^{[2]}-u^{[2m-2]} \overline {v}^{[1]}+u^{[2m-1]}\overline{v} \bigr]_{a}^{b}. \end{aligned}$$
By the BCs, we conclude
$$ \bigl[-u\overline{v}^{[2m-1]}+u^{[1]}\overline{v}^{[2m-2]}-u^{[2]\overline {v}^{[2m-3]}}+ \cdots +u^{[2m-3]}\overline{v}^{[2]}-u^{[2m-2]}\overline {v}^{[1]}+u^{[2m-1]}\overline{v} \bigr](a)=0. $$
Then
$$\begin{aligned} & \bigl(\lambda(\beta)-\lambda(\beta+h) \bigr) \int_{a}^{b}uvw\,ds \\ &\quad=-u(b)\overline {v}^{[2m-1]}(b)+u^{[1]}(b) \overline{v}^{[2m-2]}(b)-u^{[3]}(b) \overline {v}^{[2m-3]}(b)+ \cdots \\ &\qquad{}+u^{[m-1]}(b)\overline{v}^{[m]}(b) -u^{[m]}(b) \overline{v}^{[m-1]}(b) \\ &\qquad{}+\cdots+u^{[2m-3]}(b)\overline{v}^{[2]}(b)-u^{[2m-2]}(b) \overline {v}^{[1]}(b)+u^{[2m-1]}(b)\overline{v}(b) \\ &\quad=-\tan\beta u^{[1]}(b)\overline{v}^{[2m-1]}(b)+\tan(\beta +h)u^{[1]}(b)\overline{v}^{[2m-1]}(b)-\tan\beta u^{[3]}(b)\overline {v}^{[2m-3]}(b)+\cdots \\ &\qquad{}-\tan\beta u^{[m-1]}(b)\overline{v}^{[m+1]}(b)+\tan(\beta +h)u^{[m-1]}(b)\overline{v}^{[m+1]}(b)-\cdots \\ &\qquad{} +\tan(\beta+h) u^{[2m-3]}(b)\overline{v}^{[3]}(b) \\ &\qquad{}-\tan\beta u^{[2m-2]}(b)\overline{v}^{[1]}(b)+\tan(\beta +h)u^{[2m-1]}(b)\overline{v}^{[1]}(b) \\ &\quad= \bigl[\tan(\beta+h)-\tan(\beta) \bigr] \bigl(u^{[1]}(b) \overline {v}^{[2m-1]}(b)+u^{[3]}(b)\overline{v}^{[2m-3]}(b)+ \cdots +u^{[m-1]}(b)\overline{v}^{[m+1]}(b) \\ &\qquad{}+u^{[m+1]}(b)\overline{v}^{[m-1]}(b)+ \cdots+u^{[2m-3]}(b) \overline {v}^{[3]}(b)+u^{[2m-1]}(b) \overline{v}^{[1]}(b) \bigr). \end{aligned}$$
(4.57)
We divide both sides of (4.57) by h, as \(h\rightarrow0\), and we have
$$\begin{aligned} -\lambda^{\prime}(\beta)={}&\sec^{2}\beta \bigl(u^{[1]}(b) \overline {u}^{[2m-1]}(b)+u^{[2]}(b)\overline{u}^{[2m-3]}(b)+ \cdots+u^{[m-1]}(b)\overline{u}^{[m+1]}(b) \\ &{}+u^{[m+1]}(b)\overline{u}^{[m-1]}(b)+\cdots+u^{[2m-3]}(b) \overline {u}^{[3]}(b)+u^{[2m-1]}(b)\overline{u}^{[1]}(b) \bigr) \\ ={}&\tan^{2}\beta u^{[1]}(b)\overline{u}^{[2m-1]}(b)+ \tan^{2}\beta u^{[3]}(b)\overline{u}^{[2m-3]}(b)+\cdots\\ &{} + \tan^{2}\beta u^{[m-1]}(b)\overline{u}^{[m+1]}(b) \\ &{}+ \tan^{2}\beta u^{[m+1]}(b)\overline{u}^{[m-1]}(b)+ \cdots+\tan ^{2}\beta u^{[2m-3]}(b)\overline{u}^{[3]}(b)\\ &{}+ \tan^{2}\beta u^{[2m-1]}(b)\overline{u}^{[1]}(b) \\ &{}+ u^{[1]}(b)\overline{u}^{[2m-1]}(b)+u^{[3]}(b) \overline {u}^{[2m-3]}(b)+\cdots\\ &{} +u^{[m-1]}(b)\overline{u}^{[m+1]}(b)+u^{[m+1]}(b) \overline {u}^{[m-1]}(b) \\ &{}+\cdots+u^{[2m-3]}(b)\overline{u}^{[3]}(b)+u^{[2m-1]}(b) \overline {u}^{[1]}(b) \\ ={}& u(b)\overline{u}^{[2m-2]}(b)+ u^{[2]}(b)\overline {u}^{[2m-4]}(b)+\cdots+ u^{[m-2]}(b)\overline{u}^{[m]}(b)\\ &{} + u^{[m]}(b)\overline{u}^{[m-2]}(b)+\cdots \\ &{}+ u^{[2m-2]}(b)\overline{u}(b)+ u^{[1]}(b)\overline {u}^{[2m-1]}(b)+u^{[3]}(b)\overline{u}^{[2m-3]}(b)\\ &{}+\cdots +u^{[m-1]}(b)\overline{u}^{[m+1]}(b) \\ &{}+u^{[m+1]}(b)\overline{u}^{[m-1]}(b)+\cdots+u^{[2m-3]}(b) \overline {u}^{[3]}(b)+u^{[2m-1]}(b)\overline{u}^{[1]}(b) \\ ={}&2\operatorname{Re}\sum_{i=1} ^{\frac{m-1}{2}} \bigl(u^{[2i]}(b)\overline {u}^{[2m-2i]}(b)+u^{[2i+1]}(b) \overline{u}^{[2m-2i-1]}(b) \bigr), \end{aligned}$$
and (4.45) holds;
if m is odd,
$$\begin{aligned} & \bigl(\lambda(\beta)-\lambda(\beta+h) \bigr) \int_{a}^{b}uvw\,ds \\ &\quad= \bigl[u\overline{v}^{[2m-1]}-u^{[1]}\overline{v}^{[2m-2]}+u^{[2]} \overline {v}^{[2m-3]}-\cdots -u^{[2m-3]}\overline{v}^{[2]}+u^{[2m-2]} \overline {v}^{[1]}-u^{[2m-1]}\overline{v} \bigr]_{a}^{b}. \end{aligned}$$
By the BCs, we infer that
$$\bigl[u\overline{v}^{[2m-1]}-u^{[1]}\overline{v}^{[2m-2]}+u^{[2]}\overline {v}^{[2m-3]}- \cdots -u^{[2m-3]}\overline{v}^{[2]}+u^{[2m-2]}\overline {v}^{[1]}-u^{[2m-1]}\overline{v}\bigr](a)=0. $$
Then
$$\begin{aligned} & \bigl(\lambda(\beta)-\lambda(\beta+h) \bigr) \int_{a}^{b}uvw\,ds \\ &\quad=u(b)\overline {v}^{[2m-1]}(b)-u^{[1]}(b) \overline{v}^{[2m-2]}(b)+u^{[2]}(b) \overline {v}^{[2m-3]}(b)+ \cdots \\ &\qquad{}+u^{[m-1]}(b)\overline{v}^{[m]}(b) -u^{[m]}(b) \overline{v}^{[m-1]}(b) \\ &\qquad{}+\cdots-u^{[2m-3]}(b)\overline{v}^{[2]}(b)+u^{[2m-2]}(b) \overline {v}^{[1]}(b)-u^{[2m-1]}(b)\overline{v}(b) \\ &\quad=\tan\beta u^{[1]}(b)\overline{v}^{[2m-1]}(b)-\tan(\beta +h)u^{[1]}(b)\overline{v}^{[2m-1]}(b)+\tan\beta u^{[3]}(b)\overline {v}^{[2m-3]}(b)-\cdots \\ &\qquad{}+\tan\beta u^{[m]}(b)\overline{v}^{[m]}(b)-\tan(\beta +h)u^{[m]}(b)\overline{v}^{[m]}(b)+\cdots \\ &\qquad{} -\tan(\beta+h) u^{[2m-3]}(b)\overline{v}^{[3]}(b) \\ &\qquad{}+\tan\beta u^{[2m-2]}(b)\overline{v}^{[1]}(b)-\tan(\beta +h)u^{[2m-1]}(b)\overline{v}^{[1]}(b) \\ &\quad=- \bigl[\tan(\beta+h)-\tan(\beta) \bigr] \bigl(u^{[1]}(b) \overline {v}^{[2m-1]}(b)+u^{[3]}(b)\overline{v}^{[2m-3]}(b) \\ &\qquad{}+ \cdots +u^{[m-2]}(b)\overline{v}^{[m+2]}(b) \\ &\qquad{}+u^{[m]}(b)\overline{v}^{[m]}(b)+ \cdots+u^{[2m-3]}(b) \overline {v}^{[3]}(b)+u^{[2m-1]}(b) \overline{v}^{[1]}(b) \bigr). \end{aligned}$$
(4.58)
We divide both sides of (4.58) by h, as \(h\rightarrow0\), and we obtain
$$\begin{aligned} \lambda^{\prime}(\beta)={}&\sec^{2}\beta \bigl(u^{[1]}(b) \overline {u}^{[2m-1]}(b)+u^{[3]}(b)\overline{u}^{[2m-3]}(b)+ \cdots +u^{[m-2]}(b)\overline{u}^{[m+2]}(b) \\ &{}+u^{[m]}(b)\overline{u}^{[m]}(b)+\cdots+u^{[2m-3]}(b) \overline {u}^{[3]}(b)+u^{[2m-1]}(b)\overline{u}^{[1]}(b) \bigr) \\ ={}&\tan^{2}\beta u^{[1]}(b)\overline{u}^{[2m-1]}(b)+ \tan^{2}\beta u^{[3]}(b)\overline{u}^{[2m-3]}(b)+\cdots + \tan^{2}\beta u^{[m-2]}(b)\overline{u}^{[m+2]}(b) \\ &{}+ \tan^{2}\beta u^{[m]}(b)\overline{u}^{[m]}(b)+ \cdots+\tan^{2}\beta u^{[2m-3]}(b)\overline{u}^{[3]}(b) + \tan^{2}\beta u^{[2m-1]}(b)\overline {u}^{[1]}(b) \\ &{}+ u^{[1]}(b)\overline{u}^{[2m-1]}(b)+u^{[3]}(b) \overline {u}^{[2m-3]}(b)+\cdots +u^{[m-2]}(b)\overline{u}^{[m+2]}(b)+u^{[m]}(b) \overline{u}^{[m]}(b) \\ &{}+\cdots+u^{[2m-3]}(b)\overline{u}^{[3]}(b)+u^{[2m-1]}(b) \overline {u}^{[1]}(b) \\ ={}& u(b)\overline{u}^{[2m-2]}(b)+ u^{[2]}(b)\overline {u}^{[2m-4]}(b)+\cdots\\ &{}+ u^{[m-3]}(b)\overline{u}^{[m+1]}(b) + u^{[m-1]}(b)\overline{u}^{[m-1]}(b) \\ &{}+u^{[m+1]}(b)\overline{u}^{[m-3]}(b)+\cdots+ u^{[2m-2]}(b) \overline {u}(b)\\ &{}+ u^{[1]}(b)\overline{u}^{[2m-1]}(b)+u^{[3]}(b) \overline {u}^{[2m-3]}(b)+\cdots \\ &{}+u^{[m-2]}(b)\overline{u}^{[m+2]}(b)+u^{[m]}(b) \overline {u}^{[m]}(b)+\cdots\\ &{}+u^{[2m-3]}(b)\overline{u}^{[3]}(b) +u^{[2m-1]}(b)\overline{u}^{[1]}(b) \\ ={}&2\operatorname{Re}\sum_{i=0} ^{\frac{m-3}{2}} \bigl(u^{[2i]}(b)\overline {u}^{[2m-2i-2]}(b)+u^{[2i+1]}(b) \overline{u}^{[2m-2i-1]}(b) \bigr)\\ &{}+ \bigl\vert u^{[m-1]}(b) \bigr\vert ^{2}+ \bigl\vert u^{[m](b)} \bigr\vert ^{2}, \end{aligned}$$
and (4.46) holds. This finishes the proof.
Next, we prove (4.53) and (4.54). Firstly, we show (4.53) is true. Let \(\mu=\lambda(\theta), u=u(\cdot, \theta)\) and \(\nu=\lambda(\theta +h), v=u(\cdot, \theta+h)\) be the corresponding eigenvalues and eigenfunctions, respectively. When \(h\in\mathbb{R}\) is small enough, we apply (4.40) to infer
$$\begin{aligned} & \bigl(\lambda(\theta)-\lambda(\theta+h) \bigr) \int_{a}^{b} uvw\,ds \\ &\quad= \Biggl[(-1)^{m}\sum_{i=0}^{2m-1}(-1)^{2m+1-i}u^{[i]} \bar {v}^{[2m-i-1]} \Biggr]_{a}^{b} \\ &\quad=(-1)^{m+1} \begin{pmatrix} \overline{v}^{[2m-1]}, & -\overline{v}^{[2m-2]}, &\cdots, & \overline {v}^{[1]}, & -\overline{v} \end{pmatrix} (b) \begin{pmatrix} u \\ u^{[1]} \\ \vdots\\ u^{[2m-2]} \\ u^{[2m-1]} \end{pmatrix} (b) \\ &\qquad{}-(-1)^{m+1} \begin{pmatrix} \overline{v}^{[2m-1]}, & -\overline{v}^{[2m-2]}, &\cdots, & \overline {v}^{[1]}, & -\overline{v} \end{pmatrix} (a) \begin{pmatrix} u \\ u^{[1]} \\ \vdots\\ u^{[2m-2]} \\ u^{[2m-1]} \end{pmatrix}(a) \\ &\quad=(-1)^{m+1}e^{i\theta} \begin{pmatrix} \overline{v}^{[2m-1]}, & -\overline{v}^{[2m-2]}, &\cdots, & \overline {v}^{[1]}, & -\overline{v} \end{pmatrix} (b) \mathbf{K}\begin{pmatrix} u \\ u^{[1]} \\ \vdots\\ u^{[2m-2]} \\ u^{[2m-1]} \end{pmatrix} (a) \\ &\qquad{}-(-1)^{m+1}e^{i(\theta+h)} \begin{pmatrix} \overline{v}^{[2m-1]}, & -\overline{v}^{[2m-2]}, &\cdots, & \overline {v}^{[1]}, & -\overline{v} \end{pmatrix}(b) \mathbf{K}\begin{pmatrix} u \\ u^{[1]} \\ \vdots\\ u^{[2m-2]} \\ u^{[2m-1]} \end{pmatrix}(a) \\ &\quad=(-1)^{m}e^{i\theta} \begin{pmatrix} \overline{v}^{[2m-1]}, & -\overline{v}^{[2m-2]}, &\cdots, & \overline {v}^{[1]}, & -\overline{v} \end{pmatrix} (b) \mathbf{K} \\ &\qquad{}\times\begin{pmatrix} u \\ u^{[1]} \\ \vdots\\ u^{[2m-2]} \\ u^{[2m-1]} \end{pmatrix} (a) \bigl(e^{ih}-1 \bigr). \end{aligned}$$
(4.59)
We divide both sides of (4.59) by h, as \(h\rightarrow0\), and we get
$$\begin{aligned} \lambda^{\prime}(\theta)&=(-1)^{m+1}ie^{i\theta} \begin{pmatrix} \overline{u}^{[2m-1]}, & -\overline{u}^{[2m-2]}, &\cdots, & \overline {u}^{[1]}, & -\overline{u} \end{pmatrix} (b)\mathbf{K}\begin{pmatrix} u \\ u^{[1]} \\ \vdots\\ u^{[2m-2]} \\ u^{[2m-1]} \end{pmatrix} (a) \\ &=(-1)^{m+1}i \begin{pmatrix} \overline{u}^{[2m-1]}, & -\overline{u}^{[2m-2]}, &\cdots, & \overline {u}^{[1]}, & -\overline{u} \end{pmatrix} (b) \begin{pmatrix} u \\ u^{[1]} \\ \vdots\\ u^{[2m-2]} \\ u^{[2m-1]} \end{pmatrix} (b) \\ &=(-1)^{m}2\operatorname{Im}\sum_{i=0} ^{m-1}(-1)^{i}u^{[i]}(b)\overline {u}^{[2m-i-1]}(b), \end{aligned}$$
and (4.53) holds.
Then we turn to establishing (4.54). Let \(u=u(\cdot, \mathbf {K}), v=u(\cdot, \mathbf{K+H})\) for K and \(\mathbf{K+H}\) satisfy (2.16). Proceeding similar to the argument above, we have
$$\begin{aligned} & \bigl(\lambda(\mathbf{K})-\lambda(\mathbf{K+H}) \bigr) \int_{a}^{b} uvw\,ds \\ &\quad= \Biggl[(-1)^{m}\sum_{i=0}^{2m-1}(-1)^{2m+1-i}u^{[i]} \bar {v}^{[2m-i-1]} \Biggr]_{a}^{b} \\ &\quad=(-1)^{m+1} \begin{pmatrix} \overline{v}^{[2m-1]}, & -\overline{v}^{[2m-2]}, &\cdots, & \overline {v}^{[1]}, & -\overline{v} \end{pmatrix} (b) \begin{pmatrix} u \\ u^{[1]} \\ \vdots\\ u^{[2m-2]} \\ u^{[2m-1]} \end{pmatrix} (b) \\ &\qquad{}-(-1)^{m+1} \begin{pmatrix} \overline{v}^{[2m-1]}, & -\overline{v}^{[2m-2]}, &\cdots, & \overline {v}^{[1]}, & -\overline{v} \end{pmatrix} (a) \begin{pmatrix} u \\ u^{[1]} \\ \vdots\\ u^{[2m-2]} \\ u^{[2m-1]} \end{pmatrix} (a) \\ &\quad=(-1)^{m+1}e^{i\theta} \begin{pmatrix} \overline{v}^{[2m-1]}, & -\overline{v}^{[2m-2]}, &\cdots, & \overline {v}^{[1]}, & -\overline{v} \end{pmatrix} (b) \mathbf{K}\begin{pmatrix} u \\ u^{[1]} \\ \vdots\\ u^{[2m-2]} \\ u^{[2m-1]} \end{pmatrix} (a) \\ &\qquad{}-(-1)^{m+1}e^{i\theta} \begin{pmatrix} \overline{v}^{[2m-1]}, & -\overline{v}^{[2m-2]}, &\cdots, & \overline {v}^{[1]}, & -\overline{v} \end{pmatrix} (b) \mathbf{(K+H)}\begin{pmatrix} u \\ u^{[1]} \\ \vdots\\ u^{[2m-2]} \\ u^{[2m-1]} \end{pmatrix} (a) \\ &\quad=(-1)^{m}e^{i\theta} \begin{pmatrix} \overline{v}^{[2m-1]}, & -\overline{v}^{[2m-2]}, &\cdots, & \overline {v}^{[1]}, & -\overline{v} \end{pmatrix}(b) \mathbf{H}\begin{pmatrix} u \\ u^{[1]} \\ \vdots\\ u^{[2m-2]} \\ u^{[2m-1]} \end{pmatrix}(a) \\ &\quad=(-1)^{m} \begin{pmatrix} \overline{v}^{[2m-1]}, & -\overline{v}^{[2m-2]}, &\cdots, & \overline {v}^{[1]}, & -\overline{v} \end{pmatrix}(b) \mathbf{HK^{-1}}\begin{pmatrix} u \\ u^{[1]} \\ \vdots\\ u^{[2m-2]} \\ u^{[2m-1]} \end{pmatrix}(b) \\ &\quad=(-1)^{m} \begin{pmatrix} \overline{u}^{[2m-1]}, & \bar{u}^{[2m-2]}, &\cdots, & \overline {u}^{[1]}, & -\overline{u} \end{pmatrix} (b) \mathbf{HK^{-1}}\begin{pmatrix} u \\ u^{[1]} \\ \vdots\\ u^{[2m-2]} \\ u^{[2m-1]} \end{pmatrix} (b) \\ &\quad=(-1)^{m} \begin{pmatrix} \overline{v}^{[2m-1]}- \overline{u}^{[2m-1]}, & \overline {u}^{[2m-2]}-\overline{v}^{[2m-2]}, &\cdots, & \overline{v}^{[1]}- \overline{u}^{[1]}, & \overline{u}-\overline{v} \end{pmatrix} (b) \mathbf{HK^{-1}} \\ &\qquad{}\times\begin{pmatrix} u \\ u^{[1]} \\ \vdots\\ u^{[2m-2]} \\ u^{[2m-1]} \end{pmatrix} (b). \end{aligned}$$
Then
$$\begin{aligned} \lambda(\mathbf{K+H})-\lambda(\mathbf{K})={}&(-1)^{m+1} \begin{pmatrix} \overline{u}^{[2m-1]}, & -\overline{u}^{[2m-2]}, &\cdots, & \overline {u}^{[1]}, & -\overline{u} \end{pmatrix} (b)\\ &{}\times\mathbf{HK^{-1}}\begin{pmatrix} u \\ u^{[1]} \\ \vdots\\ u^{[2m-2]} \\ u^{[2m-1]} \end{pmatrix} (b) + o(\mathbf{H}), \end{aligned}$$
and (4.54) follows.
To show (4.55), we let \(u=u(\cdot, q), v=u(\cdot, q+h)\) where \(h\in \mathbf{L}^{1}(a, b)\). Using (2.1) and integration by parts, then
$$ \bigl[\lambda(p_{0})-\lambda(p_{0}+h) \bigr] \int_{a}^{b} u\bar{v}w\,ds = \Biggl[(-1)^{m} \sum_{i=0}^{2m-1}(-1)^{2m+1-i}u^{[i]} \bar {v}^{[2m-i-1]} \Biggr]_{a}^{b}- \int_{a}^{b} u\bar{v}h\,ds. $$
For all BCs, we have
$$ \Biggl[(-1)^{m}\sum_{i=0}^{2m-1}(-1)^{2m+1-i}u^{[i]} \bar {v}^{[2m-i-1]} \Biggr]_{a}^{b}=0. $$
An application of Lemma 3.1 and Theorem 3.1 implies
$$ \bigl[\lambda(p_{0}+h)-\lambda(p_{0}) \bigr] \bigl(1+o(h) \bigr)= \int_{a}^{b} \vert u \vert ^{2}h \,ds+o(h). $$
Consequently,
$$ \lambda(p_{0}+h)-\lambda(p_{0})= \biggl[ \int_{a}^{b} \vert u \vert ^{2}h \,ds+o(h) \biggr] \bigl(1+o(h) \bigr)^{-1}= \int_{a}^{b} \vert u \vert ^{2}h \,ds+o(h), $$
as \(h\rightarrow0\) in \(\mathbf{L}^{1}(a, b)\), and (4.55) is proved. □
In a similar manner, we can deduce (4.56), here we omit its details.
Remark 4.1
The continuity of the nth eigenvalue of the self-adjoint BCs for 2mth-order spectral problems is more complicated. For regular Sturm-Liouville problems, Everitt, Möller and Zettl [25] show that the nth eigenvalue is not a continuous function of the BCs with separated boundary conditions, and similar results are obtained by Kong, Wu and Zettl [26] for general BCs.