### Operator formulations

Put \(V=H_{0}^{2}(\Omega)\). Then the weak form of (1)-(2) is given as follows.

Find \(\lambda\in \mathbf {R}\), \(0\neq u \in V\), such that

$$\begin{aligned} a(u,v)=\lambda b(u,v), \quad \forall v\in V, \end{aligned}$$

(5)

where \(a(u,v)=\int_{\Omega}\triangle u \triangle v \,\mathrm{d} \mathbf{x}\), \(b(u,v)=\int_{\Omega}u v\,\mathrm{d} \mathbf{x}\).

The source problem associated with (5) is written as follows.

Find \(u \in V\) such that

$$\begin{aligned} a(u,v)=b(f,v), \quad \forall v\in V. \end{aligned}$$

(6)

It is obvious that \(a(\cdot,\cdot)\) is a continuous, symmetric, and *V*-elliptic bilinear form on \(V\times V\) and \(b(f,\cdot)\) is a continuous and linear functional on *V*. Thus, we can use \(\Vert \cdot \Vert _{a}=\sqrt{a(\cdot,\cdot)}\) as a norm in *V* which is equivalent to the norm \(\Vert \cdot \Vert _{2}\) induced by \(H^{2}(\Omega)\) and we know from the Lax-Milgram theorem that (6) has a unique solution. Thus, according to the source problem (6), we define the operator \(T: V\rightarrow V\) by

$$\begin{aligned} a(Tf,v)=b(f,v), \quad \forall v\in V. \end{aligned}$$

(7)

Thanks to [24], we know that (5) has the equivalent operator form

$$\begin{aligned} Tu=\frac{1}{\lambda}u. \end{aligned}$$

(8)

For the above operator *T*, we have the following result.

### Lemma 3.1

*The operator*
\(T:V\mapsto V\)
*is a self*-*adjoint compact one*.

### Proof

For \(\forall u,v \in V\), we have

$$\begin{aligned} a(Tu,v)=b(u,v)=b(v,u)=a(Tv,u)=a(u,Tv), \end{aligned}$$

Thus, \(T:V\mapsto V\) is self-adjoint. By taking \(f=u\), \(v=Tu\) in (7), we can obtain

$$\begin{aligned} a(Tu,Tu)=b(u,Tu). \end{aligned}$$

From the Poincaré inequality, we can derive

$$\begin{aligned} \gamma \Vert Tu\Vert _{2}^{2}&\leq a(Tu,Tu)=b(u,Tu) \leq \Vert u\Vert _{0} \Vert Tu\Vert _{0} \leq \Vert u \Vert _{0} \Vert Tu\Vert _{2}. \end{aligned}$$

Thus, we have

$$\begin{aligned} \Vert Tu\Vert _{2}\leq\frac{1}{\gamma} \Vert u\Vert _{0}, \end{aligned}$$

(9)

where *γ* is a positive constant.

Let *E* be the bounded set in *V*. Since *V* is compactly embedded in \(L_{2}(\Omega)\), so *E* is the sequentially compact set in \(L^{2}(\Omega )\). From (9), we know that *TE* is the sequentially compact set in *V*. Thus, \(T:V\rightarrow V\) is a compact operator. □

From the classical theory of abstract elliptic eigenvalue problem (see, *e.g.*, [23, 24]), we know that all eigenvalues of *T* are real and have finite algebraic multiplicity. We arrange the eigenvalues of *T* by increasing order:

$$0< \lambda_{1}\leq\lambda_{2}\leq\lambda_{3}\leq \cdots\nearrow +\infty. $$

Since the eigenfunctions corresponding to two arbitrary different eigenvalues of *T* must be orthogonal, there must exist a standard orthogonal basis in eigenspace corresponding to the same eigenvalue. Thus, by using the eigenfunctions of *T* corresponding to \(\{\lambda_{j}\}\), we can construct a complete orthonormal system of *V* as follows:

$$\begin{aligned} u_{1}, u_{2}, \ldots, u_{j},\ldots. \end{aligned}$$

Let \(X_{N}=\Pi_{N}^{d} \cap H_{0}^{2}(\Omega)\), then the spectral-Galerkin approximation of (5) is given as follows.

Find \(\lambda_{N} \in \mathbf {R}\), \(0\neq u_{N}\in X_{N}\) such that

$$\begin{aligned} a(u_{N},v_{N})=\lambda_{N} b(u_{N},v_{N}), \quad \forall v_{N} \in X_{N}. \end{aligned}$$

(10)

The source problem associated with (10) is written as follows.

Find \(u_{N}\in X_{N}\), such that

$$\begin{aligned} a(u_{N},v)=b(f,v), \quad \forall v\in X_{N}. \end{aligned}$$

(11)

Likewise, we know from the Lax-Milgram theorem that (11) has a unique solution. Thus, we can define the operator \(T_{N}: V\rightarrow X_{N}\) by

$$\begin{aligned} a(T_{N}f,v)=b(f,v), \quad \forall v\in X_{N}. \end{aligned}$$

(12)

From [24], we know that (10) has the equivalent operator form

$$\begin{aligned} T_{N}u_{N}=\frac{1}{\lambda_{N}}u_{N}. \end{aligned}$$

(13)

It is obvious that \(T_{N}: V\rightarrow X_{N}\) is a finite rank operator.

Define the projection operator \(\Pi_{N}^{2,0}: V\rightarrow X_{N}\) by

$$\begin{aligned} a\bigl(u-\Pi_{N}^{2,0}u,v\bigr)=0, \quad \text{for all } u \in V, v\in X_{N}. \end{aligned}$$

(14)

We have the following result.

### Lemma 3.2

*Let*
*T*
*and*
\(T_{N} \)
*be linear bounded operator defined by* (7) *and* (12), *respectively*. *Then we have the following equality*:

$$\begin{aligned} T_{N}=\Pi_{N}^{2,0}T. \end{aligned}$$

### Proof

For \(\forall u\in V\), \(v\in X_{N}\), we have

$$\begin{aligned} a\bigl(\Pi_{N}^{2,0}Tu-T_{N}u,v\bigr)=a\bigl( \Pi_{N}^{2,0}Tu-Tu,v\bigr)+a(Tu-T_{N}u,v)=0. \end{aligned}$$

(15)

By taking \(v=\Pi_{N}^{2,0}Tu-T_{N}u\) in (15), we can obtain

$$\begin{aligned} a\bigl(\Pi_{N}^{2,0}Tu-T_{N}u, \Pi_{N}^{2,0}Tu-T_{N}u\bigr)=0. \end{aligned}$$

Since \(a(\cdot,\cdot)\) is V-elliptic, we can obtain

$$\begin{aligned} T_{N}=\Pi_{N}^{2,0}T, \end{aligned}$$

which completes the proof of Lemma 3.2. □

It is obvious that

$$\begin{aligned} T_{N}|_{X_{N}}:X_{N}\rightarrow X_{N} \end{aligned}$$

is a self-adjoint finite rank operator and the eigenvalues of (10) can be arranged as

$$\begin{aligned} 0< \lambda_{1N}\leq\lambda_{2N}\leq\lambda_{3N}\leq \cdots\lambda _{KN},\qquad K=\dim(X_{N}). \end{aligned}$$

### Error estimates

In the following, we provide the error estimates. We first provide the following lemma.

### Lemma 3.3

*Let*
\((\lambda,u)\)
*and*
\((\lambda_{N},u_{N})\)
*be the eigenpair of* (5) *and* (10), *respectively*. *Then we have*

$$\begin{aligned} \lambda_{N}-\lambda=\frac{\Vert u_{N}-u\Vert _{a}^{2}}{\Vert u_{N}\Vert _{L^{2}(\Omega )}^{2}}-\lambda\frac{\Vert u_{N}-u\Vert _{L^{2}(\Omega)}^{2}}{\Vert u_{N}\Vert _{L^{2}(\Omega )}^{2}}. \end{aligned}$$

(16)

### Proof

From (5), we can derive

$$\begin{aligned} &a(u_{N}-u,u_{N}-u)-\lambda b(u_{N}-u,u_{N}-u) \\ &\quad =a(u_{N},u_{N})-2a(u_{N},u)+a(u,u)- \lambda b(u_{N},u_{N})+2\lambda b(u_{N},u)- \lambda b(u,u) \\ &\quad =a(u_{N},u_{N})-2\lambda b(u_{N},u)+ \lambda b(u,u)-\lambda b(u_{N},u_{N})+2\lambda b(u_{N},u)-\lambda b(u,u) \\ &\quad =a(u_{N},u_{N})-\lambda b(u_{N},u_{N}). \end{aligned}$$

By dividing \(b(u_{N},u_{N})\) and applying (10) to both sides of the above equation, we obtain

$$\begin{aligned} \lambda_{N}-\lambda=\frac{\Vert u_{N}-u\Vert _{a}^{2}}{\Vert u_{N}\Vert _{L^{2}(\Omega )}^{2}}-\lambda\frac{\Vert u_{N}-u\Vert _{L^{2}(\Omega)}^{2}}{\Vert u_{N}\Vert _{L^{2}(\Omega)}^{2}}, \end{aligned}$$

(17)

which completes the proof of Lemma 3.3. □

Put \(\eta_{N}=\sup_{u\in V,\Vert u\Vert _{a}=1}\inf_{v\in X_{N}}\Vert Tu-v\Vert _{a}\). It is clear that we have

$$\begin{aligned} \eta_{N}=\sup_{u\in V,\Vert u\Vert _{a}=1}\inf_{v\in X_{N}} \Vert Tu-v\Vert _{a}\rightarrow0 \quad (N\rightarrow \infty). \end{aligned}$$

(18)

Thus, we have the following convergence on operators (also see the proof in [25]).

### Theorem 3.1

*We have*

$$\begin{aligned} \lim_{N\rightarrow\infty} \Vert T-T_{N}\Vert _{a}= 0. \end{aligned}$$

(19)

### Proof

By the definition of the operator norm, we have

$$\begin{aligned} \Vert T-T_{N}\Vert _{a}&=\sup_{u\in V,\Vert u\Vert _{a}=1} \bigl\Vert (T-T_{N})u\bigr\Vert _{a} =\sup _{u\in V,\Vert u\Vert _{a}=1}\bigl\Vert Tu-\Pi_{N}^{2,0}Tu\bigr\Vert _{a} \\ &=\sup_{u\in V,\Vert u\Vert _{a}=1}\inf_{v\in X_{N}}\Vert Tu-v\Vert _{a}=\eta_{N}. \end{aligned}$$

Then, from (18), we obtain the desired result. □

Let \(M(\lambda)\) denote the eigenfunctions space of (5) corresponding to the eigenvalue *λ*. We have the following results (also see the proof in [25]).

### Theorem 3.2

*Let*
\((\lambda,u)\)
*and*
\((\lambda_{N},u_{N})\)
*be the eigenpair of* (5) *and* (10), *respectively*. *Then we have*

$$\begin{aligned} &\Vert u-u_{N}\Vert _{a} \leq\sup_{u\in M(\lambda),\Vert u\Vert _{a}=1} \frac {C}{\lambda} \bigl\Vert u-\Pi_{N}^{2,0}u\bigr\Vert _{a}, \end{aligned}$$

(20)

$$\begin{aligned} &\lambda_{N}-\lambda\leq\sup_{u\in M(\lambda),\Vert u\Vert _{a}=1} \frac {C}{\lambda^{2}}\frac{\Vert u-\Pi_{N}^{2,0}u\Vert _{a}^{2}}{\Vert u_{N}\Vert _{L^{2}(\Omega )}^{2}}. \end{aligned}$$

(21)

### Proof

From Theorem 3.1, we know that \(\Vert T-T_{N}\Vert _{a}\rightarrow0\) (\(N\rightarrow\infty\)). Thus, according to Theorem 7.4 in [24], we have

$$\begin{aligned} \Vert u-u_{N}\Vert _{a}\leq C\bigl\Vert (T-T_{N})|_{M(\lambda)}\bigr\Vert _{a}. \end{aligned}$$

(22)

Therefore, for any \(u\in M(\lambda)\) satisfying \(\Vert u\Vert _{a}=1\), we have

$$\begin{aligned} &\bigl\Vert (T-T_{N})u\bigr\Vert _{a}= \bigl\Vert Tu-\Pi_{N}^{2,0}Tu\bigr\Vert _{a}= \frac{1}{\lambda}\bigl\Vert u-\Pi _{N}^{2,0}u\bigr\Vert _{a}, \end{aligned}$$

(23)

$$\begin{aligned} &\bigl\Vert (T-T_{N})|_{M(\lambda)} \bigr\Vert _{a}=\sup_{u\in M(\lambda),\Vert u\Vert _{a}=1}\bigl\Vert (T-T_{N})u\bigr\Vert _{a}. \end{aligned}$$

(24)

By combining (23) and (24) with (22), we get the desired result (20). By Lemma 3.3, we obtain

$$\begin{aligned} &\lambda_{N}-\lambda\leq\frac{\Vert u_{N}-u\Vert _{a}^{2}}{\Vert u_{N}\Vert _{L^{2}(\Omega)}^{2}}, \end{aligned}$$

(25)

which together with (20) yields (21). □

### Theorem 3.3

*Let*
\((\lambda,u)\)
*and*
\((\lambda_{N},u_{N})\)
*be the eigenpair of* (5) *and* (10), *respectively*. *If*
\(u\in H_{0}^{2}(\Omega)\cap H^{r}(\Omega)\)
*with*
\(r\geq2\), *then*, *for*
\(N\geq2\),

$$\begin{aligned} &\Vert u-u_{N}\Vert _{a} \leq\frac{C}{\lambda}N^{(2-r)} \sup_{u\in M(\lambda ),\Vert u\Vert _{a}=1}\Vert u\Vert _{r}, \\ &\lambda_{N}-\lambda\leq\frac{C}{\lambda^{2}}N^{2(2-r)}\sup _{u\in M(\lambda),\Vert u\Vert _{a}=1}\frac{\Vert u\Vert _{r}^{2}}{\Vert u_{N}\Vert _{L^{2}(\Omega)}^{2}}. \end{aligned}$$

### Proof

From (14) and the continuity of \(a(u,v)\), we have

$$\begin{aligned} \bigl\Vert u-\Pi_{N}^{2,0}u\bigr\Vert _{a}^{2}&=a\bigl(u-\Pi_{N}^{2,0}u,u- \Pi_{N}^{2,0}u\bigr) \\ &=\inf_{\phi_{N}\in X_{N}}a(u-\phi_{N},u-\phi_{N}) \\ &\leq C\inf_{\phi_{N}\in X_{N}}\vert u-\phi_{N}\vert _{2}^{2} \\ & \leq C\bigl\vert u-S_{N}^{-2}u\bigr\vert _{2}^{2}. \end{aligned}$$

Thus, from Theorem 2.1, we can obtain

$$\begin{aligned} &\bigl\Vert u-\Pi_{N}^{2,0}u\bigr\Vert _{a}^{2} \leq C\bigl\vert u-S_{N}^{-2}u \bigr\vert _{2}^{2} \leq CN^{2(2-r)}\Vert u\Vert _{r}^{2}. \end{aligned}$$

(26)

By combing (26) with Theorem 3.2, we can get the desired result. □