Eigenfunction expansion / series solution technique

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In Theorem 6.12 show that h∈Dh \in \mathcal{D} if and only if h∈L2[a,b]h \in L^{2}[a,b] and ∑n=1∞λn2∣⟨h,en⟩∣2<∞\sum_{n=1}^{\infty } \lambda_{n}^{2} \left| \left\langle h, e_{n} \right\rangle \right|^{2} < \infty. If h∈Dh \in \mathcal{D}, show that h(x)=∑n=1∞⟨h,en⟩en(x)h(x) = \sum_{n=1}^{\infty } \left\langle h, e_{n} \right\rangle e_{n}(x), where this series converges uniformly and absolutely on [a,b][a,b].

Theorem 6.12 (context): Assume (6.4): if h∈Dh \in \mathcal{D} and Lh=0Lh=0, then h=0h=0. Then there is a sequence {λ1,λ2,…}\left\{ \lambda_{1}, \lambda_{2}, \ldots \right\} of real numbers and a basis {e1,e2,…}\left\{ e_{1}, e_{2}, \ldots \right\} for L2[a,b]L^{2}[a,b] such that

  1. 0<∣λ1∣<∣λ2∣<⋯0 < \left|\lambda_{1}\right| < \left|\lambda_{2}\right| < \cdots and ∣λn∣→∞\left|\lambda_{n}\right| \rightarrow \infty.

  2. en∈De_{n} \in \mathcal{D} and Len=λnenLe_{n} = \lambda_{n} e_{n} for all nn.

  3. If λ≠λn\lambda \neq \lambda_{n} for any λn\lambda_{n} and f∈L2[a,b]f \in L^{2}[a,b], then there is a unique hh in D\mathcal{D} with Lh−λh=fLh - \lambda h = f.

  4. If λ=λn\lambda = \lambda_{n} for some nn and f∈L2[a,b]f \in L^{2}[a,b], then there is an hh in D\mathcal{D} with Lh−λh=fLh - \lambda h = f if and only if ⟨f,en⟩=0\left\langle f, e_{n} \right\rangle = 0. If ⟨f,en⟩=0\left\langle f, e_{n} \right\rangle = 0, any two solutions of Lh−λh=fLh - \lambda h = f differ by a multiple of ene_{n}.

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In Theorem 6.12(c), show that h(x)=∑n=1∞(λn−λ)−1⟨f,en⟩en(x)h(x) = \sum_{n=1}^{\infty } (\lambda_{n} - \lambda )^{-1} \left\langle f, e_{n} \right\rangle e_{n}(x) and this series converges uniformly and absolutely on [a,b][a,b].

Theorem 6.12(c) (context): Assume (6.4). If λ≠λn\lambda \neq \lambda_{n} for any λn\lambda_{n} and f∈L2[a,b]f \in L^{2}[a,b], then there is a unique hh in D\mathcal{D} with Lh−λh=fLh - \lambda h = f, where {λn}\left\{ \lambda_{n}\right\} and {en}\left\{ e_{n}\right\} are as in Theorem 6.12.

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In Theorem 6.12(d), show that if f⊥enf \perp e_{n} and Lh−λnh=fLh - \lambda_{n} h = f, then h(x)=∑j≠n(λj−λn)−1⟨f,ej⟩ej(x)+αen(x)h(x) = \sum_{j \neq n} (\lambda_{j} - \lambda_{n})^{-1} \left\langle f, e_{j} \right\rangle e_{j}(x) + \alpha e_{n}(x) for some α\alpha, where the series converges uniformly and absolutely on [a,b][a,b].

Theorem 6.12(d) (context): Assume (6.4). If λ=λn\lambda = \lambda_{n} for some nn and f∈L2[a,b]f \in L^{2}[a,b], then there is an hh in D\mathcal{D} with Lh−λh=fLh - \lambda h = f if and only if ⟨f,en⟩=0\left\langle f, e_{n} \right\rangle = 0. If ⟨f,en⟩=0\left\langle f, e_{n} \right\rangle = 0, any two solutions of Lh−λh=fLh - \lambda h = f differ by a multiple of ene_{n}, where {λn}\left\{ \lambda_{n}\right\} and {en}\left\{ e_{n}\right\} are as in Theorem 6.12.

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A string of length ll is located along the axis OxOx between the points (0,0)(0,0) and (l,0)(l,0) and vibrates in the plane xOyxOy. The displacement yy at the point xx and at time tt satisfies the equation ∂2y∂t2=a2∂2y∂x2\dfrac {\partial^{2}y}{\partial t^{2}}=a^{2}\dfrac {\partial^{2}y}{\partial x^{2}}. Find the types of vibration for which y=u(x)sin⁡bty=u(x)\sin bt, where bb characterizes the frequency of vibration. The ends of the string are fixed: y=0y=0 at x=0x=0 and x=lx=l for any tt.

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References

[Con07] Conway J.B. A Course in Functional Analysis. 2nd ed., Springer, 2007.
[ГюКу03] Гюнтер Н.М., Кузьмин Р.О. Сборник задач по высшей математике. 13-е изд., Издательство «Лань», 2003.