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Chapter 7: Problem 6

Let \(\Omega\) be the square \(\\{(x, y):-1

Short Answer

Expert verified
The Neumann eigenvalues of a slitted circle \(\hat{\Omega}\) do not satisfy the inequality \(\mu_{n}(\hat{\Omega}) \geq \mu_{n}(\Omega)\) for all \(n \geq 1\), despite the slitted circle being a subset of the square \(\Omega\). This is due to the particular alterations that slitting introduces to the eigenvalue problem.

Step by step solution

01

Understand the Neumann eigenvalues

The Neumann eigenvalues are defined as the sequences of eigenvalues calculated from solutions to Neumann boundary problems. The Neumann boundary condition is a type of boundary condition, named after Carl Neumann. When used with differential equations, it specifies the values that the derivative of a solution is to take on the boundary of the domain.
02

Comparing Eigenvalues

It is generally expected that if one domain \(\hat{\Omega}\) is a subset of another domain \(\Omega\), the eigenvalues of the subset would be greater than or equal to that of the set, i.e., \(\mu_{n}(\hat{\Omega}) \geq \mu_{n}(\Omega)\) for all \(n \geq 1\). This is because eigenvalues are usually associated with the 'size' of the domain, and a smaller domain (subset) typically results in larger eigenvalues.
03

Concept of Slitting

However, in this exercise, the 'circle' \(\hat{\Omega}\) is not actually a full circle, it's a 'slitted' circle, which means it has a 'cut' or slit in it. This complicates the eigenvalue comparison - and in some particular cases, a slit can cause the eigenvalues to be smaller than those for a full, non-slit domain.
04

Conclusion

Therefore, although the circle \(\hat{\Omega}\) is a subset of the square \(\Omega\), the eigenvalues do not necessarily satisfy \(\mu_{n}(\hat{\Omega}) \geq \mu_{n}(\Omega)\) for all \(n \geq 1\). This points out a peculiar property of Neumann eigenvalues with respect to domain modifications.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Neumann boundary condition
Neumann boundary conditions play a critical role in solving differential equations, especially when the behavior at the boundaries of a given domain is essential. These conditions are named after Carl Neumann and differ from Dirichlet boundary conditions. While Dirichlet conditions set the function's value on the boundary, Neumann conditions specify the derivative's value.

Essentially, this means that for a differential equation solved over a domain, the rate of change (or slope) of the solution at the boundaries is fixed. For instance, if you're thinking about heat flow, the Neumann condition can represent a scenario where the object loses heat at a constant rate across its surface.

When applied to eigenvalue problems, Neumann boundary conditions help determine a sequence of eigenvalues based on these derivative conditions. These eigenvalues, known as Neumann eigenvalues, can offer insight into various physical, mechanical, or geometrical problems, highlighting how the system behaves at its most stable states.
eigenvalue comparison
Eigenvalue comparison is an interesting and insightful aspect when working with different domains. Even though it is generally expected that the eigenvalues of a subset domain should be greater than or equal to those of the larger domain, this is not always the case.

Intuitively, a smaller domain, being more 'compact', typically leads to higher eigenvalues due to constraints. The reasoning is that these constraints put more 'pressure' on the solution curves, forcing them to adjust more sharply.

However, this comparison doesn't hold in every situation. The exercise in this context defies the general rule due to the nature of the domains involved. This circle, being 'slitted', deviates from usual domain configurations, thus revealing surprising results like the possibility of smaller eigenvalues. Remember, eigenvalues are crucial as they determine phenomena like vibration frequencies or heat distribution patterns.
domain subset and eigenvalues
Examining how eigenvalues change with domain size is fascinating and offers much to learn. Generally, if you have one domain as a subset of another, you might anticipate its eigenvalues to be larger owing to the smaller area or volume.

This expectation stems from the idea that reducing the domain increases certain restrictions on the functions, leading to elevated eigenvalues. However, the peculiar aspect arises when you have irregular shapes or modifications鈥攍ike slits鈥攚hich make these usual rules less predictable.

Even though \( \hat{\Omega} \), the slitted circle, is entirely contained within\( \Omega \), the square, the slitting can counteract the shrinkage effect and upset usual comparisons. Essentially, this understanding challenges preconceived notions and illustrates the sensitivity of eigenvalues to structural changes in the domain. Identifying these intricacies requires a deeper dive into the geometry and boundaries of the domains involved.

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Most popular questions from this chapter

Suppose \(F\) is a \(C^{1}\) functional on \(X\) satisfying (i) \(F^{\prime}(u)=L+K(u)\), where \(L: X \rightarrow X^{*}\) is an isomorphism (i.e., a bounded linear operator that is one to one and onto) and \(K: X \rightarrow X^{*}\) is a compact map (i.e., maps bounded sets of \(X\) to precompact sets of \(X^{*}\) but \(K\) is not necessarily linear); and (ii) every Palais-Smale sequence is bounded in \(X\). Show that \(F\) satisfies the Palais- Smale condition.

Let \(X\) be a complete metric space and \(\mathcal{T}_{1}: X \rightarrow X\) and \(\mathcal{T}_{2}: X \rightarrow X\) be two contractions: \(d\left(\mathcal{T}_{j} x, \mathcal{T}_{j} y\right) \leq \alpha_{j} d(x, y)\) with \(0<\alpha_{j}<1\) for \(j=1,2\). Let \(\bar{x}_{j}\) be the (unique) fixed point of \(\mathcal{T}_{j}\). Assume \(\mathcal{T}_{1}, \mathcal{T}_{2}\) are \(\epsilon\)-close: \(d\left(\mathcal{T}_{1} x, \mathcal{T}_{2} x\right) \leq \epsilon\) for all \(x \in X\). Show that the fixed points are close: \(d\left(\bar{x}_{1}, \bar{x}_{2}\right)<\epsilon\) where \(\alpha=\min \left\\{\alpha_{1}, \alpha_{2}\right\\}\).

Suppose \(\Omega\) is a bounded domain, \(q(x)\) is a bounded function on \(\Omega\) satisfying \(q(x) \leq \eta\), and \(f \in L^{2}(\Omega)\). If \(\eta \geq 0\) is sufficiently small, then show that the Dirichlet problem \(\Delta u+q(x) u=f\) in \(\Omega, u=0\) on \(\partial \Omega\) admits a unique weak solution.

(a) If \(X\) is a Hilbert space and \(u_{j} \rightarrow u\) weakly in \(X\), then \(\|u\|_{X} \leq\) \(\liminf _{j \rightarrow \infty}\left\|u_{j}\right\|_{X}\). (b) Show that "weak upper semicontinuity" for the norm on a Hilbert space need not be true: \(u_{j} \rightarrow \neq\|u\|_{X} \geq \limsup \left\|u_{j}\right\|_{X}\).

Show that the solution \(y(t)\) of ( 40 ) depends continuously on the initial condition \(y_{0} ;\) if \(z(t)\) is the solution for the initial condition \(z_{0}\), then \(|y(t)-z(t)|<\epsilon\) for \(\left|t-t_{0}\right|<\tau\), provided \(\left|y_{0}-z_{0}\right|<\delta_{\epsilon}\).
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