Question

Consider a rectangular box with sides \(a, b\), and \(c\) located in the first octant with one corner at the origin. Let \(D\) denote the inside of this box. (a) Show that zero cannot be an eigenvalue of the Laplacian operator with the Dirichlet \(\mathrm{BCs}\) on \(\partial D\). (b) Find the GF for this Dirichlet BVP.

Short answer

For part (a), by applying the Laplacian operator and integrating, it's identified that zero cannot be an eigenvalue. For part (b), the Green's function for the Dirichlet boundary value problem is \(G(x, y, z; x', y', z') = \sum_{k, l, m = 1}^{\infty} \frac{\sin(\frac{k\pi x}{a}) \sin(\frac{k\pi x'}{a}) \sin(\frac{l\pi y}{b}) \sin(\frac{l\pi y'}{b}) \sin(\frac{m\pi z}{c}) \sin(\frac{m\pi z'}{c})}{(k/a)^2 + (l/b)^2 + (m/c)^2}\)

Step-by-step solution

Understanding the Laplacian

First, recall the definition of the Laplacian operator where \(\Delta u = \nabla^2 u = u_{xx} + u_{yy} + u_{zz}\) for cartesian coordinates.

Prove zero is not an eigenvalue

Start by assuming that there is a function \(f(x, y, z)\) such that it satisfies the eigenvalue problem \(\Del f = \lambda f\) in the solid, \(D\), and \(f = 0\) on \(\partial D\). Now, we multiply the operator equation by \(f\) and integrate over \(D\) to prove that zero cannot be an eigenvalue: \(\int_D f \; \Del f \, dxdydz = \lambda \int_D f^2 \, dxdydz\). We know that the left hand side of above equation is equivalent to \(-\int_D \nabla f \cdot \nabla f \, dxdydz\). Since all the values \(-\int_D \nabla f \cdot \nabla f \, dxdydz\) and \(\int_D f^2 \, dxdydz\) are nonnegative, for the equality to hold, \(\lambda\) cannot be zero.

Understanding the Green's Function

The Green's function, \(G\), for a given linear differential operator \(L\) in a region \(V\) satisfies the equation \(L[G(\mathbf{x},\mathbf{x'})] = -\delta(\mathbf{x}-\mathbf{x'})\), where \(L\) is the differential operator, \(\mathbf{x}\) is a point in \(V\), \(\mathbf{x'}\) is the source point in \(V\), and \(\delta\) is the Dirac delta function.

Find the Green's Function for the Dirichlet boundary value problem

The Green's function can be found by using the method of separation of variables. We can write the Green's function as a sum over the eigenfunctions of the Laplacian, i.e., \(G(x, y, z; x', y', z') = \sum_{k, l, m = 1}^{\infty} \frac{\sin(\frac{k\pi x}{a}) \sin(\frac{k\pi x'}{a}) \sin(\frac{l\pi y}{b}) \sin(\frac{l\pi y'}{b}) \sin(\frac{m\pi z}{c}) \sin(\frac{m\pi z'}{c})}{(k/a)^2 + (l/b)^2 + (m/c)^2}\). We get this by imposing that \(G = 0\) on \(\partial D\) (i.e., at \(x=0, a\), \(y=0, b\), and \(z=0, c\)) and finding the terms that satisfy this condition and the differential operator.

Key concepts

Eigenvalue Problem

An eigenvalue problem involves finding a scalar, known as an eigenvalue, and a corresponding function, known as an eigenfunction. This is typically set up with differential equations. In the context of a Laplacian operator, the eigenvalue problem can be expressed as:

• \( \Delta f = \lambda f \)

Here, \( \Delta \) is the Laplacian operator, \( f \) is the eigenfunction, and \( \lambda \) is the eigenvalue. This equation is crucial when determining the behavior of physical systems like those defined by quantum mechanics or heat conduction.
In our exercise, we are tasked with showing that zero cannot be an eigenvalue of the Laplacian with Dirichlet boundary conditions in a rectangular box. If \( \lambda = 0 \), the equation becomes \( \Delta f = 0 \), giving rise to a harmonic function. With Dirichlet boundary conditions \( f = 0 \) on the boundary, the only solution is the trivial one (\( f = 0 \) everywhere in \( D \)). Thus, zero can never be a valid eigenvalue in this setup.

Dirichlet Boundary Conditions

Dirichlet boundary conditions are a type of constraint used in differential equations where the solution is fixed to a certain value on the boundary of the domain.

• They are expressed as \( f = 0 \) on \( \partial D \).
• These conditions help in determining unique solutions by restricting the problem to only certain functions that satisfy these fixed boundary values.

In the given problem, the Dirichlet boundary conditions mean that the function \( f(x, y, z) \) must be zero on all surfaces of the box.This strict condition is essential, as it ensures that solutions within the domain \( D \) that are harmonic can only trivial, i.e., \( f(x, y, z) = 0 \) everywhere when zero is proposed as an eigenvalue.

Green's Function

A Green's function is a crucial tool used to solve differential equations with boundary conditions. For an operator \( L \), the Green's function satisfies:

• \( L[G(\mathbf{x}, \mathbf{x'})] = -\delta(\mathbf{x} - \mathbf{x'}) \)

Where \( \delta \) is the Dirac delta function.
The concept of a Green's function allows us to construct solutions for boundary value problems by acting like an influence function that links inputs to outputs in the domain.
To find the Green's function for the Dirichlet boundary value problem within the box, we apply the method of separation of variables, decomposing the function into a series of eigenfunctions. As calculated in the solution, this results in:

• \( G(x, y, z; x', y', z') = \sum_{k, l, m = 1}^{\infty} \frac{\sin(\frac{k\pi x}{a}) \sin(\frac{k\pi x'}{a}) \sin(\frac{l\pi y}{b}) \sin(\frac{l\pi y'}{b}) \sin(\frac{m\pi z}{c}) \sin(\frac{m\pi z'}{c})}{(k/a)^2 + (l/b)^2 + (m/c)^2} \)

Separation of Variables

Separation of variables is a mathematical method used to solve partial differential equations, where the solution is expressed as the product of functions, each depending on just one of the variables.

• This method is especially useful when dealing with problems with symmetrical boundaries or domains, like a rectangular box.
• In our context, this approach helps handle the Laplacian operator by turning a complex multi-variable problem into simpler many single-variable problems.

Given the boundaries of a box, we assume the potential solutions have forms like \( f(x, y, z) = X(x)Y(y)Z(z) \) which meet boundary conditions independently. Each function component, \( X, Y, \) and \( Z \), can then be solved using simple eigenvalue problems:

• For example, \( \frac{d^2 X}{dx^2} = -\left(\frac{k \pi}{a}\right)^2 X \)

Such simplifications make complex problems more approachable, allowing us to develop solutions like the Green's function for practical applications.