Question

In each exercise, a rectangle is given. Consider the Dirichlet problem $$ u_{x x}(x, y)+u_{y y}(x, y)=0, \quad 0

Short answer

Question: Determine the function v(x, y) and the solution u(x, y) for the given Dirichlet problem where the boundary values are: $$ \begin{aligned} u(2, y) &= 1, \quad 0 \leq y < 1 \\ u(x, 1) &= 1, \quad 0 < x \leq 2 \\ u(0, y) &= 1, \quad 0 < y \leq 1 \\ u(x, 0) &= \cos(2 \pi x), \quad 0 \leq x < 2 \end{aligned} $$ Answer: The function v(x, y) is given by: $$ v(x, y) = \cos(2\pi x) $$ And the solution u(x, y) is given by: $$ u(x, y) = \cos(2\pi x) + \sum_{n=1}^{\infty} b_n \sin\left(\frac{(2n-1)\pi x}{2}\right) \sinh\left(\frac{(2n-1)\pi y}{2}\right) $$ where the coefficients \(b_n\) are: $$ b_n = -\frac{2}{(2n-1)\pi}\int_{0}^{1}(\cos(2\pi x) - 1)\sin\left(\frac{(2n-1)\pi x}{2}\right) dx $$

Step-by-step solution

Determine the function v(x, y)

As this is a Dirichlet problem, we apply Laplace's equation and use separation of variables to solve it. We will find the function \(v(x, y)\) by utilizing the given boundary conditions, resulting in a Fourier series expression for \(v(x, y)\). Since the boundary values at \(y = 0\) have the form \(\cos(2\pi x)\), we start by proposing a solution of the form: $$ v(x, y) = a \cos(2\pi x) + b \sin(2\pi x) $$ where \(a\) and \(b\) are constants to be determined. Applying the boundary conditions, we have: $$ v(0, y) = 1 \quad \Rightarrow \quad a = 1 $$ $$ v(2, y) = 1 \quad \Rightarrow \quad b = 0 $$ So, the function \(v(x, y)\) is given by: $$ v(x, y) = \cos(2\pi x) $$

Form U(x, y) and solve the boundary value problem

Now, we will form the function \(U(x, y) = u(x, y) - v(x, y)\), and solve the boundary value problem for \(U(x, y)\). By subtracting the function \(v(x, y)\) from the original problem, we obtain the following boundary values for \(U(x, y)\): $$ \begin{aligned} U(x, 1) &= 0, \quad 0 < x \leq 2 \\ U(0, y) &= 0, \quad 0 < y \leq 1 \\ U(x, 0) &= 0, \quad 0 \leq x < 2 \\ \end{aligned} $$ As \(U(x, y)\) satisfies homogeneous boundary conditions, we can use separation of variables to solve for \(U(x, y)\). The solution can be expressed as a Fourier sine series: $$ U(x, y) = \sum_{n=1}^{\infty} b_n \sin\left(\frac{(2n-1)\pi x}{2}\right) \sinh\left(\frac{(2n-1)\pi y}{2}\right) $$ Using the given function \(u(2, y) = 1\) and \(v(2, y) = \cos(2\pi x)\), we can solve for the coefficients \(b_n\): $$ \begin{aligned} b_n &= -\frac{2}{(2n-1)\pi}\int_{0}^{1}(\cos(2\pi x) - 1)\sin\left(\frac{(2n-1)\pi x}{2}\right) dx \end{aligned} $$ Plugging these coefficients back into the Fourier series expression, we obtain the solution for \(U(x, y)\).

Obtain the solution u(x, y)

Finally, we can obtain the solution \(u(x, y)\) by adding the function \(v(x, y)\) to the solution for \(U(x, y)\): $$ u(x, y) = U(x, y) + v(x, y) $$ $$ u(x, y) = \cos(2\pi x) + \sum_{n=1}^{\infty} b_n \sin\left(\frac{(2n-1)\pi x}{2}\right) \sinh\left(\frac{(2n-1)\pi y}{2}\right) $$ For part (c), use computer software to display a partial sum approximation of the solution surface using the obtained expression for \(u(x, y)\).

Key concepts

Laplace's equation

Laplace's equation is a second-order partial differential equation fundamental in mathematical physics. It is expressed as \( abla^2 u = 0 \), where \( u \) is a twice-differentiable function and \( abla^2 \) is the Laplacian operator. This equation describes the behavior of scalar fields like gravitational, electric potential, and temperature, under steady-state conditions. In physical terms, solutions to Laplace's equation imply that the potential function is harmonic, meaning no change or accumulation in that field.

• It shows that any point is the average of its neighbors.
• Commonly appears in problems involving heat, light, and fluid flow.
• Underlies the Dirichlet and Neumann boundary value problems.

In this exercise, it represents the equilibrium condition for the potential function \( u(x, y) \) over a rectangular region. Solving this equation lets us model potential fields including voltage or static fluid pressure.

Fourier series

A Fourier series is a way to represent a function as a sum of periodic components, and in many cases, it makes problems easier to solve. It's particularly useful in solving boundary value problems and partial differential equations. When you need to solve a function with periodic behavior, Fourier series lets you decompose it into sine and cosine terms, each weighted with coefficients. For example, in this problem, the boundary condition \( u(x, 0) = \cos(2\pi x) \) suggests using trigonometric functions in the solution.

• Each term in a Fourier series corresponds to a frequency component of the function.
• These terms are infinite, but in practical applications, we consider a finite sum (partial sum approximation) for simplicity.
• Fourier series converge to a function as the number of terms increases.

Using Fourier series allows us to express the solution \( U(x, y) \) in terms of a series of sines, simplifying the separation of variables approach.

Separation of variables

Separation of variables is a powerful mathematical technique used to solve partial differential equations like Laplace's equation. The idea is to transform a complex problem into simpler ones by assuming the solution is a product of functions, each in a single independent variable. It allows us to find solutions by solving simpler ordinary differential equations.To apply this technique:

• Assume the form \( u(x, y) = f(x)g(y) \).
• Substitute into Laplace's equation and separate terms such that each depends on only one variable.
• Solve the resulting ordinary differential equations separately.

In this exercise, separation of variables helps compute the solution \( U(x, y) \), which we've expressed as a Fourier sine series. This approach simplifies finding solutions to complex boundary value problems.

Boundary value problem

A boundary value problem involves solving a differential equation with specific conditions applied at the boundary of the domain. These conditions can be values, derivatives, or a mix of both, and determine the uniqueness of the solution.Dirichlet boundary conditions, as seen here, specify the values the solution must take on the boundary of the domain.

• In this exercise, the boundaries are defined by \( u(2, y), u(x, 1), u(0, y) \), and \( u(x, 0) \).
• Boundary conditions are critical in defining the behavior and physical interpretation of the solution.
• Boundary value problems can model scenarios like temperature distribution or structural deflection.

Solving such problems often involves methods like separation of variables and Fourier series, as used here, where boundary conditions lead to the decomposition of \( u(x, y) \) into its components \( U(x, y) \) and \( v(x, y) \). This approach offers a comprehensive understanding of how solutions vary across the domain.