Problem 9 - Applied Partial Differential Equations

Question

Solve Laplace's equation inside a \(90^{\circ}\) sector of a circular annulus \((a

Step-by-Step Solution

Answer

The solution of the Laplace equation in the given sector with boundary conditions (a) or (b) involves the separation of variables method. A set of two ordinary differential equations is derived from this, where the solutions are determined considering the compatibility with the imposed conditions at the boundaries of the sector. In boundary conditions (a), the solution is given as a power series obtained from the zero conditions at the boundaries. In boundary conditions (b), the solution involves an integral over the function \(f(r)\), which takes into account the given non-zero conditions.

Step 1: Formulate Laplace's Equation in Polar Coordinates

Formulate Laplace's equation in polar coordinates, which reads: \(\nabla^2 u = \frac{1}{r} \frac{\partial}{\partial r} (r \frac{\partial u}{\partial r}) + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2} = 0\).

Step 2: Apply Separation of Variables

Separate the solution in radial and angular parts, \(u(r, \theta) = R(r)\Theta(\theta)\), and substitute it into Laplace's equation. This will lead to two ordinary differential equations.

Step 3: Solve the Ordinary Differential Equations

The angular equation is a second order homogeneous ordinary differential equation with constant coefficients, which solutions can be given in terms of sine and cosine functions. The radial equation is a Euler-Cauchy differential equation. The solutions to the Euler-Cauchy equation can be expressed as powers of r.

Step 4: Apply Boundary Conditions

(a) Set the boundary conditions: either the radial or angular part of the solution is zero. This will lead to a series solution with coefficients that can be determined by compatibility with the function \(f(\theta)\) at the boundary. (b) Apply the boundary conditions: since now the non-zero condition involves both r and \theta, this will require the solution to be written as an integral over the functions \(f(r)\) where the separation constant is now considered as a variable in the integral.

Key Concepts

Polar Coordinates

When dealing with problems involving circular domains, such as a sector of a circular annulus, using polar coordinates makes the mathematics simpler because they naturally describe the geometry of a circle. Instead of using Cartesian coordinates \(x, y\), polar coordinates use \(r\) for the radial distance from the origin and \(\theta\) for the angle from a reference direction. In this context, Laplace's Equation takes a different form in polar coordinates:

• It involves derivatives with respect to \(r\) and \(\theta\), offering a more intuitive approach for circular domains.
• The equation becomes: \(abla^2 u = \frac{1}{r} \frac{\partial}{\partial r} \left(r \frac{\partial u}{\partial r}\right) + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2} = 0\).

Using this form allows the problem to be attacked using methods that exploit the symmetry of circular shapes. This makes it not only easier but also more accurate to solve the equation for the given conditions.

Separation of Variables

The method of separation of variables is a powerful mathematical technique used to solve partial differential equations, like Laplace's Equation. This technique assumes that the solution can be separated into a product of functions, each depending only on one of the coordinates. For example, in the reported scenario, assume:

• The solution has the form \(u(r, \theta) = R(r)\Theta(\theta)\).
• Substituting this into Laplace's equation allows the equation to be divided into two separate ordinary differential equations, each in terms of \(r\) or \(\theta\).

The advantage of this method is that it transforms a partial differential equation such as Laplace's Equation into simpler ordinary differential equations, which are often easier to solve. Each of these equations can then be tackled independently. This is especially helpful in problems having boundary conditions defined along the radial and angular components, respectively.

Boundary Conditions

Boundary conditions are crucial for determining a unique solution to differential equations. For this exercise, specific boundary conditions were given for the limits in \(r\) and \(\theta\):

• The solution was required to be zero at some boundaries: \(u(r, 0) = 0\), \(u(a, \theta) = 0\), etc.
• At other boundaries, the solution had to match given functions, like \(u(b, \theta) = f(\theta)\) or \(u(r, \pi/2) = f(r)\).

These conditions guide us to choose appropriate parameter values and ultimately select the correct solutions and constants in the general solution of the ordinary differential equations. By satisfying these conditions, the solution can be made to behave correctly at the edges of the domain, conforming to the physical situation described.

Ordinary Differential Equations

The separation of variables technique transforms Laplace's Equation into simpler equations called ordinary differential equations (ODEs). Solving these ordinary differential equations is a key step:

• The angular part, represented by \(\Theta(\theta)\), often results in a second-order ODE with solutions in terms of sine and cosine functions, due to periodic boundary conditions.
• The radial part, represented by \(R(r)\), typically results in an Euler-Cauchy type ODE, which has solutions in the form of power laws (i.e., \(r^n\)).

These solutions are then subject to the imposed boundary conditions, which help to determine any unknown constants. By solving these ODEs and applying boundary conditions, one constructs a specific solution to the original Laplace's Equation, reflecting the scenario's physical constraints.