91Ó°ÊÓ

Boundary conditions are specifications required to solve partial differential equations (PDEs) uniquely. In the problem given, we have boundary conditions stating that the function \( u \) must be zero at the edge of the circle \( (r = a) \) and that initially, \( u = 0 \) for all points inside the circle.

These conditions are crucial because they limit the possible solutions of a differential equation to those that are physically relevant to the problem's context. For example, the condition \( u = 0 \) at the boundary \( r = a \) says that outside influences are not affecting the system at that boundary.

Furthermore, the initial condition \( u = 0 \) signifies that at the start (time \( t = 0 \)), there's no quantity of \( u \) present anywhere inside \( r < a \). This ties into the nature of the problem as it assures us of a clean starting point, allowing us to work through the problem using these constraints to find a specific solution.

Separation of variables is a method used to solve a differential equation by splitting it into two or more simpler equations. This technique assumes that the solution can be expressed as a product of functions, each depending only on one of the independent variables. For our PDE, we presumed a solution of the form \( u(r, t) = v(r)T(t) \).

By substituting this form into the original PDE, it transforms the equation into a form where each side depends on only one variable (in this case, \( r \) or \( t \)). The key result of separation is that, if it is possible, each side of the equation must equal a constant because one side depends on time while the other depends on space.

This constant, often denoted as \(-\lambda\), essentially helps in splitting the PDE into two ordinary differential equations (ODEs) which are generally simpler to solve. Each resulting ODE then describes how the system evolves over time (T equation) and how it distributes over space (v equation). By solving these equations separately, we can eventually reassemble them to find our desired solution.

Ordinary Differential Equations (ODEs) arise naturally from the separation of variables in solving Partial Differential Equations (PDEs). In this exercise, the substitution led to two ODEs: \( \frac{T'}{kT} = -\lambda \) and \( \frac{v''}{v} + \frac{f}{kv} = -\lambda \).

Solving these ODEs independently allows us to find the functions \( T(t) \) and \( v(r) \), which describe how \( u \) evolves over time and space respectively.

• The ODE \( \frac{T'}{kT} = -\lambda \) is relatively straightforward. It reduces to a standard exponential decay/growth solution, determined by integrating with respect to time.
• The second ODE \( \frac{v''}{v} + \frac{f}{kv} = -\lambda \) deals with the spatial aspect, which might be more complex due to the presence of the term \( f(r,t) \).

By applying initial and boundary conditions to these solutions, constants that arise during integration can be evaluated to satisfy all the problem's requirements. This provides a complete and specific solution tailored to the problem's unique parameters.