Initial Condition Example 1

The final part of the solution of the equation is to apply the initial condition to determine the values of the weighting constants. Let the initial condition be one of equilibrium, or f(x, 0) = 0. In this case

This solution must hold for every n value of the summation. Therefore, it is solved using the concept of orthogonality. Multiplying both sides of eq. ↓ by sin⎛⎝(nπx)/(L)⎞⎠ and integrating over the region of interest x ∈ (0, L) results in the equation

The exponential term is equal to one at t = 0, and the summations and A_n terms can be moved outside of the integral since they are independent of x.

Integrating the left hand side of the equation yields a zero (0) result if m ≠ n, but results in L ⁄ 2 if m = n, so that

So, if m ≠ n, then A_n are not known, but are trivial since the equation is zero. When n, then A_n = 0 to ensure that the initial condition is met. This too is a trivial solution.

Initial Condition Example 2

A more interesting solution is if the initial condition is non-zero. Let the initial deflection be f(x, 0) = sin(2πx ⁄ L). This is consistent with the boundary conditions at x = 0 and x = L. Once again, orthogonality is applied and results in the equation

In multiplying the left and right sides of the equation, the following solutions are found. If, n ≠ 2, then the solution is

Here, A_n = 0 to ensure that the solution remains zero. At n = 2, the solution is If n = 2, then A₂ = 1. Therefore, the equation that describes the heat equation is:

This was another relatively simple result since the initial function was one of the integral terms. If a non-trigonometric function describes the initial condition, the solution is not as simple.