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Step four Solve the ordinary differential equation in time.

The ordinary differential equation in time is

(10) (T̈(t))/(T(t)) = C²λ²

As before with the spatial term, there are different solutions for the temporal equation dependent upon the value and sign of σ:
- σ = 0. The solution has the form: T(t) = Dt + E,
- σ < 0. Let σ = − λ² to be consistent with the spatial assumption. Here, the solution will have the form: T(t) = Dsinh(λ√(C)t) + Ecosh(λ√(C)t)
- σ > 0. Let σ = λ² to be consistent with the spatial assumption. Here, the solution will have the form: T(t) = Dsin(λ√(C)t) + Ecos(λ√(C)t)
where D and E are constant coefficients that are determined by two initial conditions (boundary conditions that are functions of time).

Step five Recombine the separated variables

Having the solution for X(x) and T(t), they are recombined to form the solution for u(x, t) = X(x)T(t).

The equations obtained during steps three and four have an infinite number of possible solutions. To solve a particular problem, the boundary conditions in space and time are applied to find the correct and unique values of the solution in space and time. The application of these boundary conditions can lead to periodic solutions, as one may perceive from the use of the trigonometric functions in the assumed solutions. The boundary conditions can be applied to the spatial and the temporal terms independently.

To illustrate, let the spatial range of interest be set to x ∈ (0, 1), and that the two spatial boundary conditions are x(0, t) = x(1, t) = 0. When these are applied to the three different possible values of σ, the following outcomes are observed:

σ = 0 where the assumed solution is X(x) = Ax + B. At x = 0, X(0) = 0 = A(0) + B. Therefore, B = 0. At x = 1, X(1) = 0 = A(1). Therefore, A = 0 also, and X(x) = 0. Therefore, there is no spatial response if σ = 0, and this solution is observed to be trivial.
σ < 0 where the assumed solution is X(x) = Asinh(λx) + Bcosh(λx). At x = 0, X(0) = 0 = Asinh(0) + Bcosh(0) = A(0) + B(1). Therefore, B = 0. At x = 1, X(1) = 0 = Asinh(λ). Since, sinh(λ) is not equal to zero except when λ = 0, which is not the case here, then A = 0. Therefore, X(x) = 0, and this solution is also observed to be trivial.
σ > 0 where the assumed solution is X(x) = Asin(λx) + Bcos(λx). At x = 0, X(0) = 0 = Asin(0) + Bcos(0) = A(0) + B(1). Therefore, B = 0. At x = 1, X(1) = 0 = Asin(λ). For this equation, there is an infinite set of solutions when λ = nπ, where n = 1, 2, 3, … Therefore, A can be a finite, nonzero value, and there is a non-trivial solution to this equation.

The equation sin(λ) = 0 is the simplest form of the assumed solution that results in a non-trivial solution to the spatial equation. It is known as the characteristic equation. The solutions to this equation, λ = nπ, are known as eigenvalues, and the equation with these eigenvalues, X(x) = sin(nπx), is known as an eigenfunction.

Turning to the temporal equation, a similar analysis can be applied once two initial conditions are known. Let

(11) u(x, t) = ^∞⎲⎳_n = 1X_n(x)T_n(t) = ^∞⎲⎳_n = 1sin(nπx)(a_ncos(nCπt) + b_nsin(nCπt))

By applying the initial conditions one can find a_n and b_n.

(12) u(x, 0) = f(x) = ^∞⎲⎳_n = 1a_nsin(nπx)

and

(13) u̇(x, 0) = g(x) = ^∞⎲⎳_n = 1b_nCnπsin(nπx)

By using Fourier series one can solve for a_n and b_n from eqs. 12↑ and 13↑.

(14) a_n = 2¹⌠⌡₀f(x)sin(nπx)dx

and

(15) b_n = (2)/(Cnπ)¹⌠⌡₀g(x)sin(nπx)dx

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