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SEPARATION OF VARIABLES

Separation of variables is a mathematical technique for solving ordinary and partial differential equations, though here only the solution of linear partial differential equations will be addressed. To demonstrate this method, a one-dimensional wave equation is solved using this approach. The wave equation can be written as a function of the independent space (x) and time (t) variables as:
(1) (∂2u(x, t))/(∂t2) = C2(∂2u(x, t))/(∂x2)
where C is a constant. For simplicity, (∂u(x, t))/(∂x) = u′ denotes the spatial derivative, and (∂u(x, t))/(∂t) = u̇ denotes a temporal derivative. Hence, eqn. 1↑ can be written as
(2) ü = C2u’’
Step one  Separate the variables.
  • The dependent variable is assumed to be separable into two new dependent variables that are dependent only on one independent variable. For the example here, the variable u is separated into a product of two variables, X and T, which are functions of space and time, respectively, so that u(x, t) = X(x)T(t).
Step two  Simplify the partial differential equation.
  • The separated variable, u(x, t) = X(x)T(t), is substituted into the original differential equation in eqn. 2↑ and the derivatives employed, resulting in
    (3) T̈(t)X(x) = C2T(t)X’’(x)
    The equation can be rearranged so that each side of the equation is a function of only one independent variable. Dividing both sides of eqn. 3↑ by u = T(t)X(x) yields
    (4) (1)/(C2)(T̈(t))/(T(t)) = (X’’(x))/(X(x))
    The right hand side of eqn. 4↑ ⎛⎝(X’’(x))/(X(x))⎞⎠ is a function only of x and the left hand side ⎛⎝(1)/(C2)(T̈(t))/(T(t))⎞⎠ is a function of t only. The only way that the two sides of the equation can be equal if they both are equal to a constant. Therefore, eqn. 5↓ can be separated into two distinct ordinary differential equations
    (5) (1)/(C2)(T̈(t))/(T(t)) = σ   (X’’(x))/(X(x)) = σ
Step three  Solve the ordinary differential equations in space.
  • The spatial ordinary differential equation,
    (6) (X’’(x))/(X(x)) = σ
    will have a solution of different forms depending on the sign of σ.
    • σ = 0. In this case the ordinary differential equation is
      (7) (X’’(x))/(X(x)) = 0
      This differential equation is solved by integrating it twice. Thus, the general solution will have the form X(x) = Ax + B. Two boundary conditions are applied to find the values of A and B.
    • σ < 0. Let σ =  − λ2. In this case, the ordinary differential equation is
      (8) (X’’(x))/(X(x)) =  − λ2
      The general solution is X(x) = Asinh(λx) + Bcosh(λx). Again two boundary conditions are applied to find the values of the coefficients A and B
    • σ > 0. Let σ = λ2. In this case the ordinary differential equation is
      (9) (X’’(x))/(X(x)) = λ2
      The general solution is X(x) = Asin(λx) + Bcos(λx). Two boundary conditions are applied to find A and B.
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