SEPARATION OF VARIABLES EXAMPLE

Consider the one-dimensional heat equation, which describes the change in heat (or temperature) along the x-axis over time. Thus, it is a linear partial differential equation with independent variables in time (t) and space (x). It can be written for a dependent variable, f(x, t) as

where c² is a constant. The equation is first order in time and second order in space. If this equation is rewritten so that the time derivative is on the left hand side of the equation, while the spatial derivative is on the right hand side of the equation, it becomes:

For these two sides to be equal for all points in x and t, the equation must be equal to a new constant, α:

Now, separate the variable f into two components that are independent in space (X(x)) and time (T(t)):

Substituting eq. 4↑ into eq. 3↑ yields:

Rearranging so that the spatial terms are on one side and the temporal terms are on the other:

where the dot over the T indicates a derivative in time, and the double prime next to the X indicates a second order derivative in x. These can be separated into two first order ordinary differential equations:

The time equation can be written as T(t)̇ = c²αT. Direct integration yields the solution T(t) = Te^c2αt.

The spatial equation as written has an awkward solution, and one that is nonphysical in many applications. It is actually easier to note that there is a standard second-order equation similar to eq. 7↑ that will result a well-known solution. This equation is X’’(x) + γ²X = 0, which has an assumed solution of X(x) = Asin(γx) + Bcos(γx) if γ is a nonzero number. Comparing this equation to the second equation in eq. 7↑, it is easily seen that if α =  − γ², then the well-known solution can be used. Since α and γ are simply parameters that denote a constant that is yet to be determined, it does not matter if α =  − γ², but it does significantly simplify the problem mathematically.

Using this assumed solution, there are an infinite number of solutions to the ordinary differential equation. Two boundary conditions, typically applied at the two spatial (x) limits of the region of interest are needed to reduce the solution size.

Assume that the boundary conditions are f(0, t) = 0 and f(L, t) = 0 for a region of interest of x ∈ (0, L). When the f(0, t) = 0 boundary condition is applied, the solution has the form X(0) = A(0) + B(1) = 0, so that the constant, B is zero. The remaining solution is X(x) = Asin(γx).

When the second boundary condition at x = L is applied, the solution becomes X(L) = Asin(γL) = 0. If A = 0, then the solution has the form X(x) = 0 where there is no change of the variable in the spatial coordinate. This is known as a trivial solution, and it is not usually a solution of interest. If γ = 0, then sin(0) = 0 as well. However, the assumed solution was only valid if γ ≠ 0, so this is not an admissible solution at all since it violates the mathematical assumptions. Another way that the boundary condition can be zero is if sin(γL) = 0. sin is a trigonometric function that is periodic; that is, it will be zero whenever γL = nπ, where n = 1, 2, 3, .... Note that n cannot be zero, since that implies that γ = 0.

The solution of the equation is not X or T, but f. So, the solutions after the boundary conditions have been applied must be combined per eq. 4↑. Since there are a number of infinite solutions, the solution is summed:

Here, the γ values from the spatial boundary conditions is applied to the temporal solution. A_n will be constants that weight the importance of each n component in the final solution, and replaces the T constant from the temporal solution.