# Separation of variables

In mathematics, separation of variables is any of several methods of solving ordinary and partial differential equations.

 Contents

## Ordinary differential equations (ODE)

Occasionally an ordinary differential equation allows a separation of variables, which we here exemplify rather than define. The differential equation

[itex]\frac{df(x)}{dx}=f(x)(1-f(x))[itex]

may be written as

[itex]\frac{dy}{dx}=y(1-y).[itex]

Pretend that dy and dx are numbers, so that both sides of the equation may be multiplied by dx. Also divide both sides by y(1 − y). We get

[itex]\frac{dy}{y(1-y)}=dx.[itex]

At this point we have separated the variables x and y from each other, since x appears only on the right side of the equation and y only on the left.

Integrating both sides, we get

[itex]\int\frac{dy}{y(1-y)}=\int dx,[itex]

which, via partial fractions, becomes

[itex]\int\frac{1}{y}+\frac{1}{1-y}\,dy=\int dx,[itex]

and then

[itex]\log_e y -\log_e (1-y)=x+C.[itex]

A bit of algebra gives a solution for y:

[itex]y=\frac{1}{1+Be^{-x}}.[itex]

One may check that if B is any positive constant, this function satisfies the differential equation.

This process also exemplifies the utility of the Leibniz notation, in which dy and dx are thought of as infinitely small increments of y and x respectively.

Note that once dividing in (1 − y) and (y − 0) one must check if the solutions y(x) = 0 and y(x) = 1 solve the differential equation. See also: singular solutions.

## Partial differential equations

Given a partial differential equation of a function

[itex] F(x_1,x_2,...,x_n) [itex]

of n variables, it is sometimes useful to guess solution of the form

[itex] F = F_1(x_1) \cdot F_2(x_2) \cdots F_n(x_n) [itex]

or

[itex] F = f_1(x_1) + f_2(x_2) + \cdots + f_n(x_n) [itex]

which turns the partial differential equation (PDE) into a set of ODEs. Usually, each independent variable creates a separation constant that cannot be determined only from the equation itself.

### Example (I)

Suppose F(x,y,z) and the following PDE:

[itex] \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} + \frac{\partial F}{\partial z} = 0 [itex] (1)

We shall guess

[itex] F(x,y,z) = X(x) + Y(y) + Z(z) [itex] (2)

thus making the equation (1) to

[itex] \frac{dX}{dx} + \frac{dY}{dy} + \frac{dZ}{dz} = 0 [itex]

(since [itex]\frac{\partial F}{\partial x} = \frac{dX}{dx} [itex]).

Now, since X'(x) is dependent only at x and Y'(y) is dependent only at y (so on for Z'(z)) and that the equation (1) is true for every x,y,z it is clear that each one of the term is constant. More precisely,

[itex] \frac{dX}{dx} = c_1 \quad \frac{dY}{dy} = c_2 \quad \frac{dZ}{dz} = c_3 [itex] (3)

were the constants c1, c2, c3 satisfy

[itex] c_1 + c_2 + c_3 = 0 [itex] (4)

Eq (3) is actually a set of three ODEs. In this case they are trivial and can be solved by simple integration, giving:

[itex] F(x,y,z) = c_1 x + c_2 y + c_3 z + c_4[itex] (5)

where the integration constant c4 is determined by initial conditions.

### Example (II)

Consider the differential equation

[itex]\nabla^2 v + \lambda v = {\partial^2 v \over \partial x^2} + {\partial^2 v \over \partial y^2} + \lambda v = 0[itex]

We suppose the solution is of the form

[itex] v = X(x)Y(y)\,[itex]

substituting,

[itex] {\partial^2\over\partial x^2}(X(x)Y(y))+{\partial^2\over\partial y}(X(x)Y(y))+\lambda X(x)Y(y)= [itex]
[itex] = X''(x)Y(y)+X(x)Y''(y)+\lambda X(x)Y(y)= 0\,[itex]

Divide throughout by X(x)

[itex] = {X''(x)Y(y) \over X(x)}+{X(x)Y''(y)\over X(x)}+{\lambda X(x)Y(y)\over X(x)} ={X''(x)Y(y) \over X(x)}+Y''(y)+\lambda Y(y) = 0[itex]

and then by Y(y)

[itex] ={X''(x)\over X(x)}+{Y''(y)+\lambda Y(y)\over Y(y)} = 0[itex]

Now X′′(x)/X(x) is a function of x only, as is (Y′′(y)+λY(y))/Y(y), so there are separation constants so

[itex] {X''(x)\over X(x)} = k = {Y''(y)+\lambda Y(y)\over Y(y)}[itex]

which splits up into ordinary differential equations

[itex]{X''(x)\over X(x)} = k[itex]
[itex]X''(x) - k X(x)=0\,[itex]

and

[itex]{Y''(y)+\lambda Y(y)\over Y(y)} =k[itex]
[itex]Y''(y)+(\lambda-k) Y(y) =0\,[itex]

which we can solve accordingly. If the equation as posed originally was a boundary value problem, one would use the given boundary values. See that article for an example which uses boundary values.

## Bibliography

• A. D. Polyanin and V. F. Zaitsev, Handbook of Exact Solutions for Ordinary Differential Equations, Chapman & Hall/CRC Press, Boca Raton, 2003 (2nd edition).
• A. D. Polyanin, Handbook of Linear Partial Differential Equations for Engineers and Scientists, Chapman & Hall/CRC Press, 2002.
• A. D. Polyanin and V. F. Zaitsev, Handbook of Nonlinear Partial Differential Equations, Chapman & Hall/CRC Press, 2004.
• A. D. Polyanin, V. F. Zaitsev, and A. Moussiaux, Handbook of First Order Partial Differential Equations], Taylor & Francis, London, 2002.

• Art and Cultures
• Countries of the World (http://www.academickids.com/encyclopedia/index.php/Countries)
• Space and Astronomy