Separation of variables

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In mathematics, separation of variables is any of several methods for solving ordinary and partial differential equations, in which algebra allows one to re-write an equation so that each of two variables occurs on a different side of the equation.

Contents 1 Ordinary differential equations (ODE) 1.1 Example 2 Partial differential equations 2.1 Example (I) 2.2 Example (II) 3 External links 4 Bibliography

Ordinary differential equations (ODE)

Suppose a differential equation can be written in the form

<math>\frac{d}{dx} f(x) = g(x)h(f(x)),\qquad\qquad (1)</math>

which we can write more simply by letting <math>y = f(x)</math>:

<math>\frac{dy}{dx}=g(x)h(y)\qquad\qquad (1).</math>

As long as h(y) ≠ 0, we can rearrange terms to obtain:

<math>{dy \over h(y)} = {g(x)dx},</math>

so that the two variables x and y have been separated.

Some who dislike the Leibniz notation may prefer to write this as

<math>\frac{1}{h(y)} \frac{dy}{dx} = g(x),</math>

but that fails to make it quite as obvious why this is called "separation of variables".

Integrating both sides of the equation with respect to <math>x</math>, we have

<math>\int \frac{1}{h(y)} \frac{dy}{dx} \, dx = \int g(x) \, dx + C, \qquad\qquad (2)</math>

or equivalently,

<math>\int \frac{1}{h(y)} \, dy = \int g(x) \, dx + C</math>

because of the substitution rule for integrals.

If one can evaluate the two integrals, one can find a solution to the differential equation. Observe that this process effectively allows us to treat the derivative <math>\frac{dy}{dx}</math> as a fraction which can be separated. This allows us to solve separable differential equations more conveniently, as demonstrated in the example below.

(Note that we do not need to use two constants of integration, in equation (2) as in

<math>\int \frac{1}{h(y)} \, dy + C_1 = \int g(x) \, dx + C_2,</math>

because a single constant <math>C = C_2 - C_1</math> is equivalent.)

Example

The ordinary differential equation

<math>\frac{d}{dx}f(x)=f(x)(1-f(x))</math>

may be written as

<math>\frac{dy}{dx}=y(1-y).</math>

If we let <math>g(x) = 1</math> and <math>h(y) = y(1-y)</math>, we can write the differential equation in the form of equation (1) above. Thus, the differential equation is separable.

As shown above, we can treat <math>dy</math> and <math>dx</math> as separate values, so that both sides of the equation may be multiplied by <math>dx</math>. Subsequently dividing both sides by <math>y(1 - y)</math>, we have

<math>\frac{dy}{y(1-y)}=dx.</math>

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

<math>\int\frac{dy}{y(1-y)}=\int dx,</math>

which, via partial fractions, becomes

<math>\int\frac{1}{y}+\frac{1}{1-y}\,dy=\int dx,</math>

and then

<math>\ln y -\ln (1-y)=x+C</math>

where C is the constant of integration. A bit of algebra gives a solution for y:

<math>y=\frac{1}{1+Be^{-x}}.</math>

One may check our solution by taking the derivative with respect to x of the function we found, where B is an arbitrary constant. The result should be equal to our original problem.

Note that since we divided by <math>(1 - y)</math> and <math>(y - 0)</math> we must check to see whether the solutions <math>y(x) = 0</math> and <math>y(x) = 1</math> solve the differential equation. See also: singular solutions.

Partial differential equations

Given a partial differential equation of a function

<math> F(x_1,x_2,\dots,x_n) </math>

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

<math> F = F_1(x_1) \cdot F_2(x_2) \cdots F_n(x_n) </math>

or

<math> F = f_1(x_1) + f_2(x_2) + \cdots + f_n(x_n) </math>

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:

<math> \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} + \frac{\partial F}{\partial z} = 0 \qquad\qquad (1)</math>

We shall guess

<math> F(x,y,z) = X(x) + Y(y) + Z(z)\qquad\qquad (2) </math>

thus making the equation (1) to

<math> \frac{dX}{dx} + \frac{dY}{dy} + \frac{dZ}{dz} = 0 </math>

(since <math>\frac{\partial F}{\partial x} = \frac{dX}{dx} </math>).

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,

<math> \frac{dX}{dx} = c_1 \quad \frac{dY}{dy} = c_2 \quad \frac{dZ}{dz} = c_3\qquad\qquad (3) </math>

were the constants c₁, c₂, c₃ satisfy

<math> c_1 + c_2 + c_3 = 0\qquad\qquad (4) </math>

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

<math> F(x,y,z) = c_1 x + c_2 y + c_3 z + c_4\qquad\qquad (5)</math>

where the integration constant c₄ is determined by initial conditions.

Example (II)

Consider the differential equation

<math>\nabla^2 v + \lambda v = {\partial^2 v \over \partial x^2} + {\partial^2 v \over \partial y^2} + \lambda v = 0.</math>

First we seek solutions of the form

<math> v = X(x)Y(y).\,</math>

Most solutions are not of that form, but other solutions are sums of (generally infinitely many) solutions of that form.

Substituting,

<math> {\partial^2\over\partial x^2}(X(x)Y(y))+{\partial^2\over\partial y^2}[X(x)Y(y)]+\lambda X(x)Y(y)= </math>

<math> = X(x)Y(y)+X(x)Y(y)+\lambda X(x)Y(y)= 0\,</math>

Divide throughout by X(x)

<math> = {X(x)Y(y) \over X(x)}+{X(x)Y(y)\over X(x)}+{\lambda X(x)Y(y)\over X(x)}</math>

<math> ={X(x)Y(y) \over X(x)}+Y(y)+\lambda Y(y) = 0</math>

and then by Y(y)

<math> ={X(x)\over X(x)}+{Y(y)+\lambda Y(y)\over Y(y)} = 0</math>

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

<math> {X(x)\over X(x)} = k = {Y(y)+\lambda Y(y)\over Y(y)}</math>

which splits up into ordinary differential equations

<math>{X(x)\over X(x)} = k</math>

<math>X(x) - k X(x)=0\,</math>

and

<math>{Y(y)+\lambda Y(y)\over Y(y)} =k</math>

<math>Y(y)+(\lambda-k) Y(y) =0\,</math>

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.

External links

• Methods of Generalized and Functional Separation of Variables at EqWorld: The World of Mathematical Equations.

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). ISBN 1-58488-297-2
• A. D. Polyanin, Handbook of Linear Partial Differential Equations for Engineers and Scientists, Chapman & Hall/CRC Press, Boca Raton, 2002. ISBN 1-58488-299-9ja:変数分離

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