NationMaster - Encyclopedia: Separation of variables

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. For other meanings of mathematics or uses of math and maths, see Mathematics (disambiguation) and Math (disambiguation). ... A simulation of airflow into a duct using the Navier-Stokes equations A differential equation is a mathematical equation for an unknown function of one or several variables which relates the values of the function itself and of its derivatives of various orders. ...

1 Ordinary differential equations (ODE)
2 Partial differential equations
- 2.1 Example (I)
- 2.2 Example (II)
3 References
4 External links

Ordinary differential equations (ODE)

Suppose a differential equation can be written in the form

$frac{d}{dx} f(x) = g(x)h(f(x)),qquadqquad (1)$

which we can write more simply by letting $y = f (x)$ :

$frac{dy}{dx}=g(x)h(y)qquadqquad (1).$

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

${dy over h(y)} = {g(x)dx},$

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

Alternative method

Some who dislike Leibniz's notation may prefer to write this as In calculus, Leibnizs notation, named in honor of the 17th century German philosopher and mathematician Gottfried Wilhelm Leibniz, was originally the use of expressions such as dx and dy and to represent infinitely small (or infinitesimal) increments of quantities x and y, just as Î”x and Î”y represent finite...

$frac{1}{h(y)} frac{dy}{dx} = g(x),$

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 $x$ , we have

$int frac{1}{h(y)} frac{dy}{dx} , dx = int g(x) , dx, qquadqquad (2)$

or equivalently,

$int frac{1}{h(y)} , dy = int g(x) , dx$

because of the substitution rule for integrals. In calculus, the substitution rule is a tool for finding antiderivatives and 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 $frac{dy}{dx}$ as a fraction which can be separated. This allows us to solve separable differential equations more conveniently, as demonstrated in the example below. For a non-technical overview of the subject, see Calculus. ...

(Note that we do not need to use two constants of integration, in equation (2) as in In calculus, the indefinite integral of a given function (i. ...

$int frac{1}{h(y)} , dy + C_1 = int g(x) , dx + C_2,$

because a single constant $C = C 2 - C 1$ is equivalent.)

Example (I)

The ordinary differential equation

$frac{d}{dx}f(x)=f(x)(1-f(x))$

may be written as

$frac{dy}{dx}=y(1-y).$

If we let $g (x) = 1$ and $h (y) = y (1 - y)$ , 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 $d y$ and $d x$ as separate values, so that both sides of the equation may be multiplied by $d x$ . Subsequently dividing both sides by $y (1 - y)$ , we have

$frac{dy}{y(1-y)}=dx.$

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

$intfrac{dy}{y(1-y)}=int dx,$

which, via partial fractions, becomes In algebra, the partial fraction decomposition or (partial fraction expansion) is used to reduce the degree of either the numerator or the denominator of a rational function. ...

$intfrac{1}{y}+frac{1}{1-y},dy=int dx,$

and then

ln | y | - ln | 1 - y | = x + C

where C is the constant of integration. A bit of algebra gives a solution for y: In calculus, the indefinite integral of a given function (i. ... This article is about the branch of mathematics. ...

$y=frac{1}{1+Be^{-x}}.$

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. (One must be careful with the absolute values when solving the equation above. It turns out that the different signs of the absolute value contribute the positive and negative values for B, respectively. And the B = 0 case is contributed by the case that y = 1, as discussed below.)

Note that since we divided by $y$ and $(1 - y)$ we must check to see whether the solutions $y (x) = 0$ and $y (x) = 1$ solve the differential equation (in this case they are both solutions). See also: singular solutions. A simulation of airflow into a duct using the Navier-Stokes equations A differential equation is a mathematical equation for an unknown function of one or several variables which relates the values of the function itself and of its derivatives of various orders. ... A singular solution of a differential equation is a solution that satisfies the following conditions: It solves the original differential equation. ...

Example (II)

Population growth is often modeled by the differential equation

$frac{dP}{dt}=kPleft(1-frac{P}{K}right)$

where $P$ is the population with respect to time $t$ , $k$ is the rate of growth, and $K$ is the carrying capacity of the environment.

Separation of variables may be used to solve this differential equation.

$frac{dP}{dt}=kPleft(1-frac{P}{K}right)$

$intfrac{dP}{Pleft(1-frac{P}{K}right)}=int k,dt$

To evaluate the integral on the left side, we simplify the complex fraction:

$frac{1}{Pleft(1-frac{P}{K}right)}=frac{K}{Pleft(K-Pright)}$

Then, we decompose the fraction into partial fractions:

$frac{K}{Pleft(K-Pright)}=frac{1}{P}+frac{1}{K-P}$

Thus we have

$intleft(frac{1}{P}+frac{1}{K-P}right),dP=int k,dt$

$lnbegin{vmatrix}Pend{vmatrix}-lnbegin{vmatrix}K-Pend{vmatrix}=kt+C$

$lnbegin{vmatrix}K-Pend{vmatrix}-lnbegin{vmatrix}Pend{vmatrix}=-kt-C$

$lnbegin{vmatrix}cfrac{K-P}{P}end{vmatrix}=-kt-C$

$begin{vmatrix}cfrac{K-P}{P}end{vmatrix}=e^{-kt-C}$

$begin{vmatrix}cfrac{K-P}{P}end{vmatrix}=e^{-C}e^{-kt}$

$frac{K-P}{P}=pm e^{-C}e^{-kt}$

Let $A=pm e^{-C}$ .

$frac{K-P}{P}=Ae^{-kt}$

$frac{K}{P}-1=Ae^{-kt}$

$frac{K}{P}=1+Ae^{-kt}$

$frac{P}{K}=frac{1}{1+Ae^{-kt}}$

$P=frac{K}{1+Ae^{-kt}}$

Therefore, the solution to the logistic equation is

$Pleft(tright)=frac{K}{1+Ae^{-kt}}$

To find $A$ , let $t = 0$ and $Pleft(0right)=P_0$ . Then we have

$P_0=frac{K}{1+Ae^0}$

Noting that $e 0 = 1$ , and solving for A we get

$A=frac{K-P_0}{P_0}$

Partial differential equations

Given a partial differential equation of a function In mathematics, a partial differential equation (PDE) is a relation involving an unknown function of several independent variables and its partial derivatives with respect to those variables. ...

$F(x_1,x_2,dots,x_n)$

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

$F = F_1(x_1) cdot F_2(x_2) cdots F_n(x_n)$

$F = f_1(x_1) + f_2(x_2) + cdots + f_n(x_n)$

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. In mathematics, an ordinary differential equation (or ODE) is a relation that contains functions of only one independent variable, and one or more of its derivatives with respect to that variable. ...

When such a technique works, it is called a separable partial differential equation. A separable partial differential equation (PDE) is one that can be broken into a set of separate equations of lower dimensionality (fewer independent variables) by a method of separation of variables. ...

Example (I)

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

$frac{partial F}{partial x} + frac{partial F}{partial y} + frac{partial F}{partial z} = 0 qquadqquad (1)$

We shall guess

$F(x,y,z) = X(x) + Y(y) + Z(z)qquadqquad (2)$

thus making the equation (1) to

$frac{dX}{dx} + frac{dY}{dy} + frac{dZ}{dz} = 0$

(since $frac{partial F}{partial x} = frac{dX}{dx}$ ).

Now, since X'(x) is dependent only on x and Y'(y) is dependent only on 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,

$frac{dX}{dx} = c_1 quad frac{dY}{dy} = c_2 quad frac{dZ}{dz} = c_3qquadqquad (3)$

where the constants c₁, c₂, c₃ satisfy

$c_1 + c_2 + c_3 = 0qquadqquad (4)$

Eq. (3) is actually a set of three ODEs. In this case they are trivial and can be solved by simple integration, giving: In mathematics, an ordinary differential equation (or ODE) is a relation that contains functions of only one independent variable, and one or more of its derivatives with respect to that variable. ... This article is about the concept of integrals in calculus. ...

$F(x,y,z) = c_1 x + c_2 y + c_3 z + c_4qquadqquad (5)$

where the integration constant c₄ is determined by initial conditions.

Example (II)

Consider the differential equation

$nabla^2 v + lambda v = {partial^2 v over partial x^2} + {partial^2 v over partial y^2} + lambda v = 0.$

First we seek solutions of the form

$v = X(x)Y(y).,$

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

Substituting,

${partial^2overpartial x^2} [X(x)Y(y)]+{partial^2overpartial y^2}[X(x)Y(y)]+lambda X(x)Y(y)=$

$= X''(x)Y(y)+X(x)Y''(y)+lambda X(x)Y(y)= 0,$

Divide throughout by X(x)

$= {X''(x)Y(y) over X(x)}+{X(x)Y''(y)over X(x)}+{lambda X(x)Y(y)over X(x)}$

$={X''(x)Y(y) over X(x)}+Y''(y)+lambda Y(y) = 0$

and then by Y(y)

$={X''(x)over X(x)}+{Y''(y)+lambda Y(y)over Y(y)} = 0$

Now X′′(x)/X(x) is a function of x only, and (Y′′(y)+λY(y))/Y(y) is a function of y only, so for their sum to be equal to zero for all x and y, they must both be constant. Thus,

${X''(x)over X(x)} = k = -{Y''(y)+lambda Y(y)over Y(y)}$

where k is the separation constant. This splits up into ordinary differential equations In mathematics, an ordinary differential equation (or ODE) is a relation that contains functions of only one independent variable, and one or more of its derivatives with respect to that variable. ...

${X''(x)over X(x)} = k$

$X''(x) - k X(x)=0,$

and

${Y''(y)+lambda Y(y)over Y(y)} =-k$

$Y''(y)+(lambda+k) Y(y) =0,$

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. Shows a region where a differential equation is valid and the associated boundary values In mathematics, in the field of differential equations, a boundary value problem is a differential equation together with a set of additional restraints, called the boundary conditions. ...

References

A. D. Polyanin, Handbook of Linear Partial Differential Equations for Engineers and Scientists, Chapman & Hall/CRC Press, Boca Raton, 2002. ISBN 1-58488-299-9.

External links

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

Examples of separating variables to solve PDEs.

Categories: Ordinary differential equations | Partial differential equations

Results from FactBites:

Separation of variables - Wikipedia, the free encyclopedia (644 words)
	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.
	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.
	Methods of Generalized and Functional Separation of Variables at EqWorld: The World of Mathematical Equations.

More results at FactBites »

COMMENTARY

Share your thoughts, questions and commentary here

Want to know more?
Search encyclopedia, statistics and forums:

Feeds | Contact
The Wikipedia article included on this page is licensed under the GFDL.
Images may be subject to relevant owners' copyright.
All other elements are (c) copyright NationMaster.com 2003-5. All Rights Reserved.
Usage implies agreement with terms.

Contents

Ordinary differential equations (ODE) var ACE_AR = {Site: '756743', Size: '300250'};

Alternative method

Example (I)

Example (II)

Partial differential equations

Example (I)

Example (II)

References

External links

Ordinary differential equations (ODE)