Separation of variables

From Wikipedia, the free encyclopedia

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

[edit] Ordinary differential equations (ODE)

Suppose a differential equation can be written in the form

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

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

\frac{dy}{dx}=g(x)h(y).

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.

[edit] Alternative method

Some who dislike Leibniz's notation may prefer to write this as

\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, \qquad\qquad (2)

or equivalently,

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

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 \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.

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

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

because a single constant C = C2 − C1 is equivalent.)

[edit] 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 dy and dx as separate values, so that both sides of the equation may be multiplied by dx. 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

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

which, via partial fractions, becomes

\int\left(\frac{1}{y}+\frac{1}{1-y}\right)\,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:

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.

[edit] Example (II)

Population growth is often modeled by the differential equation

\frac{dP}{dt}=kP\left(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}=kP\left(1-\frac{P}{K}\right)
\int\frac{dP}{P\left(1-\frac{P}{K}\right)}=\int k\,dt

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

\frac{1}{P\left(1-\frac{P}{K}\right)}=\frac{K}{P\left(K-P\right)}

Then, we decompose the fraction into partial fractions:

\frac{K}{P\left(K-P\right)}=\frac{1}{P}+\frac{1}{K-P}

Thus we have

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

\ln\begin{vmatrix}P\end{vmatrix}-\ln\begin{vmatrix}K-P\end{vmatrix}=kt+C

\ln\begin{vmatrix}K-P\end{vmatrix}-\ln\begin{vmatrix}P\end{vmatrix}=-kt-C

\ln\begin{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

P\left(t\right)=\frac{K}{1+Ae^{-kt}}

To find A, let t = 0 and P\left(0\right)=P_0. Then we have

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

Noting that e0 = 1, and solving for A we get

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

[edit] Partial differential equations

Given a partial differential equation of a function

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)

or

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.

When such a technique works, it is called a separable partial differential equation.

[edit] 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 \qquad\qquad (1)

We shall guess

F(x,y,z) = X(x) + Y(y) + Z(z)\qquad\qquad (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 (and 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_3\qquad\qquad (3)

where the constants c1, c2, c3 satisfy

c_1 + c_2 + c_3 = 0\qquad\qquad (4)

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

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

where the integration constant c4 is determined by initial conditions.

[edit] 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^2\over\partial x^2} [X(x)Y(y)]+{\partial^2\over\partial 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, and so on 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

{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.

[edit] 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.

[edit] External links

  • Examples of separating variables to solve PDEs.