Template:Differential equations
In mathematics, separation of variables (also known as the Fourier method) is any of several methods for solving ordinary and partial differential equations, in which algebra allows one to rewrite an equation so that each of two variables occurs on a different side of the equation.
Ordinary differential equations (ODE)
Suppose a differential equation can be written in the form
 $\backslash frac\{d\}\{dx\}\; f(x)\; =\; g(x)h(f(x)),\backslash qquad\backslash qquad\; (1)$
which we can write more simply by letting $y\; =\; f(x)$:
 $\backslash frac\{dy\}\{dx\}=g(x)h(y).$
As long as h(y) ≠ 0, we can rearrange terms to obtain:
 $\{dy\; \backslash over\; h(y)\}\; =\; \{g(x)dx\},$
so that the two variables x and y have been separated. dx (and dy) can be viewed, at a simple level, as just a convenient notation, which provides a handy mnemonic aid for assisting with manipulations. A formal definition of dx as a differential (infinitesimal) is somewhat advanced.
Alternative notation
Some who dislike Leibniz's notation may prefer to write this as
 $\backslash frac\{1\}\{h(y)\}\; \backslash frac\{dy\}\{dx\}\; =\; g(x),$
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
 $\backslash int\; \backslash frac\{1\}\{h(y)\}\; \backslash frac\{dy\}\{dx\}\; \backslash ,\; dx\; =\; \backslash int\; g(x)\; \backslash ,\; dx,\; \backslash qquad\backslash qquad\; (2)$
or equivalently,
 $\backslash int\; \backslash frac\{1\}\{h(y)\}\; \backslash ,\; dy\; =\; \backslash int\; g(x)\; \backslash ,\; dx$
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 $\backslash frac\{dy\}\{dx\}$ 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
 $\backslash int\; \backslash frac\{1\}\{h(y)\}\; \backslash ,\; dy\; +\; C\_1\; =\; \backslash int\; g(x)\; \backslash ,\; dx\; +\; C\_2,$
because a single constant $C\; =\; C\_2\; \; C\_1$ is equivalent.)
Example (I)
The ordinary differential equation
 $\backslash frac\{d\}\{dx\}f(x)=f(x)(1f(x))$
may be written as
 $\backslash frac\{dy\}\{dx\}=y(1y).$
If we let $g(x)\; =\; 1$ and $h(y)\; =\; y(1y)$, 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
 $\backslash frac\{dy\}\{y(1y)\}=dx.$
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
 $\backslash int\backslash frac\{dy\}\{y(1y)\}=\backslash int\; dx,$
which, via partial fractions, becomes
 $\backslash int\backslash frac\{1\}\{y\}\; \backslash ,\; dy\; +\; \backslash int\backslash frac\{1\}\{1y\}\backslash ,dy=\backslash int\; 1\; \backslash ,\; dx,$
and then
 $\backslash ln\; y\; \backslash ln\; 1y=x+C$
where C is the constant of integration. A bit of algebra gives a solution for y:
 $y=\backslash frac\{1\}\{1+Be^\{x\}\}.$
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.
Example (II)
Population growth is often modeled by the differential equation
 $\backslash frac\{dP\}\{dt\}=kP\backslash left(1\backslash frac\{P\}\{K\}\backslash right)$
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.
 $\backslash frac\{dP\}\{dt\}=kP\backslash left(1\backslash frac\{P\}\{K\}\backslash right)$
 $\backslash int\backslash frac\{dP\}\{P\backslash left(1\backslash frac\{P\}\{K\}\backslash right)\}=\backslash int\; k\backslash ,dt$
To evaluate the integral on the left side, we simplify the fraction
 $\backslash frac\{1\}\{P\backslash left(1\backslash frac\{P\}\{K\}\backslash right)\}=\backslash frac\{K\}\{P\backslash left(KP\backslash right)\}$
and then, we decompose the fraction into partial fractions
 $\backslash frac\{K\}\{P\backslash left(KP\backslash right)\}=\backslash frac\{1\}\{P\}+\backslash frac\{1\}\{KP\}$
Thus we have
 $\backslash int\backslash left(\backslash frac\{1\}\{P\}+\backslash frac\{1\}\{KP\}\backslash right)\backslash ,dP=\backslash int\; k\backslash ,dt$

$\backslash ln\backslash begin\{vmatrix\}P\backslash end\{vmatrix\}\backslash ln\backslash begin\{vmatrix\}KP\backslash end\{vmatrix\}=kt+C$

$\backslash ln\backslash begin\{vmatrix\}KP\backslash end\{vmatrix\}\backslash ln\backslash begin\{vmatrix\}P\backslash end\{vmatrix\}=ktC$

$\backslash ln\backslash begin\{vmatrix\}\backslash cfrac\{KP\}\{P\}\backslash end\{vmatrix\}=ktC$

$\backslash begin\{vmatrix\}\backslash cfrac\{KP\}\{P\}\backslash end\{vmatrix\}=e^\{ktC\}$

$\backslash begin\{vmatrix\}\backslash cfrac\{KP\}\{P\}\backslash end\{vmatrix\}=e^\{C\}e^\{kt\}$

$\backslash frac\{KP\}\{P\}=\backslash pm\; e^\{C\}e^\{kt\}$

Let $A=\backslash pm\; e^\{C\}$.

$\backslash frac\{KP\}\{P\}=Ae^\{kt\}$

$\backslash frac\{K\}\{P\}1=Ae^\{kt\}$

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

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

$P=\backslash frac\{K\}\{1+Ae^\{kt\}\}$
Therefore, the solution to the logistic equation is

$P\backslash left(t\backslash right)=\backslash frac\{K\}\{1+Ae^\{kt\}\}$
To find $A$, let $t=0$ and $P\backslash left(0\backslash right)=P\_0$. Then we have
 $P\_0=\backslash frac\{K\}\{1+Ae^0\}$
Noting that $e^0=1$, and solving for A we get
 $A=\backslash frac\{KP\_0\}\{P\_0\}$
Partial differential equations
The method of separation of variables is also used to solve a wide range of linear partial differential equations with boundary and initial conditions, such as heat equation, wave equation, Laplace equation and Helmholtz equation.
Homogeneous case
Consider the onedimensional heat equation.The equation is

$\backslash frac\{\backslash partial\; u\}\{\backslash partial\; t\}\backslash alpha\backslash frac\{\backslash partial^\{2\}u\}\{\backslash partial\; x^\{2$


(})

=0
Template:EqRef}}
The boundary condition is homogeneous, that is

$u\backslash big$


(Template:EqRef)

Let us attempt to find a solution which is not identically zero satisfying the boundary conditions but with the following property: u is a product in which the dependence of u on x, t is separated, that is:

Substituting u back into equation and using the product rule,

Since the right hand side depends only on x and the left hand side only on t, both sides are equal to some constant value − λ. Thus:

and

− λ here is the eigenvalue for both differential operators, and T(t) and X(x) are corresponding eigenfunctions.
We will now show that solutions for X(x) for values of λ ≤ 0 cannot occur:
Suppose that λ < 0. Then there exist real numbers B, C such that
 $X(x)\; =\; B\; e^\{\backslash sqrt\{\backslash lambda\}\; \backslash ,\; x\}\; +\; C\; e^\{\backslash sqrt\{\backslash lambda\}\; \backslash ,\; x\}.$
From Template:EqNote we get

and therefore B = 0 = C which implies u is identically 0.
Suppose that λ = 0. Then there exist real numbers B, C such that
 $X(x)\; =\; Bx\; +\; C.$
From Template:EqNote we conclude in the same manner as in 1 that u is identically 0.
Therefore, it must be the case that λ > 0. Then there exist real numbers A, B, C such that
 $T(t)\; =\; A\; e^\{\backslash lambda\; \backslash alpha\; t\},$
and
 $X(x)\; =\; B\; \backslash sin(\backslash sqrt\{\backslash lambda\}\; \backslash ,\; x)\; +\; C\; \backslash cos(\backslash sqrt\{\backslash lambda\}\; \backslash ,\; x).$
From Template:EqNote we get C = 0 and that for some positive integer n,
 $\backslash sqrt\{\backslash lambda\}\; =\; n\; \backslash frac\{\backslash pi\}\{L\}.$
This solves the heat equation in the special case that the dependence of u has the special form of Template:EqNote.
In general, the sum of solutions to Template:EqNote which satisfy the boundary conditions Template:EqNote also satisfies Template:EqNote and Template:EqNote. Hence a complete solution can be given as
 $u(x,t)\; =\; \backslash sum\_\{n\; =\; 1\}^\{\backslash infty\}\; D\_n\; \backslash sin\; \backslash frac\{n\backslash pi\; x\}\{L\}\; \backslash exp\backslash left(\backslash frac\{n^2\; \backslash pi^2\; \backslash alpha\; t\}\{L^2\}\backslash right),$
where D_{n} are coefficients determined by initial condition.
Given the initial condition
 $u\backslash big\_\{t=0\}=f(x),$
we can get
 $f(x)\; =\; \backslash sum\_\{n\; =\; 1\}^\{\backslash infty\}\; D\_n\; \backslash sin\; \backslash frac\{n\backslash pi\; x\}\{L\}.$
This is the sine series expansion of f(x). Multiplying both sides with $\backslash sin\; \backslash frac\{n\backslash pi\; x\}\{L\}$ and integrating over [0,L] result in
 $D\_n\; =\; \backslash frac\{2\}\{L\}\; \backslash int\_0^L\; f(x)\; \backslash sin\; \backslash frac\{n\backslash pi\; x\}\{L\}\; \backslash ,\; dx.$
This method requires that the eigenfunctions of x, here $\backslash left\backslash \{\backslash sin\; \backslash frac\{n\backslash pi\; x\}\{L\}\backslash right\backslash \}\_\{n=1\}^\{\backslash infty\}$, are orthogonal and complete. In general this is guaranteed by SturmLiouville theory.
Nonhomogeneous case
Suppose the equation is nonhomogeneous,

$\backslash frac\{\backslash partial\; u\}\{\backslash partial\; t\}\backslash alpha\backslash frac\{\backslash partial^\{2\}u\}\{\backslash partial\; x^\{2$


(})

=h(x,t)
Template:EqRef}}
with the boundary condition the same as Template:EqNote.
Expand h(x,t), u(x,t) and f(x,t) into



where h_{n}(t) and b_{n} can be calculated by integration, while u_{n}(t) is to be determined.
Substitute Template:EqNote and Template:EqNote back to Template:EqNote and considering the orthogonality of sine functions we get
 $u\text{'}\_\{n\}(t)+\backslash alpha\backslash frac\{n^\{2\}\backslash pi^\{2\}\}\{L^\{2\}\}u\_\{n\}(t)=h\_\{n\}(t),$
which are a sequence of linear differential equations that can be readily solved with, for instance, Laplace transform, or Integrating factor. Finally, we can get
 $u\_\{n\}(t)=e^\{\backslash alpha\backslash frac\{n^\{2\}\backslash pi^\{2\}\}\{L^\{2\}\}\; t\}\; \backslash left\; (b\_\{n\}+\backslash int\_\{0\}^\{t\}h\_\{n\}(s)e^\{\backslash alpha\backslash frac\{n^\{2\}\backslash pi^\{2\}\}\{L^\{2\}\}\; s\}\; \backslash ,\; ds\; \backslash right).$
If the boundary condition is nonhomogeneous, then the expansion of Template:EqNote and Template:EqNote is no longer valid. One has to find a function v that satisfies the boundary condition only, and subtract it from u. The function uv then satisfies homogeneous boundary condition, and can be solved with the above method.
In orthogonal curvilinear coordinates, separation of variables can still be used, but in some details different from that in Cartesian coordinates. For instance, regularity or periodic condition may determine the eigenvalues in place of boundary conditions. See spherical harmonics for example.
Matrices
The matrix form of the separation of variables is the Kronecker sum.
As an example we consider the 2D discrete Laplacian on a regular grid:
 $L\; =\; \backslash mathbf\{D\_\{xx\}\}\backslash oplus\backslash mathbf\{D\_\{yy\}\}=\backslash mathbf\{D\_\{xx\}\}\backslash otimes\backslash mathbf\{I\}+\backslash mathbf\{I\}\backslash otimes\backslash mathbf\{D\_\{yy\}\},\; \backslash ,$
where $\backslash mathbf\{D\_\{xx\}\}$ and $\backslash mathbf\{D\_\{yy\}\}$ are 1D discrete Laplacians in the x and ydirections, correspondingly, and $\backslash mathbf\{I\}$ are the identities of appropriate sizes. See the main article Kronecker sum of discrete Laplacians for details.
References
 A. D. Polyanin, Handbook of Linear Partial Differential Equations for Engineers and Scientists, Chapman & Hall/CRC Press, Boca Raton, 2002. ISBN 1584882999.


External links
 Template:Springer
 John Renze, MathWorld
 Methods of Generalized and Functional Separation of Variables at EqWorld: The World of Mathematical Equations
 Examples of separating variables to solve PDEs
 "A Short Justification of Separation of Variables"
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