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| {{Differential equations}}
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| In [[mathematics]], '''separation of variables''' (also known as the '''Fourier method''') is any of several methods for solving ordinary and partial [[differential equation]]s, in which algebra allows one to rewrite an equation so that each of two variables occurs on a different side of the equation.
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| == Ordinary differential equations (ODE) ==
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| Suppose a differential equation can be written in the form
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| :<math>\frac{d}{dx} f(x) = g(x)h(f(x))</math>
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| which we can write more simply by letting <math>y = f(x)</math>:
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| :<math>\frac{dy}{dx}=g(x)h(y).</math>
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| As long as ''h''(''y'') ≠ 0, we can rearrange terms to obtain:
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| :<math>{dy \over h(y)} = {g(x)dx},</math>
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| 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.
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| === Alternative notation ===
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| Some who dislike [[Leibniz's notation]] may prefer to write this as
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| :<math>\frac{1}{h(y)} \frac{dy}{dx} = g(x),</math>
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| 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
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| :<math>\int \frac{1}{h(y)} \frac{dy}{dx} \, dx = \int g(x) \, dx, \qquad\qquad (1) </math>
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| or equivalently,
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| :<math>\int \frac{1}{h(y)} \, dy = \int g(x) \, dx </math>
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| because of the [[integration by substitution|substitution rule for integrals]].
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| 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.
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| (Note that we do not need to use two [[arbitrary constant of integration|constants of integration]], in equation (1) as in
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| :<math>\int \frac{1}{h(y)} \, dy + C_1 = \int g(x) \, dx + C_2,</math>
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| because a single constant <math>C = C_2 - C_1</math> is equivalent.)
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| ===Example (I)===
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| The ordinary differential equation | |
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| :<math>\frac{d}{dx}f(x)=f(x)(1-f(x))</math>
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| may be written as
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| :<math>\frac{dy}{dx}=y(1-y).</math>
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| 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.
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| 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
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| :<math>\frac{dy}{y(1-y)}=dx.</math>
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| 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.
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| Integrating both sides, we get
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| :<math>\int\frac{dy}{y(1-y)}=\int dx,</math>
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| which, via [[partial fraction]]s, becomes
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| :<math>\int\frac{1}{y} \, dy + \int\frac{1}{1-y}\,dy=\int 1 \, dx,</math>
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| and then
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| :<math>\ln |y| -\ln |1-y|=x+C</math>
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| where ''C'' is the [[arbitrary constant of integration|constant of integration]]. A bit of [[algebra]] gives a solution for ''y'':
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| :<math>y=\frac{1}{1+Be^{-x}}.</math>
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| 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.)
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| Note that since we divided by <math>y</math> and <math>(1 - y)</math> we must check to see whether the solutions <math>y(x) = 0</math> and <math>y(x) = 1</math> solve the [[differential equation]] (in this case they are both solutions). See also: [[singular solution]]s.
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| ===Example (II)===
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| Population growth is often modeled by the differential equation
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| : <math>\frac{dP}{dt}=kP\left(1-\frac{P}{K}\right)</math>
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| where <math>P</math> is the population with respect to time <math>t</math>, <math>k</math> is the rate of growth, and <math>K</math> is the [[carrying capacity]] of the environment. | |
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| Separation of variables may be used to solve this differential equation.
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| : <math>\frac{dP}{dt}=kP\left(1-\frac{P}{K}\right)</math>
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| : <math>\int\frac{dP}{P\left(1-\frac{P}{K}\right)}=\int k\,dt</math>
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| To evaluate the integral on the left side, we simplify the fraction
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| : <math>\frac{1}{P\left(1-\frac{P}{K}\right)}=\frac{K}{P\left(K-P\right)}</math>
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| and then, we decompose the fraction into partial fractions
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| : <math>\frac{K}{P\left(K-P\right)}=\frac{1}{P}+\frac{1}{K-P}</math>
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| Thus we have
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| : <math>\int\left(\frac{1}{P}+\frac{1}{K-P}\right)\,dP=\int k\,dt</math>
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| : <p><math>\ln\begin{vmatrix}P\end{vmatrix}-\ln\begin{vmatrix}K-P\end{vmatrix}=kt+C</math></p>
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| : <p><math>\ln\begin{vmatrix}K-P\end{vmatrix}-\ln\begin{vmatrix}P\end{vmatrix}=-kt-C</math></p>
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| : <p><math>\ln\begin{vmatrix}\cfrac{K-P}{P}\end{vmatrix}=-kt-C</math></p>
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| : <p><math>\begin{vmatrix}\cfrac{K-P}{P}\end{vmatrix}=e^{-kt-C}</math></p>
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| : <p><math>\begin{vmatrix}\cfrac{K-P}{P}\end{vmatrix}=e^{-C}e^{-kt}</math></p>
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| : <p><math>\frac{K-P}{P}=\pm e^{-C}e^{-kt}</math></p>
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| : <p>Let <math>A=\pm e^{-C}</math>.</p>
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| : <p><math>\frac{K-P}{P}=Ae^{-kt}</math></p>
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| : <p><math>\frac{K}{P}-1=Ae^{-kt}</math></p>
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| : <p><math>\frac{K}{P}=1+Ae^{-kt}</math></p>
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| : <p><math>\frac{P}{K}=\frac{1}{1+Ae^{-kt}}</math></p>
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| : <p><math>P=\frac{K}{1+Ae^{-kt}}</math></p>
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| Therefore, the solution to the logistic equation is
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| : <p><math>P\left(t\right)=\frac{K}{1+Ae^{-kt}}</math></p>
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| To find <math>A</math>, let <math>t=0</math> and <math>P\left(0\right)=P_0</math>. Then we have
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| : <math>P_0=\frac{K}{1+Ae^0}</math></p>
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| Noting that <math>e^0=1</math>, and solving for A we get
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| : <math>A=\frac{K-P_0}{P_0}</math>
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| == Partial differential equations{{anchor|pde}} ==
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| 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]].
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| ===Homogeneous case===
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| Consider the one-dimensional [[heat equation]].The equation is
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| {{NumBlk|:|<math>\frac{\partial u}{\partial t}-\alpha\frac{\partial^{2}u}{\partial x^{2}}=0</math>|{{EqRef|1}}}}
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| The boundary condition is homogeneous, that is
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| {{NumBlk|:|<math>u\big|_{x=0}=u\big|_{x=L}=0 </math>|{{EqRef|2}}}}
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| 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:
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| {{NumBlk|:|<math> u(x,t) = X(x) T(t).</math>|{{EqRef|3}}}}
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| Substituting ''u'' back into equation and using the [[product rule]],
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| {{NumBlk|:|<math>\frac{T'(t)}{\alpha T(t)} = \frac{X''(x)}{X(x)}.</math>|{{EqRef|4}}}}
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| 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:
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| {{NumBlk|:|<math>T'(t) = - \lambda \alpha T(t),</math>|{{EqRef|5}}}}
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| and
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| {{NumBlk|:|<math>X''(x) = - \lambda X(x).</math>|{{EqRef|6}}}}
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| − λ here is the [[eigenvalue]] for both differential operators, and ''T(t)'' and ''X(x)'' are corresponding [[eigenfunction]]s.
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| We will now show that solutions for ''X(x)'' for values of λ ≤ 0 cannot occur:
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| Suppose that λ < 0. Then there exist real numbers ''B'', ''C'' such that
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| :<math>X(x) = B e^{\sqrt{-\lambda} \, x} + C e^{-\sqrt{-\lambda} \, x}.</math>
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| From {{EqNote|2}} we get
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| {{NumBlk|:|<math>X(0) = 0 = X(L),</math>|{{EqRef|7}}}}
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| and therefore ''B'' = 0 = ''C'' which implies ''u'' is identically 0.
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| Suppose that λ = 0. Then there exist real numbers ''B'', ''C'' such that
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| :<math>X(x) = Bx + C.</math>
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| From {{EqNote|7}} we conclude in the same manner as in 1 that ''u'' is identically 0.
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| Therefore, it must be the case that λ > 0. Then there exist real numbers ''A'', ''B'', ''C'' such that
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| :<math>T(t) = A e^{-\lambda \alpha t},</math> | |
| and
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| :<math>X(x) = B \sin(\sqrt{\lambda} \, x) + C \cos(\sqrt{\lambda} \, x).</math> | |
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| From {{EqNote|7}} we get ''C'' = 0 and that for some positive integer ''n'',
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| :<math>\sqrt{\lambda} = n \frac{\pi}{L}.</math>
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| This solves the heat equation in the special case that the dependence of ''u'' has the special form of {{EqNote|3}}.
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| In general, the sum of solutions to {{EqNote|1}} which satisfy the boundary conditions {{EqNote|2}} also satisfies {{EqNote|1}} and {{EqNote|3}}. Hence a complete solution can be given as
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| :<math>u(x,t) = \sum_{n = 1}^{\infty} D_n \sin \frac{n\pi x}{L} \exp\left(-\frac{n^2 \pi^2 \alpha t}{L^2}\right),</math>
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| where ''D''<sub>''n''</sub> are coefficients determined by initial condition.
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| Given the initial condition
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| :<math>u\big|_{t=0}=f(x),</math>
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| we can get
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| :<math>f(x) = \sum_{n = 1}^{\infty} D_n \sin \frac{n\pi x}{L}.</math>
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| This is the [[sine series]] expansion of ''f(x)''. Multiplying both sides with <math>\sin \frac{n\pi x}{L}</math> and integrating over ''[0,L]'' result in
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| :<math>D_n = \frac{2}{L} \int_0^L f(x) \sin \frac{n\pi x}{L} \, dx.</math>
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| This method requires that the eigenfunctions of ''x'', here <math>\left\{\sin \frac{n\pi x}{L}\right\}_{n=1}^{\infty}</math>, are [[orthogonal]] and [[complete]]. In general this is guaranteed by [[Sturm-Liouville theory]].
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| === Nonhomogeneous case ===
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| Suppose the equation is nonhomogeneous,
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| {{NumBlk|:|<math>\frac{\partial u}{\partial t}-\alpha\frac{\partial^{2}u}{\partial x^{2}}=h(x,t)</math>|{{EqRef|8}}}}
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| with the boundary condition the same as {{EqNote|2}}.
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| Expand ''h(x,t)'', ''u(x,t)'' and ''f(x,t)'' into
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| {{NumBlk|:|<math>h(x,t)=\sum_{n=1}^{\infty}h_{n}(t)\sin\frac{n\pi x}{L},</math>|{{EqRef|9}}}}
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| {{NumBlk|:|<math>u(x,t)=\sum_{n=1}^{\infty}u_{n}(t)\sin\frac{n\pi x}{L},</math>|{{EqRef|10}}}}
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| {{NumBlk|:|<math>f(x)=\sum_{n=1}^{\infty}b_{n}\sin\frac{n\pi x}{L},</math>|{{EqRef|11}}}}
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| where ''h''<sub>''n''</sub>(''t'') and ''b''<sub>''n''</sub> can be calculated by integration, while ''u''<sub>''n''</sub>(''t'') is to be determined.
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| Substitute {{EqNote|9}} and {{EqNote|10}} back to {{EqNote|8}} and considering the orthogonality of sine functions we get
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| : <math>u'_{n}(t)+\alpha\frac{n^{2}\pi^{2}}{L^{2}}u_{n}(t)=h_{n}(t),</math>
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| 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
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| : <math>u_{n}(t)=e^{-\alpha\frac{n^{2}\pi^{2}}{L^{2}} t} \left (b_{n}+\int_{0}^{t}h_{n}(s)e^{\alpha\frac{n^{2}\pi^{2}}{L^{2}} s} \, ds \right).</math>
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| If the boundary condition is nonhomogeneous, then the expansion of {{EqNote|9}} and {{EqNote|10}} is no longer valid. One has to find a function ''v'' that satisfies the boundary condition only, and subtract it from ''u''. The function ''u-v'' then satisfies homogeneous boundary condition, and can be solved with the above method.
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| 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#Laplace's spherical harmonics|spherical harmonics]] for example.
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| == Matrices ==
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| The matrix form of the separation of variables is the [[Kronecker sum]].
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| As an example we consider the 2D [[Discrete Laplace operator|discrete Laplacian]] on a [[regular grid]]:
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| :<math>L = \mathbf{D_{xx}}\oplus\mathbf{D_{yy}}=\mathbf{D_{xx}}\otimes\mathbf{I}+\mathbf{I}\otimes\mathbf{D_{yy}}, \,</math>
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| where <math>\mathbf{D_{xx}} </math> and <math>\mathbf{D_{yy}} </math> are 1D discrete Laplacians in the ''x''- and ''y''-directions, correspondingly, and <math>\mathbf{I} </math> are the identities of appropriate sizes. See the main article [[Kronecker sum of discrete Laplacians]] for details.
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| == References ==
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| * 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.
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| *{{Cite book
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| | author = Tyn Myint-U, Lokenath Debnath
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| | isbn = 978-0-8176-4393-5
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| | title = Linear Partial Differential Equations for Scientists and Engineers
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| | location = Boston, MA
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| | accessdate = 2011-03-29
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| | year = 2007
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| | url = http://www.springerlink.com/index/10.1007/978-0-8176-4560-1
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| }}
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| * {{Cite book | last=Teschl | first=Gerald | author-link =Gerald Teschl| title=Ordinary Differential Equations and Dynamical Systems | series=Graduate Studies in Mathematics | volume=140 | publisher=[[American Mathematical Society]]| location= Providence, RI| year= 2012 | isbn=978-0-8218-8328-0| url=http://www.mat.univie.ac.at/~gerald/ftp/book-ode/}}
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| == External links==
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| * {{springer|title=Fourier method|id=p/f041070}}
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| * {{mathworld2 |urlname=SeparationofVariables |title=Separation of variables|urlname2=DifferentialEquation|title2=Differential Equation|author=John Renze, [[Eric W. Weisstein]]}}
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| * [http://eqworld.ipmnet.ru/en/education/edu-pde.htm Methods of Generalized and Functional Separation of Variables] at EqWorld: The World of Mathematical Equations
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| * [http://www.exampleproblems.com/wiki/index.php/PDE:Integration_and_Separation_of_Variables Examples] of separating variables to solve PDEs
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| * [http://www.math-cs.gordon.edu/courses/mat225/handouts/sepvar.pdf "A Short Justification of Separation of Variables"]
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| [[Category:Ordinary differential equations]]
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| [[Category:Partial differential equations]]
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