Editing Separation of variables (section)

== Partial differential equations{{anchor|pde}} ==

{{See also|Separable partial differential equation}}

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 the [[heat equation]], [[wave equation]], [[Laplace equation]], [[Helmholtz equation]] and [[biharmonic equation]].

The analytical method of separation of variables for solving partial differential equations has also been generalized into a computational method of decomposition in invariant structures that can be used to solve systems of partial differential equations.<ref>{{Cite book|url=https://books.google.com/books?id=kxVDDwAAQBAJ|title = Harmonic Wave Systems: Partial Differential Equations of the Helmholtz Decomposition|isbn = 9781618964069|last1 = Miroshnikov|first1 = Victor A.|date = 15 December 2017| publisher=Scientific Research Publishing, Inc. USA }}</ref>

=== Example: homogeneous case ===

Consider the one-dimensional [[heat equation]]. The equation is

{{NumBlk|:|<math>\frac{\partial u}{\partial t} - \alpha\frac{\partial^{2}u}{\partial x^{2}} = 0</math>|{{EqRef|1}}}}

The variable ''u'' denotes temperature. The boundary condition is homogeneous, that is
{{NumBlk|:|<math>u\big|_{x=0}=u\big|_{x=L}=0 </math>|{{EqRef|2}}}}

Let us attempt to find a [[Triviality_(mathematics)#Trivial_and_nontrivial_solutions|nontrivial solution]] 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:

{{NumBlk|:|<math> u(x,t) = X(x) T(t).</math>|{{EqRef|3}}}}

Substituting ''u'' back into equation {{EqNote|1}} and using the [[product rule]],

{{NumBlk|:|<math>\frac{T'(t)}{\alpha T(t)} = \frac{X''(x)}{X(x)}= -\lambda,</math>|{{EqRef|4}}}}

where ''λ'' must be constant since the right hand side depends only on ''x'' and the left hand side only on ''t''. Thus:

{{NumBlk|:|<math>T'(t) = - \lambda \alpha T(t),</math>|{{EqRef|5}}}}

and

{{NumBlk|:|<math>X''(x) = - \lambda X(x).</math>|{{EqRef|6}}}}

−''λ'' here is the [[eigenvalue]] for both differential operators, and ''T''(''t'') and ''X''(''x'') are corresponding [[eigenfunction]]s.

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

:<math>X(x) = B e^{\sqrt{-\lambda} \, x} + C e^{-\sqrt{-\lambda} \, x}.</math>

From {{EqNote|2}} we get

{{NumBlk|:|<math>X(0) = 0 = X(L),</math>|{{EqRef|7}}}}

and therefore ''B'' = 0 = ''C'' which implies ''u'' is identically 0.

Suppose that ''λ'' = 0. Then there exist real numbers ''B'', ''C'' such that

:<math>X(x) = Bx + C.</math>

From {{EqNote|7}} 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
:<math>T(t) = A e^{-\lambda \alpha t},</math>
and
:<math>X(x) = B \sin(\sqrt{\lambda} \, x) + C \cos(\sqrt{\lambda} \, x).</math>

From {{EqNote|7}} we get ''C'' = 0 and that for some positive integer ''n'',

:<math>\sqrt{\lambda} = n \frac{\pi}{L}.</math>

This solves the heat equation in the special case that the dependence of ''u'' has the special form of {{EqNote|3}}.

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

:<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>

where ''D''<sub>''n''</sub> are coefficients determined by initial condition.

Given the initial condition

:<math>u\big|_{t=0}=f(x),</math>

we can get

:<math>f(x) = \sum_{n = 1}^{\infty} D_n \sin \frac{n\pi x}{L}.</math>

This is the [[Fourier sine series]] expansion of ''f''(''x'') which is amenable to [[Fourier analysis]]. Multiplying both sides with <math display="inline">\sin \frac{n\pi x}{L}</math> and integrating over {{closed-closed|0, ''L''}} results in

:<math>D_n = \frac{2}{L} \int_0^L f(x) \sin \frac{n\pi x}{L} \, dx.</math>

This method requires that the eigenfunctions ''X'', here <math display="inline">\left\{\sin \frac{n\pi x}{L}\right\}_{n=1}^{\infty}</math>, are [[orthogonal]] and [[Schauder basis|complete]]. In general this is guaranteed by [[Sturm–Liouville theory]].

=== Example: nonhomogeneous case ===

Suppose the equation is nonhomogeneous,

{{NumBlk|:|<math>\frac{\partial u}{\partial t}-\alpha\frac{\partial^{2}u}{\partial x^{2}}=h(x,t)</math>|{{EqRef|8}}}}

with the boundary condition the same as {{EqNote|2}}.

Expand ''h''(''x,t''), ''u''(''x'',''t'') and ''f''(''x'') into

{{NumBlk|:|<math>h(x,t)=\sum_{n=1}^{\infty}h_{n}(t)\sin\frac{n\pi x}{L},</math>|{{EqRef|9}}}}

{{NumBlk|:|<math>u(x,t)=\sum_{n=1}^{\infty}u_{n}(t)\sin\frac{n\pi x}{L},</math>|{{EqRef|10}}}}

{{NumBlk|:|<math>f(x)=\sum_{n=1}^{\infty}b_{n}\sin\frac{n\pi x}{L},</math>|{{EqRef|11}}}}

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.

Substitute {{EqNote|9}} and {{EqNote|10}} back to {{EqNote|8}} and considering the orthogonality of sine functions we get

: <math>u'_{n}(t)+\alpha\frac{n^{2}\pi^{2}}{L^{2}}u_{n}(t)=h_{n}(t),</math>

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
: <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>

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.

=== Example: mixed derivatives ===

For some equations involving mixed derivatives, the equation does not separate as easily as the heat equation did in the first example above, but nonetheless separation of variables may still be applied. Consider the two-dimensional [[biharmonic equation]]

:<math>\frac{\partial^4 u}{\partial x^4} + 2\frac{\partial^4 u}{\partial x^2\partial y^2} + \frac{\partial^4 u}{\partial y^4} = 0.</math>

Proceeding in the usual manner, we look for solutions of the form

:<math>u(x,y) = X(x)Y(y)</math>

and we obtain the equation

:<math>\frac{X^{(4)}(x)}{X(x)} + 2\frac{X''(x)}{X(x)}\frac{Y''(y)}{Y(y)} + \frac{Y^{(4)}(y)}{Y(y)} = 0.</math>

Writing this equation in the form

:<math>E(x) + F(x)G(y) + H(y) = 0,</math>

Taking the derivative of this expression with respect to <math> x </math> gives <math> E'(x)+F'(x)G(y)=0 </math> which means <math> G(y)=const. </math> or <math> F'(x)=0 </math> and likewise, taking derivative with respect to <math> y </math> leads to <math> F(x)G'(y)+H'(y)=0 </math> and thus <math> F(x)=const. </math> or <math> G'(y)=0 </math>, hence either ''F''(''x'') or ''G''(''y'') must be a constant, say −λ. This further implies that either <math>-E(x)=F(x)G(y)+H(y)</math> or <math>-H(y)=E(x)+F(x)G(y)</math> are constant. Returning to the equation for ''X'' and ''Y'', we have two cases

:<math>\begin{align}
    X''(x) &= -\lambda_1X(x) \\
    X^{(4)}(x) &= \mu_1X(x) \\
    Y^{(4)}(y) - 2\lambda_1Y''(y) &= -\mu_1Y(y)
\end{align}</math>

and

:<math>\begin{align}
    Y''(y) &= -\lambda_2Y(y) \\
    Y^{(4)}(y) &= \mu_2Y(y) \\
    X^{(4)}(x) - 2\lambda_2X''(x) &= -\mu_2X(x)
\end{align}</math>

which can each be solved by considering the separate cases for <math>\lambda_i<0, \lambda_i=0, \lambda_i>0</math> and noting that <math>\mu_i=\lambda_i^2</math>.

=== Curvilinear coordinates ===

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.