Editing Heat equation (section)

== Solving the heat equation using Fourier series ==
[[Image:Temp Rod homobc.svg|right|thumb|300px|Idealized physical setting for heat conduction in a rod with homogeneous boundary conditions.]]
The following solution technique for the heat equation was proposed by [[Joseph Fourier]] in his treatise ''Théorie analytique de la chaleur'', published in 1822. Consider the heat equation for one space variable. This could be used to model heat conduction in a rod. The equation is
{{NumBlk|:|<math>\displaystyle u_t = \alpha u_{xx}</math>|{{EqRef|1}}}}
where ''u'' = ''u''(''x'', ''t'') is a function of two variables ''x'' and ''t''. Here 
* ''x'' is the space variable, so ''x'' ∈ [0, ''L''], where ''L'' is the length of the rod.
* ''t'' is the time variable, so ''t'' ≥ 0.

We assume the initial condition
{{NumBlk|:|<math>u(x,0) = f(x) \quad \forall x \in [0,L]</math> |{{EqRef|2}}}}
where the function ''f'' is given, and the boundary conditions
{{NumBlk|:|<math>u(0,t) = 0 = u(L,t) \quad \forall t > 0 </math>.|{{EqRef|3}}}}

Let us attempt to find a solution of {{EqNote|1}} that is not identically zero satisfying the boundary conditions {{EqNote|3}} 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|4}}}}

This solution technique is called [[separation of variables]]. Substituting ''u'' back into equation {{EqNote|1}},
: <math>\frac{T'(t)}{\alpha T(t)} = \frac{X''(x)}{X(x)}.</math>

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:
{{NumBlk|:|<math>T'(t) = - \lambda \alpha T(t)</math>|{{EqRef|5}}}}
and
{{NumBlk|:|<math>X''(x) = - \lambda X(x).</math>|{{EqRef|6}}}}

We will now show that nontrivial solutions for {{EqNote|6}} for values of ''λ'' ≤ 0 cannot occur:
# Suppose that ''λ'' < 0. Then there exist real numbers ''B'', ''C'' such that <math display="block">X(x) = B e^{\sqrt{-\lambda} \, x} + C e^{-\sqrt{-\lambda} \, x}.</math> From {{EqNote|3}} we get  ''X''(0) = 0 = ''X''(''L'') 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 equation {{EqNote|3}} 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 display="block">T(t) = A e^{-\lambda \alpha t}</math> and <math display="block">X(x) = B \sin\left(\sqrt{\lambda} \, x\right) + C \cos\left(\sqrt{\lambda} \, x\right).</math> From {{EqNote|3}} we get ''C'' = 0 and that for some positive integer ''n'', <math display="block">\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 {{EqNote|4}}.

In general, the sum of solutions to {{EqNote|1}} that satisfy the boundary conditions {{EqNote|3}} also satisfies {{EqNote|1}} and {{EqNote|3}}. We can show that the solution to {{EqNote|1}}, {{EqNote|2}} and {{EqNote|3}} is given by
: <math>u(x,t) = \sum_{n = 1}^{\infty} D_n \sin \left(\frac{n\pi x}{L}\right) e^{-\frac{n^2 \pi^2 \alpha t}{L^2}}</math>
where
: <math>D_n = \frac{2}{L} \int_0^L f(x) \sin \left(\frac{n\pi x}{L}\right ) \, dx.</math>

=== Generalizing the solution technique ===
The solution technique used above can be greatly extended to many other types of equations. The idea is that the operator ''u<sub>xx</sub>'' with the zero boundary conditions can be represented in terms of its [[eigenfunction]]s. This leads naturally to one of the basic ideas of the [[spectral theory]] of linear [[self-adjoint operator]]s.

Consider the [[linear operator]] Δ''u'' = ''u<sub>xx</sub>''. The infinite sequence of functions
: <math> e_n(x) = \sqrt{\frac{2}{L}}\sin \left(\frac{n\pi x}{L}\right)</math>
for ''n'' ≥ 1 are eigenfunctions of Δ. Indeed,
: <math> \Delta e_n = -\frac{n^2 \pi^2}{L^2} e_n. </math>

Moreover, any eigenfunction ''f'' of Δ with the boundary conditions ''f''(0) = ''f''(''L'') = 0 is of the form ''e''<sub>''n''</sub> for some ''n'' ≥ 1. The functions ''e''<sub>''n''</sub> for ''n'' ≥ 1 form an [[orthonormal]] sequence with respect to a certain [[inner product]] on the space of real-valued functions on [0, ''L'']. This means
: <math> \langle e_n, e_m \rangle = \int_0^L e_n(x) e^*_m(x) dx = \delta_{mn}</math>

Finally, the sequence {''e''<sub>''n''</sub>}<sub>''n'' ∈ '''N'''</sub> spans a dense linear subspace of ''L''<sup>2</sup>((0, ''L'')). This shows that in effect we have [[diagonal matrix|diagonalized]] the operator Δ.

=== Mean-value property ===
Solutions of the heat equations 
: <math>(\partial_t -\Delta)u=0</math>
satisfy a mean-value property analogous to the [[Harmonic_function#The_mean_value_property|mean-value properties of harmonic functions]], solutions of 
: <math>\Delta u = 0,</math>
though a bit more complicated. Precisely, if ''u'' solves 
: <math>(\partial_t -\Delta)u=0</math>
and 
: <math>(x,t)+E_\lambda\subset\mathrm{dom}(u)</math>
then
: <math>u(x,t)=\frac{\lambda}{4}\int_{E_\lambda}u(x-y,t-s)\frac{|y|^2}{s^2}ds\,dy,</math>
where ''E<sub>λ</sub>'' is a ''heat-ball'', that is a super-level set of the fundamental solution of the heat equation:
: <math>E_\lambda := \{(y,s) : \Phi(y,s) > \lambda\},</math>
: <math>\Phi(x,t) := (4t\pi)^{-\frac{n}{2}}\exp\left(-\frac{|x|^2}{4t}\right).</math>
Notice that 
: <math>\mathrm{diam}(E_\lambda)=o(1)</math>
as ''λ'' → ∞ so the above formula holds for any (''x'', ''t'') in the (open) set dom(''u'') for ''λ'' large enough.<ref>Conversely, any function ''u'' satisfying the above mean-value property on an open domain of '''R'''<sup>''n''</sup> × '''R''' is a solution of the heat equation</ref>