Editing Heat equation (section)

== Fundamental solutions ==
{{see also|Weierstrass transform}}
A [[fundamental solution]] of the heat equation is a solution that corresponds to the initial condition of an initial point source of heat at a known position. These can be used to find a general solution of the heat equation over certain domains (see, for instance, {{harv|Evans|2010}}).

In one variable, the [[Green's function]] is a solution of the initial value problem (by [[Duhamel's principle]], equivalent to the definition of Green's function as one with a delta function as solution to the first equation)
: <math>\begin{cases}
u_t(x,t) - k u_{xx}(x,t) = 0& (x, t) \in \R \times (0, \infty)\\
u(x,0)=\delta(x)&
\end{cases}</math>
where ''<math>\delta</math>'' is the [[Dirac delta function]]. The fundamental solution to this problem is given by the [[heat kernel]]
: <math>\Phi(x,t)=\frac{1}{\sqrt{4\pi kt}}\exp\left(-\frac{x^2}{4kt}\right).</math>

One can obtain the general solution of the one variable heat equation with initial condition ''u''(''x'', 0) = ''g''(''x'') for −∞ < ''x'' < ∞ and 0 < ''t'' < ∞ by applying a [[convolution]]:
: <math>u(x,t) = \int \Phi(x-y,t) g(y) dy.</math>

In several spatial variables, the fundamental solution solves the analogous problem
: <math>\begin{cases}
u_t(\mathbf{x},t) - k \sum_{i=1}^nu_{x_ix_i}(\mathbf{x},t) = 0 & (\mathbf{x}, t) \in \R^n \times (0, \infty)\\
u(\mathbf{x},0)=\delta(\mathbf{x})
\end{cases}</math>

The ''n''-variable fundamental solution is the product of the fundamental solutions in each variable; i.e.,
: <math>\Phi(\mathbf{x},t) = \Phi(x_1,t) \Phi(x_2,t) \cdots \Phi(x_n,t) = \frac{1}{\sqrt{(4\pi k t)^n}} \exp \left (-\frac{\mathbf{x}\cdot\mathbf{x}}{4kt} \right).</math>

The general solution of the heat equation on '''R'''<sup>''n''</sup> is then obtained by a convolution, so that to solve the initial value problem with ''u''('''x''', 0) = ''g''('''x'''), one has
: <math>u(\mathbf{x},t) = \int_{\R^n}\Phi(\mathbf{x}-\mathbf{y},t)g(\mathbf{y})d\mathbf{y}.</math>

The general problem on a domain Ω in '''R'''<sup>''n''</sup> is
: <math> \begin{cases}
u_t(\mathbf{x},t) - k \sum_{i=1}^nu_{x_ix_i}(\mathbf{x},t) = 0& (\mathbf{x}, t) \in \Omega\times (0, \infty)\\
u(\mathbf{x},0)=g(\mathbf{x})&\mathbf{x}\in\Omega
\end{cases}</math>
with either [[Dirichlet problem|Dirichlet]] or [[Neumann problem|Neumann]] boundary data.  A [[Green's function]] always exists, but unless the domain Ω can be readily decomposed into one-variable problems (see below), it may not be possible to write it down explicitly.  Other methods for obtaining Green's functions include the [[method of images]], [[separation of variables]], and [[Laplace transform]]s (Cole, 2011). 

=== Some Green's function solutions in 1D ===
A variety of elementary Green's function solutions in one-dimension are recorded here; many others are available elsewhere.<ref>The [https://www.engr.unl.edu/~glibrary/home/index.html Green's Function Library] contains a variety of fundamental solutions to the heat equation.</ref>  In some of these, the spatial domain is (−∞,∞).  In others, it is the semi-infinite interval (0,∞) with either [[Neumann problem|Neumann]] or [[Dirichlet problem|Dirichlet]] boundary conditions.  One further variation is that some of these solve the inhomogeneous equation
: <math>u_{t}=ku_{xx}+f.</math>
where ''f'' is some given function of ''x'' and ''t''.

==== Homogeneous heat equation ====
; Initial value problem on (−∞,∞) :
: <math>\begin{cases}
u_{t}=ku_{xx} & (x, t) \in \R \times (0, \infty) \\ 
u(x,0)=g(x) & \text{Initial condition}
\end{cases} </math>
: <math>u(x,t) = \frac{1}{\sqrt{4\pi kt}} \int_{-\infty}^{\infty} \exp\left(-\frac{(x-y)^2}{4kt}\right)g(y)\,dy </math>

[[Image:Fundamental solution to the heat equation.gif|right|thumb|upright=2|Fundamental solution of the one-dimensional heat equation. Red: time course of <math>\Phi(x,t)</math>. Blue: time courses of <math>\Phi(x_0,t)</math> for two selected points x<sub>0</sub> = 0.2 and x<sub>0</sub> = 1. Note the different rise times/delays and amplitudes.<br/> [https://www.geogebra.org/m/SV6PruXx Interactive version.]]]

''Comment''. This solution is the [[convolution]] with respect to the variable ''x'' of the fundamental solution
: <math>\Phi(x,t) := \frac{1}{\sqrt{4\pi kt}} \exp\left(-\frac{x^2}{4kt}\right),</math>
and the function ''g''(''x''). (The [[Green's function number]] of the fundamental solution is X00.)

Therefore, according to the general properties of the convolution with respect to differentiation, ''u'' = ''g'' ∗ Φ is a solution of the same heat equation, for
: <math>\left (\partial_t-k\partial_x^2 \right )(\Phi*g)=\left [\left (\partial_t-k\partial_x^2 \right )\Phi \right ]*g=0.</math>

Moreover,
: <math>\Phi(x,t)=\frac{1}{\sqrt{t}}\,\Phi\left(\frac{x}{\sqrt{t}},1\right)</math>
: <math>\int_{-\infty}^{\infty}\Phi(x,t)\,dx=1,</math>
so that, by general facts about [[mollifier|approximation to the identity]], Φ(⋅, ''t'') ∗ ''g'' → ''g'' as ''t'' → 0 in various senses, according to the specific ''g''. For instance, if ''g'' is assumed bounded and continuous on '''R''' then {{nowrap|Φ(⋅, ''t'') ∗ ''g''}} converges uniformly to ''g'' as ''t'' → 0, meaning that ''u''(''x'', ''t'') is continuous on {{nowrap|'''R''' × [0, ∞)}} with {{nowrap|1=''u''(''x'', 0) = ''g''(''x'').}}

; Initial value problem on (0,∞) with homogeneous Dirichlet boundary conditions :
: <math>\begin{cases}
u_{t}=ku_{xx} & (x, t) \in [0, \infty) \times (0, \infty) \\ 
u(x,0)=g(x) & \text{IC} \\
u(0,t)=0 & \text{BC}
\end{cases} </math>
: <math>u(x,t)=\frac{1}{\sqrt{4\pi kt}} \int_{0}^{\infty} \left[\exp\left(-\frac{(x-y)^2}{4kt}\right)-\exp\left(-\frac{(x+y)^2}{4kt}\right)\right] g(y)\,dy </math>

''Comment.'' This solution is obtained from the preceding formula as applied to the data ''g''(''x'') suitably extended to '''R''', so as to be an [[odd function]], that is, letting ''g''(−''x'') := −''g''(''x'') for all ''x''. Correspondingly, the solution of the initial value problem on (−∞,∞) is an odd function with respect to the variable ''x'' for all values of ''t'', and in particular it satisfies the homogeneous Dirichlet boundary conditions ''u''(0, ''t'') = 0.
The [[Green's function number]] of this solution is X10.

; Initial value problem on (0,∞) with homogeneous Neumann boundary conditions :
: <math>\begin{cases}
u_{t}=ku_{xx} & (x, t) \in [0, \infty) \times (0, \infty) \\
u(x,0)=g(x) & \text{IC} \\
u_{x}(0,t)=0 & \text{BC}
\end{cases} </math>
: <math>u(x,t)=\frac{1}{\sqrt{4\pi kt}} \int_{0}^{\infty} \left[\exp\left(-\frac{(x-y)^2}{4kt}\right)+\exp\left(-\frac{(x+y)^2}{4kt}\right)\right]g(y)\,dy </math>

''Comment.'' This solution is obtained from the first solution formula as applied to the data ''g''(''x'') suitably extended to '''R''' so as to be an [[even function]], that is, letting ''g''(−''x'') := ''g''(''x'') for all ''x''. Correspondingly, the solution of the initial value problem on '''R''' is an even  function with respect to the variable ''x'' for all values of ''t'' > 0, and in particular, being smooth, it satisfies the homogeneous Neumann boundary conditions ''u<sub>x</sub>''(0, ''t'') = 0.  The [[Green's function number]] of this solution is X20.

; Problem on (0,∞) with homogeneous initial conditions and non-homogeneous Dirichlet boundary conditions :
: <math>\begin{cases}
u_{t}=ku_{xx} & (x, t) \in [0, \infty) \times (0, \infty) \\
u(x,0)=0 & \text{IC} \\
u(0,t)=h(t) & \text{BC}
\end{cases} </math>
: <math>u(x,t)=\int_{0}^{t} \frac{x}{\sqrt{4\pi k(t-s)^3}} \exp\left(-\frac{x^2}{4k(t-s)}\right)h(s)\,ds, \qquad\forall x>0</math>

''Comment''. This solution is the [[convolution]] with respect to the variable ''t'' of
: <math>\psi(x,t):=-2k \partial_x \Phi(x,t) = \frac{x}{\sqrt{4\pi kt^3}} \exp\left(-\frac{x^2}{4kt}\right)</math>
and the function ''h''(''t''). Since Φ(''x'', ''t'') is the fundamental solution of
: <math>\partial_t-k\partial^2_x,</math>
the function ''ψ''(''x'', ''t'') is also a solution of the same heat equation, and so is ''u'' := ''ψ'' ∗ ''h'', thanks to general properties of the convolution with respect to differentiation. Moreover,
: <math>\psi(x,t)=\frac{1}{x^2}\,\psi\left(1,\frac{t}{x^2}\right)</math>
: <math>\int_0^{\infty}\psi(x,t)\,dt=1,</math>
so that, by general facts about [[mollifier|approximation to the identity]], ''ψ''(''x'', ⋅) ∗ ''h'' → ''h'' as ''x'' → 0 in various senses, according to the specific ''h''. For instance, if ''h'' is assumed continuous on '''R''' with support in [0, ∞) then ''ψ''(''x'', ⋅) ∗ ''h'' converges uniformly on compacta to ''h'' as ''x'' → 0, meaning that ''u''(''x'', ''t'') is continuous on {{nowrap|[0, ∞) × [0, ∞)}} with {{nowrap|1=''u''(0, ''t'') = ''h''(''t'').}}
[[File:2D Nonhomogeneous heat equation .gif|thumb|Depicted is a numerical solution of the non-homogeneous heat equation. The equation has been solved with 0 initial and boundary conditions and a source term representing a stove top burner.]]

==== Inhomogeneous heat equation  ====
; Problem on (-∞,∞) homogeneous initial conditions :
: 
: 

''Comment''. This solution is the convolution in '''R'''<sup>2</sup>, that is with respect to both the variables ''x'' and ''t'', of the fundamental solution
: <math>\Phi(x,t) := \frac{1}{\sqrt{4\pi kt}} \exp\left(-\frac{x^2}{4 kt}\right)</math>
and the function ''f''(''x'', ''t''), both meant as defined on the whole '''R'''<sup>2</sup> and identically 0 for all ''t'' → 0. One verifies that
: <math>\left (\partial_t-k \partial_x^2 \right )(\Phi*f)=f,</math>
which expressed in the language of distributions becomes
: <math>\left (\partial_t-k \partial_x^2 \right )\Phi=\delta,</math>
where the distribution δ is the [[Dirac's delta function]], that is the evaluation at 0.

; Problem on (0,∞) with homogeneous Dirichlet boundary conditions and initial conditions:
: <math>\begin{cases}
u_{t}=ku_{xx}+f(x,t) & (x, t) \in [0, \infty) \times (0, \infty)  \\
u(x,0)=0 & \text{IC} \\
u(0,t)=0 & \text{BC}
\end{cases} </math>
: <math>u(x,t)=\int_{0}^{t}\int_{0}^{\infty} \frac{1}{\sqrt{4\pi k(t-s)}} \left(\exp\left(-\frac{(x-y)^2}{4k(t-s)}\right)-\exp\left(-\frac{(x+y)^2}{4k(t-s)}\right)\right)
f(y,s)\,dy\,ds </math>

''Comment''. This solution is obtained from the preceding formula as applied to the data ''f''(''x'', ''t'') suitably extended to '''R''' × [0,∞), so as to be an odd function of the variable ''x'', that is, letting ''f''(−''x'', ''t'') := −''f''(''x'', ''t'') for all ''x'' and ''t''. Correspondingly, the solution of the inhomogeneous problem on (−∞,∞) is an odd function with respect to the variable ''x'' for all values of ''t'', and in particular it satisfies the homogeneous Dirichlet boundary conditions ''u''(0, ''t'') = 0.

; Problem on (0,∞) with homogeneous Neumann boundary conditions and initial conditions :
: <math>\begin{cases}
u_{t} = ku_{xx}+f(x,t) & (x, t) \in [0, \infty) \times (0, \infty)  \\
u(x,0)=0 & \text{IC} \\
u_x(0,t)=0 & \text{BC}
\end{cases} </math>
: <math>u(x,t)=\int_{0}^{t}\int_{0}^{\infty} \frac{1}{\sqrt{4\pi k(t-s)}} \left(\exp\left(-\frac{(x-y)^2}{4k(t-s)}\right)+\exp\left(-\frac{(x+y)^2}{4k(t-s)}\right)\right)
f(y,s)\,dy\,ds </math>

''Comment''. This solution is obtained from the first formula as applied to the data ''f''(''x'', ''t'') suitably extended to '''R''' × [0,∞), so as to be an even function of the variable ''x'', that is, letting ''f''(−''x'', ''t'') := ''f''(''x'', ''t'') for all ''x'' and ''t''. Correspondingly, the solution of the inhomogeneous problem on (−∞,∞) is an even function with respect to the variable ''x'' for all values of ''t'', and in particular, being a smooth function, it satisfies the homogeneous Neumann boundary conditions ''u<sub>x</sub>''(0, ''t'') = 0.

==== Examples ====
Since the heat equation is linear, solutions of other combinations of boundary conditions, inhomogeneous term, and initial conditions can be found by taking an appropriate [[linear combination]] of the above Green's function solutions.

For example, to solve
: <math>\begin{cases}
u_{t}=ku_{xx}+f & (x, t) \in \R \times (0, \infty) \\
u(x,0)=g(x) & \text{IC}
\end{cases} </math>
let ''u'' = ''w'' + ''v'' where ''w'' and ''v'' solve the problems
: <math>\begin{cases} 
v_{t}=kv_{xx}+f, \, w_{t}=kw_{xx} \, & (x, t) \in \R \times (0, \infty) \\ 
v(x,0)=0,\, w(x,0)=g(x) \, & \text{IC}
\end{cases} </math>

Similarly, to solve
: <math>\begin{cases}
u_{t}=ku_{xx}+f & (x, t) \in [0, \infty) \times (0, \infty) \\
u(x,0)=g(x) & \text{IC} \\
u(0,t)=h(t) & \text{BC}
\end{cases} </math>
let ''u'' = ''w'' + ''v'' + ''r'' where ''w'', ''v'', and ''r'' solve the problems
: <math>\begin{cases} 
v_{t}=kv_{xx}+f, \, w_{t}=kw_{xx}, \, r_{t}=kr_{xx} & (x, t) \in [0, \infty) \times (0, \infty)  \\ 
v(x,0)=0, \; w(x,0)=g(x), \; r(x,0)=0 & \text{IC} \\ 
v(0,t)=0, \; w(0,t)=0, \; r(0,t)=h(t) & \text{BC}
\end{cases}</math>