Editing Heat equation (section)

==== 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.]]