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, ). 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) : \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} where
\delta is the
Dirac delta function. The fundamental solution to this problem is given by the
heat kernel : \Phi(x,t)=\frac{1}{\sqrt{4\pi kt}}\exp\left(-\frac{x^2}{4kt}\right). One can obtain the general solution of the one variable heat equation with initial condition
u(
x, 0) =
g(
x) for −∞ u(x,t) = \int \Phi(x-y,t) g(y) dy. In several spatial variables, the fundamental solution solves the analogous problem : \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} The
n-variable fundamental solution is the product of the fundamental solutions in each variable; i.e., : \Phi(\mathbf{x},t) = \Phi(x_1,t) \Phi(x_2,t) \cdots \Phi(x_n,t) = \frac{1}{(4\pi k t)^{n/2}} \exp \left (-\frac{\mathbf{x}\cdot\mathbf{x}}{4kt} \right). The general solution of the heat equation on
Rn is then obtained by a convolution, so that to solve the initial value problem with
u(
x, 0) =
g(
x), one has : u(\mathbf{x},t) = \int_{\R^n}\Phi(\mathbf{x}-\mathbf{y},t)g(\mathbf{y})d\mathbf{y}. The general problem on a domain Ω in
Rn is : \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} with either
Dirichlet or
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 transforms (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. In some of these, the spatial domain is (−∞,∞). In others, it is the semi-infinite interval (0,∞) with either
Neumann or
Dirichlet boundary conditions. One further variation is that some of these solve the inhomogeneous equation : u_{t}=ku_{xx}+f. where
f is some given function of
x and
t.
Homogeneous heat equation ; Initial value problem on (−∞,∞) : : \begin{cases} u_{t}=ku_{xx} & (x, t) \in \R \times (0, \infty) \\ u(x,0)=g(x) & \text{Initial condition} \end{cases} : u(x,t) = \frac{1}{\sqrt{4\pi kt}} \int_{-\infty}^{\infty} \exp\left(-\frac{(x-y)^2}{4kt}\right)g(y)\,dy ]
Comment. This solution is the
convolution with respect to the variable
x of the fundamental solution : \Phi(x,t) := \frac{1}{\sqrt{4\pi kt}} \exp\left(-\frac{x^2}{4kt}\right), 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 : \left (\partial_t-k\partial_x^2 \right )(\Phi*g)=\left [\left (\partial_t-k\partial_x^2 \right )\Phi \right ]*g=0. Moreover, : \Phi(x,t)=\frac{1}{\sqrt{t}}\,\Phi\left(\frac{x}{\sqrt{t}},1\right) : \int_{-\infty}^{\infty}\Phi(x,t)\,dx=1, so that, by general facts about
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 converges uniformly to
g as
t → 0, meaning that
u(
x,
t) is continuous on with ; Initial value problem on (0,∞) with homogeneous Dirichlet boundary conditions : : \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} : 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
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 : : \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} : 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
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
ux(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 : : \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} : 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
Comment. This solution is the
convolution with respect to the variable
t of : \psi(x,t):=-2k \partial_x \Phi(x,t) = \frac{x}{\sqrt{4\pi kt^3}} \exp\left(-\frac{x^2}{4kt}\right) and the function
h(
t). Since Φ(
x,
t) is the fundamental solution of : \partial_t-k\partial^2_x, 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, : \psi(x,t)=\frac{1}{x^2}\,\psi\left(1,\frac{t}{x^2}\right) : \int_0^{\infty}\psi(x,t)\,dt=1, so that, by general facts about
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 with
Inhomogeneous heat equation ; Problem on (-∞,∞) homogeneous initial conditions : : :
Comment. This solution is the convolution in
R2, that is with respect to both the variables
x and
t, of the fundamental solution : \Phi(x,t) := \frac{1}{\sqrt{4\pi kt}} \exp\left(-\frac{x^2}{4 kt}\right) and the function
f(
x,
t), both meant as defined on the whole
R2 and identically 0 for all
t → 0. One verifies that : \left (\partial_t-k \partial_x^2 \right )(\Phi*f)=f, which expressed in the language of distributions becomes : \left (\partial_t-k \partial_x^2 \right )\Phi=\delta, 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: : \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} : 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
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 : : \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} : 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
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
ux(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 : \begin{cases} u_{t}=ku_{xx}+f & (x, t) \in \R \times (0, \infty) \\ u(x,0)=g(x) & \text{IC} \end{cases} let
u =
w +
v where
w and
v solve the problems : \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} Similarly, to solve : \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} let
u =
w +
v +
r where
w,
v, and
r solve the problems : \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} == Applications ==