Consider the one-dimensional heat equation given by
![{\displaystyle -{\cfrac {d^{2}u}{dx^{2}}}=f(x)\qquad x\in (0,1),~~u(0)=0,u(1)=0~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5194ae76be7e19881257697598671cc5018501be)
This equation has Green's function
which satisfies
![{\displaystyle -{\frac {\partial ^{2}g}{\partial x^{2}}}=\delta (x-y)\qquad x,y\in (0,1)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ab9a9b0e65ae532d76678b59d83f9b43f205edbe)
The Green's function is given by
![{\displaystyle g(x,y)={\begin{cases}(1-y)~x&0\leq x<y\\(1-x)~y&y<x\leq 1\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fe2858bc76a0b342137b6f51f5b29a49719e7f7f)
or,
![{\displaystyle g(x,y)=(1-y)~x+(y-x)~H(x-y)~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8b5a401eaf117f3927383c66ca3e409be12c9382)
Therefore,
![{\displaystyle g_{x}(x,y)=(1-y)-H(x-y)+(y-x){\frac {\partial }{\partial x}}[H(x-y)]=(1-y)-H(x-y)+(y-x)\delta (x-y)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8a627600faea8113280a3fb7bab8f51412b200f8)
Now,
![{\displaystyle \left\langle x~\delta (x),\phi \right\rangle =\left\langle \delta (x),x~\phi \right\rangle =(x~\phi )(0)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/963bd7d79e8641200bfa0d0545dc0958f63dff3e)
Hence,
![{\displaystyle g_{x}(x,y)=(1-y)-H(x-y)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/02d30fcf08840980636478bb42b7113a8999d6ca)
And,
![{\displaystyle g_{xx}(x,y)=-\delta (x-y)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/62d86ed1feef72439f93f14d4f64fc344070cfc7)
Note that the second derivative of
is a delta function.
We can use this observation to arrive at a more general description of the
Green's function for a particular differential equation. That is, for a
th order linear differential operator we would want the
th derivative of
to be like a delta function. Thus the
th derivative of
should be like a Heaviside function and all lower derivatives should
be continuous.
In particular, consider the operator
acting on
such that
![{\displaystyle {\mathcal {L}}~g=\delta (x-y)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/024bfe384ceccf8e3990eb0eafff89f8933e5d61)
with
![{\displaystyle {\mathcal {L}}=a_{n}(x)~{\cfrac {d^{n}}{dx^{n}}}+a_{n-1}(x)~{\cfrac {d^{(n-1)}}{dx^{(n-1)}}}+\dots a_{0}(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fda2dc6279231019ddc83d1b30c87fa3493efe41)
where
. If we integrate this equation across the
point
from
to
we get
![{\displaystyle \left.a_{n}(x){\cfrac {d^{n-1}g}{dx^{n-1}}}\right|_{x=y^{-}}^{x=y^{+}}=\int _{y^{-1}}^{y^{+}}\delta (x)~{\text{d}}x=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/41b50e60e7d3161b7312e7faddc81a2ee0c4e3ac)
This condition is called a Jump condition.
This suggests that the Green's function
satisfying
![{\displaystyle {\mathcal {L}}_{x}g(x,y)=\delta (x-y)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4428f043ce85a9d8aeafd092495d842b63385dbb)
in the sense of distributions has the properties that
for all
.
is continuous at
for
.
.
must satisfy all appropriate homogeneous boundary conditions.
If the Green's function exists then
![{\displaystyle u(x)=\int _{a}^{b}g(x,y)~f(y)~{\text{d}}y~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/12a297befe32b1265eecee98e8433398da3754e2)
Let us consider a second order differentiable linear operator, i.e.,
on
with separated boundary conditions,
![{\displaystyle {\begin{aligned}\alpha _{1}~u(a)+\beta _{1}~u'(a)&=0\qquad \implies {\mathcal {B}}_{1}[u(a)]=0\\\alpha _{2}~u(b)+\beta _{2}~u'(b)&=0\qquad \implies {\mathcal {B}}_{2}[u(b)]=0\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/47dcc66153a1f620975375dd4bbf4d77e7da7040)
where
and
are boundary operators.
The differential equation
![{\displaystyle {\mathcal {L}}_{x}g(x,y)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bcaa221f0cb06ac2685d7d167ce7464076216783)
and its required continuity conditions are satisfied is
![{\displaystyle g(x,y)={\begin{cases}A_{1}~u_{1}(x)~u_{2}(y)&a\leq y<x\\A_{2}~u_{2}(x)~u_{1}(y)&y<x\leq b\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0ea1bf278237f8bddfce69c579095fc2927d18f4)
The first two terms above are to satisfy the boundary conditions while the
third terms gives us continuity.
The jump condition is that
![{\displaystyle \left.{\cfrac {dg(x,y)}{dx}}\right|_{x=y^{-}}^{x=y^{+}}={\cfrac {1}{a_{2}(y)}}=A[u_{2}'(y)~u_{1}(y)-u_{1}'(y)~u_{2}(y)]=A~W(y)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/592ee730e49d5ef6d3fee9d8670f1e90e32533c5)
where
is the Wronskian.
Therefore,
![{\displaystyle A={\cfrac {1}{a_{2}(y)~W(y)}},~~~W(y)\neq 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e35a1e792d401df77acaa51c0fa492b9af875b73)
For the heat equation
![{\displaystyle {\mathcal {L}}u=-u''}](https://wikimedia.org/api/rest_v1/media/math/render/svg/45ed0f1b59d47edf66c21c247ae800bb631d3185)
we can take
![{\displaystyle u_{1}(x)=x,~~u_{2}(x)=(1-x),~~{\text{and}}~~W(y)=-1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a4989dd35bf804e2b97a98bd71c9f523333d76ff)
That gives us
and we recover the same
as before.
In the general case, the solution is
![{\displaystyle u(x)=\int _{a}^{x}{\cfrac {u_{2}(x)~u_{1}(y)~f(y)}{a_{2}(y)~W(y)}}~{\text{d}}y+\int _{x}^{b}{\cfrac {u_{1}(x)~u_{2}(y)~f(y)}{a_{2}(y)~W(y)}}~{\text{d}}y}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cfa7e514bc4cff5f532bce22793e6b91a56b1c39)
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