We can also generalize the notion of a differential equation.
Definition:
The differential equation
is a differential equation in the sense
of a distribution (i.e., in the weak sense) if
and
are distributions
and all the derivatives are interpreted in the weak sense.
Suppose
is the generalized differential operator
![{\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/1f18657f7abad657252a4cdf9f06fe031d9dbad1)
where
is infinitely differentiable.
We seek a
such that
![{\displaystyle \left\langle {\mathcal {L}}u,\phi \right\rangle =\left\langle f,\phi \right\rangle \qquad \forall \phi \in D}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3db2551a8a4db40c36534564f43dd96d858d2e7d)
which is taken to mean that
![{\displaystyle \left\langle u,{\mathcal {L}}^{*}\phi \right\rangle =\left\langle f,\phi \right\rangle \qquad \forall \phi \in D~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e77b571964bdc75508420713eca4f1eeee1d2410)
Note that
![{\displaystyle \left\langle a_{k}(x){\cfrac {d^{k}u}{dx^{k}}},\phi \right\rangle =\left\langle {\cfrac {d^{k}u}{dx^{k}}},a_{k}(x)~\phi \right\rangle =(-1)^{k}\left\langle u,{\cfrac {d}{dx^{k}}}[a_{k}(x)~\phi ]\right\rangle }](https://wikimedia.org/api/rest_v1/media/math/render/svg/36da68c0911e98d38799efa059bb945c0215d346)
Therefore,
![{\displaystyle {\mathcal {L}}^{*}\phi =(-1)^{n}~{\cfrac {d^{n}(a_{n}\phi )}{dx^{n}}}+(-1)^{n-1}~{\cfrac {d^{(n-1)}(a_{n-1}\phi )}{dx^{(n-1)}}}+\dots +a_{0}~\phi ~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc0a37a2cb4163a64144ef9ab0e31006d4b51a46)
Here
is the formal adjoint of
. We can check that
. If
we say that
is
formally self adjoint.
For example, if
then
![{\displaystyle {\mathcal {L}}=a_{2}(x){\cfrac {d^{2}}{dx^{2}}}+a_{1}(x){\cfrac {d}{dx}}+a_{0}(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0c0d3b51b0de7208db6fd2f23b757efcfb0f68f4)
Then
![{\displaystyle {\mathcal {L}}^{*}\phi ={\cfrac {d^{2}(a_{2}\phi )}{dx^{2}}}-{\cfrac {d(a_{1}\phi )}{dx}}+a_{0}~\phi }](https://wikimedia.org/api/rest_v1/media/math/render/svg/80047d20b02cefe5fe2776f48210fb8dfa14871c)
or,
![{\displaystyle {\mathcal {L}}^{*}\phi =a_{2}~\phi ''+(2a'_{2}-a_{1})\phi '+(a''_{2}-a'_{1}+a_{0})\phi ~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cc1be12fa8cf572cb5bc834d67f648442e4eedc4)
Therefore, for
to be self adjoint,
![{\displaystyle {\mathcal {L}}^{*}\phi ={\mathcal {L}}\phi =a_{2}\phi ''+a_{1}\phi '+a_{0}\phi }](https://wikimedia.org/api/rest_v1/media/math/render/svg/2242173b77348ccc92446641740dd0fd8284781e)
Hence
![{\displaystyle a'_{2}=a_{1}\implies a''_{2}=a_{1}'~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/db22889428402c2bf060fd5470aecf5079e0681c)
In such a case,
is called a Sturm-Liouville operator.
To solve the differential equation
![{\displaystyle x{\cfrac {du}{dx}}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/276d83c4ebb6a01c8dbcbfc13216471feece8a09)
we seek a distribution
which satisfies
![{\displaystyle \left\langle xu',\phi \right\rangle =-\left\langle u,(x\phi )'\right\rangle =0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4acdc7bf5624a29dec624d2e25fb42d9cb4c6139)
Define
. Then
must be a test function. We can
show that
is a test function if and only if
![{\displaystyle {\text{(2)}}\qquad \int _{\infty }^{\infty }\psi (x){\text{d}}x=0\qquad {\text{and}}\qquad \int _{0}^{\infty }\psi (x){\text{d}}x=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0cca5a7a6d6f7a014dcd6fc6bee2c40f5cc13118)
Now let us pick two test functions
and
satisfying
![{\displaystyle \int _{-\infty }^{\infty }\phi _{0}(x){\text{d}}x=0\qquad {\text{and}}\qquad \int _{0}^{\infty }\phi _{0}(x){\text{d}}x=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1a9a58e09ebce9e6bb75fcb8990aac731706e13e)
and
![{\displaystyle \int _{-\infty }^{\infty }\phi _{1}(x){\text{d}}x=1\qquad {\text{and}}\qquad \int _{0}^{\infty }\phi _{1}(x){\text{d}}x=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b9ee05c861dae383dac184fad362122abfa37d79)
Then we can write any arbitrary test function
as a linear
combination of
and
plus a terms which has the form of
:
![{\displaystyle \phi (x)=\phi _{0}(x)\int _{0}^{\infty }\phi (s)~ds+\phi _{1}(x)\int _{-\infty }^{\infty }\phi (s)~ds+\psi (x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9888d92357d90a137f8e178273f1d6841865450a)
which serves to define
. Note that
satisfies
equation (2).
Since
, the action of
on
is given by
![{\displaystyle \left\langle u,\phi \right\rangle =\left\langle u,\phi _{0}\right\rangle \int _{\infty }^{\infty }H(x)\phi (s)ds+\left\langle u,\phi _{1}\right\rangle \int _{\infty }^{\infty }\phi (s)ds}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6f471f0079a6258da735cbd85991c0d628123ae5)
Therefore the solution is
![{\displaystyle u=C_{1}~H(x)+C_{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5e26e9a2ff127bf7a8513ce160a17132640e1a99)
where
and
.
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