# Applied linear operators and spectral methods/Differential equations of distributions

## Differential equations in the sense of distributions

We can also generalize the notion of a differential equation.

Definition:

The differential equation ${\mathcal {L}}u=f$  is a differential equation in the sense of a distribution (i.e., in the weak sense) if $f$  and $u$  are distributions and all the derivatives are interpreted in the weak sense.

Suppose ${\mathcal {L}}$  is the generalized differential operator

${\mathcal {L}}=a_{n}(x){\cfrac {d^{n}}{dx^{n}}}+a_{n-1}(x){\cfrac {d^{(n-1)}}{dx^{(n-1)}}}+\dots +a_{0}(x)$

where $a_{i}(x)$  is infinitely differentiable.

We seek a $u$  such that

$\left\langle {\mathcal {L}}u,\phi \right\rangle =\left\langle f,\phi \right\rangle \qquad \forall \phi \in D$

which is taken to mean that

$\left\langle u,{\mathcal {L}}^{*}\phi \right\rangle =\left\langle f,\phi \right\rangle \qquad \forall \phi \in D~.$

Note that

$\left\langle a_{k}(x){\cfrac {d^{k}u}{dx^{k}}},\phi \right\rangle =\left\langle {\cfrac {d^{k}u}{dx^{k}}},a_{k}(x)~\phi \right\rangle =(-1)^{k}\left\langle u,{\cfrac {d}{dx^{k}}}[a_{k}(x)~\phi ]\right\rangle$

Therefore,

${\mathcal {L}}^{*}\phi =(-1)^{n}~{\cfrac {d^{n}(a_{n}\phi )}{dx^{n}}}+(-1)^{n-1}~{\cfrac {d^{(n-1)}(a_{n-1}\phi )}{dx^{(n-1)}}}+\dots +a_{0}~\phi ~.$

Here ${\mathcal {L}}^{*}$  is the formal adjoint of ${\mathcal {L}}$ . We can check that $({\mathcal {L}}^{*})^{*}={\mathcal {L}}$ . If ${\mathcal {L}}={\mathcal {L}}^{*}$  we say that ${\mathcal {L}}$  is formally self adjoint.

For example, if $n=2$  then

${\mathcal {L}}=a_{2}(x){\cfrac {d^{2}}{dx^{2}}}+a_{1}(x){\cfrac {d}{dx}}+a_{0}(x)$

Then

${\mathcal {L}}^{*}\phi ={\cfrac {d^{2}(a_{2}\phi )}{dx^{2}}}-{\cfrac {d(a_{1}\phi )}{dx}}+a_{0}~\phi$

or,

${\mathcal {L}}^{*}\phi =a_{2}~\phi ''+(2a'_{2}-a_{1})\phi '+(a''_{2}-a'_{1}+a_{0})\phi ~.$

Therefore, for ${\mathcal {L}}^{*}$  to be self adjoint,

${\mathcal {L}}^{*}\phi ={\mathcal {L}}\phi =a_{2}\phi ''+a_{1}\phi '+a_{0}\phi$

Hence

$a'_{2}=a_{1}\implies a''_{2}=a_{1}'~.$

In such a case, ${\mathcal {L}}$  is called a Sturm-Liouville operator.

### Example

To solve the differential equation

$x{\cfrac {du}{dx}}=0$

we seek a distribution $u$  which satisfies

$\left\langle xu',\phi \right\rangle =-\left\langle u,(x\phi )'\right\rangle =0$

Define $\psi :=(x\phi )'$ . Then $\psi$  must be a test function. We can show that $\psi$  is a test function if and only if

${\text{(2)}}\qquad \int _{\infty }^{\infty }\psi (x){\text{d}}x=0\qquad {\text{and}}\qquad \int _{0}^{\infty }\psi (x){\text{d}}x=0$

Now let us pick two test functions $\phi _{0}$  and $\phi _{1}$  satisfying

$\int _{-\infty }^{\infty }\phi _{0}(x){\text{d}}x=0\qquad {\text{and}}\qquad \int _{0}^{\infty }\phi _{0}(x){\text{d}}x=1$

and

$\int _{-\infty }^{\infty }\phi _{1}(x){\text{d}}x=1\qquad {\text{and}}\qquad \int _{0}^{\infty }\phi _{1}(x){\text{d}}x=0$

Then we can write any arbitrary test function $\phi (x)$  as a linear combination of $\phi _{0}$  and $\phi _{1}$  plus a terms which has the form of $\psi$ :

$\phi (x)=\phi _{0}(x)\int _{0}^{\infty }\phi (s)~ds+\phi _{1}(x)\int _{-\infty }^{\infty }\phi (s)~ds+\psi (x)$

which serves to define $\psi (x)$ . Note that $\psi$  satisfies equation (2).

Since $\left\langle u,\psi \right\rangle =0$ , the action of $u$  on $\phi$  is given by

$\left\langle u,\phi \right\rangle =\left\langle u,\phi _{0}\right\rangle \int _{\infty }^{\infty }H(x)\phi (s)ds+\left\langle u,\phi _{1}\right\rangle \int _{\infty }^{\infty }\phi (s)ds$

Therefore the solution is

$u=C_{1}~H(x)+C_{2}$

where $C_{1}:=\left\langle u,\phi _{0}\right\rangle$  and $C_{2}:=\left\langle u,\phi _{1}\right\rangle$ . Template:Lectures