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

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

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

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

We seek a ${\displaystyle u}$  such that

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

which is taken to mean that

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

Note that

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

Therefore,

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

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

For example, if ${\displaystyle n=2}$  then

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

Then

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

or,

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

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

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

Hence

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

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

Example

To solve the differential equation

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

we seek a distribution ${\displaystyle u}$  which satisfies

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

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

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

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

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

and

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

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

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

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

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

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

Therefore the solution is

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

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