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{d u}{d x} = 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$$.