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 $ℒu=f$ 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 $ℒ$ is the generalized differential operator

$ℒ=a_n(x)\frac{d^n}{d{x}^n}+a_{n-1}(x)\frac{d^{(n-1)}}{d{x}^{(n-1)}}+\dots +a_0(x)$

where $a_i(x)$ is infinitely differentiable.

We seek a $u$ such that

$⟨ℒu,\varphi ⟩=⟨f,\varphi ⟩\mathrm{\forall }\varphi \in D$

which is taken to mean that

$⟨u,{ℒ}^{*}\varphi ⟩=⟨f,\varphi ⟩\mathrm{\forall }\varphi \in D.$

Note that

$⟨a_k(x)\frac{d^{k}u}{d{x}^k},\varphi ⟩=⟨\frac{d^{k}u}{d{x}^k},a_k(x)\varphi ⟩=(-1{)}^k⟨u,\frac{d}{d{x}^k}[a_k(x)\varphi ]⟩$

Therefore,

${ℒ}^{*}\varphi =(-1{)}^n\frac{d^n(a_n\varphi )}{d{x}^n}+(-1{)}^{n-1}\frac{d^{(n-1)}(a_{n-1}\varphi )}{d{x}^{(n-1)}}+\dots +a_0\varphi .$

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

For example, if $n=2$ then

$ℒ=a_2(x)\frac{d^2}{d{x}^2}+a_1(x)\frac{d}{dx}+a_0(x)$

Then

${ℒ}^{*}\varphi =\frac{d^2(a_2\varphi )}{d{x}^2}-\frac{d(a_1\varphi )}{dx}+a_0\varphi$

or,

${ℒ}^{*}\varphi =a_2{\varphi }^{″}+(2a{\text{'}}_2-a_1){\varphi }^{\prime }+(a^{\prime }{\text{'}}_2-a{\text{'}}_1+a_0)\varphi .$

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

${ℒ}^{*}\varphi =ℒ\varphi =a_2{\varphi }^{″}+a_1{\varphi }^{\prime }+a_0\varphi$

Hence

$a{\text{'}}_2=a_1⟹a^{\prime }{\text{'}}_2={a_1}^{\prime }.$

In such a case, $ℒ$ is called a Sturm-Liouville operator.

Example

To solve the differential equation

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

we seek a distribution $u$ which satisfies

$⟨x{u}^{\prime },\varphi ⟩=-⟨u,(x\varphi {)}^{\prime }⟩=0$

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

$\text{(2)}\underset{\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\psi (x)\text{d}x=0\text{and}\underset{0}{\overset{\mathrm{\infty }}{\int }}\psi (x)\text{d}x=0$

Now let us pick two test functions ${\varphi }_0$ and ${\varphi }_1$ satisfying

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\varphi }_0(x)\text{d}x=0\text{and}\underset{0}{\overset{\mathrm{\infty }}{\int }}{\varphi }_0(x)\text{d}x=1$

and

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\varphi }_1(x)\text{d}x=1\text{and}\underset{0}{\overset{\mathrm{\infty }}{\int }}{\varphi }_1(x)\text{d}x=0$

Then we can write any arbitrary test function $\varphi (x)$ as a linear combination of ${\varphi }_0$ and ${\varphi }_1$ plus a terms which has the form of $\psi$:

$\varphi (x)={\varphi }_0(x)\underset{0}{\overset{\mathrm{\infty }}{\int }}\varphi (s)ds+{\varphi }_1(x)\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\varphi (s)ds+\psi (x)$

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

Since $⟨u,\psi ⟩=0$, the action of $u$ on $\varphi$ is given by

$⟨u,\varphi ⟩=⟨u,{\varphi }_0⟩\underset{\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}H(x)\varphi (s)ds+⟨u,{\varphi }_1⟩\underset{\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\varphi (s)ds$

Therefore the solution is

$u=C_{1}H(x)+C_2$

where $C_1:=⟨u,{\varphi }_0⟩$ and $C_2:=⟨u,{\varphi }_1⟩$. Template:Lectures