Inverse Operators

Foreword

Recall that a general solution is the sum of the complentary function and a particular solution i.e. $y=y_c+y_p$.

The complentary function of the differntial equation:

$$\begin{array}{cc}(D^2-1)y=x^2& \text{(a)}\end{array}$$

is

$$\begin{array}{cc}{y}_c=c_1{e}^x+c_2{e}^{-x}& \text{(b)}\end{array}$$

and a particular solution of $(a)$ is:

$$\begin{array}{cc}{y}_p=-x^2-2& \text{(c)}\end{array}$$

Therefore the general solution is:

$$\begin{array}{cc}y=-x^2-2+c_1{e}^x+c_2{e}^{-x}& \text{(d)}\end{array}$$

Note that $(b)$ can be obtained by assigning the value of 0 to both arbitrary constants in $(d)$, and so can infinitely other particular solutions.

In this lesson and the next, whenver we refer to a particular solution $y_p$ of a differential equation, we shall mean that particular solution from which all constant multiples of terms in $y_c$ have been eliminated.

(jmh The implication here is that there is only one such particular solution in which all arbitrary constants have been set to zero - the infinite multitude of particular solutions arises from those arbitrary constants being values other than zero).

Meaning of an Inverse Operator

Given the $n$th order linear differential equation:

$$\begin{array}{cc}P(D)y=Q(x),& \text{(25.1)}\end{array}$$

Recall that when we say $P(D)$ we mean the polynomial operator:

$$\begin{array}{cc}P(D)=a_n{D}^n+\cdots +a_{1}D+a_0,a_n\ne 0& \text{(25.11)}\end{array}$$

Definition 25.2 Let $P(D)y=Q(x)$ where $P(D)$ is the polynomial operator $(25.11)$ and $Q(x)$ is the special function consisting only of such terms as $b,x^k,e^{ax},\sin ax,\cos ax$, and a finite number of combinations of such terms, where $a,b$ are constants and $k$ is a positive integer. Then the inverse operator of $P(D)$, written as $P^{-1}(D)$ or $1/P(D)$, is defined as an operator which, when operating on $Q(x)$, will give the particular solution $y_p$ of $(25.1)$ that contains no constant multiples of a term in the complementary function $y_c$, i.e.,

$$\begin{array}{cc}{P}^{-1}(D)Q(x)=y_p\text{or}\frac{1}{P(D)}Q(x)=y_p,& \text{(25.21)}\end{array}$$

where $y_p$ is the particular solution of $P(D)y=Q(x)$ that containts no constant multiple of a term in $y_c$.

By Definition 25.2, we can conclude that:

$$\begin{array}{cc}{D}^{-n}Q(x)\equiv \text{integrating}Q(x)n\text{times, ignoring constants of integration}& \text{(25.25)}\end{array}$$

Solution by Means of Inverse Operators

As stated above, in general we can find $y_p$ by:

$$y_p=\frac{1}{P(D)}{Q}_x$$

This is often a much easier method of finding $y_p$ than previous methods.

We can find some formulas/shortcuts/rules to handle the possible cases.

We shall consider each of the functions mentioned in Definition 25.2, case by case.

(jmh Note that the book proves each scenario but I'll just capture the recipe for each scenario here.)

1. If $Q(x)=b{x}^k$ and $P(D)=D-a_0$:

Given:

$$\begin{array}{cc}(D-a_0)y=b{x}^k& \text{(a)}\end{array}$$

By definition 25.2 we have:

$$\begin{array}{cc}{y}_p=\frac{1}{D-a_0}(b{x}^k)=\frac{1}{-a_0(1-\frac{D}{a_0})}(b{x}^k)& \text{(b)}\end{array}$$

Which, via geometric series expansion can be written as:

$$\begin{array}{cc}-\frac{1}{a_0}(1+\frac{D}{a_0}+\frac{D^2}{{a_0}^2}+\cdots +\frac{D^k}{{a_0}^k})(b{x}^k),a_0\ne 0& \text{(c)}\end{array}$$

Note that we can stop the series expansion at $D^k$ because we're differentiating a $k$th degree polynomial, and higher derivatives will just be $0$.

Also note that for the special case $k=0$, we have:

$$\begin{array}{cc}{y}_p=\frac{b}{a_0}& \text{(d)}\end{array}$$

2. If $Q(x)=b{x}^k$ and $P(D)=a_n{D}^n+\cdots +a_{1}D$:

We have here that $D$ is a factor of $P(D)$d and can write $P(D)=D(a_n{D}^{n-1}+\cdots +a_{2}D+a_1$, where $a_1\ne 0$. We can also have higher powered factors of $D$ in $P(D)$. In general, let $D^r$ be a factor of $P(D)$. Then $P(D)y=b{x}^k$ can be written as:

$$\begin{array}{cc}P(D)y=D^r(a_n{D}^{n-r}+\cdots +a_{r+1}D+a_r)y=b{x}^k,a_r\ne 0& \text{(a)}\end{array}$$

Therefore, by Definition 25.2:

$$\begin{array}{cc}{y}_p=\frac{1}{D^r}(\frac{1}{a_n{D}^{n-r}+\cdots +a_{r+1}D+a_r}(b{x}^k)),a_r\ne 0& \text{(25.34)}\end{array}$$

From there, we first perform series expansion to get a non-inverse differential operator (see above from the first $b{x}^k$ recipe), apply differntiation, and then take $r$ successive integrals against the result, ignoring arbitrary constants of differentiation.

3. If $Q(x)=b{e}^{ax}$:

Then a particular solution of this equation is:

$$\begin{array}{cc}{y}_p=\frac{1}{P(D)}b{e}^{ax}=\frac{b{e}^{ax}}{P(a)},P(a)\ne 0& \text{(25.4)}\end{array}$$

4. If $Q(x)=b\sin ax$ or $b\cos ax$

Here we can take advantage of the exponential form of these functions apply the method given for the exponential form above. Even easier, we can use the method given in $(25.4)$.

That is, for $P(D)y=b\sin ax$, we use the imaginary part of the particular solution of:

$$\begin{array}{cc}P(D)y=b{e}^{aix}& \text{(a)}\end{array}$$

and for $P(D)y=b\cos ax$, we use the real part of the particular solution of $(a)$. In both cases, we use the method given above for $Q(x)=b{e}^{ax}$.

5. If $Q(x)=u{e}^{ax}$ where $u$ is a polynomial in $x$ then:

$$\begin{array}{cc}{y}_p=\frac{1}{P(D)}u{e}^{ax}=e^{ax}\frac{1}{P(D+a)}u& \text{(25.51)}\end{array}$$

From there, we use the methods given above for $Q(x)=b{x}^k$ to convert the inverse polynomial operator into a polynomial operator.

6. If $Q(x)=b{e}^{ax}$ and $P(a)=0$:

Given $P(D)y=b{e}^{ax}$, if $P(a)=0$ then $(D-a)$ is a factor of $P(D)$. Assume $(D-a{)}^r$ is a factor of $P(D)$. Then we can write:

$$\begin{array}{cc}P(D)=(D-a{)}^{r}F(D),F(a)\ne 0.& \text{(a)}\end{array}$$

Then a particular solution is:

$$\begin{array}{cc}{y}_p=\frac{1}{P(D)}(b{e}^{ax})\equiv \frac{1}{(D-a{)}^{r}F(D)}(b{e}^{ax})=\frac{b{x}^r{e}^{ax}}{r!F(a)},F(a)\ne 0& \text{(25.6)}\end{array}$$

7. If $Q(x)=Q_1(x)+Q_2(x)+\cdots +Q_n(x)$:

Then $P(D)y=Q(x)=Q_1(x)+Q_2(x)+\cdots +Q_n(x)$ and a particular solution $y_p$ of $P(D)y=Q(x)$ is the sum of the respective solutions of $P(D)y=Q_1$, $P(D)y=Q_2$, $\cdots$, $P(D)y=Q_n$.

Therefore,

$$\begin{array}{cc}{y}_p=\frac{1}{P(D)}Q\equiv \frac{1}{P(D)}{Q}_1+\frac{1}{P(D)}{Q}_2+\cdots +\frac{1}{P(D)}{Q}_n& \text{(25.6)}\end{array}$$

Here, we find particular solutions for each part using the methods developed in this lesson, then sum them to find a particular solution for the whole $P(D)y=Q(x)$.