Differential Operators
Solved Problems
Example 1.
Check whether the commutative law of multiplication is valid for the operators \[L = {D^2} + 1, M = 2D +3.\]
Solution.
Calculate \(LMy:\)
\[My = \left( {2D + 3} \right)y = 2y' + 3y.\]
We obtain the following differential expression:
\[LMy = L\left( {My} \right)
= \left( {{D^2} + 1} \right)\left( {2y' + 3y} \right)
= 2y^{\prime\prime\prime} + 3y^{\prime\prime} + 2y' + 3y
= \left( {2{D^3} + 3{D^2} + 2D + 3} \right)y.\]
Now we compute \(MLy:\)
\[Ly = \left( {{D^2} + 1} \right)y = y^{\prime\prime} + y.\]
Hence,
\[MLy = M\left( {Ly} \right)
= \left( {2D + 3} \right)\left( {y^{\prime\prime} + y} \right)
= 2y^{\prime\prime\prime} + 3y^{\prime\prime} + 2y' + 3y
= \left( {2{D^3} + 3{D^2} + 2D + 3} \right)y.\]
We see that in this case, the commutative law of multiplication is valid (this is true for any operator \(L\left( D \right)\) with constant coefficients).
Example 2.
Check whether the commutative law of multiplication is valid for the operators \[L = xD - 1, M = {D^2} + {x^2}.\]
Solution.
We first calculate the differential expression \(LMy:\)
\[My = \left( {{D^2} + {x^2}} \right)y = y^{\prime\prime} + {x^2}y,\]
\[LMy = L\left( {My} \right)
= \left( {xD - 1} \right)\left( {y^{\prime\prime} + {x^2}y} \right)
= xy^{\prime\prime\prime} - y^{\prime\prime} + x\left( {2xy + {x^2}y'} \right) - {x^2}y
= xy^{\prime\prime\prime} - y^{\prime\prime} + 2{x^2}y + {x^3}y' - {x^2}y
= xy^{\prime\prime\prime} - y^{\prime\prime} + {x^3}y' + {x^2}y
= \left( {x{D^3} - {D^2} + {x^3}D + {x^2}} \right)y.\]
Similarly, we calculate \(MLy\) and compare the results
\[Ly = \left( {xD - 1} \right)y = xy' - y,\]
\[MLy = M\left( {Ly} \right)
= \left( {{D^2} + {x^2}} \right)\left( {xy' - y} \right)
= D\left( {y' + xy^{\prime\prime}} \right) + {x^3}y' - y^{\prime\prime} - {x^2}y
= y^{\prime\prime} + \cancel{y^{\prime\prime}} + xy^{\prime\prime\prime} - \cancel{y^{\prime\prime}} + {x^3}y' - {x^2}y
= xy^{\prime\prime\prime} + y^{\prime\prime} + {x^3}y' - {x^2}y = \left( {x{D^3} + {D^2} + {x^3}D - {x^2}} \right)y.\]
Thus, we get different expressions for a different order of the operators \(L\) and \(M.\) This property is typical for differential operators with variable coefficients.
Example 3.
Find a particular solution of the differential equation \[y^{\prime\prime\prime} + 3y = {e^{2x}}\] using the operator method.
Solution.
Consider the action of an arbitrary operator \(L\left( D \right)\) with constant coefficients to the exponential function \({e^{kx}}:\)
\[L\left( D \right){e^{kx}} = \left( {{k^n} + {a_1}{k^{n - 1}} + \cdots + {a_n}} \right){e^{kx}}
= L\left( k \right){e^{kx}}.\]
It follows that
\[L\left( D \right)\left[ {\frac{{{e^{kx}}}}{{L\left( k \right)}}} \right] = {e^{kx}}.\]
Since the differential equation is written in operator form as
\[L\left( D \right)y = {e^{kx}},\]
then one of the solutions of this equation is the function
\[{y_1} = \frac{{{e^{kx}}}}{{L\left( k \right)}}.\]
In this example, the operator is
\[L\left( D \right) = {D^3} + 3.\]
Hence, the particular solution of the differential equation is given by
\[{y_1} = \frac{{{e^{kx}}}}{{L\left( k \right)}} = \frac{{{e^{kx}}}}{{{k^3} + 3}}
= \frac{{{e^{2x}}}}{{{2^3} + 3}} = \frac{{{e^{2x}}}}{{11}}.\]
Example 4.
Find a particular solution of the differential equation \[{y^{IV}} - y^{\prime\prime} + y = 2\sin x\] using the operator method.
Solution.
The equation can be written as
\[L\left( D \right)y = 2\sin x\;\; \text{or}\;\;\left( {{D^4} - {D^2} + 1} \right)y = 2\sin x.\]
This differential polynomial contains only even powers of \(D.\) You may notice that the operator \({D^2}\) acting on the function \(A\sin kx\) produces
\[{D^2}y = {D^2}\left( {A\sin kx} \right) = - {k^2}A\sin kx = - {k^2}y.\]
It is clear that for an arbitrary differential polynomial \(L\left( {{D^2}} \right)\) with constant coefficients the following formula holds:
\[L\left( {{D^2}} \right)y = L\left( {{D^2}} \right)\left( {A\sin kx} \right)
= L\left( { - {k^2}} \right)A\sin kx = L\left( { - {k^2}} \right)y.\]
Then the particular solution is given by
\[{y_1} = \frac{{A\sin kx}}{{L\left( { - {k^2}} \right)}}.\]
In our case, the right-hand side of the equation is equal to \(2\sin x,\) and the differential operator can be written as
\[L\left( {{D^2}} \right) = {\left( {{D^2}} \right)^2} - {D^2} + 1.\]
As a result, we obtain:
\[{y_1} = \frac{{A\sin kx}}{{L\left( { - {k^2}} \right)}}
= \frac{{A\sin kx}}{{{{\left( { - {k^2}} \right)}^2} - \left( { - {k^2}} \right) + 1}}
= \frac{{2\sin x}}{{{{\left( { - {1^2}} \right)}^2} - \left( { - {1^2}} \right) + 1}}
= \frac{{2\sin x}}{3}.\]
