Q39E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Let A, B and C represent three linear differential operators with constant coefficients; for example,

$$\begin{gathered} \mathbf{A}\mathbf{:}\mathbf{=}{\mathbf{a}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{a}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{a}}_{\mathbf{0}}\mathbf{,}\mathbf{B}\mathbf{:}\mathbf{=}{\mathbf{b}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{b}}_{\mathbf{0}}\mathbf{,} \\ \mathbf{C}\mathbf{:}\mathbf{=}{\mathbf{c}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{c}}_{\mathbf{0}}\mathbf{,} \end{gathered}$$

 Where the a’s, b’s, and c’s are constants. Verify the following properties:

 

(a) Commutative laws: 

      $$\begin{gathered} \mathit{A}\mathbf{ }\mathbf{+}\mathbf{ }\mathit{B}\mathbf{ }\mathbf{=}\mathbf{ }\mathit{B}\mathbf{ }\mathbf{+}\mathbf{ }\mathit{A}\mathbf{,}\mathbf{ } \\ \mathit{A}\mathit{B}\mathbf{ }\mathbf{=}\mathbf{ }\mathit{B}\mathit{A} \end{gathered}$$

 (b) Associative laws:

       $$\begin{gathered} \left(\mathbf{A}\mathbf{+}\mathbf{B}\right)\mathbf{+}\mathbf{C}\mathbf{=}\mathbf{A}\mathbf{+}\left(\mathbf{B}\mathbf{+}\mathbf{C}\right)\mathbf{,} \\ \left(\mathbf{AB}\right)\mathbf{C}\mathbf{=}\mathbf{A}\left(\mathbf{BC}\right)\mathbf{.} \end{gathered}$$

 (c) Distributive law: $\mathbf{A}\left(\mathbf{B}\mathbf{+}\mathbf{C}\right)\mathbf{=}\mathbf{AB}\mathbf{+}\mathbf{AC}$

       

Step-by-Step Solution

Answer

1. Therefore, the given equations satisfy the commutative law.
2. Hence, the given equations satisfy the associative law.
3. Consequently, the given equations satisfy the distributive law.

Step 1: General form

Commutative laws: 

 $$\begin{gathered} \mathbf{A}\mathbf{+}\mathbf{B}\mathbf{=}\mathbf{B}\mathbf{+}\mathbf{A}\mathbf{,} \\ \mathbf{AB}\mathbf{=}\mathbf{BA}\mathbf{.} \end{gathered}$$

Associative laws:

 $$\begin{gathered} \left(\mathbf{A}\mathbf{+}\mathbf{B}\right)\mathbf{+}\mathbf{C}\mathbf{=}\mathbf{A}\mathbf{+}\left(\mathbf{B}\mathbf{+}\mathbf{C}\right)\mathbf{,} \\ \left(\mathbf{AB}\right)\mathbf{C}\mathbf{=}\mathbf{A}\left(\mathbf{BC}\right)\mathbf{.} \end{gathered}$$

Distributive law:

 $\mathbf{A}\left(\mathbf{B}\mathbf{+}\mathbf{C}\right)\mathbf{=}\mathbf{AB}\mathbf{+}\mathbf{AC}$.

Step 2: Demonstrating the given equation

Given that,

 

$\mathbf{A}\mathbf{=}{\mathbf{a}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{a}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{a}}_{\mathbf{0}}$  …… (1)

   $\mathbf{B}\mathbf{=}{\mathbf{b}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{b}}_{\mathbf{0}}$…… (2)

  $\mathbf{C}\mathbf{=}{\mathbf{c}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{c}}_{\mathbf{0}}$…… (3)

 

Let us prove the commutative property.

 

Case (1):

 

Then, find the L.H.S.

 $$\begin{gathered} \mathbf{A}\mathbf{+}\mathbf{B}\mathbf{=}{\mathbf{a}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{a}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{a}}_{\mathbf{0}}\mathbf{+}{\mathbf{b}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{b}}_{\mathbf{0}} \\ \mathbf{=}\left({\mathbf{a}}_{\mathbf{2}}\mathbf{+}{\mathbf{b}}_{\mathbf{2}}\right){\mathbf{D}}^{\mathbf{2}}\mathbf{+}\left({\mathbf{a}}_{\mathbf{1}}\mathbf{+}{\mathbf{b}}_{\mathbf{1}}\right)\mathbf{D}\mathbf{+}\left({\mathbf{a}}_{\mathbf{0}}\mathbf{+}{\mathbf{b}}_{\mathbf{0}}\right) \end{gathered}$$

 

Now, R.H.S.

$\mathbf{B}\mathbf{+}\mathbf{A}\mathbf{=}\left({\mathbf{b}}_{\mathbf{2}}\mathbf{+}{\mathbf{a}}_{\mathbf{2}}\right){\mathbf{D}}^{\mathbf{2}}\mathbf{+}\left({\mathbf{b}}_{\mathbf{1}}\mathbf{+}{\mathbf{a}}_{\mathbf{1}}\right)\mathbf{D}\mathbf{+}\left({\mathbf{b}}_{\mathbf{0}}\mathbf{+}{\mathbf{a}}_{\mathbf{0}}\right)$ 

.So, A + B = B + A.

 Case (2):

$$\begin{gathered} \mathbf{AB}\mathbf{=}\left({\mathbf{a}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{a}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{a}}_{\mathbf{0}}\right)\left({\mathbf{b}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{b}}_{\mathbf{0}}\right) \\ \mathbf{BA}\mathbf{=}\left({\mathbf{b}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{b}}_{\mathbf{0}}\right)\left({\mathbf{a}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{a}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{a}}_{\mathbf{0}}\right) \end{gathered}$$ 

Therefore, AB = BA

Step 3: Showing the given equation

To prove: 

 $$\begin{gathered} \left(\mathbf{A}\mathbf{+}\mathbf{B}\right)\mathbf{+}\mathbf{C}\mathbf{=}\mathbf{A}\mathbf{+}\left(\mathbf{B}\mathbf{+}\mathbf{C}\right)\mathbf{,} \\ \left(\mathbf{AB}\right)\mathbf{C}\mathbf{=}\mathbf{A}\left(\mathbf{BC}\right)\mathbf{.} \end{gathered}$$

 Case (1):

 $$\begin{gathered} \left(\mathbf{A}\mathbf{+}\mathbf{B}\right)\mathbf{+}\mathbf{C}\mathbf{=}\left({\mathbf{a}}_{\mathbf{2}}\mathbf{+}{\mathbf{b}}_{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\right){\mathbf{D}}^{\mathbf{2}}\mathbf{+}\left({\mathbf{a}}_{\mathbf{1}}\mathbf{+}{\mathbf{b}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{1}}\right)\mathbf{D}\mathbf{+}\left({\mathbf{a}}_{\mathbf{0}}\mathbf{+}{\mathbf{b}}_{\mathbf{0}}\mathbf{+}{\mathbf{c}}_{\mathbf{0}}\right) \\ \mathbf{A}\mathbf{+}\left(\mathbf{B}\mathbf{+}\mathbf{C}\right)\mathbf{=}\left({\mathbf{a}}_{\mathbf{2}}\mathbf{+}{\mathbf{b}}_{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\right){\mathbf{D}}^{\mathbf{2}}\mathbf{+}\left({\mathbf{a}}_{\mathbf{1}}\mathbf{+}{\mathbf{b}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{1}}\right)\mathbf{D}\mathbf{+}\left({\mathbf{a}}_{\mathbf{0}}\mathbf{+}{\mathbf{b}}_{\mathbf{0}}\mathbf{+}{\mathbf{c}}_{\mathbf{0}}\right) \end{gathered}$$

Henceforth, (A + B) + C = A + (B + C).

 

Case (2):

 $$\begin{gathered} \left(\mathbf{AB}\right)\mathbf{C}\mathbf{=}\left({\mathbf{a}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{a}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{a}}_{\mathbf{0}}\right)\left({\mathbf{b}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{b}}_{\mathbf{0}}\right)\left({\mathbf{c}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{c}}_{\mathbf{0}}\right) \\ \mathbf{A}\left(\mathbf{BC}\right)\mathbf{=}\left({\mathbf{a}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{a}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{a}}_{\mathbf{0}}\right)\left({\mathbf{b}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{b}}_{\mathbf{0}}\right)\left({\mathbf{c}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{c}}_{\mathbf{0}}\right) \end{gathered}$$

Hence, (AB) C = A (BC).

 

To prove: $\mathbf{A}\left(\mathbf{B}\mathbf{+}\mathbf{C}\right)\mathbf{=}\mathbf{AB}\mathbf{+}\mathbf{AC}$ .

 

Then,

$$\begin{gathered} \mathbf{A}\mathbf{+}\left(\mathbf{B}\mathbf{+}\mathbf{C}\right)\mathbf{=}\left({\mathbf{a}}_{\mathbf{2}}\mathbf{+}{\mathbf{b}}_{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\right){\mathbf{D}}^{\mathbf{2}}\mathbf{+}\left({\mathbf{a}}_{\mathbf{1}}\mathbf{+}{\mathbf{b}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{1}}\right)\mathbf{D}\mathbf{+}\left({\mathbf{a}}_{\mathbf{0}}\mathbf{+}{\mathbf{b}}_{\mathbf{0}}\mathbf{+}{\mathbf{c}}_{\mathbf{0}}\right) \\ \mathbf{AB}\mathbf{+}\mathbf{AC}\mathbf{=}\left({\mathbf{a}}_{\mathbf{2}}\mathbf{+}{\mathbf{b}}_{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\right){\mathbf{D}}^{\mathbf{2}}\mathbf{+}\left({\mathbf{a}}_{\mathbf{1}}\mathbf{+}{\mathbf{b}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{1}}\right)\mathbf{D}\mathbf{+}\left({\mathbf{a}}_{\mathbf{0}}\mathbf{+}{\mathbf{b}}_{\mathbf{0}}\mathbf{+}{\mathbf{c}}_{\mathbf{0}}\right) \end{gathered}$$ 

Consequently, A (B + C) = AB + AC.