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What conditions must the functions \(f(t)\) and \(g(t)\) satisfy in order for the following operator pairs to be commutative? (a) \(\mathrm{L}=\frac{\mathrm{d}}{\mathrm{d} t}+f(t), \quad \mathrm{M}=\frac{\mathrm{d}}{\mathrm{d} t}+g(t)\) (b) \(\mathrm{L}=\frac{\mathrm{d}^{2}}{\mathrm{~d} t^{2}}+f(t), \quad \mathrm{M}=\frac{\mathrm{d}^{2}}{\mathrm{~d} t^{2}}+g(t)\)

Short Answer

Expert verified
For both (a) and (b), commutativity requires that \( f(t) = g(t) \).

Step by step solution

01

Define Commutativity

For the operators \(L\) and \(M\) to be commutative, their product in any order must be equal: \(LM = ML\). This means applying \(L\) after \(M\) should give the same result as applying \(M\) after \(L\).
02

Express Product of Operators in (a)

Compute \(LM\) and \(ML\) for the operators in part (a): - \(L = \frac{d}{dt} + f(t)\) - \(M = \frac{d}{dt} + g(t)\)Calculate:\(LM = \left(\frac{d}{dt} + f(t)\right)\left(\frac{d}{dt} + g(t)\right) = \frac{d^2}{dt^2} + g(t)\frac{d}{dt} + f(t)\frac{d}{dt} + f(t)g(t)\).Similarly, calculate:\(ML = \left(\frac{d}{dt} + g(t)\right)\left(\frac{d}{dt} + f(t)\right) = \frac{d^2}{dt^2} + f(t)\frac{d}{dt} + g(t)\frac{d}{dt} + g(t)f(t)\).
03

Equate LM and ML in (a)

Equate the expressions from step 2:For \(LM = ML\), we need:\[g(t)\frac{d}{dt} + f(t)g(t) = f(t)\frac{d}{dt} + g(t)f(t)\]Simplify to:\[g(t) = f(t)\]Hence, for commutativity, \(f(t)\) must equal \(g(t)\).
04

Express Product of Operators in (b)

Compute \(LM\) and \(ML\) for part (b): - \(L = \frac{d^2}{dt^2} + f(t)\) - \(M = \frac{d^2}{dt^2} + g(t)\)Calculate:\(LM = \left(\frac{d^2}{dt^2} + f(t)\right)\left(\frac{d^2}{dt^2} + g(t)\right) = \frac{d^4}{dt^4} + g(t)\frac{d^2}{dt^2} + f(t)\frac{d^2}{dt^2} + f(t)g(t)\).Similarly, calculate:\(ML = \left(\frac{d^2}{dt^2} + g(t)\right)\left(\frac{d^2}{dt^2} + f(t)\right) = \frac{d^4}{dt^4} + f(t)\frac{d^2}{dt^2} + g(t)\frac{d^2}{dt^2} + g(t)f(t)\).
05

Equate LM and ML in (b)

For commutativity, equate expressions from step 4:We need:\[g(t)\frac{d^2}{dt^2} + f(t)g(t) = f(t)\frac{d^2}{dt^2} + g(t)f(t)\]Which simplifies to:\[g(t) = f(t)\]Thus, \(f(t)\) must equal \(g(t)\) for commutativity.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Operators
Differential operators are fundamental in calculus and analysis. They are used to execute differentiation, a key mathematical process.
In essence, a differential operator is any function or expression that involves derivatives. A common example is the derivative operator \(\frac{d}{dt}\), which signifies differentiation with respect to time \(t\).
However, differential operators are not limited to basic derivatives. They can include higher-order derivatives such as \(\frac{d^2}{dt^2}\), which represent the second derivative, and so on.
In our context, operators like \(L = \frac{d}{dt} + f(t)\) combine the basic derivative with a function \(f(t)\). Such compositions are quite common in solving differential equations. The role of \(f(t)\) here is to adjust or modify the operator to fit specific scenarios in analysis or physics.
Commutativity Conditions
Commutativity is a key concept in mathematics, especially when dealing with operators or matrices.
The term refers to the ability to swap operations without changing the outcome. For example, multiplying numbers or matrices might or might not be commutative depending on the specific case.
When it comes to operators, such as our differential operators \(L\) and \(M\), commutativity means performing them in any sequence yields the same result. Mathematically, this is expressed as \(LM = ML\).
To determine if two operators \(L = \frac{d}{dt} + f(t)\) and \(M = \frac{d}{dt} + g(t)\) are commutative, we look for conditions where \(g(t) = f(t)\).
This equality ensures that applying one operator after another or vice versa does not change the result. It's critical in ensuring consistent results in calculations and analyses.
Operator Algebra
Operator algebra involves the study and manipulation of operators. Operators, particularly linear ones, are vital in mathematical analysis and quantum mechanics.
Their algebra is much about how operators interact in a set of functions or equations. In operator algebra, one is likely to encounter concepts such as commutators, which measure the degree of non-commutativity between two operators.
In our exercise, identifying commutative conditions for operators \(L\) and \(M\) involves exploring their equations or expressions and balancing them algebraically. The step-by-step process usually involves expanding their multiplication \(LM\) and \(ML\), followed by simplification to match terms.
This type of algebraic manipulation is almost like solving puzzles. It involves determining what needs to be true for everything to fit together accurately, a crucial skill in many scientific and engineering fields.

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Most popular questions from this chapter

Sketch the direction field of the differential equation $$ \frac{\mathrm{d} x}{\mathrm{~d} t}=t-x $$ Verify that \(x=t-1+C \mathrm{e}^{-t}\) is the solution of the equation. Sketch the solution curve for which \(x(0)=2\), and that for which \(x(4)=0\), and check that these are consistent with your direction field.

A chemical \(A\) is formed by an irreversible reaction from chemicals \(B\) and \(C\). Assuming that the amounts of \(B\) and \(C\) are adequate to sustain the reaction, the amount of \(A\) formed at time \(t\) is governed by the differential equation $$ \frac{\mathrm{d} A}{\mathrm{~d} t}=K(1-\alpha A)^{7} $$ If no \(A\) is present at time \(t=0\), find an expression for the amount of \(A\) present at time \(t\).

Show that the characteristic equation of the differential equation $$ \frac{\mathrm{d}^{4} x}{\mathrm{~d} t^{4}}-2 \frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}+3 \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}-2 \frac{\mathrm{d} x}{\mathrm{~d} t}+x=0 $$ is $$ \left(m^{2}-m+1\right)^{2}=0 $$ and hence find the general solution of the equation.

An open vessel is in the shape of a right circular cone of semi-vertical angle \(45^{\circ}\) with axis vertical and apex downwards. At time \(t=0\) the vessel is empty. Water is pumped in at a constant rate \(p \mathrm{~m}^{3} \mathrm{~s}^{-1}\) and escapes through a small hole at the vertex at a rate \(k y \mathrm{~m}^{3} \mathrm{~s}^{-1}\), where \(k\) is a positive constant and \(y\) is the depth of water in the cone. Given that the volume of a circular cone is \(\pi r^{2} h / 3\), where \(r\) is the radius of the base and \(h\) its vertical height, show that $$ \pi y^{2} \frac{\mathrm{d} y}{\mathrm{~d} t}=p-k y $$ Deduce that the water level reaches the value \(y=p /(2 k)\) at time $$ t=\frac{\pi p^{2}}{k^{3}}\left(\ln 2-\frac{5}{8}\right) $$

Show that the characteristic equation of the differential equation $$ \frac{\mathrm{d}^{4} x}{\mathrm{~d} t^{4}}-3 \frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}-5 \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+9 \frac{\mathrm{d} x}{\mathrm{~d} t}-2 x=0 $$ is $$ \left(m^{2}+m-2\right)\left(m^{2}-4 m+1\right)=0 $$ and hence find the general solutions of the equations (a) \(\frac{\mathrm{d}^{4} x}{\mathrm{~d} t^{4}}-3 \frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}-5 \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+9 \frac{\mathrm{d} x}{\mathrm{~d} t}-2 x=\cos 2 t\) (b) \(\frac{\mathrm{d}^{4} x}{\mathrm{~d} t^{4}}-3 \frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}-5 \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+9 \frac{\mathrm{d} x}{\mathrm{~d} t}-2 x=\mathrm{e}^{2 t}+\mathrm{e}^{-2 t}\) (c) \(\frac{\mathrm{d}^{4} x}{\mathrm{~d} t^{4}}-3 \frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}-5 \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+9 \frac{\mathrm{d} x}{\mathrm{~d} t}-2 x=t^{2}-1+\mathrm{e}^{-t}\)

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