Chapter 6: Problem 27
Let \((x(t), y(t))\) be a nontrivial solution of the nonautonomous system $$ \frac{d x}{d t}=y, \quad \frac{d y}{d t}=t x $$ Suppose that \(\phi(t)=x(t+\gamma)\) and \(\psi(t)=y(t+\gamma)\), where \(\gamma \neq 0\). Show that \((\phi(t), \psi(t))\) is not a solution of the system.
Short Answer
Expert verified
\((\phi(t), \psi(t))\) is not a solution, as it leads to a contradiction with \(\gamma \neq 0\).
Step by step solution
01
Differentiate \(\phi(t)\) and \(\psi(t)\)
Find the derivatives of \(\phi(t) = x(t+\gamma)\) and \(\psi(t) = y(t+\gamma)\) with respect to \(t\). Using the chain rule, the derivative of \(\phi(t)\) is \(\frac{d\phi}{dt} = \frac{dx}{dt}\bigg|_{t+\gamma}\cdot \frac{d}{dt}(t+\gamma) = y(t+\gamma)\). Similarly, \(\frac{d\psi}{dt} = \frac{dy}{dt}\bigg|_{t+\gamma}\cdot \frac{d}{dt}(t+\gamma) = (t+\gamma)x(t+\gamma)\).
02
Compare with the Original System
We want to check if \((\phi(t), \psi(t))\) satisfies the original system equations: \(\frac{d\phi}{dt} = \psi(t)\) and \(\frac{d\psi}{dt} = tx(t+\gamma)\). Substituting the derivatives from Step 1, we have \(y(t+\gamma) = \psi(t)\) and \((t+\gamma)x(t+\gamma) = t\phi(t)\).
03
Analyze for Contradictions
Notice the second condition is \((t+\gamma)x(t+\gamma) = t\phi(t)\). Since \(\phi(t) = x(t+\gamma)\), the equation becomes \((t+\gamma)\phi(t) = t\phi(t)\), which is only true if \(\phi(t) = 0\) or \(\gamma = 0\). However, since \(x(t+\gamma)\) is nontrivial, \(\phi(t)\) is not identically zero and \(\gamma eq 0\). So this gives a contradiction.
04
Conclusion: \( (\phi(t), \psi(t)) \) Does Not Satisfy the System
Since the assumption leads to a contradictory statement that cannot hold for nontrivial solutions given that \(\gamma eq 0\), \((\phi(t), \psi(t))\) does not satisfy the system.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Nontrivial Solutions
In the world of differential equations, a **nontrivial solution** refers to a solution that is not identically zero. This means that for every time point, the solution vector components are not all zero. In the given system,
- \( (x(t), y(t)) \) is nontrivial
- This implies that the functions \( x(t) \) and \( y(t) \) are not zero across their domain.
Nonautonomous Systems
Differential equations can be classified into two categories: autonomous and nonautonomous. A **nonautonomous system** is one where the rate of change is not solely dependent on the current state but also explicitly on time itself.The system given by:\[ \frac{d x}{d t}=y, \quad \frac{d y}{d t}=t x \],is clearly nonautonomous, as we see from the presence of the time variable \( t \) in the equation for \( \frac{d y}{d t} \). This time dependence introduces a level of dynamism; it means the system’s behavior is influenced by how time progresses.
- Unlike autonomous systems, where behavior is constant over time,
- nonautonomous systems echo the idea that historical changes can impact future behavior significantly.
Chain Rule
The **chain rule** is a fundamental tool in calculus used to differentiate compositions of functions. In our problem, it helps when differentiating functions of the form \( x(t+\gamma) \) and \( y(t+\gamma) \) with respect to \( t \).Let's examine how the chain rule is applied:
- For \( \phi(t) = x(t+\gamma) \), applying the chain rule gives:\[ \frac{d\phi}{dt} = \frac{dx}{dt}\bigg|_{t+\gamma}\cdot \frac{d}{dt}(t+\gamma) = y(t+\gamma) \]
- Similarly, for \( \psi(t) = y(t+\gamma) \), we get: \[ \frac{d\psi}{dt} = \frac{dy}{dt}\bigg|_{t+\gamma}\cdot \frac{d}{dt}(t+\gamma) = (t+\gamma)x(t+\gamma) \]
System of Differential Equations
A **system of differential equations** involves more than one interdependent differential equation, often to describe systems in which multiple factors influence each other. The exercise offers a classic example of such a system by giving two equations:
- \( \frac{d x}{d t}=y \)
- \( \frac{d y}{d t}=t x \)