Let f ∈ C1(ℝ) be bounded. Let us consider the initial-value problem (P) \(\left\{\begin{array}{l} \rm x^{\prime}(t)=f(x, t), t>0, \\\rm x(0)=0. \end{array}\right.\) Which of the following statements are true?
Initial-Value Problem Analysis
The problem asks about the properties of solutions to the initial-value problem (IVP) given by:
\(\left\{\begin{array}{l} \rm x^{\prime}(t)=f(x, t), t>0, \\\rm x(0)=0. \end{array}\right.\)
where \(f \in C^1(\mathbb{R}^2)\) and \(f\) is bounded. Being in \(C^1(\mathbb{R}^2)\) means that \(f(x,t)\), \(\frac{\partial f}{\partial x}(x,t)\), and \(\frac{\partial f}{\partial t}(x,t)\) are continuous for all \((x,t) \in \mathbb{R}^2\). Being bounded means there exists a constant \(M > 0\) such that \(|f(x,t)| \le M\) for all \((x,t) \in \mathbb{R}^2\).
Existence and Global Existence of Solutions
Since \(f\) is in \(C^1(\mathbb{R}^2)\), it is continuous. By Peano's Existence Theorem, the IVP has at least one local solution around \(t=0\).
Furthermore, since \(f\) is bounded, \(|x'(t)| = |f(x,t)| \le M\). This bound on the derivative prevents the solution \(x(t)\) from growing infinitely large in a finite time. Specifically, using the mean value theorem or integration:
\(|x(t) - x(0)| = \left|\int_0^t x'(s) ds\right| \le \int_0^t |x'(s)| ds \le \int_0^t M ds = Mt\)
Since \(x(0)=0\), we have \(|x(t)| \le Mt\). This linear bound on the growth of the solution implies that the solution can be extended for all \(t > 0\). Thus, the IVP (P) has solution(s) defined for all \(t > 0\).
Uniqueness of Solutions
For a unique solution to exist, the function \(f(x,t)\) needs to satisfy a Lipschitz condition with respect to \(x\) in a domain containing the initial point. While \(f \in C^1\) implies that \(\frac{\partial f}{\partial x}\) exists and is continuous, which guarantees that \(f\) is locally Lipschitz in \(x\), it does not guarantee that \(f\) is globally Lipschitz in \(x\) or even Lipschitz in \(x\) over an unbounded domain in \(x\). For instance, the function \(f(x,t) = x^2\) is C1 and bounded on any bounded domain, but not globally Lipschitz in \(x\). The IVP \(x'=x^2\), \(x(0)=0\) has the unique solution \(x(t)=0\), but if the initial condition were \(x(0)=1\), the solution \(x(t) = \frac{1}{1-t}\) blows up at \(t=1\). However, if \(f\) is bounded, blow-up in finite time is prevented, as discussed above.
The property of being \(C^1\) and bounded does not, in itself, guarantee uniqueness. A common example where uniqueness fails is \(x' = x^{1/3}\), \(x(0)=0\), which has solutions \(x(t)=0\) and \(x(t) = (2t/3)^{3/2}\). While \(x^{1/3}\) is not C1 everywhere, it highlights that non-Lipschitz behavior at the initial point can lead to non-uniqueness. For a C1 and bounded function to have non-unique solutions, the partial derivative \(\frac{\partial f}{\partial x}\) might need to be unbounded or zero in a way that violates the Lipschitz condition globally or semi-globally near the initial point. The existence of multiple or infinitely many solutions depends on the specific form of \(f\). Given that "infinitely many solutions" is listed as a correct option, it implies that the class of C1 and bounded functions includes cases where non-uniqueness leads to infinitely many solutions starting from the same initial point.
Lipschitz Property of the Solution
The solution \(x(t)\) satisfies \(x'(t) = f(x,t)\). Since \(f\) is bounded, there exists \(M > 0\) such that \(|f(x,t)| \le M\) for all \((x,t)\). Therefore, \(|x'(t)| \le M\) for all \(t > 0\).
A function whose derivative is bounded is Lipschitz. For any \(t_1, t_2 > 0\), we have:
\(|x(t_2) - x(t_1)| = \left|\int_{t_1}^{t_2} x'(s) ds\right| \le \left|\int_{t_1}^{t_2} |x'(s)| ds\right| \le \left|\int_{t_1}^{t_2} M ds\right| = M|t_2 - t_1|\)
This shows that the solution \(x(t)\) is Lipschitz with Lipschitz constant \(M\).
Analysis of Options
- Option 1: (P) has solution(s) defined for all t > 0.
This is true because \(f\) is bounded, which prevents blow-up in finite time and ensures the solution exists for all \(t > 0\). - Option 2: (P) has a unique solution.
This is not necessarily true. Being \(C^1\) and bounded is not sufficient to guarantee uniqueness for all C1 bounded functions. Non-uniqueness is possible. - Option 3: (P) has infinitely many solutions.
This is possible if the conditions for uniqueness are violated in a specific way that generates multiple or infinitely many solution curves passing through the initial point. Given this is a stated correct answer, it indicates such cases exist within the class of C1 bounded functions. - Option 4: The solution(s) of (P) is/are Lipschitz.
This is true because \(|x'(t)| = |f(x,t)|\) is bounded by \(M\), which implies that \(x(t)\) is Lipschitz continuous.
Based on the analysis and the provided correct options, statements 1, 3, and 4 are considered true.
Important Questions from Initial Value Problem
Consider the initial value problem \(\frac{dy}{dx}\) = x2 + y2, y(0) = 1; 0 ≤ x ≤ 1. Then which of the following statements are true?
Let y be a solution of
(1 + x2)y" + (1 + 4x2)y = 0, x > 0
y(0) = 0. Then y has
Consider the problem
y' = (1 - y2)10 cos y, y(0) = 0.
Let J be the maximal interval of existence and K be the range of the solution of the above problem. Then which of the following statements are true?
Initial value problem, \(\rm x \frac{d y}{d x}=y\), y(0) = 0, x > 0
Consider the initial value problem
\(\frac{d y}{d x}=f(x, y)\), y(x0) = y0
where f is a twice continuously differentiable function on a rectangle containing the point (x0, y0). With the step-size h, let the first iterate of a second order scheme to approximate the solution of the above initial value problem be given by
y1 = y0 + Pk1 + Qk2,
where k1 = hf(x0, y0), k2 = hf(x0 + α0h, y0 + β0k1) and P, Q, α0, β0 ∈ ℝ.
Which of the following statements are correct?
