Consider the initial value problem \(\frac{dy}{dx}\) = x2 + y2, y(0) = 1; 0 ≤ x ≤ 1. Then which of the following statements are true?
Initial Value Problem Analysis
We are given the initial value problem (IVP):
\(\frac{dy}{dx}\) = x2 + y2, with the initial condition y(0) = 1, for the interval 0 ≤ x ≤ 1.
This is a first-order ordinary differential equation (ODE).
Existence and Uniqueness
The function \(f(x, y) = x^2 + y^2\) and its partial derivative with respect to y, \(\frac{\partial f}{\partial y} = 2y\), are continuous for all real x and y. According to the Existence and Uniqueness Theorem for ODEs, for any initial condition \((x_0, y_0)\), there exists a unique solution in some interval containing \(x_0\). For our initial condition (0, 1), a unique solution exists in a maximal interval \((T_{min}, T_{max})\) containing 0.
Analyzing the Statements
Uniqueness in \([0, \pi/4]\)
Since \(f(x, y)\) and \(\frac{\partial f}{\partial y}\) are continuous, a unique solution exists in a maximal interval. We need to determine if this interval includes \([0, \pi/4]\).
Consider the comparison equation \(\frac{dz}{dx} = (\pi/4)^2 + z^2\) with initial condition z(0) = 1. For \(x \in [0, \pi/4]\), we have \(x^2 \le (\pi/4)^2\). Thus, \(\frac{dy}{dx} = x^2 + y^2 \le (\pi/4)^2 + y^2\). By the comparison principle, since y(0)=z(0)=1, we have \(y(x) \le z(x)\) as long as both solutions exist.
Solving \(\frac{dz}{dx} = (\pi/4)^2 + z^2\):
\(\int \frac{dz}{(\pi/4)^2 + z^2} = \int dx\)
\(\frac{1}{\pi/4} \arctan\left(\frac{z}{\pi/4}\right) = x + C\)
Using z(0)=1:
\(\frac{4}{\pi} \arctan\left(\frac{1}{\pi/4}\right) = 0 + C \implies C = \frac{4}{\pi} \arctan\left(\frac{4}{\pi}\right)\)
So, \(\frac{4}{\pi} \arctan\left(\frac{z(x)}{\pi/4}\right) = x + \frac{4}{\pi} \arctan\left(\frac{4}{\pi}\right)\).
The solution z(x) becomes unbounded when \(\frac{\pi}{4} x + \arctan\left(\frac{4}{\pi}\right) = \frac{\pi}{2}\). This happens at \(x_s = \frac{4}{\pi}\left(\frac{\pi}{2} - \arctan\left(\frac{4}{\pi}\right)\right)\). Numerically, \(x_s \approx 2 - \frac{4}{3.14} \arctan(1.27) \approx 2 - 1.27 \times 0.907 \approx 2 - 1.15 = 0.85\). Since \(\pi/4 \approx 0.785\), we have \(x_s > \pi/4\). The solution z(x) exists uniquely and is bounded on \([0, \pi/4]\). Because \(y(x) \le z(x)\) on \([0, \pi/4]\), the unique solution y(x) also exists on \([0, \pi/4]\).
Thus, there exists a unique solution in \(\left[ 0, \frac{\pi}{4} \right]\). This statement is true.
Boundedness in \([0, \pi/4]\)
As shown above, the comparison solution \(z(x)\) defined on \([0, \pi/4]\) is bounded because its singularity occurs at \(x_s > \pi/4\). Since \(y(x) \le z(x)\) for \(x \in [0, \pi/4]\), \(y(x)\) is also bounded on \([0, \pi/4]\).
Thus, every solution is bounded in \(\left[ 0, \frac{\pi}{4} \right]\). This statement is true.
Singularity in [0, 1]
Consider the comparison equation \(\frac{dv}{dx} = v^2\) with initial condition v(0) = 1. The solution is \(v(x) = \frac{1}{1-x}\), which is defined on \([0, 1)\) and becomes unbounded as \(x \to 1^-\).
For \(x \in [0, 1]\), we have \(x^2 \ge 0\), so \(\frac{dy}{dx} = x^2 + y^2 \ge y^2 = \frac{dv}{dx}\). By the comparison principle, since y(0)=v(0)=1, we have \(y(x) \ge v(x)\) as long as both solutions exist. For \(x \in [0, 1)\), \(y(x) \ge \frac{1}{1-x}\). Since \(\frac{1}{1-x} \to \infty\) as \(x \to 1^-\), \(y(x)\) must also become unbounded at some point \(x_0 \le 1\). This point \(x_0\) is where the solution exhibits a singularity.
We know the solution exists and is bounded on \([0, \pi/4]\), so the singularity point \(x_0\) must be \(x_0 \ge \pi/4\). Combined with \(x_0 \le 1\), the singularity must occur at some point \(x_0 \in [\pi/4, 1]\), which is within [0, 1].
Thus, the solution exhibits a singularity at some point in [0, 1]. This statement is true.
Unbounded in a subinterval of \([\pi/4, 1]\)
From the analysis above, the solution exists on \([0, x_0)\), where \(x_0 \in [\pi/4, 1]\) is the point of singularity. As \(x \to x_0^-\), the solution \(y(x)\) becomes unbounded.
If \(x_0 \in [\pi/4, 1]\), then the interval \([x_0 - \epsilon, x_0)\) for small \(\epsilon > 0\) (such that \(x_0 - \epsilon \ge \pi/4\)) is a subinterval of \([\pi/4, 1]\), or the interval \([\pi/4, x_0)\) is a subinterval of \([\pi/4, 1]\). The solution becomes unbounded as x approaches \(x_0\) within such subintervals.
Thus, the solution becomes unbounded in some subinterval of \(\left[ \frac{\pi}{4},1 \right]\). This statement is true.
All the given statements are true.
Important Questions from Initial Value Problem
Initial value problem, \(\rm x \frac{d y}{d x}=y\), y(0) = 0, x > 0
Consider the eigenvalue problem
((1 + x4)y')' + λy = 0, x ∈ (0, 1),
y(0) = 0, y(1) + 2y'(1) = 0.
Then which of the following statements are true?
Consider the following two initial value ODEs
(A) \(\frac{dx}{dt}=x^3,x(0)=1;\)
(B) \(\frac{dx}{dt}=x\sin x^2,x(0)=2.\)
Related to these ODEs, we make the following assertions.
I. The solution to (A) blows up in finite time.
II. The solution to (B) blows up in finite time.
Which of the following statements is true?
Let y0 > 0, z0 > 0 and α > 1.
Consider the following two differential equations:
\(\begin{aligned} &(*)\left\{\begin{array}{l} \frac{d y}{d t}=y^\alpha \quad \text { for } t>0, \\ y(0)=y_0 \end{array}\right. \\ &(* *)\left\{\begin{array}{l} \frac{d z}{d t}=-z^\alpha \quad \text { for } t>0, \\ z(0)=z_0 \end{array}\right. \end{aligned}\)
We say that the solution to a differential equation exists globally if it exists for all t > 0.Which of the following statements is true?
Let f ∶ ℝ2 → ℝ be a locally Lipschitz function. Consider the initial value problem
ẋ = f(t, x), x(t0) = x0
for (t0, x0) ∈ ℝ2. Suppose that J(t0, x0) represents the maximal interval of existence for the initial value problem. Which of the following statements is true?
