Initial Value Problems Analysis

We are given two initial value problems (IVPs) and asked to determine the nature of their solutions (unique or infinitely many) on the interval $[0, \infty)$. Let's analyze each problem separately.

Problem (P): \( \rm x^{\prime}(t)=\sqrt{x(t)}, t>0, x(0)=0 \)

This is a first-order ordinary differential equation with an initial condition. The equation is of the form \(x'(t) = f(x(t))\), where \(f(x) = \sqrt{x}\). The initial condition is \(x(0)=0\).

To analyze the uniqueness of solutions for such an IVP, we often use the Picard-Lindelöf theorem. This theorem guarantees a unique solution in a neighborhood of the initial point if the function \(f(x)\) is continuous and Lipschitz continuous with respect to \(x\) in that neighborhood.

• The function \(f(x) = \sqrt{x}\) is continuous for \(x \ge 0\). Since the initial condition is \(x(0)=0\), we need to check continuity near \(x=0\). It is continuous at \(x=0\).
• Now let's check for Lipschitz continuity at \(x=0\). A function \(f\) is Lipschitz continuous at \(x_0\) if there exists a constant \(L > 0\) such that \(|f(x) - f(x_0)| \le L|x - x_0|\) for all \(x\) in some neighborhood of \(x_0\). At \(x_0=0\), this condition becomes \(|\sqrt{x} - \sqrt{0}| \le L|x - 0|\) for \(x\) near 0, i.e., \(\sqrt{x} \le L|x|\). For \(x > 0\) near 0, this is \(\sqrt{x} \le Lx\). Dividing by \(x\) (assuming \(x>0\)), we get \(\frac{\sqrt{x}}{x} \le L\), which simplifies to \(\frac{1}{\sqrt{x}} \le L\). As \(x \to 0^+\), \(\frac{1}{\sqrt{x}}\) approaches infinity. Thus, no such finite constant \(L\) exists. The function \(f(x)=\sqrt{x}\) is continuous but not Lipschitz continuous at \(x=0\).

Since the Lipschitz condition is not met at the initial point \(x=0\), the Picard-Lindelöf theorem does not guarantee uniqueness. Let's look for solutions.

• One obvious solution is the trivial solution \(x(t) = 0\) for all \(t \ge 0\). We can check this: \(x'(t) = 0\), and \(\sqrt{x(t)} = \sqrt{0} = 0\). So \(x'(t) = \sqrt{x(t)}\) holds. The initial condition \(x(0)=0\) is also satisfied.
• Let's look for non-trivial solutions. For \(x(t) > 0\), we can separate variables: \[ \frac{dx}{\sqrt{x}} = dt \] Integrating both sides: \[ \int \frac{1}{\sqrt{x}} dx = \int dt \] \[ 2\sqrt{x} = t + C \] Using the initial condition \(x(0)=0\), we get \(2\sqrt{0} = 0 + C\), which implies \(C=0\). So, \(2\sqrt{x} = t\), which gives \(\sqrt{x} = \frac{t}{2}\). Squaring both sides, we get \(x(t) = \left(\frac{t}{2}\right)^2 = \frac{t^2}{4}\). This solution is valid for \(t \ge 0\). Let's check: \(x'(t) = \frac{2t}{4} = \frac{t}{2}\), and \(\sqrt{x(t)} = \sqrt{\frac{t^2}{4}} = \frac{|t|}{2}\). Since we consider \(t \ge 0\), \(\sqrt{x(t)} = \frac{t}{2}\). So \(x'(t) = \sqrt{x(t)}\) holds for \(t \ge 0\), and \(x(0) = 0^2/4 = 0\). This is another solution.
• We can construct even more solutions by combining the trivial solution with the non-trivial one. Consider a solution that stays at 0 for some time \(t_0 \ge 0\) and then follows the quadratic form. Let's define a family of functions: \[ x_{t_0}(t) = \begin{cases} 0 & 0 \le t \le t_0 \\ \frac{1}{4}(t-t_0)^2 & t > t_0 \end{cases} \] For \(t_0=0\), this gives \(x_0(t) = \frac{1}{4}t^2\) for \(t > 0\) and \(x_0(0)=0\), which is just \(x(t)=\frac{t^2}{4}\) for \(t \ge 0\). For \(t_0 > 0\), let's check if \(x_{t_0}(t)\) is a solution.
  • If \(0 \le t < t_0\), \(x_{t_0}(t)=0\), so \(x'_{t_0}(t)=0\) and \(\sqrt{x_{t_0}(t)}=\sqrt{0}=0\). The ODE holds.
  • If \(t > t_0\), \(x_{t_0}(t)=\frac{1}{4}(t-t_0)^2\), so \(x'_{t_0}(t) = \frac{1}{4} \cdot 2(t-t_0) = \frac{1}{2}(t-t_0)\). Also, \(\sqrt{x_{t_0}(t)} = \sqrt{\frac{1}{4}(t-t_0)^2} = \sqrt{\frac{1}{4}}|t-t_0| = \frac{1}{2}(t-t_0)\) since \(t > t_0\). The ODE holds.
  • At \(t=t_0\), \(x_{t_0}(t_0)=0\). The limits of \(x'_{t_0}(t)\) from the left and right are \(\lim_{t \to t_0^-} 0 = 0\) and \(\lim_{t \to t_0^+} \frac{1}{2}(t-t_0) = 0\). So \(x'_{t_0}(t_0)=0\). Also \(\sqrt{x_{t_0}(t_0)} = \sqrt{0} = 0\). The ODE holds at \(t=t_0\).
  • The initial condition \(x_{t_0}(0)=0\) is satisfied for any \(t_0 \ge 0\).
  Since \(t_0\) can be any non-negative real number, this construction gives infinitely many distinct solutions for problem (P) in $[0, \infty)$. This includes the solution \(x(t)=0\) (corresponding to \(t_0 \to \infty\)).

Problem (Q): \( \rm y^{\prime}(t)=-\sqrt{y(t)}, t>0, y(0)=0 \)

This is another first-order ordinary differential equation with an initial condition. The equation is of the form \(y'(t) = g(y(t))\), where \(g(y) = -\sqrt{y}\). The initial condition is \(y(0)=0\).

• The function \(g(y) = -\sqrt{y}\) is continuous for \(y \ge 0\). At the initial point \(y=0\), it is continuous.
• Now let's check for Lipschitz continuity at \(y=0\). We need \(|g(y) - g(0)| \le L|y - 0|\) for \(y\) near 0. This becomes \(|-\sqrt{y} - (-\sqrt{0})| \le L|y|\) for \(y\) near 0, i.e., \(\sqrt{y} \le L|y|\). For \(y > 0\) near 0, this is \(\sqrt{y} \le Ly\), which means \(\frac{1}{\sqrt{y}} \le L\). As \(y \to 0^+\), \(\frac{1}{\sqrt{y}}\) approaches infinity. Thus, \(g(y)=-\sqrt{y}\) is continuous but not Lipschitz continuous at \(y=0\).

Again, Picard-Lindelöf does not guarantee uniqueness. Let's look for solutions.

• The trivial solution \(y(t) = 0\) for all \(t \ge 0\) is a solution: \(y'(t)=0\), \(-\sqrt{y(t)} = -\sqrt{0}=0\), so \(y'(t) = -\sqrt{y(t)}\) holds. The initial condition \(y(0)=0\) is satisfied.
• Let's look for non-trivial solutions for \(t > 0\). For \(\sqrt{y(t)}\) to be defined, we must have \(y(t) \ge 0\). The ODE is \(y'(t) = -\sqrt{y(t)}\). Since \(\sqrt{y(t)} \ge 0\), \(-\sqrt{y(t)} \le 0\). Thus, \(y'(t) \le 0\), which means \(y(t)\) must be a non-increasing function.
• Given the initial condition \(y(0)=0\), and the requirement that \(y(t) \ge 0\) for all \(t \ge 0\) (because \(\sqrt{y(t)}\) is in the equation), the only non-increasing function that starts at 0 and stays non-negative is \(y(t) = 0\) for all \(t \ge 0\).
• Suppose there was some \(t_1 > 0\) such that \(y(t_1) > 0\). Since \(y(t)\) is non-increasing and \(y(0)=0\), this is not possible. A non-increasing function starting at 0 can only be positive if it was positive at an earlier point, which contradicts the starting point being 0. More formally, if \(y(t_1) > 0\) for some \(t_1 > 0\), then since \(y(t)\) is non-increasing, \(y(t) \ge y(t_1) > 0\) for all \(t \in [0, t_1]\). But this contradicts \(y(0)=0\). Therefore, \(y(t)\) must be 0 for all \(t \ge 0\).

So, for problem (Q), the only solution is \(y(t)=0\) for all \(t \ge 0\). This means problem (Q) has a unique solution.

Summary of Findings

• Problem (P) \(x'(t) = \sqrt{x(t)}\), \(x(0)=0\) has infinitely many solutions in $[0, \infty)$.
• Problem (Q) \(y'(t) = -\sqrt{y(t)}\), \(y(0)=0\) has a unique solution in $[0, \infty)$.

Let's review the statements:

1. (P) has a unique solution in [0, ∞). (False)
2. (Q) has a unique solution in [0, ∞). (True)
3. (P) has infinitely many solutions in [0, ∞). (True)
4. (Q) has infinitely many solutions in [0, ∞). (False)

Based on our analysis, statements 2 and 3 are true.