Let y0 0, z0 0 and α 1.Consider the following two differential equations:$\begin{aligned} &(*)\left\{\begin{array}{l} \frac{d y}{d t}=y^\alpha \quad \text { for } t0, \ y(0)=y_0 \end{array}\right. \ &(* *)\left\{\begin{array}{l} \frac{d z}{d t}=-z^\alpha \quad \text { for } t0, \ z(0)=z_0 \end{array}\right. \end{aligned}$ 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?

Question

Let y0 > 0, z0 > 0 and &alpha; > 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?

Prepp Team · Accepted Answer

Differential Equations Analysis

We are given two initial value problems (IVPs) for differential equations:

Equation (*): $\frac{dy}{dt} = y^\alpha$ with $y(0) = y_0$, where $y_0 > 0$ and $\alpha > 1$.
Equation (**): $\frac{dz}{dt} = -z^\alpha$ with $z(0) = z_0$, where $z_0 > 0$ and $\alpha > 1$.

We need to determine if the solutions to these differential equations exist globally for all $t > 0$ or if they exhibit finite-time blow-up (reaching infinity at a finite time T).

Equation (*) Solution Existence

The first differential equation is $\frac{dy}{dt} = y^\alpha$. This is a separable ordinary differential equation (ODE). We can separate the variables:

$\frac{dy}{y^\alpha} = dt$

Integrating both sides:

$\int y^{-\alpha} dy = \int dt$

For $\alpha > 1$, the integral of $y^{-\alpha}$ is $\frac{y^{-\alpha+1}}{-\alpha+1}$ or $\frac{y^{1-\alpha}}{1-\alpha}$. So,

$\frac{y^{1-\alpha}}{1-\alpha} = t + C$

Using the initial condition $y(0) = y_0$, we find the constant $C$:

$\frac{y_0^{1-\alpha}}{1-\alpha} = 0 + C \implies C = \frac{y_0^{1-\alpha}}{1-\alpha}$

Substitute C back into the solution:

$\frac{y^{1-\alpha}}{1-\alpha} = t + \frac{y_0^{1-\alpha}}{1-\alpha}$

Multiply by $(1-\alpha)$:

$y^{1-\alpha} = (1-\alpha)t + y_0^{1-\alpha}$

Since $\alpha > 1$, $1-\alpha$ is negative. Let $\beta = \alpha-1 > 0$. Then $1-\alpha = -\beta$. So,

$y^{-\beta} = -\beta t + y_0^{-\beta}$

$\frac{1}{y^{\beta}} = \frac{- \beta t y_0^{\beta} + 1}{y_0^{\beta}}$

$y^{\beta} = \frac{y_0^{\beta}}{1 - \beta t y_0^{\beta}}$

$y(t) = \left( \frac{y_0^{\beta}}{1 - \beta t y_0^{\beta}} \right)^{1/\beta} = \left( \frac{y_0^{\alpha-1}}{1 - (\alpha-1)t y_0^{\alpha-1}} \right)^{\frac{1}{\alpha-1}}$

The solution $y(t)$ exists as long as the denominator $1 - (\alpha-1)t y_0^{\alpha-1}$ is positive. Since $y_0 > 0$ and $\alpha > 1$, $(\alpha-1)y_0^{\alpha-1} > 0$.

The denominator becomes zero when $1 - (\alpha-1)t y_0^{\alpha-1} = 0$, which occurs at $t = \frac{1}{(\alpha-1)y_0^{\alpha-1}}$. Let $T = \frac{1}{(\alpha-1)y_0^{\alpha-1}}$. Since $y_0 > 0$ and $\alpha > 1$, $T$ is a finite positive value.

As $t \rightarrow T^-$, the denominator $1 - (\alpha-1)t y_0^{\alpha-1} \rightarrow 0^+$.

Therefore, $\lim_{t \rightarrow T^-} y(t) = \lim_{t \rightarrow T^-} \left( \frac{y_0^{\alpha-1}}{1 - (\alpha-1)t y_0^{\alpha-1}} \right)^{\frac{1}{\alpha-1}} = \left( \frac{y_0^{\alpha-1}}{0^+} \right)^{\frac{1}{\alpha-1}} = (+\infty)^{\frac{1}{\alpha-1}} = +\infty$.

This shows that the solution to equation (*) exhibits finite-time blow-up at $t = T$. Thus, it does not have a global solution for all $t > 0$.

Equation (**) Solution Existence

The second differential equation is $\frac{dz}{dt} = -z^\alpha$. This is also a separable ODE. Separating variables:

$\frac{dz}{-z^\alpha} = dt$

Integrating both sides:

$\int -z^{-\alpha} dz = \int dt$

For $\alpha > 1$, the integral of $-z^{-\alpha}$ is $\frac{-z^{-\alpha+1}}{-\alpha+1}$ or $\frac{z^{1-\alpha}}{\alpha-1}$. So,

$\frac{z^{1-\alpha}}{\alpha-1} = t + C$

Using the initial condition $z(0) = z_0$, we find the constant $C$:

$\frac{z_0^{1-\alpha}}{\alpha-1} = 0 + C \implies C = \frac{z_0^{1-\alpha}}{\alpha-1}$

Substitute C back into the solution:

$\frac{z^{1-\alpha}}{\alpha-1} = t + \frac{z_0^{1-\alpha}}{\alpha-1}$

Multiply by $(\alpha-1)$:

$z^{1-\alpha} = (\alpha-1)t + z_0^{1-\alpha}$

Since $\alpha > 1$, $1-\alpha$ is negative. Let $\beta = \alpha-1 > 0$. Then $1-\alpha = -\beta$. So,

$z^{-\beta} = \beta t + z_0^{-\beta}$

$\frac{1}{z^{\beta}} = \frac{\beta t z_0^{\beta} + 1}{z_0^{\beta}}$

$z^{\beta} = \frac{z_0^{\beta}}{1 + \beta t z_0^{\beta}}$

$z(t) = \left( \frac{z_0^{\beta}}{1 + \beta t z_0^{\beta}} \right)^{1/\beta} = \left( \frac{z_0^{\alpha-1}}{1 + (\alpha-1)t z_0^{\alpha-1}} \right)^{\frac{1}{\alpha-1}}$

The solution $z(t)$ exists as long as the denominator $1 + (\alpha-1)t z_0^{\alpha-1}$ is positive. Since $y_0 > 0$, $\alpha > 1$, and $t \ge 0$, the term $(\alpha-1)t z_0^{\alpha-1}$ is always non-negative. Thus, the denominator $1 + (\alpha-1)t z_0^{\alpha-1}$ is always greater than or equal to 1 for $t \ge 0$.

This means the solution $z(t)$ is well-defined for all $t \ge 0$.

As $t \rightarrow \infty$, the denominator $1 + (\alpha-1)t z_0^{\alpha-1} \rightarrow \infty$.

Therefore, $\lim_{t \rightarrow \infty} z(t) = \lim_{t \rightarrow \infty} \left( \frac{z_0^{\alpha-1}}{1 + (\alpha-1)t z_0^{\alpha-1}} \right)^{\frac{1}{\alpha-1}} = \left( \frac{z_0^{\alpha-1}}{\infty} \right)^{\frac{1}{\alpha-1}} = (0)^{\frac{1}{\alpha-1}} = 0$ (since $\alpha-1 > 0$).

The solution to equation (**) exists globally for all $t > 0$ and decays to 0 as $t \rightarrow \infty$.

Summary of Findings

Equation (*), $\frac{dy}{dt} = y^\alpha$, has a solution that blows up in finite time $T = \frac{1}{(\alpha-1)y_0^{\alpha-1}} < \infty$. It does not have a global solution.
Equation (**), $\frac{dz}{dt} = -z^\alpha$, has a solution that exists globally for all $t > 0$.

Evaluating the Options

Based on our analysis, let's look at the given options:

Both (*) and (**) have global solutions. This is false, as (*) does not have a global solution.
None of (*) and (**) have global solutions. This is false, as (**) has a global solution.
There exists a global solution for (*) and there exists a T < $\infty$ such that $\lim _{t \rightarrow T}|z(t)|=+∞$. This is false, as (*) does not have a global solution, and $z(t)$ decays to 0, not blowing up.
There exists a global solution for (**) and there exists a T < $\infty$ such that $\lim _{t \rightarrow T}|y(t)|=+∞$. This statement is true. We found that (**) has a global solution, and (*) has a finite-time blow-up at $T = \frac{1}{(\alpha-1)y_0^{\alpha-1}} < \infty$, where $\lim_{t \rightarrow T}|y(t)|=+∞$.

Therefore, the statement in option 4 is the true statement.