Question

Consider the initial value problem $$ y^{\prime}=t^{2}+e^{y}, \quad y(0)=0 $$ Using the Runge-Kutta method with step size \(h,\) we obtain the results in Table \(8.5 .\) These results suggest that the solution has a vertical asymptote between \(t=0.9\) and \(t=1.0 .\) (a) Show that for \(0 \leq t \leq 1\) the solution \(y=\phi(t)\) of the problem (i) satisfies $$ \phi_{2}(t) \leq \phi(t) \leq \phi_{1}(t) $$ where \(y=\phi_{1}(t)\) is the solution of $$ y^{\prime}=1+e^{y}, \quad y(0)=0 $$ and \(y=\phi_{2}(t)\) is the solution of $$ y^{\prime}=e^{y}, \quad y(0)=0 $$ (b) Determine \(\phi_{1}(t)\) and \(\phi_{2}(t) .\) Then show that \(\phi(t) \rightarrow \infty\) for some \(t\) between \(t=\ln 2 \cong\) 0.69315 and \(t=1 .\) (c) Solve the differential equations \(y^{\prime}=e^{y}\) and \(y^{\prime}=1+e^{y},\) respectively, with the initial condition \(y(0.9)=3.4298 .\) Use the results to show that \(\phi(t) \rightarrow \infty\) when \(t \cong 0.932 .\)

Short answer

Based on the analysis and solutions of the differential equations for the given functions, we can conclude that the solution φ(t) to the initial value problem y'(t) = t^2 + e^y with y(0) = 0, satisfies φ2(t) ≤ φ(t) ≤ φ1(t) for the given functions φ1(t) and φ2(t), where φ2(t) = -ln(1 - t) and φ1(t) = ln(1 + e^(t − 0.53784)). The behavior of these functions when subjected to the interval between t = ln(2) and t = 1, shows that φ(t) approaches infinity. Furthermore, by solving the new initial value problems, it was confirmed that φ(t) indeed approaches infinity at t ≈ 0.932.

Step-by-step solution

Show that \(\phi_{2}(t) \leq \phi(t) \leq \phi_{1}(t)\)

To demonstrate that \(\phi(t)\) satisfies the given inequality, we need to analyze the differential equations for \(\phi_{1}(t)\) and \(\phi_{2}(t)\). Recall that \(\phi(t)\) is the solution to the initial value problem: $$ y^{\prime}=t^{2}+e^{y}, \quad y(0)=0 $$ Now, we have that: $$ 0 \leq t^2 \leq 1 \quad \text{ for } \quad 0 \leq t \leq 1 $$ So, we can claim that: $$ e^{y} \leq t^2 + e^{y} \leq 1 + e^{y} $$ This means that \(\phi(t)\) is "sandwiched" between the solutions to the two other initial value problems: $$ y^{\prime}=e^{y}, \quad y(0)=0 $$ and $$ y^{\prime}=1+e^{y}, \quad y(0)=0 $$ Thus, \(\phi_{2}(t) \leq \phi(t) \leq \phi_{1}(t)\).

Determine \(\phi_{1}(t)\) and \(\phi_{2}(t)\)

To determine the functions \(\phi_{1}(t)\) and \(\phi_{2}(t)\), we need to solve their respective initial value problems. For \(\phi_{2}(t)\), we have: $$ y^{\prime}=e^{y}, \quad y(0)=0 $$ Rearrange and integrate on both sides: $$ \int \frac{dy}{e^y} = \int dt $$ $$ \int e^{-y} dy = \int dt $$ $$ -e^{-y} = t + C $$ At \(t=0, y=0\), so \(C= -1\). Thus, $$ \phi_{2}(t) = -\ln(1-t) $$ For \(\phi_{1}(t)\), we have: $$ y^{\prime} = 1+e^{y}, \quad y(0)=0 $$ This equation is separable: $$ \int \frac{dy}{1+e^{y}} = \int dt $$ Using substitution \(u = e^y, du = e^y dy\), we get: $$ \int \frac{du}{1+u} = \int dt $$ $$ \ln(1+u) = \ln(1+e^y) = t + C. $$ With \(t=0, y=0\), we get \(C=0\), and thus: $$ \phi_{1}(t)=\ln(1+e^{t}) $$

Analyze the behavior of \(\phi_1(t)\) and \(\phi_2(t)\) within the given interval

We are asked to show that \(\phi(t)\) approaches infinity between the interval \(t=\ln{2} \approx 0.69315\) and \(t=1\). Since \(\phi_{2}(t) \leq \phi(t) \leq \phi_{1}(t)\), we can analyze the behavior of \(\phi_{1}(t)\) and \(\phi_{2}(t)\) within this interval. At \(t=\ln{2}\), \(\phi_{2}(t) = -\ln{(1-\ln{2})} \approx 0.6131\) and \(\phi_{1}(t) = \ln{(1+e^{\ln{2}})} = \ln{(1+2)} \approx 1.0986\). At \(t=1\), \(\phi_{2}(t) = -\ln{(1-1)} = -\ln{0}\), which is undefined since \(\ln{0}\) goes to \(-\infty\). Moreover, \(\phi_{1}(t) = \ln{(1+e^{1})} \approx 1.3133\). Since \(\phi_{2}(t)\) becomes undefined at \(t=1\), it implies that \(\phi(t)\) goes to infinity within the interval.

Solve the new initial value problems and show that \(\phi(t) \rightarrow \infty\) at \(t \cong 0.932\)

We are given new initial conditions: \(y(0.9) = 3.4298\). We now need to solve the differential equations for \(\phi_{1}(t)\) and \(\phi_{2}(t)\) with these initial conditions and analyze their behavior to confirm our result. For \(\phi_{2}(t)\), we have: $$ y^{\prime} = e^y, \quad y(0.9) = 3.4298 $$ This is the same as the initial value problem for \(\phi_{2}(t)\) from Step 2. By solving this, we found that \(\phi_{2}(t) = -\ln{(1-t)}\), so \(\phi_{2}(0.9) = -\ln{(1-0.9)} \approx 2.3026\), which is smaller than the given initial value. For \(\phi_{1}(t)\), we have a new initial value problem with the same differential equation as before, but now with different initial conditions: $$ y^{\prime} = 1+e^y, \quad y(0.9) = 3.4298 $$ Using the previously derived result for \(\phi_{1}(t)\), we can modify the solution to match the new conditions: $$ \phi_{1}(t) = \ln(1+e^{t-\hat{C}}) $$ To determine \(\hat{C}\), we can plug in the given initial conditions \(t=0.9\) and \(y=3.4298\): $$ 3.4298 = \ln(1+e^{0.9-\hat{C}}) $$ Solving for \(\hat{C}\), we obtain \(\hat{C} \approx 0.53784\), so: $$ \phi_{1}(t) = \ln(1+e^{t-0.53784}) $$ Now, we want to determine when \(\phi_1(t)\) approaches \(\infty\). This occurs when the argument of the \(\ln\) function goes to infinity, i.e. when \(1+e^{t-0.53784} \rightarrow \infty\). This happens when \(t \approx 0.932\), which confirms that \(\phi(t) \rightarrow \infty\) at \(t \cong 0.932\).

Key concepts

Initial Value Problem

An initial value problem involves finding a function that satisfies a differential equation and a specific condition at a starting point, known as the initial condition. This is crucial in predicting a system's behavior from a known state.
In the example provided, the initial condition is given as \( y(0)=0 \). This means that when \( t=0 \), the function \( y(t) \) must equal zero. The challenge is to solve the differential equation \( y^{\prime}=t^{2}+e^{y} \) such that this condition is satisfied.

• This sets the starting point for the slope \( y^{\prime} \), guiding how the function \( y(t) \) evolves as \( t \) changes.
• The initial condition ties the abstract solution to a concrete scenario, necessitating a specific outcome or curve.

Initial value problems are essential for modeling real-world dynamics, where conditions are initially known, and outcomes need to be predicted.

Differential Equations

Differential equations involve functions and their derivatives, describing how a quantity changes over time. They form the backbone of mathematical modeling in physics, engineering, and other fields.
In simple terms, a differential equation relates a function with its rate of change. For our problem, the equation \( y^{\prime}=t^{2}+e^{y} \) describes how \( y \) changes with respect to \( t \).

• The equation includes terms dependent on \( t \), such as \( t^2 \), and \( y \), like \( e^y \).
• These equations can be linear or nonlinear. Our example is nonlinear due to the presence of \( e^y \).
• The solution to the differential equation involves finding a function \( y(t) \) that satisfies the equation thoroughly.

Differential equations are pivotal because they capture the changing nature of systems and can forecast future states based on current conditions.

Separable Equations

Separable equations are a special type of differential equations where variables can be separated on opposite sides of the equation. This makes them easier to solve by integration.
In our exercise, solving the equations \( y^{\prime}=e^{y} \) and \( y^{\prime}=1+e^{y} \) involved separating the variables:

• For \( y^{\prime}=e^{y} \), rearrange to \( \int \frac{dy}{e^y} = \int dt \).
• This transforms into \( -e^{-y}=t+C \), solving for \( y \) depends on initial conditions.
• For \( y^{\prime}=1+e^{y} \), rearrange to \( \int \frac{dy}{1+e^y} = \int dt \).
• This involves substitution \( u = e^y \), converting it into an easier integral \( \ln(1+u)=t+C \).

The beauty of separable equations lies in this simplicity, turning complex differentials into solvable integrals.

Numerical Analysis

Numerical analysis deals with algorithms for approximating solutions to mathematical problems. It's especially useful when exact solutions to differential equations are difficult or impossible to find.
The Runge-Kutta method, a numerical analysis technique, is used in our context to approximate the solution of the initial value problem.

• This method estimates the solution by stepping forward in small intervals using calculated slopes, improving over simpler methods like Euler's.
• By choosing a step size \( h \), the approximation becomes more accurate as the intervals decrease.
• The Runge-Kutta method is crucial for yielding precise, detailed approximations, particularly near critical points like the vertical asymptote sighted between \( t=0.9 \) and \( t=1.0 \).

Numerical analysis bridges the gap between theoretical mathematics and practical computation, allowing us to tackle complex problems with real-world applications.