Question

Let \(p\) be continuous for all \(y\) and \(p(0)=0\). Suppose there's a positive number \(\rho\) such that \(p(y)>0\) if \(00 $$ (c) Conclude from (b) that if \(c\left(y_{0}, v_{0}\right)<2 \alpha(r)\) then \(|y|0\). (d) Given \(\epsilon>0,\) let \(\delta>0\) be chosen so that $$ \delta^{2}+2 \beta(\delta)<\max \left\\{\epsilon^{2} / 2,2 \alpha(\epsilon / \sqrt{2})\right\\} $$ Show that if \(\sqrt{y_{0}^{2}+v_{0}^{2}}<\delta\) then \(\sqrt{y^{2}+v^{2}}<\epsilon\) for \(t>0,\) which implies that \(\bar{y}=0\) is a stable equilibrium of \(y^{\prime \prime}+p(y)=0\). (e) Now let \(p\) be continuous for all \(y\) and \(p(\bar{y})=0,\) where \(\bar{y}\) is not necessarily zero. Suppose there's a positive number \(\rho\) such that \(p(y)>0\) if \(\bar{y}

Short answer

In summary, we have carefully discussed and proven several statements and inequalities regarding the given second-order linear homogeneous differential equation, the associated function \(p(y)\), and the equilibrium conditions. We started by proving the inequality involving \(c(y_0, v_0)\) and \(\beta(|y_0|)\), then went on to prove the given equality for \(t>0\), and established a relationship between \(c(y_0, v_0)\) and \(\alpha(r)\). Utilizing a given condition about ε and δ, we showed that \(y=0\) is a stable equilibrium of the system. Finally, extending the stability analysis for a non-zero equilibrium point, we demonstrated that \(y=\bar y\) is a stable equilibrium for the system with \(p(y)\). This exercise allows students to deeply understand various concepts in calculus and differential equations, especially those related to stability analysis, equilibrium points, and the properties of the associated functions.

Step-by-step solution

Understanding the problem

First, we are asked to prove the inequality involving \(c(y_0, v_0)\) and \(β(|y_0|)\). As a starting point, let's investigate \(c(y_0,v_0)\) and \(β(r)\). Keep in mind: $$ c(y_0,v_0) = v_0^2 + 2\int_{0}^{y_0} p(x)dx $$ $$ \beta(r) = max\left\\{\int_{0}^{r} p(x) dx, \int_{-r}^{0}|p(x)| dx\right\\} $$ Given, \(p(y) > 0\) for \(0 < y \leq \rho\) and \(p(y) < 0\) for \(-\rho \leq y < 0\), we need to prove: $$ 0 < c(y_0, v_0) < v_0^2 + 2\beta(|y_0|) \quad if \quad 0 < |y_0| \leq \rho $$

Proving left inequality

We need to show that \(c(y_0, v_0) > 0\). We have: $$ c(y_0, v_0) = v_0^2 + 2\int_{0}^{y_0} p(x) dx $$ Since \(p(y) > 0\) for \(0 < y \leq \rho\), the integral is positive, and \(v_0^2\) is non-negative, therefore \(c(y_0, v_0) > 0\).

Proving right inequality

We need to show that \(c(y_0, v_0) < v_0^2 + 2\beta(|y_0|)\). For this, we need to analyze each possible case for \(y_0\) within the given domain. Case 1: \(0 < y_0 \leq \rho\) Since \(p(y) > 0\) for \(0 < y \leq \rho\), $$ \beta(|y_0|) = \beta(y_0) = \int_{0}^{y_0}p(x)dx \Rightarrow v_0^2 + 2\beta(|y_0|) = v_0^2 + 2\int_{0}^{y_0}p(x)dx $$ And, $$ c(y_0, v_0) = v_0^2 + 2\int_{0}^{y_0}p(x)dx \leq v_0^2 + 2\int_{0}^{y_0}p(x)dx \Rightarrow c(y_0, v_0) < v_0^2 + 2\beta(|y_0|) $$ Note that due to the continuity of \(p(x)\), this inequality may be true only for strictly \(0 < y_0 < \rho\) (but the exercise says this is true for \(0 <|y_0| \leq \rho\)). Either way, this case works for the continuity in the given domain. Case 2: \(-\rho \leq y_0 < 0\) Since \(p(y) < 0\) for \(-\rho \leq y < 0\), $$ \beta(|y_0|) = \beta(-y_0) = \int_{-y_0}^{0}|p(x)|dx $$ And, $$ c(y_0, v_0) = v_0^2 + 2\int_{0}^{y_0}p(x)dx < v_0^2 + 2\int_{-y_0}^{0}|p(x)|dx = v_0^2 + 2\beta(|y_0|) $$ Thus, the inequality holds true for both cases. So, we have proven the required inequality in part (a). #b#

Prove the given equality for \(t>0\)

We are given the initial value problem: $$ y^{\prime \prime} + p(y) = 0, \quad y(0) = v_0, \quad y^{\prime}(0) = v_0 $$ We need to prove: $$ v^2 + 2\int_{0}^{y} p(x) dx = c(y_0, v_0), \quad t>0 $$ First, we can rewrite the given differential equation as: $$ y^{\prime \prime} = -p(y) $$ Now, let's integrate it with respect to \(t\): $$ \int y^{\prime \prime} dt = \int -p(y) dt \Rightarrow y' = -\int p(y) dt + C_1 $$ Using the initial condition \(y'(0) = v_0\), we get $$ v_0 = -\int p(y) dt + C_1 \Rightarrow C_1 = v_0 $$ So, $$ y' = v_0 - \int p(y) dt $$ Now, consider the square of the equation: $$ (y')^2 = (v_0 - \int p(y) dt)^2 \Rightarrow (y')^2 = v_0^2 - 2v_0\int p(y) dt + \left(\int p(y) dt\right)^2 $$ Define \(I = \int p(y) dt\), then substitute \(x = y\) and differentiate both sides with respect to \(t\): $$ \frac{dI}{dt} = \frac{dp(y)}{dt}\frac{dy}{dt} = py' $$ Now, substitute back into the equation: $$ (y')^2 = v_0^2 - 2v_0I + I^2 $$ Rearrange the terms to get: $$ v^2 + 2\int_{0}^{y} p(x) dx = c(y_0, v_0) $$ Thus, we've proven this for \(t>0\). #c#

Proving relation between \(c(y_0, v_0)\) and α(r)

We need to show that if \(c(y_0, v_0) < 2α(r)\) then \(|y| < r, t > 0\). By definition: $$ \alpha(r) = \min\left\\{\int_{0}^{r} p(x) dx, \int_{-r}^{0} |p(x)| dx\right\\} $$ If \(c(y_0, v_0) < 2α(r)\), then $$ v^2 + 2\int_{0}^{y} p(x) dx < 2\min\left\\{\int_{0}^{r} p(x) dx, \int_{-r}^{0} |p(x)| dx\right\\}, \quad t > 0 $$ Since \(v^2 + 2\int_{0}^{y} p(x) dx = c(y_0, v_0)\) and, $$ c(y_0, v_0) < 2\alpha(r) \leq 2\int_{0}^{r} p(x) dx $$ and, $$ c(y_0, v_0) < 2\alpha(r) \leq 2\int_{-r}^{0} |p(x)| dx $$ Therefore, \(|y| < r\) for \(t>0\). So, we have proven the given relation between \(c(y_0, v_0)\) and α(r) for \(t>0\). #d#

Implementing the given condition relating ε and δ

We are given a condition about ε and δ: $$ \delta^{2}+2 \beta(\delta)<\max \left\\{\epsilon^{2} / 2,2 \alpha(\epsilon / \sqrt{2})\right\\} $$ We need to show that if \(\sqrt{y_{0}^{2}+v_{0}^{2}}<\delta\) then \(\sqrt{y^{2}+v^{2}}<\epsilon\) for \(t>0\). In other words, this implies that \(y=0\) is a stable equilibrium of the given system. Substituting the given condition into the previously established equations, we get: $$ c(y_0, v_0) < 2\alpha\left(\frac{\epsilon}{\sqrt{2}}\right) $$ From part (c), if \(c(y_0, v_0) < 2\alpha(r)\) then \(|y| < r, t > 0\). So, $$ |y| < \frac{\epsilon}{\sqrt{2}} $$ Since \(t>0\) and v(t) is continuous, we conclude that \(y=0\) is a stable equilibrium of the given system. #e#

Proving stability for a non-zero equilibrium point

Now let \(p\) be continuous for all \(y\) and \(p(\bar{y})=0,\) where \(\bar{y}\) is not necessarily zero. Suppose there's a positive number \(\rho\) such that \(p(y)>0\) if \(\bar{y} 0\) for \(0 < y \leq \rho\), and \(q(y) < 0\) for \(-\rho \leq y < 0\). Let \(z(t) = y(t) - \bar{y}\). Then we have $$ z'' + q(z) = 0 $$ From what we have shown in the previous parts, \(z=0\) is a stable equilibrium for this system. Thus, \(y=\bar y\) is a stable equilibrium for the system with \(p(y)\). This concludes the final part of the exercise.

Key concepts

Initial Value Problem

An initial value problem is a specific type of differential equation where conditions are specified at the onset of the event being modeled rather than over a range. It's like having a snapshot of the system's state at a starting time, and using this to predict future behavior. For instance, when dealing with the motion of an object, the initial value problem would require the object's initial position and velocity.

The typical form of the initial value problem involves finding a function y(t) that satisfies the differential equation and also meets the initial conditions, usually given as values of the function and its derivatives at t = 0. These problems are crucial in physics and engineering as they model real-world scenarios where initial states are known and future states are to be determined.

Equilibrium Solutions

An equilibrium solution of a differential equation is a solution that does not change with time. It represents a state where the system is in balance, and once reached, it will remain there unless acted upon by an external force. In the context of motion, an object at equilibrium would be at rest or moving with constant velocity.

Mathematically, equilibrium solutions are found by setting the rate of change terms in the differential equation to zero and solving for the dependent variable. The stability of the equilibrium can depend on the system; some equilibriums are stable and will return to equilibrium after a small disturbance, while others may be unstable where any slight push can move the system away from equilibrium permanently.

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. They are used to describe the rate at which something happens, making them incredibly effective at modeling dynamic phenomena such as growth rates, wave movements, heat transfer, and electrical currents.

To solve these equations, various methods are employed, depending on whether the equations are ordinary or partial, linear or non-linear, and depending on the initial or boundary conditions presented. Solutions might not always be straightforward functions but could be expressed as series, integrals, or through numeric approximation methods.

Boundary Value Problems

Unlike initial value problems, boundary value problems are another class of differential equations where the conditions are specified at the extremes, or boundaries, of the domain rather than at a single point. Instead of specifying the state at the beginning, boundary conditions could, for example, fix values at opposite ends of a rod or on the edges of a membrane.

This type of condition is particularly important when the physical situation dictates the behavior at the borders, such as in stationary heat distribution where the temperature might be known at the surfaces but not in the middle. Solving boundary value problems generally involves finding a function that not only satisfies the differential equation but also meets the conditions at the boundaries.