Question

Consider the initial value problem $$ \frac{d}{d t}\left[\begin{array}{l} y_{1} \\ y_{2} \end{array}\right]=\left[\begin{array}{c} \frac{5}{4} y_{1}^{1 / 5}+y_{2}^{2} \\ 3 y_{1} y_{2} \end{array}\right], \quad\left[\begin{array}{l} y_{1}(0) \\ y_{2}(0) \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \end{array}\right] $$ For the given autonomous system, the two functions \(f_{1}\left(y_{1}, y_{2}\right)=\frac{5}{4} y_{1}^{1 / 5}+y_{2}^{2}\) and \(f_{2}\left(y_{1}, y_{2}\right)=3 y_{1} y_{2}\) are continuous functions for all \(\left(y_{1}, y_{2}\right)\). (a) Show by direct substitution that $$ y_{1}(t)=\left\\{\begin{array}{lr} 0, & -\infty

Short answer

In summary, the given solution of the two-dimensional autonomous system of ordinary differential equations satisfies both the initial conditions and the differential equations for any positive constant c. However, this solution is not unique because c can take any arbitrary positive value, which does not contradict Theorem 6.1. The exception in this case is due to the lack of continuity of the partial derivative of the function with respect to y1 at y1 = 0. Thus, Theorem 6.1 is still valid, and this example serves as a specific exception.

Step-by-step solution

Check if the given solution satisfies the initial conditions

To show that the given functions \(y_1(t)\) and \(y_2(t)\) satisfy the initial conditions, we need to evaluate them at the given initial values: $$ y_1(0)=0,\quad y_2(0)=0 $$ Let's check for both \(t\leq c\) and \(c<t<\infty\) cases: $$ y_1(t) = \begin{cases} 0, & -\infty<t \leq c, \\ (t-c)^{5/4}, & c<t<\infty, \end{cases} \quad y_2(t) = 0 $$ At \(t = 0\), we have \(y_1(0) = 0\) and \(y_2(0) = 0\), which satisfy the initial conditions.

Check if the given solution satisfies the differential equations

Now, we need to verify that the given functions \(y_1(t)\) and \(y_2(t)\) satisfy the differential equations: $$ \frac{d}{dt} \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} = \begin{bmatrix} \frac{5}{4} y_1^{1/5} + y_2^2 \\ 3 y_1 y_2 \end{bmatrix} $$ Firstly, compute the derivatives of the given functions \(y_1(t)\) and \(y_2(t)\) with respect to \(t\): $$ \frac{d y_1}{dt} = \begin{cases} 0, & -\infty<t \leq c, \\ \frac{5}{4}(t-c)^{1/4}, & c<t<\infty, \end{cases} \quad \frac{d y_2}{dt} = 0 $$ Now, substitute the given functions and their derivatives into the differential equations separately for \(t\leq c\) and \(c<t<\infty\): For \(-\infty<t \leq c\): $$ \begin{bmatrix} 0 \\ 0 \end{bmatrix} = \begin{bmatrix} \frac{5}{4} 0^{1/5} + 0^2 \\ 3\cdot 0\cdot 0 \end{bmatrix} \Rightarrow \begin{bmatrix} 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$ For \(c<t<\infty\): $$ \begin{bmatrix} \frac{5}{4}(t-c)^{1/4} \\ 0 \end{bmatrix} = \begin{bmatrix} \frac{5}{4} [(t-c)^{5/4}]^{1/5} + 0^2 \\ 3\cdot (t-c)^{5/4}\cdot 0 \end{bmatrix} \Rightarrow \begin{bmatrix} \frac{5}{4}(t-c)^{1/4} \\ 0 \end{bmatrix} = \begin{bmatrix} \frac{5}{4}(t-c)^{1/4} \\ 0 \end{bmatrix} $$ In both cases, the given solution satisfies the differential equations.

Discuss the uniqueness of the solution and Theorem 6.1

The given solution is clearly not unique since the positive constant \(c\) can be any arbitrary value. However, this does not contradict Theorem 6.1, as the theorem states that the solution is unique as long as \(\mathbf{F}(\mathbf{y})\) (in our case \(\begin{bmatrix} \frac{5}{4} y_1^{1/5} + y_2^2 \\ 3 y_1 y_2 \end{bmatrix}\)) is continuous and its partial derivatives are continuous in some open set containing the initial point \((t_0, \mathbf{y}_0)\). In this particular case, \(\mathbf{F}(\mathbf{y})\) is continuous for all \((y_1, y_2)\). However, the partial derivative \(\frac{\partial}{\partial y_1} (y_1^{1/5})\) does not exist when \(y_1 = 0\), making this an exception to Theorem 6.1. Hence, we don't have a contradiction with Theorem 6.1.

Key concepts

Autonomous System

An autonomous system in the context of differential equations is a set of equations where the rate of change depends solely on the variables themselves rather than explicitly on an independent variable like time. In simpler terms, the system equations describe how each state variable changes based on the current state itself, independent of external influences as time changes.

This concept is particularly useful in modeling real-world situations where the future state of a system depends only on its current state. Its advantage lies in the simplification it introduces; only the current state impacts future behavior, making the equations seem 'autonomous' or self-contained.

In this initial value problem, you are given a two-variable autonomous system where the change in each variable over time is exclusively dependent on itself and the other variable's current state. This helps in analyzing and solving the system without any direct time dependency.

Differential Equations

Differential equations play a crucial role in modeling various natural phenomena and describe how a particular quantity changes over time. They come in different forms, such as ordinary differential equations (ODEs) where changes depend on a single independent variable, usually time. In this initial value problem, we deal with a system of ODEs.

A unique feature of differential equations is that they require both a function and its derivatives to define relationships between quantities. In simpler terms, they are equations involving a variable and its rate of change. The order of a differential equation indicates the highest derivative in the equation. Our equations \(rac{d}{dt}\begin{bmatrix} y_1(t) \ y_2(t) \end{bmatrix} = \begin{bmatrix} \frac{5}{4} y_1^{1/5} + y_2^2 \ 3 y_1 y_2 \end{bmatrix}\) have various complexities due to the presence of non-linear terms which often make analytical solutions challenging.

Solving these involves techniques that integrate functions into specific forms, leveraging initial conditions \(y_1(0) = 0, y_2(0) = 0\) to find unique or general solutions to predict system behavior over time.

Uniqueness of Solutions

In the world of differential equations, the uniqueness of solutions is a fascinating topic. It deals with whether a given initial value problem has one and only one solution that satisfies the conditions provided. Many times, it helps ensure that predictions made by mathematical models remain consistent and reliable.

For an initial value problem like the one mentioned, the existence and uniqueness of solutions are often dictated by specific theorems, such as the Picard-Lindelöf theorem. This theorem implies that if all functions involved and their partial derivatives are continuous, then a solution exists and is unique in a neighborhood of that initial point.

However, the uniqueness may lag if conditions of those theorems aren't fully met, like in our example, where the solution isn't unique due to discontinuities at \(y_1 = 0\). Here, differentiability isn't met due to non-existence of certain derivatives, thus multiple solutions can satisfy the same initial conditions.

Theorem 6.1 in Differential Equations

Theorem 6.1 in the context of differential equations typically addresses the existence and uniqueness of solutions under certain conditions. It asserts that a solution is unique if a set of criteria are satisfied, primarily focusing on the continuity of the functions and their partial derivatives.

The essence of Theorem 6.1 rests on ensuring that the rate of change in the system is smoothly predictable from the start, relying on initial conditions. However, if a function or its derivatives lack continuity at the initial point, uniqueness might not hold, allowing multiple trajectories or solutions that fulfill the same initial values.

In the given problem, although the function governing changes is continuous, the broken differentiability at \(y_1 = 0\) creates an exception. This illustrates a vital aspect of applied mathematics where even small discontinuities can lead to multiple valid solutions, thus showcasing a scenario where uniqueness of solutions doesn't apply despite adhering to Theorem 6.1.