Question

Show that the functions \(f(t)=t^{2}|t|\) and \(g(t)=t^{3}\) are linearly dependent on \(0< t< 1\) and on \(-1< t<0,\) but are linearly independent on \(-1< t< 1 .\) Although \(f\) and \(g\) are linearly independent there, show that \(W(f, g)\) is zero for all \(t\) in \(-1< t< 1 .\) Hence \(f\) and \(g\) cannot be solutions of an equation \(y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0\) with \(p\) and \(q\) continuous on \(-1< t< 1 .\)

Short answer

#Short_Answer# The functions \(f(t) = t^2|t|\) and \(g(t) = t^3\) are linearly dependent on the intervals \(0<t<1\) and \(-1<t<0\), and linearly independent on the interval \(-1<t<1\). Their Wronskian is zero for all \(t\) in the interval \(-1<t<1\). This means that \(f(t)\) and \(g(t)\) cannot be solutions of an equation of the form \(y^{\prime\prime} + p(t)y^{\prime} + q(t)y = 0\) with continuous \(p\) and \(q\) on \(-1<t<1\).

Step-by-step solution

Linear dependence on 0 < t < 1 and -1 < t < 0 intervals

To determine if the given functions are linearly dependent, we need to check whether there exists a non-trivial combination of the functions that is equal to zero. Let C1 and C2 be constants. For \(0<t<1\), \(|t| = t\) so \(f(t) = t^2|t| = t^3\). Now let's check if we can find constants C1 and C2 such that \(C_1f(t) + C_2g(t) = C_1t^3 + C_2t^3 = 0\) for \(0<t<1\). If we choose C1 = 1 and C2 = -1, then: \(C_1t^3 + C_2t^3 = t^3 - t^3 = 0\) As we can find non-zero constants, the functions are linearly dependent on the interval \(0<t<1\). Now, let's evaluate linear dependence on the interval \(-1<t<0\). On this interval, \(|t| = -t\), so \(f(t) = t^2|t| = -t^3\). Now let's check whether \(C_1f(t) + C_2g(t) = C_1(-t^3) + C_2t^3 = 0\) for \(-1<t<0\). If we choose C1 = 1 and C2 = 1, then: \(C_1(-t^3) + C_2t^3 = -t^3 + t^3 = 0\) As we can find no zero constants, the functions are linearly dependent on the interval \(-1<t<0\).

Linear independence on -1 < t < 1 interval

To check for linear independence, we need to determine if the only solution for \(C_1f(t) + C_2g(t) = 0\) is \(C_1 = C_2 = 0\) for the given interval. Let \(C_1f(t) + C_2g(t) = C_1t^2|t| + C_2t^3 = 0\) for \(-1<t<1\). For \(0<t<1\), the equation becomes \(C_1t^3 + C_2t^3 = 0\). However, this relation does not hold for all values of t on the interval \(-1<t<1\) due to its dependence on t itself. Thus, the only solution to this equation is \(C_1=C_2=0\). Therefore, f(t) and g(t) are linearly independent on this interval.

Calculate Wronskian of f(t) and g(t) on -1 < t < 1 interval

Secondly, we must find the Wronskian and show it's equal to zero. The Wronskian is defined as \(W(f, g) = f(t)g'(t) - f'(t)g(t)\). In this case: - \(f(t) = t^2|t|\), so \(f'(t) = 2t|t| + t^2\mathrm{sgn}(t)\) - \(g(t) = t^3\), so \(g'(t) = 3t^2\) Now, let's calculate \(W(f, g)\) for \(-1<t<1\): \(W(f, g) = (t^2|t|)(3t^2) - (2t|t| + t^2\mathrm{sgn}(t))(t^3)\) \(= 3t^4|t| - (2t^3|t| + t^4\mathrm{sgn}(t))t\) \(= (3t^4|t| - 2t^4|t| - t^5\mathrm{sgn}(t))t\) \(= 0\) Thus, \(W(f, g)\) is zero for all \(t\) in the interval \(-1<t<1\).

Conclude that f(t) and g(t) cannot be solutions of a specific equation

Since \(W(f, g) = 0\) for all \(t\) in the interval \(-1<t<1\), \(f(t)\) and \(g(t)\) cannot be solutions of an equation \(y^{\prime\prime} + p(t)y^{\prime} + q(t)y = 0\) with \(p\) and \(q\) continuous on \(-1<t<1\). This is because the Wronskian of linearly independent solutions to such an equation is nonzero. However, our calculation of \(W(f, g)\) showed that it is equal to zero, which means these functions cannot be solutions of an equation with the given form.

Key concepts

Wronskian

Understanding the Wronskian is crucial when dealing with differential equations and the concept of linear independence. Essentially, the Wronskian is a determinant used to determine whether two or more functions are linearly independent. If you're considering functions in the context of solutions to differential equations, the Wronskian can be particularly revealing.

Let's define it more formally. For two functions, say, \( f(t) \) and \( g(t) \), the Wronskian, denoted as \( W(f, g) \), is given by the formula:
\[ W(f, g) = f(t)g'(t) - f'(t)g(t) \].

When the Wronskian is equal to zero at a point, it hints that the functions may be dependent, but it doesn't conclusively prove it because the Wronskian can be zero at some points for independent functions. However, if the Wronskian is non-zero at least at one point, the functions are guaranteed to be independent. This small but crucial distinction can sometimes lead to confusion. It's important to check the value of the Wronskian not just at a single point, but over the entire interval of interest.

Linear Independence

The concept of linear independence is at the heart of understanding solutions to differential equations. In simple terms, functions are considered to be linearly independent if no function in the set can be written as a linear combination of the others. For those who might not be familiar, a linear combination involves multiplying functions by constants and adding them together.

If we're speaking mathematically, for functions \( f(t) \) and \( g(t) \) to be linearly independent, the only solution to \( C_1f(t)+C_2g(t)=0 \) for all \( t \) should be the trivial solution \( C_1 = C_2 = 0 \). This is equivalent to saying that you can't scale and add \( f(t) \) and \( g(t) \) together in any way to get the zero function unless you scale them both by zero.

In the context of differential equations, having a set of linearly independent solutions is vital because it implies that the solutions form a basis for the solution space, which is a fancy way of saying that any possible solution can be expressed as a linear combination of these basic, independent solutions.

Differential Equations

Differential equations are the bread and butter of mathematical modeling, describing relationships involving rates of change. They are used extensively to model phenomena in physics, engineering, economics, and other fields. A differential equation expresses a relationship between a function and its derivatives.

An important category of differential equations is the second-order linear homogeneous differential equation, which has the form:\[ y'' + p(t)y' + q(t)y = 0 \]. When dealing with such an equation, two solutions are of particular interest: they should be both non-trivial and linearly independent. This requirement is crucial because such solutions provide the foundation to construct the general solution of the differential equation.

The previous exercise mentions functions that cannot be solutions to a specific differential equation. This ties back to our concepts of the Wronskian and linear independence. Understanding why certain functions don't qualify as solutions to differential equations often involves checking their Wronskian and establishing their independence or dependence in the context of the given interval. The crux is that the structure of the solutions is tightly woven with the equation's coefficients being continuous and the Wronskian not being identically zero.