Determinant - Revision history

Khanh at 05:53, 19 November 2021

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Khanh at 05:50, 19 November 2021

2021-11-19T05:50:18Z

Khanh: Created page with "== Wronskian Determinants of Two Functions == We are going to look more into second order linear homogenous differential equations, but before we do, we need to first learn a..."

2021-11-19T05:37:12Z

Created page with "== Wronskian Determinants of Two Functions == We are going to look more into second order linear homogenous differential equations, but before we do, we need to first learn a..."

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== Wronskian Determinants of Two Functions ==

We are going to look more into second order linear homogenous differential equations, but before we do, we need to first learn about a type of determinant known as a '''Wronskian Determinant''' which we define below.

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
:'''Definition:''' Let <math>f</math> and <math>g</math> be two differentiable functions. Then the '''Wronskian Determinant''' of <math>f</math> and <math>g</math> is the <math>2 \times 2</math> determinant <math>W(f, g) = \begin{vmatrix} f(x) & g(x) \\ f'(x) & g'(x) \end{vmatrix} = f(x)g'(x) + f'(x)g(x)</math>.</td>
</blockquote>

''Sometimes the term "Wronskian" by itself is used to mean the same thing as "Wronskian Determinant". Furthermore, sometimes we can just write "<math>W</math>", or "<math>W(x)</math>" instead of <math>W(f, g)</math> to represent the Wronskian of <math>f</math> and <math>g</math>.''

Let's look at some examples of computing the Wronskian determinant of two differentiable functions.

===Example 1===
'''Determine the Wronskian of the functions <math>f(x) = x^2</math> and <math>g(x) = 3x^2</math>. For what values of <math>x</math> is the Wronskian equal to zero?'''

We note that <math>f</math> and <math>g</math> are both differentiable functions and that <math>f'(x) = 2x</math> and <math>g'(x) = 6x</math>. Therefore the Wronskian of <math>f</math> and <math>g</math> is:
<div style="text-align: center;"><math>\begin{align} \quad \begin{vmatrix} x^2 & 3x^2\\ 2x & 6x \end{vmatrix} = x^2 \cdot 6x - 3x^2 \cdot 2x = 6x^3 - 6x^3 = 0 \end{align}</math></div>

Therefore the Wronskian of <math>f</math> and <math>g</math> is equal to zero for all <math>x \in \mathbb{R}</math>.

===Example 2===
'''Determine the Wronskian of the functions <math>f(x) = e^x \sin t</math> and <math>g(x) = e^x \cos x</math>. For what values of <math>x</math> is the Wronskian equal to zero?'''

We note that <math>f</math> and <math>g</math> are both differentiable functions and that <math>f'(x) = e^x \sin x + e^x \cos x</math> and <math>g'(x) = e^x \cos x - e^x \sin x</math>. Therefore the Wronskian of <math>f</math> and <math>g</math> is:
<div style="text-align: center;"><math>\begin{align} \quad \begin{vmatrix} e^x \sin x & e^x \cos x \\ e^x \sin x + e^x \cos x & e^x \cos x - e^x \sin x \end{vmatrix} = \left ( e^x\sin x \right ) \cdot \left (e^x \cos x - e^x \sin x \right ) - \left ( e^x \cos x \right ) \cdot \left ( e^x \sin x + e^x \cos x \right ) \\ \quad = e^{2x} \sin x \cos x - e^{2x} \sin^2 x - e^{2x} \sin x \cos x - e^{2x} \cos^2 x = -e^{2x}(\cos^2 x + \sin^2 x) = -e^{2x} \end{align}</math></div>

Note that <math>-e^{2x} < 0</math> for all <math>x \in \mathbb{R}</math> so the Wronskian of <math>f</math> and <math>g</math> is zero nowhere.

← Older revision		Revision as of 05:53, 19 November 2021
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	: '''Theorem 1:''' Let <math>\frac{d^2 y}{dt^2} + p(t) \frac{dy}{dt} + q(t) y = 0</math> be a second order linear homogenous differential equation where <math>p</math> and <math>q</math> are continuous functions on an open interval <math>I</math> such that <math>t_0 \in I</math> and with the initial conditions <math>y(t_0) = y_0</math> and <math>y'(t_0) = y'_0</math>. If <math>y = y_1(t)</math> and <math>y = y_2(t)</math> are solutions to this differential equation then there exists constants <math>C</math> and <math>D</math> for which <math>y = Cy_1(t) + Dy_2(t)</math> is a solution to the initial value problem if and only if the Wronskian at <math>t_0</math> is nonzero, that is <math>W(y_1, y_2) \bigg\|_{t_0} = y_1(t_0)y_2'(t_0) - y_1'(t_0)y_2(t_0) \neq 0</math>.		: '''Theorem 1:''' Let <math>\frac{d^2 y}{dt^2} + p(t) \frac{dy}{dt} + q(t) y = 0</math> be a second order linear homogenous differential equation where <math>p</math> and <math>q</math> are continuous functions on an open interval <math>I</math> such that <math>t_0 \in I</math> and with the initial conditions <math>y(t_0) = y_0</math> and <math>y'(t_0) = y'_0</math>. If <math>y = y_1(t)</math> and <math>y = y_2(t)</math> are solutions to this differential equation then there exists constants <math>C</math> and <math>D</math> for which <math>y = Cy_1(t) + Dy_2(t)</math> is a solution to the initial value problem if and only if the Wronskian at <math>t_0</math> is nonzero, that is <math>W(y_1, y_2) \bigg\|_{t_0} = y_1(t_0)y_2'(t_0) - y_1'(t_0)y_2(t_0) \neq 0</math>.
	</blockquote>		</blockquote>
		+
		+	== Licensing ==
		+	Content obtained and/or adapted from:
		+	* [http://mathonline.wikidot.com/wronskian-determinants-of-two-functions Wronskian Determinants of Two Functions, mathonline.wikidot.com] under a CC BY-SA license
		+	* [http://mathonline.wikidot.com/wronskian-determinants-and-linear-homogenous-differential-eq Wronskian Determinants and Linear Homogenous Differential Equations, mathonline.wikidot.com] under a CC BY-SA license

← Older revision		Revision as of 05:50, 19 November 2021
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	Note that <math>-e^{2x} < 0</math> for all <math>x \in \mathbb{R}</math> so the Wronskian of <math>f</math> and <math>g</math> is zero nowhere.		Note that <math>-e^{2x} < 0</math> for all <math>x \in \mathbb{R}</math> so the Wronskian of <math>f</math> and <math>g</math> is zero nowhere.
		+
		+	== Wronskian Determinants and Linear Homogenous Differential Equations ==
		+
		+	Recall that if we have the second order linear homogenous differential equation <math>\frac{d^2 y}{dt^2} + p(t) \frac{dy}{dt} + q(t)y = 0</math> and <math>y = y_1(t)</math> and <math>y = y_2(t)</math> are solutions to this differential equation, then for constants <math>C</math> and <math>D</math>, the linear combination <math>y = Cy_1(t) + Dy_2(t)</math> is also a solution to this differential equation. One question to ask if whether or not <em>all</em> of the solutions to this differential equation are in this form as we do not want to miss any other potential solutions.
		+
		+	Suppose that we are given an initial value problem to a second order linear homogenous differential equation in this form alongside the initial conditions <math>y(t_0) = y_0</math> and <math>y'(t_0) = y'_0</math>. Applying the first initial condition <math>y(t_0) = y_0</math> to our solution <math>y = Cy_1(t) + Dy_2(t)</math> and we have that:
		+	<div style="text-align: center;"><math>\begin{align} \quad y_0 = Cy_1(t_0) + Dy_2(t_0) \end{align}</math></div>
		+
		+	Now the derivative of <math>y = Cy_1(t) + Dy_2(t)</math> is <math>y' = Cy_1'(t) + Dy_2'(t)</math>. Applying the second initial condition <math>y'(t_0) = y'_0</math> and we have that:
		+	<div style="text-align: center;"><math>\begin{align} \quad y'_0 = Cy_1'(t_0) + Dy_2'(t_0) \end{align}</math></div>
		+
		+	We need the unknown constants <math>C</math> and <math>D</math> to satisfy both of these equations, that is, we want to solve the system <math>\left\{\begin{matrix} Cy_1(t_0) + Dy_2(t_0) = y_0 \\ Cy_1'(t_0) + Dy_2'(t_0) = y'_0 \end{matrix}\right.</math> for the constants <math>C</math> and <math>D</math>. We note that we have a system of two equations with two unknowns and thus, a unique solution exists if the determinant of the augmented matrix for this system is nonzero, that is:
		+	<div style="text-align: center;"><math>\begin{align} W = \begin{vmatrix} y_1(t_0) & y_2(t_0)\\ y_1'(t_0) & y_2'(t_0) \end{vmatrix} = y_1(t_0)y_2'(t_0) - y_1'(t_0)y_2(t_0) \neq 0 \end{align}</math></div>
		+
		+	Notice though that this determinant <math>W</math> is simple the Wronskian of the functions <math>y_1</math> and <math>y_2</math> evaluated at <math>t_0</math>. If <math>W \neq 0</math>, then we can apply Cramer's Rule from linear algebra to find the values of the constants <math>C</math> and <math>D</math>. We have that:
		+
		+	<div style="text-align: center;"><math>\begin{align} \quad C = \frac{\begin{vmatrix} y_0 & y_2(t_0)\\ y'_0) & y_2'(t_0) \end{vmatrix}}{\begin{vmatrix} y_1(t_0) & y_2(t_0)\\ y_1'(t_0) & y_2'(t_0) \end{vmatrix}} = \frac{y_0y_2'(t_0) - y'_0y_2(t_0)}{y_1(t_0)y_2'(t_0) - y_1'(t_0)y_2(t_0)} \quad \quad D = \frac{\begin{vmatrix} y_1(t_0) & y_0 \\ y_1'(t_0) & y'_0\end{vmatrix}}{\begin{vmatrix} y_1(t_0) & y_2(t_0)\\ y_1'(t_0) & y_2'(t_0) \end{vmatrix}} = \frac{y'_0y_1(t_0) - y_0y_1'(t_0)}{y_1(t_0)y_2'(t_0) - y_1'(t_0)y_2(t_0)} \end{align}</math></div>
		+	If this determinant <math>W</math> equals zero, then either no solutions for <math>C</math> and <math>D</math> exist, or infinitely many solutions for the values of <math>C</math> and <math>D</math> may exist. The following theorem summarizes what we have just found.
		+
		+	<blockquote style="background: white; border: 1px solid black; padding: 1em;">
		+	: '''Theorem 1:''' Let <math>\frac{d^2 y}{dt^2} + p(t) \frac{dy}{dt} + q(t) y = 0</math> be a second order linear homogenous differential equation where <math>p</math> and <math>q</math> are continuous functions on an open interval <math>I</math> such that <math>t_0 \in I</math> and with the initial conditions <math>y(t_0) = y_0</math> and <math>y'(t_0) = y'_0</math>. If <math>y = y_1(t)</math> and <math>y = y_2(t)</math> are solutions to this differential equation then there exists constants <math>C</math> and <math>D</math> for which <math>y = Cy_1(t) + Dy_2(t)</math> is a solution to the initial value problem if and only if the Wronskian at <math>t_0</math> is nonzero, that is <math>W(y_1, y_2) \bigg\|_{t_0} = y_1(t_0)y_2'(t_0) - y_1'(t_0)y_2(t_0) \neq 0</math>.
		+	</blockquote>