Solutions of Linear Systems - Revision history

Khanh at 06:29, 19 November 2021

2021-11-19T06:29:54Z

Khanh: Created page with "==Fundamental Solutions to Linear Homogenous Differential Equations==
:'''Theorem 1:''' Let
2021-11-19T06:27:44Z

Created page with "==Fundamental Solutions to Linear Homogenous Differential Equations== <blockquote style="background: white; border: 1px solid black; padding: 1em;"> :'''Theorem 1:''' Let <m..."

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==Fundamental Solutions to Linear Homogenous Differential Equations==

<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 on an open interval <math>I</math> such that <math>t_0 \in I</math>, and let <math>y = y_1(t)</math> and <math>y = y_2(t)</math> be two solutions to this differential equation. The set of all linear combinations of these two solutions, <math>y = Cy_1(t) + Dy_2(t)</math> where <math>C</math> and <math>D</math> are constants contains all solutions to this differential equation if and only if there exists a point <math>t_0</math> for which the Wronksian of <math>y_1</math> and <math>y_2</math> at <math>t_0</math> is nonzero, that is <math>W(y_1, y_2) \bigg|_{t_0} \neq 0</math>.
</blockquote>

* '''Proof:''' Let <math>y = y_1(t)</math> and <math>y = y_2(t)</math> both be solutions to the differential equation <math>\frac{d^2 y}{dt^2} + p(t) \frac{dy}{dt} + q(t)y = 0</math> and suppose that <math>y = \phi (t)</math> is any arbitrary solution as well. We want to show that <math>\phi (t)</math> is a linear combination of <math>y_1</math> and <math>y_2</math> for some constants <math>C</math> and <math>D</math>.

<math>\Leftarrow</math> Let <math>t_0</math> be such that the Wronskian of <math>y_1</math> and <math>y_2</math> evaluated at <math>t_0</math> is nonzero, that is:

<div style="text-align: center;"><math>\begin{align} \quad W(y_1, y_2) \bigg|_{t_0} \neq 0 \end{align}</math></div>

Take this value <math>t_0</math> and evaluate both <math>\phi</math> and <math>\phi'</math> at this point. Then <math>\phi (t_0) = y_0</math> and <math>\phi'(t_0) = y'_0</math> (since <math>y = \phi(t)</math> is a solution to our differential equation). Now consider the initial value problem <math>\frac{d^2 y}{dt^2} + p(t) \frac{dy}{dt} + q(t) y = 0</math> with the initial conditions <math>y(t_0) = y_0</math> and <math>y'(t_0) = y'_0</math>. The function <math>\phi</math> satisfies this differential equation. Since <math>W(y_1, y_2) \bigg|_{t_0} \neq 0</math> then we have that there exists constants <math>C</math> and <math>D</math> such that <math>y = Cy_1(t) + Dy_2(t)</math> satisfies this initial value problem. But since <math>p</math> and <math>q</math> are continuous on the open interval <math>I</math> containing <math>t_0</math> then this implies that a unique solution exists, and so:

<div style="text-align: center;"><math>\begin{align} \phi(t) = Cy_1(t) + Dy_2(t) \end{align}</math></div>

* So all solutions for this differential equation are a linear combination of the solutions <math>y = y_1(t)</math> and <math>y = y_2(t)</math>.

* <math>\Rightarrow</math> Suppose that every point <math>t_0 \in I</math> is such that <math>W(y_1, y_2) \bigg|_{t_0} = 0</math>, that is, there exists no point <math>t_0</math> on <math>I</math> where the Wronskian of <math>y_1</math> and <math>y_2</math> evaluated at <math>t_0</math> is nonzero. Let <math>y_0</math> and <math>y'_0</math> be values for which 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> has no solutions for a set of constants <math>C</math> and <math>D</math>.

Now since <math>p</math> and <math>q</math> are continuous on an open interval <math>I</math> containing <math>t_0</math>, such a solution <math>\phi(t)</math> satisfies the initial conditions <math>y(t_0) = y_0</math> and <math>y'(t_0) = y'_0</math>. Note though this solution is not a linear combination of <math>y_1</math> and <math>y_2</math> though which completes our proof. <math>\blacksquare</math>

Theorem 1 above implies that if we can find two solutions <math>y = y_1(t)</math> and <math>y = y_2(t)</math> for which the Wronskian <math>W(y_1, y_2) \neq 0</math>, then for constants <math>C</math> and <math>D</math>, all solutions of the second order linear homogenous differential equation <math>\frac{d^2 y}{dt^2} + p(t) \frac{dy}{dt} + q(t) y = 0</math> are given by:
<div style="text-align: center;"><math>\begin{align} \quad y = Cy_1(t) + Dy_2(t) \end{align}</math></div>

Also note that thus far we have not said that <math>y = y_1(t)</math> and <math>y = y_2(t)</math> need to be distinct. However, with the Theorem above, we see that if <math>y_1(t) = y_2(t)</math> then the Wronskian <math>W(y_1, y_2) = W(y_1, y_1) = W(y_2, y_2)</math> is zero (as you should verify) and so not all solutions to a second order linear homogenous differential are given by the linear combination of just <math>y_1</math>.

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
:'''Definition:''' Let <math>\frac{d^2 y}{dt^2} + p(t) \frac{dy}{dt} + q(t)y = 0</math> where <math>p</math> and <math>q</math> are continuous on an open interval <math>I</math> such that <math>t_0 \in I</math> and let <math>y = y_1(t)</math> and <math>y = y_2(t)</math> be solutions to this differential equation. If the Wronskian <math>W(y_1, y_2) \neq 0</math> then the set of linear combinations of <math>y_1</math> and <math>y_2</math> is known as the '''Fundamental Set of Solutions''' to this differential equation.
</blockquote>

From the definition above, we see that if we can find two solutions <math>y = y_1(t)</math> and <math>y = y_2(t)</math> for which the Wronskian <math>W(y_1, y_2)</math> is nonzero, then <math>y_1</math> and <math>y_2</math> form a fundamental set of solutions. The next question that we might pose is whether or not a second order linear homogenous differential equation always has a fundamental set of solutions.

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
:'''Theorem 2:''' Let <math>\frac{d^2 y}{dt} + 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 on an open interval <math>I</math> such that <math>t_0 \in I</math>. If <math>y = y_1(t)</math> is a solution to this differential equation that satisfies the initial conditions <math>y_1(t_0) = 1</math> and <math>y_1'(t_0) = 0</math>, and if <math>y = y_2(t)</math> be a solution to this differential equation that satisfies the initial conditions <math>y_2(t_0) = 0</math> and <math>y_2'(t_0) = 1</math>. Then <math>y_1</math> and <math>y_2</math> form a fundamental set of solutions for this differential equation.
</blockquote>

'''Proof:''' We note that:

<div style="text-align: center;"><math>\begin{align} \quad W(y_1, y_2) \bigg|_{t_0} = \begin{vmatrix} y_1(t_0) & y_2(t_0) \\ y_1'(t_0) & y_2'(t_0)\end{vmatrix} = \begin{vmatrix} 1 &a 0\\ 0 & 1 \end{vmatrix} = 1 \neq 0 \end{align}</math></div>

* Thus Theorem 1 implies that ALL solutions to this differential equation are given by <math>y = Cy_1(t) + Dy_2(t)</math> where <math>C</math> and <math>D</math> are constants. Thus <math>y_1</math> and <math>y_2</math> form a fundamental set of solutions for this differential equation. <math>\blacksquare</math>

== Licensing ==
Content obtained and/or adapted from:
* [] under a CC BY-SA license

@@ Line 16: / Line 16: @@
 <div style="text-align: center;"><math>\begin{align} \phi(t) = Cy_1(t) + Dy_2(t) \end{align}</math></div>
 * So all solutions for this differential equation are a linear combination of the solutions <math>y = y_1(t)</math> and <math>y = y_2(t)</math>.
 * <math>\Rightarrow</math> Suppose that every point <math>t_0 \in I</math> is such that <math>W(y_1, y_2) \bigg|_{t_0} = 0</math>, that is, there exists no point <math>t_0</math> on <math>I</math> where the Wronskian of <math>y_1</math> and <math>y_2</math> evaluated at <math>t_0</math> is nonzero. Let <math>y_0</math> and <math>y'_0</math> be values for which 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> has no solutions for a set of constants <math>C</math> and <math>D</math>.
 *Now since <math>p</math> and <math>q</math> are continuous on an open interval <math>I</math> containing <math>t_0</math>, such a solution <math>\phi(t)</math> satisfies the initial conditions <math>y(t_0) = y_0</math> and <math>y'(t_0) = y'_0</math>. Note though this solution is not a linear combination of <math>y_1</math> and <math>y_2</math> though which completes our proof. <math>\blacksquare</math>
 Theorem 1 above implies that if we can find two solutions <math>y = y_1(t)</math> and <math>y = y_2(t)</math> for which the Wronskian <math>W(y_1, y_2) \neq 0</math>, then for constants <math>C</math> and <math>D</math>, all solutions of the second order linear homogenous differential equation <math>\frac{d^2 y}{dt^2} + p(t) \frac{dy}{dt} + q(t) y = 0</math> are given by:
 <div style="text-align: center;"><math>\begin{align} \quad y = Cy_1(t) + Dy_2(t) \end{align}</math></div>
 Also note that thus far we have not said that <math>y = y_1(t)</math> and <math>y = y_2(t)</math> need to be distinct. However, with the Theorem above, we see that if <math>y_1(t) = y_2(t)</math> then the Wronskian <math>W(y_1, y_2) = W(y_1, y_1) = W(y_2, y_2)</math> is zero (as you should verify) and so not all solutions to a second order linear homogenous differential are given by the linear combination of just <math>y_1</math>.
@@ Line 41: / Line 45: @@
 <div style="text-align: center;"><math>\begin{align} \quad W(y_1, y_2) \bigg|_{t_0} = \begin{vmatrix} y_1(t_0) & y_2(t_0) \\ y_1'(t_0) & y_2'(t_0)\end{vmatrix} = \begin{vmatrix} 1 &a 0\\ 0 & 1 \end{vmatrix} = 1 \neq 0 \end{align}</math></div>
 * Thus Theorem 1 implies that ALL solutions to this differential equation are given by <math>y = Cy_1(t) + Dy_2(t)</math> where <math>C</math> and <math>D</math> are constants. Thus <math>y_1</math> and <math>y_2</math> form a fundamental set of solutions for this differential equation. <math>\blacksquare</math>
@@ Line 47: / Line 52: @@
 == Licensing ==
 Content obtained and/or adapted from:
-* [] under a CC BY-SA license
+* [http://mathonline.wikidot.com/fundamental-solutions-to-linear-homogenous-differential-equa Fundamental Solutions to Linear Homogenous Differential Equations, mathonline.wikidot.com] under a CC BY-SA license