Difference between revisions of "Linear Independence of Functions"

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<p>If we have an <math>n^{\mathrm{th}}</math> order linear homogenous differential equation <math>\frac{d^ny}{dt^n} + p_1(t) \frac{d^{n-1}y}{dt^{n-1}} + ... + p_{n-1}(t) \frac{dy}{dt} + p_n(t)y = 0</math> where <math>p_1</math>, <math>p_2</math>, &#8230;, <math>p_n</math> are continuous on an open interval <math>I</math> and if <math>y = y_1(t)</math>, <math>y = y_2(t)</math>, &#8230;, <math>y = y_n(t)</math> are solutions to this differential equation, then provided that <math>W(y_1, y_2, ..., y_n) \neq 0</math> for at least one point <math>t \in I</math>, then <math>y_1</math>, <math>y_2</math>, &#8230;, <math>y_n</math> form a fundamental set of solutions to this differential equation - that is, for constants <math>C_1</math>, <math>C_2</math>, &#8230;, <math>C_n</math>, then every solution to this differential equation can be written in the form:</p>
 
<p>If we have an <math>n^{\mathrm{th}}</math> order linear homogenous differential equation <math>\frac{d^ny}{dt^n} + p_1(t) \frac{d^{n-1}y}{dt^{n-1}} + ... + p_{n-1}(t) \frac{dy}{dt} + p_n(t)y = 0</math> where <math>p_1</math>, <math>p_2</math>, &#8230;, <math>p_n</math> are continuous on an open interval <math>I</math> and if <math>y = y_1(t)</math>, <math>y = y_2(t)</math>, &#8230;, <math>y = y_n(t)</math> are solutions to this differential equation, then provided that <math>W(y_1, y_2, ..., y_n) \neq 0</math> for at least one point <math>t \in I</math>, then <math>y_1</math>, <math>y_2</math>, &#8230;, <math>y_n</math> form a fundamental set of solutions to this differential equation - that is, for constants <math>C_1</math>, <math>C_2</math>, &#8230;, <math>C_n</math>, then every solution to this differential equation can be written in the form:</p>
  
<math> \quad y = C_1y_1(t) + C_2y_2(t) + ... + C_ny_n(t) </math>
+
<math>\begin{align} \quad y = C_1y_1(t) + C_2y_2(t) + ... + C_ny_n(t) \end{align}</math>
 
<p>We will now look at the connection between the solutions <math>y_1</math>, <math>y_2</math>, &#8230;, <math>y_n</math> forming a fundamental set of solutions and the linear independence/dependence of such solutions. We first define linear independence and linear dependence below.</p>
 
<p>We will now look at the connection between the solutions <math>y_1</math>, <math>y_2</math>, &#8230;, <math>y_n</math> forming a fundamental set of solutions and the linear independence/dependence of such solutions. We first define linear independence and linear dependence below.</p>
  
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<p>Perhaps the simplest linearly independent sets of functions is that set that contains <math>f_1(t) = 1</math>, <math>f_2(t) = t</math>, and <math>f_3(t) = t^2</math>. Let <math>k_1</math>, <math>k_2</math>, and <math>k_3</math> be constants and consider the following equation:</p>
 
<p>Perhaps the simplest linearly independent sets of functions is that set that contains <math>f_1(t) = 1</math>, <math>f_2(t) = t</math>, and <math>f_3(t) = t^2</math>. Let <math>k_1</math>, <math>k_2</math>, and <math>k_3</math> be constants and consider the following equation:</p>
  
<math> \quad k_1f_1(t) + k_2f_2(t) + k_3f_3(t) = 0 \\ \quad k_1 + k_2t + k_3t^2 = 0 </math>
+
<math>\begin{align} \quad k_1f_1(t) + k_2f_2(t) + k_3f_3(t) = 0 \\ \quad k_1 + k_2t + k_3t^2 = 0 \end{align}</math>
 
<p>It's not hard to see that equation above is satisfied if and only if the constants <math>k_1 = k_2 = k_3 = 0</math>.</p>
 
<p>It's not hard to see that equation above is satisfied if and only if the constants <math>k_1 = k_2 = k_3 = 0</math>.</p>
 
<p>For another example, consider the functions <math>f_1(t) = \sin t</math> and <math>f_2(t) = \sin (t + \pi)</math> defined on all of <math>\mathbb{R}</math>. This set of functions is not linearly independent. To show this, let <math>k_1</math> and <math>k_2</math> be constants and consider the following equation:</p>
 
<p>For another example, consider the functions <math>f_1(t) = \sin t</math> and <math>f_2(t) = \sin (t + \pi)</math> defined on all of <math>\mathbb{R}</math>. This set of functions is not linearly independent. To show this, let <math>k_1</math> and <math>k_2</math> be constants and consider the following equation:</p>
  
<math> \quad k_1f_1(t) + k_2f_2(t) = 0 \\ \quad k_1 \sin t + k_2 \sin (t + \pi) = 0 </math>
+
<math>\begin{align} \quad k_1f_1(t) + k_2f_2(t) = 0 \\ \quad k_1 \sin t + k_2 \sin (t + \pi) = 0 \end{align}</math>
 
<p>Now choose <math>t = \pi</math>. Then we have that:</p>
 
<p>Now choose <math>t = \pi</math>. Then we have that:</p>
  
<math> \quad k_1 \sin \pi + k_2 \sin 2\pi = 0 \\ </math>
+
<math>\begin{align} \quad k_1 \sin \pi + k_2 \sin 2\pi = 0 \\ \end{align}</math>
 
<p>But the above equation is true for any choice of constants <math>k_1</math> and <math>k_2</math> since <math>\sin \pi = \sin 2\pi = 0</math>, and thus <math>f_1</math> and <math>f_2</math> do not form a linearly independent set on all of <math>\mathbb{R}</math>.</p>
 
<p>But the above equation is true for any choice of constants <math>k_1</math> and <math>k_2</math> since <math>\sin \pi = \sin 2\pi = 0</math>, and thus <math>f_1</math> and <math>f_2</math> do not form a linearly independent set on all of <math>\mathbb{R}</math>.</p>
 
<p>From the concept of linear independence/dependence, we obtain the following theorem on fundamental sets of solutions for <math>n^{\mathrm{th}}</math> order linear homogenous differential equations.</p>
 
<p>From the concept of linear independence/dependence, we obtain the following theorem on fundamental sets of solutions for <math>n^{\mathrm{th}}</math> order linear homogenous differential equations.</p>
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</ul>
 
</ul>
  
<math>\begin{align} \quad W(y_1, y_2, ..., y_n)  \right|_{t_0} \neq 0 \end{align}</math>
+
<math>\begin{align} \quad W(y_1, y_2, ..., y_n)  \bigg|_{t_0} \neq 0 \end{align}</math>
 
<ul>
 
<ul>
 
<li>Thus this implies that following system of equations have only trivial solution <math>k_1 = k_2 = ... = k_n = 0</math>:</li>
 
<li>Thus this implies that following system of equations have only trivial solution <math>k_1 = k_2 = ... = k_n = 0</math>:</li>
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</ul>
 
</ul>
 
<ul>
 
<ul>
<li><math>\Leftarrow</math> We will prove the converse of Theorem 1 by contradiction. Suppose that <math>y_1</math>, <math>y_2</math>, &#8230;, <math>y_n</math> are linearly independent on <math>I</math>, and assume that instead <math>y_1</math>, <math>y_2</math>, &#8230;, <math>y_n</math> do NOT form a fundamental set of solutions on <math>I</math>. Then for some <math>t_0 \in I</math>, the Wronskian <math>W(y_1, y_2, ..., y_n)  \right|_{t_0} = 0</math>. Thus the system of equations above does not have only the trivial solution. Let the constants <math>k_1^*</math>, <math>k_2^*</math>, &#8230;, <math>k_n^*</math> be a nontrivial solution to this system. Define <math>\phi(t)</math> as:</li>
+
<li><math>\Leftarrow</math> We will prove the converse of Theorem 1 by contradiction. Suppose that <math>y_1</math>, <math>y_2</math>, &#8230;, <math>y_n</math> are linearly independent on <math>I</math>, and assume that instead <math>y_1</math>, <math>y_2</math>, &#8230;, <math>y_n</math> do NOT form a fundamental set of solutions on <math>I</math>. Then for some <math>t_0 \in I</math>, the Wronskian <math>W(y_1, y_2, ..., y_n)  \bigg|_{t_0} = 0</math>. Thus the system of equations above does not have only the trivial solution. Let the constants <math>k_1^*</math>, <math>k_2^*</math>, &#8230;, <math>k_n^*</math> be a nontrivial solution to this system. Define <math>\phi(t)</math> as:</li>
 
</ul>
 
</ul>
  
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<li>But <math>y_1</math>, <math>y_2</math>, &#8230;, <math>y_n</math> are linearly independent which implies that <math>k_1^* = k_2^* = ... = k_n^* = 0</math>. Thus <math>k_1^*</math>, <math>k_2^*</math>, &#8230;, <math>k_n^*</math> is a trivial solution to the system above, which is a contradiction. Therefore our assumption that <math>y_1</math>, <math>y_2</math>, &#8230;, <math>y_n</math> do not form a fundamental set of solutions was false. <math>\blacksquare</math></li>
 
<li>But <math>y_1</math>, <math>y_2</math>, &#8230;, <math>y_n</math> are linearly independent which implies that <math>k_1^* = k_2^* = ... = k_n^* = 0</math>. Thus <math>k_1^*</math>, <math>k_2^*</math>, &#8230;, <math>k_n^*</math> is a trivial solution to the system above, which is a contradiction. Therefore our assumption that <math>y_1</math>, <math>y_2</math>, &#8230;, <math>y_n</math> do not form a fundamental set of solutions was false. <math>\blacksquare</math></li>
 
</ul>
 
</ul>
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 +
==Licensing==
 +
Content obtained and/or adapted from:
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* [http://mathonline.wikidot.com/linear-independence-dependence-of-a-set-of-functions Linear Independence Dependence of a Set of Functions, http://mathonline.wikidot.com/] under a CC BY-SA license

Latest revision as of 18:04, 28 October 2021

If we have an order linear homogenous differential equation where , , …, are continuous on an open interval and if , , …, are solutions to this differential equation, then provided that for at least one point , then , , …, form a fundamental set of solutions to this differential equation - that is, for constants , , …, , then every solution to this differential equation can be written in the form:

We will now look at the connection between the solutions , , …, forming a fundamental set of solutions and the linear independence/dependence of such solutions. We first define linear independence and linear dependence below.

Definition: The functions , , …, are said to be Linearly Independent on an interval if for constants , , …, we have that implies that for all . This set of functions is said to be Linearly Dependent if where , , …, are not all zero for all .

Perhaps the simplest linearly independent sets of functions is that set that contains , , and . Let , , and be constants and consider the following equation:

It's not hard to see that equation above is satisfied if and only if the constants .

For another example, consider the functions and defined on all of . This set of functions is not linearly independent. To show this, let and be constants and consider the following equation:

Now choose . Then we have that:

But the above equation is true for any choice of constants and since , and thus and do not form a linearly independent set on all of .

From the concept of linear independence/dependence, we obtain the following theorem on fundamental sets of solutions for order linear homogenous differential equations.

Theorem 1: Let be an order linear homogenous differential equation. If , , …, are solutions to this differential equation then , , …, form a fundamental set of solutions to this differential equation on the open interval if and only if , , …, are linearly dependent on .

  • Proof: Consider the following order linear homogenous differential equation:

  • Suppose that , , …, form a fundamental set of solutions to this differential equation on the open interval . Then this implies that for all we have that :

  • Thus this implies that following system of equations have only trivial solution :

  • Thus the equation implies that . Thus , , …, are linearly independent on .
  • We will prove the converse of Theorem 1 by contradiction. Suppose that , , …, are linearly independent on , and assume that instead , , …, do NOT form a fundamental set of solutions on . Then for some , the Wronskian . Thus the system of equations above does not have only the trivial solution. Let the constants , , …, be a nontrivial solution to this system. Define as:

  • Note that satisfies the initial conditions , , …, , and satisfies our order linear homogenous differential equation because is a linear combination of the solutions , , …, .
  • Now note that the function also satisfies the differential equation and the initial conditions. By the existence/uniqueness theorem for order linear homogenous differential equations, this implies that for all , so:

  • But , , …, are linearly independent which implies that . Thus , , …, is a trivial solution to the system above, which is a contradiction. Therefore our assumption that , , …, do not form a fundamental set of solutions was false.

Licensing

Content obtained and/or adapted from: