Difference between revisions of "Solutions of Linear Systems"
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<div style="text-align: center;"><math>\begin{align} \phi(t) = Cy_1(t) + Dy_2(t) \end{align}</math></div> | <div style="text-align: center;"><math>\begin{align} \phi(t) = Cy_1(t) + Dy_2(t) \end{align}</math></div> | ||
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* 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>. | * 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>. | * <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>. | ||
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*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> | *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: | 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: | ||
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<div style="text-align: center;"><math>\begin{align} \quad y = Cy_1(t) + Dy_2(t) \end{align}</math></div> | <div style="text-align: center;"><math>\begin{align} \quad y = Cy_1(t) + Dy_2(t) \end{align}</math></div> | ||
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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>. | 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>. | ||
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<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> | <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> | ||
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* 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> | * 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> | ||
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== Licensing == | == Licensing == | ||
Content obtained and/or adapted from: | 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 |
Latest revision as of 00:29, 19 November 2021
Fundamental Solutions to Linear Homogenous Differential Equations
- Theorem 1: Let be a second order linear homogenous differential equation where and are continuous on an open interval such that , and let and be two solutions to this differential equation. The set of all linear combinations of these two solutions, where and are constants contains all solutions to this differential equation if and only if there exists a point for which the Wronksian of and at is nonzero, that is .
- Proof: Let and both be solutions to the differential equation and suppose that is any arbitrary solution as well. We want to show that is a linear combination of and for some constants and .
- Let be such that the Wronskian of and evaluated at is nonzero, that is:
- Take this value and evaluate both and at this point. Then and (since is a solution to our differential equation). Now consider the initial value problem with the initial conditions and . The function satisfies this differential equation. Since then we have that there exists constants and such that satisfies this initial value problem. But since and are continuous on the open interval containing then this implies that a unique solution exists, and so:
- So all solutions for this differential equation are a linear combination of the solutions and .
- Suppose that every point is such that , that is, there exists no point on where the Wronskian of and evaluated at is nonzero. Let and be values for which the system has no solutions for a set of constants and .
- Now since and are continuous on an open interval containing , such a solution satisfies the initial conditions and . Note though this solution is not a linear combination of and though which completes our proof.
Theorem 1 above implies that if we can find two solutions and for which the Wronskian , then for constants and , all solutions of the second order linear homogenous differential equation are given by:
Also note that thus far we have not said that and need to be distinct. However, with the Theorem above, we see that if then the Wronskian 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 .
- Definition: Let where and are continuous on an open interval such that and let and be solutions to this differential equation. If the Wronskian then the set of linear combinations of and is known as the Fundamental Set of Solutions to this differential equation.
From the definition above, we see that if we can find two solutions and for which the Wronskian is nonzero, then and 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.
- Theorem 2: Let be a second order linear homogenous differential equation where and are continuous on an open interval such that . If is a solution to this differential equation that satisfies the initial conditions and , and if be a solution to this differential equation that satisfies the initial conditions and . Then and form a fundamental set of solutions for this differential equation.
- Proof: We note that:
- Thus Theorem 1 implies that ALL solutions to this differential equation are given by where and are constants. Thus and form a fundamental set of solutions for this differential equation.
Licensing
Content obtained and/or adapted from:
- Fundamental Solutions to Linear Homogenous Differential Equations, mathonline.wikidot.com under a CC BY-SA license