differential equations in linear algebra

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• differential equations in linear algebra
  • 1.Idea
  • 2.differential eqautions in algebra
  • 3.first order differential equations in linear algebra
  • 4.second order equations
    • 4.1 (1)algebra approach:
    • 4.2 (2)linear algebra approach:
  • 5.imaginary eigenvalue involved

1.Idea

The explanation of differnetial equation in linear algobra has everything to do with what the same part in algebra does. So the point where we find the solution of them here is to observe how differential eqaution does in algebra!

algebra -> linear algebra

2.differential eqautions in algebra

case 1 : \(\frac{dy(x)}{dx}=y(x) \implies y(x) = C \cdot e^x\)
case 2: $\frac{dy(x)}{dx}=\lambda \implies y(x) = C \cdot e^{\lambda x} $

Question:how to map question from algebra to linear algebra?

=> Linear equations! form a matrix by collecting coefficients!

3.first order differential equations in linear algebra

solve \(\frac{du}{dt}=Au\)
=>

solution1:
\(Ax=\lambda x \iff Ae^{\lambda x}x=\lambda e^{\lambda x}x \iff \frac{du}{dt}=Au=Ae^{\lambda x}x=\lambda e^{\lambda x}x\)

Tech: eigenvector and eigenvalue remove all the complexity of matrix A.

solution2: : (integral then obtain new equations! )

Assume A is 2 by 2 : \(\frac{du}{dt}=Au\)

\( \begin{equation} \left\{ \begin{array}{r1} \frac{du_1}{dt} = a_{11}u_1 + a_{12}u_2\\ \frac{du_2}{dt} = a_{21}u_1 + a_{22}u_2 \end{array} \right. \end{equation} \)

Obtain the solution bu solving the linear equation above.^[1]

Example:

solution2:
\( \begin{Bmatrix} 0 & 1\\ 1 & 0 \end{Bmatrix}\) => \(\frac{du}{dt}=Au \)=>\( \begin{equation} \left\{ \begin{array}{r1} \frac{du_1}{dt} = u_2\\ \frac{du_2}{dt} = u_1 \end{array} \right. \end{equation} \)=>\( \begin{equation} \left\{ \begin{array}{r1} \frac{du_1+u_2}{dt} =u_1 + u_2 \\ \frac{du_1 - u_2}{dt} = -(u_1 - u_2) \end{array} \right. \end{equation} \)=>\( \begin{equation} \left\{ \begin{array}{r1} u_1 + u_2 = e^{t}\\ u_1 - u_2 = e^{-t} \end{array} \right. \end{equation} \)=>\( \begin{equation} \left\{ \begin{array}{r1} u_1 = \frac{e^{t}+e^{-t}}{2}\\ u_2 = \frac{e^{t}-e^{-t}}{2} \end{array} \right. \end{equation} \)

solution1:
\(\frac{du}{dt}=Au=A\cdot e^{\lambda t}x=\lambda e^{\lambda t}x\)=>\( \begin{equation} \left\{ \begin{array}{r1} v(t)= e^{\lambda_1 t}x_1\\ w(t)= e^{\lambda_2 t}x_2\\ \end{array} \right. \end{equation}\) Then the value will come from the \(u(0)\).
And the complete solution is \(u(t) = C\cdot v(t) + D\cdot w(t)\).

4.second order equations

Question rises : \(my\prime\prime + ky\prime+by=0\)

4.1 (1)algebra approach:

key point:we want to use \(e^{\lambda t}\) to solve this equation!

\(my\prime\prime + ky\prime+by=0 \implies (m\lambda^2+k\lambda+b)e^{\lambda t}=0\)-----equation\((a)\)

So obtaining the value of \(\lambda\) from equation(1) can get the value of \(\lambda\).

4.2 (2)linear algebra approach:

\(my\prime\prime + ky\prime+by=0\)=>(assume m = 1)\( \begin{equation} \left\{ \begin{array}{r1} \frac{dy}{dt} = y\prime\\ \frac{dy\prime}{dt} = -ky\prime - by \end{array} \right. \end{equation} \)=>\( \frac{du}{dt}=\begin{Bmatrix} 0 & 1\\ -b & -k \end{Bmatrix}u \)
\(\det{A-\lambda I}=0\)=>\((-\lambda)\cdot(-k-\lambda)+b=0\)=>\(m\lambda^2+k\lambda+b=0\) ,it is identical to \((a)\).
here convert second order equation to first order equation by solving 1 by 1 equation discussed above.

5.imaginary eigenvalue involved

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