Examples from Linear ODE systems All examples from Linear ODE systems.
[ EXAMPLE ] Consider
x ′ = ( − 3 3 3 − 3 ) x , x ( 0 ) = ( 4 0 ) . \mathbf{x}'=
\begin{pmatrix}
-3&3\\
3&-3
\end{pmatrix}\mathbf{x},
\qquad
\mathbf{x}(0)=
\begin{pmatrix}4\\0\end{pmatrix}. x ′ = ( − 3 3 3 − 3 ) x , x ( 0 ) = ( 4 0 ) . The eigenvalues are 0 0 0 − 6 -6 − 6
v 1 = ( 1 1 ) , v 2 = ( 1 − 1 ) . v_1=\begin{pmatrix}1\\1\end{pmatrix},
\qquad
v_2=\begin{pmatrix}1\\-1\end{pmatrix}. v 1 = ( 1 1 ) , v 2 = ( 1 − 1 ) . Since
( 4 0 ) = 2 ( 1 1 ) + 2 ( 1 − 1 ) , \begin{pmatrix}4\\0\end{pmatrix}
=2\begin{pmatrix}1\\1\end{pmatrix}
+2\begin{pmatrix}1\\-1\end{pmatrix}, ( 4 0 ) = 2 ( 1 1 ) + 2 ( 1 − 1 ) , the solution is
x ( t ) = 2 ( 1 1 ) + 2 e − 6 t ( 1 − 1 ) = ( 2 + 2 e − 6 t 2 − 2 e − 6 t ) . \mathbf{x}(t)
=2\begin{pmatrix}1\\1\end{pmatrix}
+2e^{-6t}\begin{pmatrix}1\\-1\end{pmatrix}
=
\begin{pmatrix}
2+2e^{-6t}\\
2-2e^{-6t}
\end{pmatrix}. x ( t ) = 2 ( 1 1 ) + 2 e − 6 t ( 1 − 1 ) = ( 2 + 2 e − 6 t 2 − 2 e − 6 t ) . The e − 6 t e^{-6t} e − 6 t x ( t ) \mathbf{x}(t) x ( t ) ( 2 , 2 ) T (2,2)^{\mathsf T} ( 2 , 2 ) T λ = 0 \lambda=0 λ = 0
[RIGHT CLICK] [ASK AI] [ EXAMPLE ] For
x ′ = ( 1 − 2 2 1 ) x , x ( 0 ) = ( 1 0 ) , \mathbf{x}'=
\begin{pmatrix}
1&-2\\
2&1
\end{pmatrix}\mathbf{x},
\qquad
\mathbf{x}(0)=
\begin{pmatrix}1\\0\end{pmatrix}, x ′ = ( 1 2 − 2 1 ) x , x ( 0 ) = ( 1 0 ) , the eigenvalues are 1 + 2 i 1+2i 1 + 2 i 1 − 2 i 1-2i 1 − 2 i
v + = ( i 1 ) , v − = ( − i 1 ) . v_+=\begin{pmatrix}i\\1\end{pmatrix},
\qquad
v_-=\begin{pmatrix}-i\\1\end{pmatrix}. v + = ( i 1 ) , v − = ( − i 1 ) . At t = 0 t=0 t = 0
c + ( i 1 ) + c − ( − i 1 ) = ( 1 0 ) . c_+\begin{pmatrix}i\\1\end{pmatrix}
+c_-\begin{pmatrix}-i\\1\end{pmatrix}
=\begin{pmatrix}1\\0\end{pmatrix}. c + ( i 1 ) + c − ( − i 1 ) = ( 1 0 ) . Solving these two scalar equations gives c + = − i / 2 c_+=-i/2 c + = − i /2 c − = i / 2 c_-=i/2 c − = i /2
x ( t ) = − i 2 e ( 1 + 2 i ) t ( i 1 ) + i 2 e ( 1 − 2 i ) t ( − i 1 ) . \mathbf{x}(t)
=
-\frac{i}{2}e^{(1+2i)t}\begin{pmatrix}i\\1\end{pmatrix}
+\frac{i}{2}e^{(1-2i)t}\begin{pmatrix}-i\\1\end{pmatrix}. x ( t ) = − 2 i e ( 1 + 2 i ) t ( i 1 ) + 2 i e ( 1 − 2 i ) t ( − i 1 ) . Using Euler's formula e i θ = cos θ + i sin θ e^{i\theta}=\cos\theta+i\sin\theta e i θ = cos θ + i sin θ θ = 2 t \theta=2t θ = 2 t
x ( t ) = e t ( cos ( 2 t ) sin ( 2 t ) ) . \mathbf{x}(t)
=e^t
\begin{pmatrix}
\cos(2t)\\
\sin(2t)
\end{pmatrix}. x ( t ) = e t ( cos ( 2 t ) sin ( 2 t ) ) . Direct differentiation checks that x ′ = A x \mathbf{x}'=A\mathbf{x} x ′ = A x x ( 0 ) = ( 1 , 0 ) T \mathbf{x}(0)=(1,0)^{\mathsf T} x ( 0 ) = ( 1 , 0 ) T
The factor e t e^t e t 2 2 2
[RIGHT CLICK] [ASK AI] [ EXAMPLE ] The third-order equation
y ′ ′ ′ + a y ′ ′ + b y ′ + c y = 0 y'''+ay''+by'+cy=0 y ′′′ + a y ′′ + b y ′ + cy = 0 becomes a first-order system after setting
x 1 = y , x 2 = y ′ , x 3 = y ′ ′ . x_1=y,\qquad x_2=y',\qquad x_3=y''. x 1 = y , x 2 = y ′ , x 3 = y ′′ . Then
d d t ( x 1 x 2 x 3 ) = ( 0 1 0 0 0 1 − c − b − a ) ( x 1 x 2 x 3 ) . \frac{d}{dt}
\begin{pmatrix}
x_1\\x_2\\x_3
\end{pmatrix}
=
\begin{pmatrix}
0&1&0\\
0&0&1\\
-c&-b&-a
\end{pmatrix}
\begin{pmatrix}
x_1\\x_2\\x_3
\end{pmatrix}. d t d x 1 x 2 x 3 = 0 0 − c 1 0 − b 0 1 − a x 1 x 2 x 3 . The first two rows record x 1 ′ = x 2 x_1'=x_2 x 1 ′ = x 2 x 2 ′ = x 3 x_2'=x_3 x 2 ′ = x 3 y ′ ′ ′ y''' y ′′′
[RIGHT CLICK] [ASK AI] [ EXAMPLE ] Let y 1 y_1 y 1 y 2 y_2 y 2
d y 1 d t = 1 2 y 1 − 1 5 y 2 , d y 2 d t = 3 5 y 1 − 2 5 y 2 , \frac{dy_1}{dt}=\frac12y_1-\frac15y_2,
\qquad
\frac{dy_2}{dt}=\frac35y_1-\frac25y_2, d t d y 1 = 2 1 y 1 − 5 1 y 2 , d t d y 2 = 5 3 y 1 − 5 2 y 2 , with
y ( 0 ) = ( 4 / 100 6 / 100 ) , A = ( 1 / 2 − 1 / 5 3 / 5 − 2 / 5 ) . \mathbf{y}(0)=
\begin{pmatrix}4/100\\6/100\end{pmatrix},
\qquad
A=
\begin{pmatrix}
1/2&-1/5\\
3/5&-2/5
\end{pmatrix}. y ( 0 ) = ( 4/100 6/100 ) , A = ( 1/2 3/5 − 1/5 − 2/5 ) . The characteristic polynomial is
det ( A − λ I ) = λ 2 − 1 10 λ − 2 25 , \det(A-\lambda I)
=\lambda^2-\frac1{10}\lambda-\frac2{25}, det ( A − λ I ) = λ 2 − 10 1 λ − 25 2 , so
λ 1 = 1 + 33 20 , λ 2 = 1 − 33 20 . \lambda_1=\frac{1+\sqrt{33}}{20},
\qquad
\lambda_2=\frac{1-\sqrt{33}}{20}. λ 1 = 20 1 + 33 , λ 2 = 20 1 − 33 . Eigenvectors with second component 1 1 1
v 1 = ( ( 9 + 33 ) / 12 1 ) , v 2 = ( ( 9 − 33 ) / 12 1 ) . v_1=
\begin{pmatrix}
(9+\sqrt{33})/12\\1
\end{pmatrix},
\qquad
v_2=
\begin{pmatrix}
(9-\sqrt{33})/12\\1
\end{pmatrix}. v 1 = ( ( 9 + 33 ) /12 1 ) , v 2 = ( ( 9 − 33 ) /12 1 ) . Writing the initial vector as y ( 0 ) = c 1 v 1 + c 2 v 2 \mathbf{y}(0)=c_1v_1+c_2v_2 y ( 0 ) = c 1 v 1 + c 2 v 2
c 1 = 33 − 33 1100 , c 2 = 33 + 33 1100 . c_1=\frac{33-\sqrt{33}}{1100},
\qquad
c_2=\frac{33+\sqrt{33}}{1100}. c 1 = 1100 33 − 33 , c 2 = 1100 33 + 33 . Hence
y ( t ) = 33 − 33 1100 e ( 1 + 33 ) t / 20 v 1 + 33 + 33 1100 e ( 1 − 33 ) t / 20 v 2 . \mathbf{y}(t)
=\frac{33-\sqrt{33}}{1100}
e^{(1+\sqrt{33})t/20}v_1
+\frac{33+\sqrt{33}}{1100}
e^{(1-\sqrt{33})t/20}v_2. y ( t ) = 1100 33 − 33 e ( 1 + 33 ) t /20 v 1 + 1100 33 + 33 e ( 1 − 33 ) t /20 v 2 . The first exponent is positive and the second is negative, so one eigenmode grows and the other decays.
The growing mode is the model's unstable direction: any non-zero component
of the initial state along v 1 v_1 v 1 v 2 v_2 v 2
[RIGHT CLICK] [ASK AI] Linear ODE systems Builds on Eigenvalues and eigenvectors and 5 more.