7.7
( \newcommand{\kernel}{\mathrm{null}\,}\)
Solve x′(t)=[1345]x(t)
Solution
Step 1: Find eigenvalues:
A−λI=|1−λ345−λ|=(1−λ)(5−λ)−12=λ2−6λ+5−12=λ2−6λ−7=(λ−7)(λ+1)=0
Thus λ=7,−1
Step 2: Find eigenvectors:
λ=7:A−7I=[1−7345−7]=[−634−2]
We then see that [−634−2][12]=[00]. So the dimension of the nullspace of [−634−2] is 1 .
Or in other words, solution space for [−634−2][x1x2]=[00] is 1-dimensional.
Thus a basis for the eigenspace for λ=7 is {[12]}
λ=−1A−(−1)I=[1+1345+1]=[2346]
Similar to above [2346][3−2]=[00]. Thus a basis for the eigenspace for λ=−1 is {[3−2]}
Thus a basis for the solution space to x′=[1345]x is
{[12]e7t,[3−2]e−t}
Hence the general solution is
x(t)=c1[12]e7t+c2[3−2]e−t
Note we can take any basis for the solution space to create the general solution
Alternate basis: {[24]e7t,[−96]e−t}
Alternate format of general solution:
x(t)=c1[24]e7t+c2[−96]e−t
Solve the initial value problem x′=[1345]x,x(0)=[ef]
Solution
IVP: x′=[1345]x,x(0)=[ef][ef]=c1[12]+c2[3−2]=[c1+3c22c1−2c2]
Solve using any method you like. We will use matrix form:
[ef]=[132−2][c1c2]
Solution exists if Wronskian evaluated at 0 is not zero.
W([12]e7t,[3−2]e−t)=|e7t3e−t2e7t−2e−t|=−2e6t−6e6t=−8e6t≠0
Fundamental matrix: Ψ(t)=[e7t3e−t2e7t−2e−t]
Back to IVP: [ef]=[132−2][c1c2]
[132−2]−1[ef]=[132−2]−1[32−2][c1c2]
Thus [c1c2]=[132−2]−1[ef]
But I would prefer a fundamental matrix whose inverse is easier to calculate, at least when t0=0.
Thus we will find another basis for the solution set to x′=Ax so that the corresponding fundamental matrix Φ has the property that Φ(0)=[1001], the 2x2 identity matrix via long method:
Step 1: Solve IVP: x′=Ax,x(0)=[10]
[10]=c1[12]e0+c2[3−2]e0=[132−2][c1c2][10]=[132−2][c1c2] implies [c1c2]=[132−2]−1[10](−18)[−2−3−21][132−2]=[1001]
[132−2]−1=[143814−18]&[c1c2]=[143814−18][10]=[1414]
Thus IVP solution where x(0)=[10] is
x(t)=14[12]e7t+14[3−2]e−t=[14e7t+34e−t12e7t−12e−t]
Step 2: Solve IVP: x′=Ax,x(0)=[01]
[01]=c1[12]e0+c2[3−2]e0=[132−2][c1c2]
Thus [c1c2]=[143814−18][01]=[38−18]
Thus IVP solution where x(0)=[01] is
x(t)=38[12]e7t−18[3−2]e−t=[38e7t−38e−t34e7t+14e−t]
Thus another basis for the solution space to x′=[1345]x
is {[14e7t+34e−t12e7t−12e−t],[38e7t−38e−t34e7t+14e−t]}
Its corresponding fundamental matrix is
Φ(t)=[14e7t+34e−t38e7t−38e−t12e7t−12e−t34e7t+14e−t]
Thus to solve IVP where x(t0)=[ef], we solve
[ef]=[14e7t0+34e−t038e7t0−38e−t012e7t0−12e−t034e7t0+14e−t0][c1c2]
When t0=0. I.e., we have an IVP where x(0)=[ef]
[ef]=[14e0+34e038e0−38e012e0−12e034e0+14e0][c1c2]
which simplifies to
[ef]=[1001][c1c2]=[c1c2]
In other words, c1=e and c2=f.
The Matrix Exponential
Let A be an n×n matrix. Then the matrix exponential of A is
exp(At)=eAt=I+Σ∞n=1Aktkk!
where I is the n×n identity matrix.
eA(0)=I+Σ∞n=1Ak0kk!=I
[eAt]′=Σ∞n=1kAktk−1k!=Σ∞n=1Aktk−1(k−1)!=Σ∞n=0Ak+1tkk!
=A+Σ∞n=1Ak+1tkk!=A(I+Σ∞n=1Aktkk!)=AeAt
Thus [eAt]′=AeAt and eA(0)=I. Thus eAt is the solution to the IVP M′=AM,M(0)=I where M is an n×n matrix.
Let Ψ be a fundamental matrix for x′=Ax
Suppose Ψ is the 2×2 matrix [f1f2].
Solution
Thus f1 and f2 are solutions to x′=Ax.
Hence f1′=Af1 and f2′=Af2
Thus A[f1f2]=
Since [f1f2]′=A[f1f2],Ψ(t)=[f1(t)f2(t)] is the general solution to M′=AM.
Let Φ(t)=Ψ(t)[Ψ(0)]−1. Then Φ(0)=Ψ(0)[Ψ(0)]−1=I.
Note Φ(t)=Ψ(t)[Ψ(0)]−1 is also a solution to M′=AM :
(Ψ(t)[Ψ(0)]−1)′=Ψ′(t)[Ψ(0)]−1=AΨ(t)[Ψ(0)]−1
Thus Φ is the solution to the IVP
M′=AM,M(0)=I
Since solution to IVP is unique (assuming entries of A are continuous functions), Φ=eAt
eAt=Φ(t)=Ψ(t)[Ψ(0)]−1 where Ψ is a fundamental matrix for x′=Ax
Calculate exp([1345]t):=e[1345]t
Solution
Solve x′(t)=[1345]x(t)
From previous work, the general solution is
x(t)=c1[12]e7t+c2[3−2]e−t
Thus a fundamental matrix is Ψ(t)=[e7t3e−t2e7t−2e−t]
A "better" fundamental matrix is Φ(t)=Ψ(t)[Ψ(0)]−1
Ψ(0)=[132−2]. Thus [Ψ(0)]−1=[132−2]−1=(−18)[−2−3−21]Φ(t)=Ψ(t)[Ψ(0)]−1=(−18)[e7t3e−t2e7t−2e−t][−2−3−21]=[14e7t+34e−t38e7t−38e−t12e7t−12e−t34e7t+14e−t]
Thus exp([1345]t)=e[1345]t=[14e7t+34e−t38e7t−38e−t12e7t−12e−t34e7t+14e−t]
Solve IVP:x′=Ax,x(0)=e
Solution
Observe if Φ(t)=[f1f2], then general solution is
x(t)=c1f1(t)+c2f2(t)=[f1(t)f2(t)][c1c2]=Φ(t)[c1c2]x(0)=c1f1(0)+c2f2(0)=[f1(0)f2(0)][c1c2]=Φ(0)[c1c2]=[1001][c1c2]=[c1c2]
If eAt=Φ(t)=Ψ(t)[Ψ(0)]−1 where Ψ is a fundamental matrix for x′=Ax
or equivalently, if Φ is the solution to IVP M′=AM,M(0)=I, Then solution to IVP x′=Ax,x(0)=e is x=Φe
Solve IVP: x′(t)=[1345]x(t),x(0)=[1792]
Solution
That is, x=17[14e7t+34e−t12e7t−12e−t]+92[38e7t−38e−t34e7t+14e−t]
Note: This only works if you use the fundamental matrix Φ(t) where Φ(0)=I.