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Mathematics LibreTexts

7.7

( \newcommand{\kernel}{\mathrm{null}\,}\)

Example 1

Solve x(t)=[1345]x(t)

Solution

Step 1: Find eigenvalues:
AλI=|1λ345λ|=(1λ)(5λ)12=λ26λ+512=λ26λ7=(λ7)(λ+1)=0

Thus λ=7,1

Step 2: Find eigenvectors:
λ=7:A7I=[173457]=[6342]

We then see that [6342][12]=[00]. So the dimension of the nullspace of [6342] is 1 .
Or in other words, solution space for [6342][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][32]=[00]. Thus a basis for the eigenspace for λ=1 is {[32]}

Thus a basis for the solution space to x=[1345]x is
{[12]e7t,[32]et}

Hence the general solution is
x(t)=c1[12]e7t+c2[32]et

Note we can take any basis for the solution space to create the general solution

Alternate basis: {[24]e7t,[96]et}
Alternate format of general solution:
x(t)=c1[24]e7t+c2[96]et


Example 2

Solve the initial value problem x=[1345]x,x(0)=[ef]

Solution

 IVP: x=[1345]x,x(0)=[ef][ef]=c1[12]+c2[32]=[c1+3c22c12c2]

Solve using any method you like. We will use matrix form:
[ef]=[1322][c1c2]

Solution exists if Wronskian evaluated at 0 is not zero.
W([12]e7t,[32]et)=|e7t3et2e7t2et|=2e6t6e6t=8e6t0

Fundamental matrix: Ψ(t)=[e7t3et2e7t2et]
Back to IVP: [ef]=[1322][c1c2]
[1322]1[ef]=[1322]1[322][c1c2]

Thus [c1c2]=[1322]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[32]e0=[1322][c1c2][10]=[1322][c1c2] implies [c1c2]=[1322]1[10](18)[2321][1322]=[1001]

[1322]1=[14381418]&[c1c2]=[14381418][10]=[1414]

Thus IVP solution where x(0)=[10] is
x(t)=14[12]e7t+14[32]et=[14e7t+34et12e7t12et]

Step 2: Solve IVP: x=Ax,x(0)=[01]
[01]=c1[12]e0+c2[32]e0=[1322][c1c2]

Thus [c1c2]=[14381418][01]=[3818]
Thus IVP solution where x(0)=[01] is
x(t)=38[12]e7t18[32]et=[38e7t38et34e7t+14et]

Thus another basis for the solution space to x=[1345]x
 is {[14e7t+34et12e7t12et],[38e7t38et34e7t+14et]}

Its corresponding fundamental matrix is
Φ(t)=[14e7t+34et38e7t38et12e7t12et34e7t+14et]

Thus to solve IVP where x(t0)=[ef], we solve
[ef]=[14e7t0+34et038e7t038et012e7t012et034e7t0+14et0][c1c2]

When t0=0. I.e., we have an IVP where x(0)=[ef]
[ef]=[14e0+34e038e038e012e012e034e0+14e0][c1c2]
which simplifies to
[ef]=[1001][c1c2]=[c1c2]

In other words, c1=e and c2=f.

 

The Matrix Exponential

Definition: Matrix exponential

Let A be an n×n matrix. Then the matrix exponential of A is
exp(At)=eAt=I+Σn=1Aktkk!
where I is the n×n identity matrix.

Note

eA(0)=I+Σn=1Ak0kk!=I

Note

[eAt]=Σn=1kAktk1k!=Σn=1Aktk1(k1)!=Σn=0Ak+1tkk!
=A+Σn=1Ak+1tkk!=A(I+Σn=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

Example 3

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

Theorem 1

eAt=Φ(t)=Ψ(t)[Ψ(0)]1 where Ψ is a fundamental matrix for x=Ax

Example 4

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[32]et

Thus a fundamental matrix is Ψ(t)=[e7t3et2e7t2et]

A "better" fundamental matrix is Φ(t)=Ψ(t)[Ψ(0)]1
Ψ(0)=[1322]. Thus [Ψ(0)]1=[1322]1=(18)[2321]Φ(t)=Ψ(t)[Ψ(0)]1=(18)[e7t3et2e7t2et][2321]=[14e7t+34et38e7t38et12e7t12et34e7t+14et]

Thus exp([1345]t)=e[1345]t=[14e7t+34et38e7t38et12e7t12et34e7t+14et]

Example 5

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]

 

 

 

Theorem 1

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

 

Example 6

Solve IVP: x(t)=[1345]x(t),x(0)=[1792]

Solution

That is, x=17[14e7t+34et12e7t12et]+92[38e7t38et34e7t+14et]
Note: This only works if you use the fundamental matrix Φ(t) where Φ(0)=I.


This page titled 7.7 is shared under a not declared license and was authored, remixed, and/or curated by Isabel K. Darcy.

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