How to write coupled linear equations as one matrix equation, solve it with eigenvalues, eigenvectors and the matrix exponential, and read off whether solutions decay, grow or circle.
Two tanks of brine drain into each other, or a mass on a spring has both a position and a velocity. Each quantity changes at a rate that depends on both, so neither equation can be solved alone. Stacking the quantities into one vector turns the pair into a single equation x′=Ax, and the eigenvalues of the matrix A decide almost everything about how the solutions behave.
Key terms
First-order vector systems
Definition A first-order linear system for n unknown functions x1(t),…,xn(t) is
x′(t)=A(t)x(t)+g(t),
where x=⟨x1,…,xn⟩ is written as a column, A(t) is an n×n matrix of functions and g(t) is a vector of functions. It is homogeneous when g=0 and has constant coefficients when A does not depend on t. If the entries of A and g are continuous on an open interval I containing t0, then for every vector x0 the initial-value problem with x(t0)=x0 has exactly one solution, and it exists on all of I.
In plain words List all the changing quantities in one vector. Each row of the system says how fast one of them changes, in terms of all of them.
Example The equation y′′+3y′+2y=0 becomes a system once you name the state components first. Set x1=y and x2=y′. Then x1′=x2 and x2′=y′′=−2x1−3x2, so
x′=[0−21−3]x.
Matrices and the solution space
Definition For an n×n real matrix A and a vector x in Rn, the product Ax is the vector whose ith entry is ∑jaijxj. The map x↦Ax is linear. For the homogeneous system x′=Ax with A constant, every linear combination of solutions is again a solution, and the solutions form a vector space of dimension n. Solutions x1,…,xn form a fundamental set when their Wronskian
W(t)=det[x1(t)⋯xn(t)]
is nonzero. For solutions of the same system on an interval, W is either zero everywhere or zero nowhere, and for a fundamental set every solution equals c1x1+⋯+cnxn for unique constants.
In plain words The matrix is the rule that turns the current state into its velocity. Because that rule is linear, you only need n independent solutions, and every other solution is a mix of them.
Example For the matrix A above and x=⟨1,−1⟩,
Ax=[0⋅1+1⋅(−1)−2⋅1−3⋅(−1)]=[−11]=−x.
The matrix sends this vector to −1 times itself, which is the idea of the next section.
Eigenvalues and eigenvectors
Definition A number λ, real or complex, is an eigenvalue of the n×n matrix A if Av=λv for some nonzero vector v, called an eigenvector for λ. The eigenvalues are the roots of the characteristic equation det(A−λI)=0. For a 2×2 matrix with trace T and determinant D this equation is
λ2−Tλ+D=0.
If Av=λv, then x(t)=eλtv solves x′=Ax, since both sides equal λeλtv.
In plain words An eigenvector is a direction the matrix only stretches. A solution that starts along it stays on that line and grows or shrinks by the factor eλt.
Example For A=[0−21−3] the trace is −3 and the determinant is 2, so λ2+3λ+2=0 and λ=−1 or λ=−2. Solving (A+I)v=0 gives v1=⟨1,−1⟩, and solving (A+2I)v=0 gives v2=⟨1,−2⟩.
Diagonalizable systems
Definition The matrix A is diagonalizable if it has n linearly independent eigenvectors v1,…,vn, with eigenvalues λ1,…,λn. This holds in particular when the n eigenvalues are distinct. Then A=PΛP−1 with P=[v1⋯vn] and Λ diagonal, and the general solution of x′=Ax is
x(t)=c1eλ1tv1+⋯+cneλntvn.
When A is real and λ=α+iβ with β=0 has eigenvector v, the real and imaginary parts of eλtv are two linearly independent real solutions, and they replace the pair for λ and λˉ.
In plain words Split the initial state into eigenvector pieces. Each piece evolves on its own with its own exponential, and you add them back up.
Example Solve x′=Ax with the matrix above and x(0)=⟨1,0⟩. The conditions c1+c2=1 and −c1−2c2=0 give c1=2 and c2=−1, so
x(t)=2e−t[1−1]−e−2t[1−2].
The first component y=2e−t−e−2t solves y′′+3y′+2y=0 with y(0)=1 and y′(0)=0. For an oscillatory case take B=[−1−22−1], with λ=−1+2i and v=⟨1,i⟩. Taking real and imaginary parts of e(−1+2i)tv gives
e−t[cos2t−sin2t],e−t[sin2tcos2t].
Repeated eigenvalues
Definition Suppose the real 2×2 matrix A has a double eigenvalue λ, which is then real. There are two cases. If A=λI, every nonzero vector is an eigenvector and every solution is eλtx0. Otherwise A is defective, with only one independent eigenvector v. In that case (A−λI)2=0, so the column space of A−λI is the line spanned by v, and there is always a vector w with (A−λI)w=v, called a generalized eigenvector. Then
x1=eλtv,x2=eλt(tv+w)
form a fundamental set of solutions. For an n×n matrix, an eigenvalue of multiplicity m with fewer than m independent eigenvectors produces solutions of the form eλt times a vector polynomial in t of degree less than m.
In plain words With only one eigen direction, the second solution needs a factor of t, much like the repeated root case of a second-order equation.
Example For C=[1−113] the trace is 4 and the determinant is 4, so λ=2 twice. Here C−2I=[−1−111], so v=⟨1,1⟩, and w=⟨0,1⟩ works. The second solution is x2=e2t⟨t,t+1⟩. Checking the first row, x1′=(2t+1)e2t and x1+x2=(t+t+1)e2t, which match.
Matrix exponential
Definition For any n×n matrix A and real t,
eAt=k=0∑∞k!(At)k=I+At+2!(At)2+⋯,
a series that converges for every A and t. It satisfies dtdeAt=AeAt and eA⋅0=I, so the unique solution of x′=Ax with x(0)=x0 is x(t)=eAtx0. If A=PΛP−1 with Λ diagonal, then eAt=PeΛtP−1. The rule eA+B=eAeB holds when AB=BA and can fail otherwise.
In plain words The matrix exponential packs all solutions into one matrix. Multiply it by the starting state and you get the state at time t.
Example For the defective matrix C above, write C=2I+N with N=C−2I. The matrices 2I and N commute and N2=0, so the series for eNt stops after two terms and
eCt=e2t(I+tN)=e2t[1−t−tt1+t].
Phase portraits
Definition For a 2×2 real system x′=Ax, the phase plane is the x1x2 plane, and the phase portrait is the family of solution curves drawn there with arrows for increasing t. If detA=0, the origin is the only equilibrium, and the eigenvalues classify it. A node, star node or spiral is stable when the real parts are negative and unstable when they are positive.
Eigenvalues
Origin
real, opposite signs
saddle
real, distinct, same sign
node
real, equal, A=λI
proper (star) node
real, equal, A=λI
degenerate node
α±iβ, β=0, α=0
spiral
±iβ, β=0
center
In plain words Each solution is a path in the plane traced over time. Real eigenvalues give straight-line solutions along the eigenvectors, and complex ones make paths turn around the origin.
Example For B=[−1−22−1] the eigenvalues −1±2i give a spiral. To find the turning direction, evaluate the velocity at ⟨1,0⟩. It is B⟨1,0⟩=⟨−1,−2⟩, which points downward, so paths turn clockwise while the factor e−t pulls them in.
Stability
Definition The equilibrium 0 of x′=Ax is stable if for every ε>0 there is δ>0 such that ∥x(0)∥<δ implies ∥x(t)∥<ε for all t≥0. It is asymptotically stable if it is stable and every solution tends to 0 as t→∞, and unstable if it is not stable. For constant real A,
if every eigenvalue has negative real part, 0 is asymptotically stable,
if some eigenvalue has positive real part, 0 is unstable,
if every real part is at most 0 and some is 0, then 0 is stable but not asymptotically stable exactly when each eigenvalue with zero real part has as many independent eigenvectors as its multiplicity, and unstable otherwise.
In plain words Negative real parts mean every mode decays. A neutral mode with zero real part neither grows nor decays by itself, but if its eigenvalue is defective the extra factor t makes it grow slowly.
Example The matrix N=[0010] has the double eigenvalue 0, so no mode grows exponentially. Yet eNt=I+tN and the solution from ⟨a,b⟩ is ⟨a+bt,b⟩, which is unbounded when b=0. The origin is unstable. By contrast, [0−110] has eigenvalues ±i, its solutions are circles, and the origin is a stable center that is not asymptotically stable.
Check yourself
Solve x′=[1221]x with x(0)=⟨2,0⟩ and classify the origin.
Show the answer
The eigenvalues are 3 with ⟨1,1⟩ and −1 with ⟨1,−1⟩. Matching x(0) gives c1=c2=1, so
x(t)=e3t[11]+e−t[1−1].
The eigenvalues have opposite signs, so the origin is a saddle and it is unstable.
Write y′′+2y′+5y=0 as a system with x1=y and x2=y′, then classify the origin and its stability.
Show the answer
The system is x1′=x2 and x2′=−5x1−2x2. The matrix has trace −2 and determinant 5, so λ2+2λ+5=0 and λ=−1±2i. The origin is a spiral sink, asymptotically stable. At ⟨1,0⟩ the velocity is ⟨0,−5⟩, so paths turn clockwise.
Compute eAt for A=[3013] and solve x′=Ax with x(0)=⟨1,2⟩.
Show the answer
Write A=3I+N with N=[0010] and N2=0. Then
eAt=e3t[10t1],
and the solution is x(t)=e3t⟨1+2t,2⟩.
How this lesson connects
It also uses Matrices and linear maps, Scalar linear differential equations and Complex numbers and exponentials.
May we count page visits to understand which courses and learning features are useful? Cloudflare collects visit and performance statistics, without your account details or answers. Your choice does not affect access. Details