Solving a Coupled System with Eigenvalues

Guess a straight-line solution

Take the pair $\frac{dx}{dt}=x+y$dx/dt = x + y, $\frac{dy}{dt}=4x+y$dy/dt = 4x + y, written $\frac{dX}{dt}=MX$dX/dt = M X with M = (1 1; 4 1). For one equation, $\frac{dx}{dt}=kx$dx/dt = kx, the solution is $e^{kt}$e^(kt) times a constant. So try the same shape here: $X=e^{\lambda t}v$X = e^(λt) v, a fixed column v multiplied by one exponential.

Such a solution keeps one direction, the direction of v, and only changes its length, at the single rate $\lambda$λ. Its derivative is $\frac{dX}{dt}=\lambda e^{\lambda t}v$dX/dt = λ e^(λt) v.

The guess needs an eigenvector

Put the guess into both sides. The left is $\lambda e^{\lambda t}v$λ e^(λt) v. The right is $M{e}^{\lambda t}v$M e^(λt) v, and since $e^{\lambda t}$e^(λt) is a number it can be taken out: $e^{\lambda t}Mv$e^(λt) M v. The exponential is never 0, so dividing by it leaves $Mv=\lambda v$M v = λ v.

That is the eigenvector equation. The guess works exactly when v is an eigenvector of M and $\lambda$λ is its eigenvalue: a direction M only stretches, by the factor $\lambda$λ. Along that direction the rate of X is $\lambda$λ times X, so X grows or shrinks like $e^{\lambda t}$e^(λt) and never turns.

The eigenvalues

v must not be the zero column, so $M-\lambda I$M − λI must send some non-zero column to 0, which means $\det \left(M-\lambda I\right)=0$det(M − λI) = 0. Here $M-\lambda I$M − λI has rows $\left(1-\lambda ,1\right)$(1 − λ, 1) and $\left(4,1-\lambda \right)$(4, 1 − λ), and its determinant is ${\left(1-\lambda \right)}^2-4={\lambda }^2-2\lambda -3=\left(\lambda -3\right)\left(\lambda +1\right)$(1 − λ)² − 4 = λ² − 2λ − 3 = (λ − 3)(λ + 1). So $\lambda =3$λ = 3 or $\lambda =-1$λ = −1.

Two checks: the eigenvalues add to the diagonal total, 1 + 1 = 2 = 3 + (−1), and multiply to the determinant, 1 × 1 − 1 × 4 = −3 = 3 × (−1).

The eigenvectors

For $\lambda =3$λ = 3, M − 3I = (−2 1; 4 −2). Its first row says −2x + y = 0, so y = 2x, and the second row says the same. Take $v_1=\left(1,2\right)$v₁ = (1, 2). Check: M sends (1; 2) to (1 + 2; 4 + 2) = (3; 6), which is 3 times (1; 2).

For $\lambda =-1$λ = −1, M + I = (2 1; 4 2). Its rows say 2x + y = 0, so y = −2x. Take $v_2=\left(1,-2\right)$v₂ = (1, −2). Check: M sends (1; −2) to (1 − 2; 4 − 2) = (−1; 2), which is −1 times (1; −2).

So $e^{3t}\left(1,2\right)$e^(3t)(1, 2) and $e^{-t}\left(1,-2\right)$e^(−t)(1, −2) are both solutions. The first runs out along the line y = 2x, tripling its rate as it goes; the second runs in along y = −2x toward the origin.

The general solution

The equation is linear: $M\left(X_1+X_2\right)=M{X}_1+M{X}_2$M(X₁ + X₂) = M X₁ + M X₂, and the derivative of a sum is the sum of the derivatives. So two solutions added are a solution, and so is any multiple of one. Every solution of this pair is $X=A{e}^{3t}\left(1,2\right)+B{e}^{-t}\left(1,-2\right)$X = A e^(3t)(1, 2) + B e^(−t)(1, −2), with one constant for each term.

Written out, $x=A{e}^{3t}+B{e}^{-t}$x = A e^(3t) + B e^(−t) and $y=2A{e}^{3t}-2B{e}^{-t}$y = 2A e^(3t) − 2B e^(−t). Substitute back: $\frac{dx}{dt}=3A{e}^{3t}-B{e}^{-t}$dx/dt = 3A e^(3t) − B e^(−t), and $x+y=3A{e}^{3t}-B{e}^{-t}$x + y = 3A e^(3t) − B e^(−t), the same. $\frac{dy}{dt}=6A{e}^{3t}+2B{e}^{-t}$dy/dt = 6A e^(3t) + 2B e^(−t), and $4x+y=6A{e}^{3t}+2B{e}^{-t}$4x + y = 6A e^(3t) + 2B e^(−t), the same.

The starting values fix A and B

At t = 0 both exponentials are 1, so X(0) = A(1, 2) + B(1, −2). With X(0) = (2, 0) this says A + B = 2 and 2A − 2B = 0. The second gives A = B, and the first then gives A = B = 1. So $x=e^{3t}+e^{-t}$x = e^(3t) + e^(−t) and $y=2e^{3t}-2e^{-t}$y = 2e^(3t) − 2e^(−t).

Check it at t = 0.5: $x=e^{1.5}+e^{-0.5}\approx 5.088$x = e^1.5 + e^(−0.5) ≈ 5.088 and $y\approx 7.750$y ≈ 7.750. A difference quotient gives $\frac{dx}{dt}\approx 12.839$dx/dt ≈ 12.839 there, and $x+y\approx 12.839$x + y ≈ 12.839 too.

A different start gives different constants. From X(0) = (1, 6): A + B = 1 and 2A − 2B = 6, so A − B = 3, A = 2 and B = −1. A start on an eigenvector needs only one term: from (1, −2), A = 0 and B = 1, and the point runs straight in along y = −2x, halving its distance every $\ln 2\approx 0.693$ln 2 ≈ 0.693 units of time.

The long run

As t grows, $e^{3t}$e^(3t) grows without limit and $e^{-t}$e^(−t) shrinks to 0. So once $B{e}^{-t}\left(1,-2\right)$B e^(−t)(1, −2) is small, X is almost $A{e}^{3t}\left(1,2\right)$A e^(3t)(1, 2), and the point lines up along (1, 2). On the path from (2, 0), $\frac{y}{x}$y/x is 0.924 at t = 0.25, 1.523 at t = 0.5, 1.928 at t = 1 and 1.999 at t = 2, closing on 2.

What decides it is the larger eigenvalue, 3, not the larger size: with eigenvalues 3 and −5, the $e^{-5t}$e^(−5t) term dies faster still and $e^{3t}$e^(3t) wins all the same. The one exception is a start with A = 0, on the eigenvector of the other eigenvalue.

The eigenvector lines $y=2x\left(\lambda =3\right)$y = 2x (λ = 3) and $y=-2x\left(\lambda =-1\right)$y = −2x (λ = −1). The gold path is the solution through (2, 0), with A = B = 1: it comes in from below beside y = −2x, passes (2, 0) at t = 0 and (2.896, 2.676) at t = 0.25, and turns toward the direction of y = 2x as it leaves the frame at (4.283, 5.965), where $\frac{y}{x}$y/x is already 1.39.

A second pair

Take x' = y and y' = x, so M = (0 1; 1 0). Its characteristic polynomial is ${\lambda }^2-1$λ² − 1, so $\lambda =1$λ = 1 with eigenvector (1, 1) and $\lambda =-1$λ = −1 with eigenvector (1, −1). The eigenvalues multiply to −1, which is det M = 0 × 0 − 1 × 1.

The general solution is $A{e}^t\left(1,1\right)+B{e}^{-t}\left(1,-1\right)$A e^t(1, 1) + B e^(−t)(1, −1). From X(0) = (3, 1): A + B = 3 and A − B = 1, so A = 2 and B = 1, giving $x=2e^t+e^{-t}$x = 2eᵗ + e^(−t) and $y=2e^t-e^{-t}$y = 2eᵗ − e^(−t). Check: the derivative of x is $2e^t-e^{-t}$2eᵗ − e^(−t), which is y, and the derivative of y is $2e^t+e^{-t}$2eᵗ + e^(−t), which is x.

The usual mistakes

Pairing an exponential with the wrong direction. $e^{3t}$e^(3t) goes with the eigenvector of 3: $A{e}^{3t}\left(1,-2\right)+B{e}^{-t}\left(1,2\right)$A e^(3t)(1, −2) + B e^(−t)(1, 2) does not solve the pair.

Leaving out the vectors. $A{e}^{3t}+B{e}^{-t}$A e^(3t) + B e^(−t) is one function; the solution has two, x and y, and the eigenvectors say how they are related.

Fitting the constants to an eigenvector instead of to the start. From (2, 0), A = B = 1, not A = 2 and B = 0: that would need the start to lie on y = 2x.

Choosing the dominant term by size. In the long run the term with the larger eigenvalue wins; a large negative eigenvalue makes its term vanish fastest.

A drug in the body, and an enzyme

In the applications below, a dose divides between the blood and the tissues, and the eigenvector of the eigenvalue nearer zero gives the split that lasts. In the second, one eigenvalue is 0, so one combination of the two sugars never changes while the other dies away.