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The Ordinary Differential Equations Project

Thomas W. Judson

Section 3.8 Linear Systems in Higher Dimensions

Suppose that we have two masses on a table, m1 and m2, connected by three springs with the outside springs connected to two walls (Figure 3.8.1), and the masses are free to move horizontally. We will assume that the springs are uniform and all have the same spring constant k. The horizontal displacements of the springs are denoted by x1(t) and x2(t) for the masses m1 and m2, respectively. Assuming that there is no damping, the only forces acting on mass m1 at time t are those of left and middle springs. The force from the left spring will be −kx1 while the force from middle spring will be k(x2−x1). By Newton’s Second Law of motion, we have
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m1x1″=−kx1+k(x2−x1).
Similarly, the only forces acting on the second mass, m2, will come from middle and right springs. Again using Newton’s Second Law,
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m2x2″=−k(x2−x1)−kx2.
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Figure 3.8.1. A double spring-mass system
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We now have a system of two second-order linear equations
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m1x1″=−2kx1+kx2m2x2″=kx1−2kx2.
If we define x3 and x4 by x3=x1′ and x4=x2′, we now have first-order linear system of four equations,
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x1′=x3x2′=x4x3′=−2km1x1+km1x2x4′=km2x1−2km2x2
We can represent this system in the matrix form x′=Ax, where
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A=(00100001−2k/m1k/m100k/m2−2k/m200).
We will learn how to analyze and solve such systems in the next two sections.
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Subsection 3.8.1 Higher-Order Linear Systems

We can write the system
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x1′=a11x1+⋯a1nxnx2′=a21x1+⋯a2nxn⋮xn′=an1x1+⋯annxn
in matrix form x′=Ax, where
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A=(a11a12⋯a1na21a22⋯a2n⋮⋮⋱⋮an1an2⋯ann)andx=(x1x2⋮xn).
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The strategy for finding solutions to the system x′=Ax is the same as for systems of two equations. If λ is an eigenvalue of A with eigenvector v,
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x(t)=eλtv
is a solution for our system. Indeed,
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x′(t)=λeλtv=eλt(λv)=eλtAv=A(eλtv)=Ax(t).
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Example 3.8.2.

The system
x′=−5x−8y−2zy′=5x+12y+4zz′=−11x−19y−5z
can be rewritten as
x′=(x′y′z′)=(−5−8−25124−11−19−5)(xyz)=Ax.
We can compute the eigenvalues of A by finding the roots of its characteristic polynomial
det(A−λI)=−λ3−2λ2−λ+2=(λ−1)(λ+1)(λ−2).
Thus, the eigenvalues of A are λ=2, λ=1, and λ=−1. To find an eigenvector for λ=2, we must find a nontrivial solution for system of equations (A−2I)x=0,
−7x−8y−2z=0−5x+10y+4z=0−11x−19y−7z=0.
It is easy to check that (2,−3,5) is a solution. Similarly, we can determine that (5,−7,13) is an eigenvector for λ=1 and (1,−1,2) is an eigenvector for λ=−1. Thus, we have found three solutions for the system x′=Ax,
x1(t)=e2t(2−35),x2(t)=et(5−713),andx3(t)=e−t(1−12).
The Principle of Superposition also holds for higher-order systems. If x1(t) and x2(t) are solutions for x′=Ax, then
c1x1(t)+c2x2(t)
is a solution for the system, since
ddt[c1x1(t)+c2x2(t)]=c1ddtx1(t)+c2ddtx2(t)=c1Ax1(t)+c2Ax2(t)=A[c1x1(t)+c2x2(t)].
Consequently,
x(t)=c1e2t(2−35)+et(5−713)+c3e−t(1−12)
is a solution for our system. This is, in fact, the general solution for the system.
Although we shall not cover the notions of linear independence, canonical matrices, and change of coordinates for Rn, the same ideas that we used for systems of first-order linear differential equations in R2 carry over to Rn. The necessary linear algebra is covered in any good linear algebra course. Also, see Chapters 5 and 6 in [12]. In addition, finding eigenvalues for matrices greater than 2×2, we will need to find the roots of a characteristic polynomial of degree greater than two, which can be difficult. A good course in linear algebra will cover techniques of finding eigenvalues for larger matrices.
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Example 3.8.3.

Let us see how the linear algebra works in the previous example. If we form the matrix
T=(251−3−7−15132)
from the eigenvectors of A, we can convert the system x′=Ax to the system
y′=(T−1AT)y(1−3−2−11141−1)(−5−8−25124−11−19−5)(251−3−7−15132)y,=(20001000−1)y,
which we can immediately solve:
y(t)=c1e2t(100)+c2et(010)+c3e−t(001).
Multiplying our solution by T yields the general solution
x(t)=Ty=c1e2t(2−35)+et(5−713)+c3e−t(1−12),
the solution to the system x′Ax.
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Subsection 3.8.2 The Geometry of Solutions

In Section 3.6, we classified all of the geometry of the solutions for planar systems using the trace-determinant plane. The geometry for linear systems in three variables is a bit more complicated. For a system
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x′=a1x+a1y+a3zy′=b1x+b2y+b3zz′=c1x+c2y+c3z,
our solution curves live in R3, and there is simply a lot more room to move around in three dimensions than in two dimensions. The origin is still an equilibrium solution for a system of linear differential equations in three variables. The origin is a stable equilibrium solution if any solution x(t) approaches 0=(0,0,0) as t→∞; otherwise, 0 is an unstable equilibrium solution. In the case of planar systems, an unstable solution is a nodal saddle, a nodal source, a spiral source, or a source with a single unstable line. In the case of R3, we could have a stable line of solutions and an unstable plane of solutions. In this case, all solutions of the system with initial condition lying on the stable line would approach the origin as t→∞, but all solutions with initial conditions that are a nonzero point on the unstable plane would move away from the origin.
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Example 3.8.4.

In Example 3.8.2, we had solutions
x1(t)=e2t(2−35),x2(t)=et(5−713),andx3(t)=e−t(1−12).
The straight line through the origin and the point (1,−1,2) is a stable line. That is, for any initial condition x(0)=(x0,y0,z0) lying on this line, our solution will tend toward the origin as t→∞. On the other hand, the plane spanned by (2,−3,5) and (5,−7,13) is unstable plane. Solutions on this plane move away from the origin as t→∞. Of course, (0,0,0) is an equlibrium solution for our system. We say that the origin is a saddle in this example (Figure 3.8.5).
a stable line and an unstable plane in three dimensions with a transformation to another unstable plane and stable line
Figure 3.8.5. A saddle in R3

Example 3.8.6.

For the system
x′=(010−10000−1)x
we have a very different type of unstable equilibrium solution. The eigenvalues of this matrix are λ=±i and λ=−1. Thus, a solution satisfying the initial condition x(0)=(x0,y0,z0) is given by
x(t)=x0(cos⁡t−sin⁡t0)+y0(sin⁡tcos⁡t0)+z0e−t(001).
If z0=0, then our initial condition is in the xy-plane and all of the solutions lie on circles centered at the origin. On the other hand, if x0=0 and y0=0, we have a stable line of solutions lying along the z-axis. In fact, each solution that does not lie on the stable line lies on a cylinder in R3 given by x2+y2=r2 for some constant r>0. These solutions spiral towards the circular solution of radius r in the xy-plane if z0≠0 (Figure 3.8.7).
Figure 3.8.7. A spiral center in R3

Example 3.8.8.

For an example of a stable plane and an unstable line, let us consider the system
x′=(−110−1−10001)x=Ax.
The characteristic equation of the matrix A is
λ3+λ2−2=(λ−1)(λ2+2λ+2)=0.
Thus, the eigenvalues of A are λ=1 and λ=−1±i. Solving
[A−(−1+i)]x=0
gives us an eigenvector (1,i,0) for λ=−1+i. Since
(1i0)=(100)+i(010),
we will let v1=(1,0,0) and v2=(0,1,0). Since v3=(0,0,1) is an eigenvector for λ=1, our system has solution
x(t)=x0e−t(cos⁡t−sin⁡t0)+y0e−t(sin⁡tcos⁡t0)+z0et(001),
where x(0)=(x0,y0,z0). If z0=0, our initial condition lies in the xy-plane and solution curves spiral in towards the origin. Thus, we have a stable plane. On the other hand, if x0=0 and y0=0 but z0≠0, then our solution approaches ±∞ as t→∞. In this case, the z-axis is an unstable line (Figure 3.8.9).
Figure 3.8.9. A spiral saddle in R3
For an example of a stable equilibrium solution at the origin, consider the system
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x′=(λ1000λ2000λ3)x,
where λ3<λ2<λ1<0. For an initial condition (x0,y0,z0) with at least one coordinate nonzero, the corresponding solution tends towards the origin tangentially to the x-axis as t→∞ (Figure 3.8.10).
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Figure 3.8.10. A sink in R3
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Changing the system that in ExampleExample 3.8.8 to be
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x′=(−110−1−1000−1)x,
we obtain the solution satisfying the initial condition x(0)=(x0,y0,z0) to be
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x(t)=x0e−t(cos⁡t−sin⁡t0)+y0e−t(sin⁡tcos⁡t0)+z0e−t(001),
where x(0)=(x0,y0,z0). In this case, all solutions will approach the origin as t→∞.
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Activity 3.8.1. Solving Higher Order Systems.

  1. Find the eigenvalues of A. You may find Sage useful.
  2. Find the eigenvectors for each eigenvalue of A.
  3. What is the general solution of dx/dt=Ax?
  4. Describe the nature of the nature of the solution space in R3. Are there stable lines or planes?
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Subsection 3.8.3 The Double Spring-Mass Systems Revisited

Let us return to our spring-mass system x′=Ax, where
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A=(00100001−2k/m1k/m100k/m2−2k/m200).
To keep matters simple, we will assume that m1=m2=k=1. Thus, our matrix now becomes
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A=(00100001−21001−200).
The characteristic polynomial of A is
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det(A−λI)=λ4+4λ2+3=(λ2+1)(λ2+3).
Thus, the eigenvalues of A are λ=±i and λ=±i3. We can find eigenvectors
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v1=(i/3−i/3−11),v2=(−i/3i/3−11),v3=(−i−i11),v4=(ii11),
corresponding to the eigenvalues λ1=i3, λ2=−i3, λ3=i, λ4=−i, respectively. Consequently, the general solution to our system is
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x(t)=c1ei3tv1+c1e−i3tv2+c1eitv3+c1e−itv4;
however, this form of the solution is not very useful. By examining real and imaginary parts of ei3tv1 and c1eitv3, we can rewrite the solution as
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x(t)=c1(cos⁡3t−cos⁡3t−3sin⁡3t3sin⁡3t)+c2(−sin⁡3tsin⁡3t−3cos⁡3t3cos⁡3t)+c3(cos⁡tcos⁡t−sin⁡t−sin⁡t)+c4(sin⁡tsin⁡tcos⁡tcos⁡t).
If we have the following initial conditions,
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x1(0)=0x2(0)=0x1′(0)=x3(0)=2x2′(0)=x4(0)=2,
we can determine c1=c2=c3=0 and c4=2. Thus, the solution to our initial value problem is
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x(t)=(2sin⁡t2sin⁡t2cos⁡t2cos⁡t),
and the two masses will oscillate with a frequency of one and an amplitude of two. We leave the details as an exercise.
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Subsection 3.8.4 Important Lessons

  • x1′=a11x1+⋯a1nxnx2′=a21x1+⋯a2nxn⋮xn′=an1x1+⋯annxn
    can be written in matrix form x′=Ax, where
    A=(a11a12⋯a1na21a22⋯a2n⋮⋮⋱⋮an1an2⋯ann)andx=(x1x2⋮xn).
  • As in the case of R2, we can solve the system x′=Ax by finding eigenvalues and eigenvectors for A.
  • The Principle of Superposition holds for higher-order systems. If x1(t) and x2(t) are solutions for x′=Ax, then
    c1x1(t)+c2x2(t)
    is a solution for the system.
  • The geometry for a system in R3 is more complicated than the planar case. However, the solutions are usually characterized by stable lines or stable planes.
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Reading Questions 3.8.5 Reading Questions

1.

Is it possible for a 3×3 to have a line of stable solutions and a plane of unstable solutions? Explain.
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Exercises 3.8.6 Exercises

Finding Solutions of 3×3 Systems.

For each of the linear systems dx/dt=Ax in Exercise Group 3.8.6.1–8
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  1. Find the eigenvalues of A. You may find Sage useful.
  2. Find the eigenvectors for each eigenvalue of A.
  3. What is the general solution of dx/dt=Ax?
  4. Describe the nature of the nature of the solution space in R3. Are there stable lines or planes?
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Solving Initial-Value Problems 3×3 Systems.

Solve each of initial-value problems in Exercise Group 3.8.6.9–16
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9.
x′=4x+3y−zy′=−x−zz′=x+y+2zx(0)=−1y(0)=1z(0)=0
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10.
x′=−2x−3y+4zy′=4x+5y−4zz′=5x+5y−3zx(0)=2y(0)=−1z(0)=−1
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11.
x′=4x+2yy′=2yz′=2x+2y+2zx(0)=1y(0)=1z(0)=1
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12.
x′=−2y−2zy′=4x+6y−zz′=6x+6y−2zx(0)=1y(0)=2z(0)=3
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13.
x′=11x+14y−9zy′=−5x−8y+7zz′=4x+4y+2zx(0)=−1y(0)=−2z(0)=−3
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14.
x′=11x+13y−6zy′=−6x−8y+6zz′=x−2y+4zx(0)=−2y(0)=2z(0)=0
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15.
x′=5y+2zy′=4x−10y−5zz′=−6x+21y+10zx(0)=1y(0)=1z(0)=3
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16.
x′=5x+7y−zy′=−x−3y+zz′=6x+6y−2zx(0)=−1y(0)=0z(0)=2
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