9.5 Solving Linear Systems

Many engineering problems involve two or more unknown functions that influence each other: the temperatures of two rooms that exchange heat, the currents in two loops of a circuit, the positions of two masses joined by springs. In Chapter 7 such systems were solved with operators and with eigenvalues. The Laplace transform offers a third way, and it is often the most direct one when initial conditions are given: each differential equation turns into a linear algebraic equation, and we are left with an ordinary $2\times 2$ linear system for the transforms.

The method

Consider the system

with constant coefficients $a_{ij}$ and forcing functions $f_1,f_2$ that are piecewise continuous and of exponential order. Write $X(s)=\mathcal{L}\{x\}$, $Y(s)=\mathcal{L}\{y\}$, $F_1=\mathcal{L}\{f_1\}$, $F_2=\mathcal{L}\{f_2\}$.

Transform each equation. With $\mathcal{L}\{x^{\prime }\}=sX-x_0$ and $\mathcal{L}\{y^{\prime }\}=sY-y_0$:

Collect $X$ and $Y$ on the left.

This is a linear system of two equations in two unknowns, $X$ and $Y$, whose coefficients happen to depend on $s$. Any method from algebra will do. For $2\times 2$ systems, Cramer's rule is the cleanest.

In our case the coefficient determinant is

This is exactly $\det (sI-A)$, the characteristic polynomial of the coefficient matrix $A=\left(\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right)$. Its roots are the eigenvalues of $A$ from Chapter 7. So, once again, the poles of the transforms are the characteristic roots, and they decide whether the solution decays, grows or oscillates. Since $\Delta (s)$ is a nonzero polynomial, it vanishes only at finitely many $s$, so Cramer's rule applies for all large $s$.

Compare with Chapter 7: the eigenvalues of $A$ are $4$ and $-1$, and the vectors multiplying $e^{4t}$ and $e^{-t}$, namely $(2,3)$ and $(-1,1)$, are eigenvectors. The Laplace method found them without computing any eigenvectors and fitted the initial conditions automatically.

A forced system: two rooms and a heater

Two adjacent rooms exchange heat with each other and with the outdoors. Let $x(t)$ and $y(t)$ be their temperatures measured above the outdoor temperature. With suitable time units, each room loses heat to the outdoors at a rate proportional to its own temperature and exchanges heat with the other room at a rate proportional to the difference. A heater in room 2 supplies heat at a constant rate. A simple model is

Try the system solver on your own problems. Enter the matrix $A$ row by row (rows separated by a semicolon), the two forcing functions and the initial values. It shows the transformed equations, the determinant, Cramer's rule and the inverse transforms. The first preset is Example 9.5.1, the second is Example 9.5.2. Change the heater input $f_2$ from $6$ to $6e^{-t}$ (a heater that fades out) and predict the long-term temperatures before you run it.

Complex eigenvalues

Systems not in normal form

The method does not need the equations to be solved for $x^{\prime }$ and $y^{\prime }$. If both derivatives appear in both equations, we still transform term by term and get two linear equations for $X$ and $Y$. Only the determinant changes.

Second-order systems: coupled oscillators

Newton's law for several masses gives systems of second-order equations. The method is the same; we use $\mathcal{L}\{x^{\prime \prime }\}=s^{2}X-sx(0)-x^{\prime }(0)$.

Two equal masses $m=1$ slide without friction between two walls, connected to the walls and to each other by three equal springs with $k=1$ (Figure 9.5.3). Let $x$ and $y$ be the displacements of the masses from equilibrium (positive to the right). The left mass feels the left spring, $-x$, and the middle spring, which is stretched by $y-x$, so the force is $-x+(y-x)$. The right mass feels $-(y-x)-y$. Therefore

Laplace transform or eigenvalues?

For a $2\times 2$ system with constant coefficients you now have three methods: the operator method, the matrix (eigenvalue) method of Chapter 7, and the Laplace transform. They always give the same answer, so choose by convenience.

• The Laplace transform is most efficient when initial values are given and the forcing is an exponential, a polynomial, a sinusoid, a step or an impulse. The initial values are built in, nonhomogeneous terms need no extra method, and systems not in normal form or with second derivatives are handled in the same way.
• The eigenvalue method is better when you want the general solution with arbitrary constants, or when you want to understand the geometry of all solutions at once (the phase portrait).
• For larger systems the Laplace method still works, but Cramer's rule becomes heavy; then one solves the $s$-domain linear system by elimination, and the determinant $\det (sI-A)$ still gives the poles.