8 Diagonalisation and Systems of differential equations.
With a linear first order system of differential equations with constant coefficients we mean a system of the form \begin{aligned} x_1'(t) &= a_{11}x_1(t) + a_{12}x_2(t) + \cdots + a_{1n}x_n(t) \\ x_2'(t) &= a_{21}x_1(t) + a_{22}x_2(t) + \cdots + a_{2n}x_n(t) \\ \vdots & \\ x_n'(t) &= a_{n1}x_1(t) + a_{n2}x_2(t) + \cdots + a_{nn}x_n(t), \end{aligned} \tag{8.1} where x_1(t),x_2(t),...,x_n(t) are unknown differentiable functions. If we write \begin{align*} \mathbf{x}(t)&=(x_1(t),x_2(t),...,x_n(t)) & \mathbf{x}'(t)&=(x_1'(t),x_2'(t),...,x_n'(t)), \end{align*} and define the matrix A to have components a_{ij}, then this can be written more compactly as \mathbf{x}'(t)=A\mathbf{x}(t). Diagonalisation is very useful when it comes to solving such systems:
Proposition 8.1 Suppose A is diagonalisable, with A=PDP^{-1}. Then we have that \mathbf{x}(t) solves the system \mathbf{x}'(t) = A\mathbf{x}(t) if and only if \mathbf{y}(t)=P^{-1}\mathbf{x}(t) solves the system \mathbf{y}'(t) = D\mathbf{y}(t).
Proof. Suppose first that \mathbf{x}(t) solves \mathbf{x}'(t) = A\mathbf{x}(t). With \mathbf{y}(t)=P^{-1}\mathbf{x}(t) we get that \mathbf{y}'(t) = (P^{-1}\mathbf{x})' = P^{-1}\mathbf{x}' = P^{-1}A\mathbf{x}(t) = P^{-1}AP\mathbf{y}(t) = D\mathbf{y}(t), so that \mathbf{y}(t) løser \mathbf{y}'(t) = D\mathbf{y}(t). Suppose then that \mathbf{y}(t) løser \mathbf{y}'(t) = D\mathbf{y}(t). We then get \mathbf{x}'(t) = (P\mathbf{y}(t))' = P\mathbf{y}'(t) = PD\mathbf{y}(t) = PDP^{-1}\mathbf{x}(t) = A\mathbf{x}(t), so that \mathbf{x}(t) solves \mathbf{x}'(t) = A\mathbf{x}(t).
The point is that the system \mathbf{y}'(t) = D\mathbf{y}(t) can be split into \begin{align*} y_1'(t)&=\lambda_1y_1(t) & y_2'(t)&=\lambda_2y_2(t) & & \cdots & y_n'(t)&=\lambda_ny_n(t) \end{align*} so that the component functions can be found independently of one-another as y_i(t)=C_ie^{\lambda_i t}, where C_i is a constant that can be complex, and \mathbf{y}(t)=\left( C_1e^{\lambda_1 t} , C_2e^{\lambda_2 t},..., C_ne^{\lambda_n t}\right).
It is interesting to interpret solutions of different systems when A is 2\times 2, since we then can plot x_1 against x_2 in the plane. The general solution when n=2 is \mathbf{x}(t)=P\mathbf{y}(t) = \begin{pmatrix} \mathbf{p}_1 & \mathbf{p}_2 \end{pmatrix} \begin{pmatrix} C_1e^{\lambda_1 t} \\ C_2e^{\lambda_2 t} \end{pmatrix} = C_1e^{\lambda_1 t}\mathbf{p}_1 + C_2e^{\lambda_2 t}\mathbf{p}_2, Let us first consider the case where A has real eigenvalues. Recall that A may not be diagonalisable when A has equal eigenvalues. We have the following result.
Proposition 8.2 Suppose that A has real eigenvalues and that \mathbf{x}(t_0) is known, and real. Then there exist unique real constants C_1 and C_2 so that \mathbf{x}(t) = C_1e^{\lambda_1 t}\mathbf{p}_1 + C_2e^{\lambda_2 t}\mathbf{p}_2. \tag{8.2}
The other way it is clear that Equation 8.2 is real when C_1 and C_2 are real.
Proof. The system \mathbf{x}(t_0)=\begin{pmatrix} \mathbf{p}_1 & \mathbf{p}_2 \end{pmatrix} \begin{pmatrix} C_1e^{\lambda_1 t_0} \\ C_2e^{\lambda_2 t_0} \end{pmatrix} is in this case real. Since P is invertible it follows that also the solution \begin{pmatrix} C_1e^{\lambda_1 t_0} \\ C_2e^{\lambda_2 t_0} \end{pmatrix} is unique and real. But then C_1 and C_2 are also unique and real.
Example 8.1 (real eigenvalues) Set A=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}. A has eigenvalues \lambda_1=-1 and \lambda_2=1, with corresponding eigenvectors \mathbf{p}_1=(-1,1) and \mathbf{p}_2=(1,1). The general solution Equation 8.2 is thus C_1e^{-t}\begin{pmatrix} -1 \\ 1 \end{pmatrix} + C_2e^t \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} -C_1e^{-t} + C_2e^t \\ C_1e^{-t} + C_2e^t \end{pmatrix}, and it is clear that these have f(x)=x and f(x)=-x as asymptotes when t\to\infty and t\to -\infty. We can plot more solutions C_1e^{\lambda_1t}\mathbf{p}_1+C_2e^{\lambda_2t}\mathbf{p}_2 in the same systems by varying C_1 and C_2. The solutions we get by choosing C_i\in\{1.5,2,2.5\} (9 solutions) are shown in Figure 8.1.
\clubsuit
Let us now consider the case where A has complex eigenvalues. We then know that A is diagonalisable, and that eigenvalues and eigenvectors come in conjugate pairs.
Proposition 8.3 Suppose that A has complex eigenvalues \lambda and \overline{\lambda} (with corresponding eigenvectors \mathbf{p} and \overline{\mathbf{p}}). Suppose also that \mathbf{x}(t_0) is known, and real. Then there exists a constant C (which can be complex) so that
\mathbf{x}(t) = 2\Re(Ce^{\lambda t}\mathbf{p} ). \tag{8.3}
The other way it is clear that Equation 8.3 is real for all values of C.
Proof. We have that \begin{align*} \mathbf{x}(t)=P\mathbf{y}(t) = \begin{pmatrix} \mathbf{p} & \overline{\mathbf{p}} \end{pmatrix} \begin{pmatrix} C_1e^{\lambda t} \\ C_2e^{\overline{\lambda} t} \end{pmatrix} \end{align*} for constants C_1 and C_2. If we set t=t_0 we get \begin{aligned} \mathbf{x}(t_0) &= \begin{pmatrix} \mathbf{p} & \overline{\mathbf{p}} \end{pmatrix} \begin{pmatrix} C_1e^{\lambda t_0} \\ C_2e^{\overline{\lambda} t_0} \end{pmatrix} \\ \mathbf{x}(t_0)&=\overline{\mathbf{x}(t_0)} = \begin{pmatrix} \overline{\mathbf{p}} & \mathbf{p} \end{pmatrix} \begin{pmatrix} \overline{C_1}e^{\overline{\lambda} t_0} \\ \overline{C_2}e^{\lambda t_0} \end{pmatrix} =\begin{pmatrix} \mathbf{p} & \overline{\mathbf{p}} \end{pmatrix} \begin{pmatrix} \overline{C_2}e^{\lambda t_0} \\ \overline{C_1}e^{\overline{\lambda} t_0} \end{pmatrix}, \end{aligned} \tag{8.4} where we changed the order of the first two components. Since P is invertible we get by comparing the two lines in Equation 8.4 that C_2=\overline{C_1}, and the solution becomes \mathbf{x}(t)=\begin{pmatrix} \mathbf{p} & \overline{\mathbf{p}} \end{pmatrix}\begin{pmatrix} C_1e^{\lambda t} \\ \overline{C_1}e^{\overline{\lambda} t}\end{pmatrix}= C_1e^{\lambda t}\mathbf{p} + \overline{C_1}e^{\overline{\lambda} t}\overline{\mathbf{p}} = 2\Re(C_1e^{\lambda t}\mathbf{p} ), so that we can set C=C_1.
The expression 2\Re(C_1e^{\lambda t}\mathbf{p} ) involves complex numbers, even thought the answer is real. But it is easy to rewrite this to a form not involving complex arithmetic. Let \mathbf{c} and \mathbf{d} be the real and imaginary parts of the complex vector C\mathbf{p}, and let \lambda=a+ib. Then C\mathbf{p}=\mathbf{c}+i\mathbf{d} and e^{\lambda t}=e^{at}(\cos(bt)+i\sin(bt)), and we get \Re(Ce^{\lambda t}\mathbf{p} ) = \Re( e^{at} (\mathbf{c}+i\mathbf{d})(\cos(bt)+i\sin(bt))= e^{at}\left( \cos(bt)\mathbf{c} - \sin(bt)\mathbf{d} \right), so that \mathbf{x}(t)=2e^{at}\left( \cos(bt)\mathbf{c} - \sin(bt)\mathbf{d} \right). \tag{8.5} From this form it is clear that
- For a>0: x(t)\to\infty when t\to\infty (spiral away from the origin),
- For a<0: x(t)\to 0 when t\to\infty (spiral towards the origin),
- For a=0: x(t) is periodic.
Since \cos(bt) and \sin(bt) are periodic with period 2\pi/b, a plot over [0,2\pi/b] should show one arm of a spiral in 1. and 2., or one whole period in 3..
The following function takes a real matrix A with complex eigenvalues, together with an initial value x(t_0)=x_0, as parameters. The code computes the eigenvalues and eigenvectors of A, solves the system \begin{pmatrix} \mathbf{p} & \overline{\mathbf{p}} \end{pmatrix} \begin{pmatrix} Ce^{\lambda t_0} \\ \overline{C}e^{\overline{\lambda} t_0} \end{pmatrix} = \mathbf{x}_0, finds C, computes C\mathbf{p}=\mathbf{c}+i\mathbf{d}, and then the solution Equation 8.5, which is returned:
The interval [t_0,t_0+4\pi/b] includes two arms of the spiral.
Example 8.2 (Complex eigenvalues) Consider the three matrices \begin{align*} A_1 &=\begin{pmatrix} 0.1 & 1 \\ -1 & 0.1 \end{pmatrix} & A_2 &=\begin{pmatrix} -0.1 & 1 \\ -1 & -0.1 \end{pmatrix} & A_3 &=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \end{align*} These have eigenvalues 0.1\pm i, -0.1\pm i, and \pm i, respectively. We also have that a>0, a<0, and a=0, respectively, so that the solutions to the systems will demonstrate the three possibilities above. The first solution will in particular show a spiral moving away from the origin, the second a spiral towards the origin. The solutions are plotted in Figure 8.2, Figure 8.3, Figure 8.4. We have used the initial value \mathbf{x}(2)=(2,3), so that all solutions start in the point (2,3).
One can also plot t against x_i(t). Doing this one can visualise all components in the same plot, also when n>2. In Figure 8.5 we plot only the first component x_1(t), over the same time interval, and with the same initial condition as above. Since c_1\cos(bt)-d_1\sin(bt) has amplitude \sqrt{c_1^2+d_1^2} (Exercise 8.1) we have upper and lower bounds -2\sqrt{c_1^2+d_1^2}e^{at}\leq x_1(t) \leq 2\sqrt{c_1^2+d_1^2}e^{at} \tag{8.6} These are also included in the plot.
\clubsuit
It turns out that one can write down solutions for systems with n>2, based on what we did above. In the general case we have a mix of real and complex eigenvalues and eigenvectors: \lambda_1,\overline{\lambda_1},...,\lambda_m,\overline{\lambda_m},\lambda_{2m+1},...,\lambda_n \mathbf{p}_1,\overline{\mathbf{p}_1},...,\mathbf{p}_m,\overline{\mathbf{p}_m},\mathbf{p}_{2m+1},...,\mathbf{p}_n, where the real ones were listed at the end. Exercise 8.2 explains how one can solve the case with two complex and one real eigenvalue.
8.1 Higher order linear differential equations
A type of system related to the systems given in Equation 8.1 are higher order linear differential equations with constant coefficients . These are on the form y^{(n)}(t) + a_1y^{(n-1)}(t) + a_2y^{(n-2)}(t) + \cdots + a_ny(t) = 0, \tag{8.7} where y(t) is the unknown function. We shall limit ourselves to second order equations, and write these on the form y''+py'+qy=0 where p and q are constants. By setting x_0=y, x_1=y' we can write this as the first order linear system \begin{array}{rrr} x_0'=& &x_1 \\ x_1'=&-qx_0&-px_1 \end{array} which has coefficient matrix A=\begin{pmatrix} 0 & 1 \\ -q & -p \end{pmatrix}. We have that \det(A-\lambda I)=\lambda(\lambda+p)+q=\lambda^2+p\lambda + q. Assume that the eigenvalues \lambda_1 and \lambda_2 are different (this is the same as that p^2\neq 4q). Then A is diagonalisable, and we have that A-\lambda I_n = \begin{pmatrix} -\lambda & 1 \\ -q & -\lambda -p\end{pmatrix} \sim \begin{pmatrix} -\lambda & 1 \\ 0 & 0 \end{pmatrix}, so that (1,\lambda_1) and (1,\lambda_2) are two particular eigenvectors. We can therefore set P=\begin{pmatrix} 1 & 1 \\ \lambda_1 & \lambda_2 \end{pmatrix}.
Let us first assume real eigenvalues. We have shown that the general solution is \begin{pmatrix} y(t) \\ y'(t) \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ \lambda_1 & \lambda_2 \end{pmatrix}\begin{pmatrix} C_1e^{\lambda_1t} \\ C_2e^{\lambda_2t}\end{pmatrix}, \tag{8.8} where one can find C_1 and C_2 with help of initial values y(t_0), y'(t_0).
Example 8.3 (Distinct real eigenvalues) Consider the equation y''-5y'+6y=0, with initial values y(0)=1, y'(0)=0. The eigenvalues of A are found by solving the characteristic equation \lambda^2-5\lambda + 6 = 0, which gives \lambda_1=2, \lambda_2=3. If we insert t=0 in Equation 8.8 we get \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix}\begin{pmatrix} C_1 \\ C_2 \end{pmatrix}, which has the solution C_1=3, C_2=-2. The solution is thus y(t)=C_1e^{\lambda_1t} + C_2e^{\lambda_2t}=3e^{2t}-2e^{3t}. \clubsuit
The case with complex eigenvalues is a bit more complicated. We shall first prove the following.
Proposition 8.4 y(t) = 2e^{at}\left( \cos(bt)c_1 - \sin(bt)d_1 \right), \tag{8.9} where c_1 and d_1 are arbitrarily chosen real numbers (recall that \lambda=a+ib).
Proof. Equation 8.5 said that the general solution is \begin{pmatrix} y(t) \\ y'(t) \end{pmatrix} = 2e^{at}\left( \cos(bt)\mathbf{c} - \sin(bt)\mathbf{d} \right), \tag{8.10} where \mathbf{c}=\Re(C\mathbf{p}), \mathbf{d}=\Im(C\mathbf{p}), and where C is an arbitrarily chosen complex number.
Let now c_1, d_1 be given numbers, and define C=c_1+id_1. Then \begin{align*} C\mathbf{p}&=\begin{pmatrix} c_1+id_1 \\ \cdot \end{pmatrix} & \mathbf{c}&=\Re(C\mathbf{p})=\begin{pmatrix} c_1 \\ \cdot \end{pmatrix} & \mathbf{d}&=\Im(C\mathbf{p})=\begin{pmatrix} d_1 \\ \cdot \end{pmatrix}, \end{align*} where \cdot indicate a concrete number. Evaluating the first row in Equation 8.10 with this we get that Equation 8.9 is a solution for all choices of c_1 and d_1, and because of Equation 8.10 all solutions must be on this form.
Example 8.4 (Complex eigenvalues) Consider the equation y''+4y'+5y=0. The characteristic equation now becomes \lambda^2+4\lambda+5=0, which has the solution \frac{-4\pm\sqrt{16-20}}{2}=-2\pm i. The general solution thus becomes y(t) = 2e^{-2t}\left( c_1\cos t - d_1\sin t \right). \clubsuit
So far we have avoided the case where \lambda_1=\lambda_2. In that case we have that \lambda_1=\lambda_2=-p/2, and that p^2=4q. We then get A-\lambda I = \begin{pmatrix} -\lambda & 1 \\ -q & -\lambda-p \end{pmatrix} = \begin{pmatrix} p/2 & 1 \\ -p^2/4 & -p/2\end{pmatrix} \sim\begin{pmatrix} p/2 & 1 \\ 0 & 0\end{pmatrix}. It follows that the eigenspace has dimension one, so that A is not diagonalisable. We can therefore not use what we have done to find the solutions, but it is an exercise to show by direct substitution that y(t)=e^{\lambda t} is a solution, and also that y(t)=te^{\lambda t} is a solution. It follows that all functions on the form y(t)=Ce^{\lambda t}+Dte^{\lambda t} also are solutions, and one can show that there is a unique such solution satisfying the initial conditions for y(t_0) and y'(t_0). We will not go into how we “guessed” the solution te^{\lambda t}, and we will also not prove that all functions can be expressed in terms of the functions e^{\lambda t} and te^{\lambda t}.
Example 8.5 (Double root) Consider the equation y''+6y'+9y=0 with initial conditions y(0)=4, y'(0)=-13. The characteristic equation becomes \lambda^2+6\lambda+9=0, which has the solution \lambda_1=\lambda_2=-3. The general solution thus becomes y(t)=Ce^{-3t}+Dte^{-3t}, and we also have that y'(t)=-3Ce^{-3t}+D(1-3t)e^{-3t}. The initial conditions give the equations \begin{align*} C &= 4 \\ -3C+D &= -13, \end{align*} which have the soltion C=4, D=-1. The solution is thus y(t)=4e^{-3t}-te^{-3t}. \clubsuit
Quiz
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Exercises
Exercise 8.1 Let b, c, and d be real numbers. Show that |c\cos(bt)-d\sin(bt)|\leq \sqrt{c^2+d^2} for all t. For which t do you have equality here?
Exercise 8.2 Consider the system \begin{array}{rrrrr} x_1'(t) &=& x_1(t) &+x_2(t) & \\ x_2'(t) &=& -x_1(t) &+x_2(t) & \\ x_3'(t) &=& & & x_3(t) \end{array}
- Find the general solution of the system.
- Find the solution of the system, given the initial condition \mathbf{x}(0)=(1,1,1). What happens with the solution when t\to-\infty, and when t\to\infty?
Exercise 8.3 Find the solution of
- y''-5y'+4y=0 with initial conditions y(0)=2, y'(0)=5.
- y''+2y'+2y=0.
- y''-4y'+4y=0 with initial conditions y(0)=3, y'(0)=8.
Exercise 8.4 Consider the equation y''+py'+qy, and assume that the characteristic equation of this has a double root \lambda.
- Show that y(t)=e^{\lambda t} solves the equation.
- Show that y(t)=te^{\lambda t} solves the equation (you will here need that \lambda_1=\lambda_2=-p/2, and that p^2=4q when we have a double root).
- Show that there is a unique choice of C and D so that y(t)=Ce^{\lambda t}+Dte^{\lambda t} satisfies the initial values for y(t_0) and y'(t_0).