Exercise 1: A Homogeneous, Complex System of Linear Differential Equations
We are given the following complex system of differential equations:
\[\begin{split}
\left\{
\begin{array}{rcl}
f_1'(t) & = & 2f_1(t)-5f_2(t)\\
f_2'(t) & = & f_1(t)-2f_2(t)
\end{array}.
\right.
\end{split}\]
Question a
The system can be written in the form
\[\begin{split}
\left[\begin{array}{c} f_1'(t)\\ f_2'(t)\end{array}\right] = {\mathbf A} \cdot \left[\begin{array}{c} f_1(t)\\ f_2(t)\end{array}\right].
\end{split}\]
What is the matrix \(\mathbf A\)? Also determine the characteristic polynomial of the matrix.
Answer
The matrix \({\mathbf A}\) is the coefficient matrix of the system. This means that:
\[\begin{split}
{\mathbf A}=\left[\begin{array}{rr} 2 & -5\\ 1 & -2\end{array}\right].
\end{split}\]
We have
\[\begin{split}
p_{\mathbf A}(Z)=\mathrm{det}\left(\left[\begin{array}{cc} 2-Z & -5\\ 1 & -2-Z\end{array}\right]\right)=Z^2+1.
\end{split}\]
Question b
We are being informed that over the complex numbers the eigenspace \(E_i\) of the matrix \(\mathbf A\) from Question a is given by
\[\begin{split}
E_i=\mathrm{span}_{\mathbb C}\left( \left[\begin{array}{c} 2+i\\ 1\end{array}\right] \right).\end{split}\]
State without further calculations the eigenspace \(E_{-i}\).
Hint
Can complex conjugation be of use here?
Answer
Because the matrix \(\mathbf A\) is real (contains only real entries) we have that \({\mathbf v} \in E_i\) if and only if \(\overline{\mathbf v} \in E_{-i}\). With \(\overline{\mathbf v}\) we mean the vector that is achieved by taking the complex conjugate of each entry in \(\mathbf v\). Hence we get:
\[\begin{split}
E_{-i}=\mathrm{span}_{\mathbb C}\left( \left[\begin{array}{c} 2-i\\ 1\end{array}\right] \right).
\end{split}\]
Question c
Provide the general, complex solution to the given system of differential equations.
Answer
\[\begin{split}
\left[\begin{array}{c} f_1(t)\\ f_2(t)\end{array}\right] =c_1 \cdot \left[\begin{array}{c} 2+i\\ 1\end{array}\right]\cdot \mathrm e^{it}+c_2 \cdot \left[\begin{array}{c} 2-i\\ 1\end{array}\right] \cdot \mathrm e^{-it}=\left[\begin{array}{c} c_1\cdot (2+i)\cdot \mathrm e^{it}+c_2\cdot (2-i)\cdot \mathrm e^{-it}\\ c_1\cdot \mathrm e^{it}+c_2\cdot \mathrm e^{-it}\end{array}\right]\, , \quad \text{ where } c_1,c_2 \in \mathbb{C}.
\end{split}\]
Exercise 2: From Complex to Real Solutions
In this exercise we consider the same system of differential equations as in Exercise 1:
\[\begin{split}
\left\{
\begin{array}{rcl}
f_1'(t) & = & 2f_1(t)-5f_2(t)\\
f_2'(t) & = & f_1(t)-2f_2(t)
\end{array}.
\right.
\end{split}\]
The general, complex solution to this system is:
\[\begin{split}
\left[\begin{array}{c} f_1(t)\\ f_2(t)\end{array}\right] =c_1 \cdot \left[\begin{array}{c} 2+i\\ 1\end{array}\right]\cdot \mathrm e^{it}+c_2 \cdot \left[\begin{array}{c} 2-i\\ 1\end{array}\right] \cdot \mathrm e^{-it}=\left[\begin{array}{c} c_1\cdot (2+i)\cdot \mathrm e^{it}+c_2\cdot (2-i)\cdot \mathrm e^{-it}\\ c_1\cdot \mathrm e^{it}+c_2\cdot \mathrm e^{-it}\end{array}\right]\, , \quad \text{ where } c_1,c_2 \in \mathbb{C}.
\end{split}\]
The goal with this exercise is to exemplify how we from complex solutions can extract real solutions. The general approach is described in Corollary 13.2.6 in the textbook.
Question a
In the given general solution we choose the constants \(c_1\) and \(c_2\) such that \(c_1=c_2=a\), where \(a\) is a real number. Show that the solution in this case can be rewritten to:
\[\begin{split}
\left[\begin{array}{c} f_1(t)\\ f_2(t)\end{array}\right] =a \cdot \left[\begin{array}{c} 2(\mathrm e^{it}+\mathrm e^{-it})\\ \mathrm e^{it}+\mathrm e^{-it}\end{array}\right] +a \cdot \left[\begin{array}{c} i(\mathrm e^{it}-\mathrm e^{-it})\\ 0\end{array}\right].
\end{split}\]
Now use Euler’s formulas (equations (4.7) and (4.8) in the textbook) to express \(\mathrm e^{it}\) and \(\mathrm e^{-it}\) in terms of \(\cos(t)\) and \(\sin(t)\). What is the result?
Answer
\[\begin{split}
\left[\begin{array}{c} f_1(t)\\ f_2(t)\end{array}\right] =a \cdot \left[\begin{array}{c} 4\cos(t)\\ 2\cos(t)\end{array}\right] +a \cdot \left[\begin{array}{c} -2\sin(t)\\ 0\end{array}\right]=a \cdot \left[\begin{array}{c} 4\cos(t)-2 \sin(t)\\ 2\cos(t)\end{array}\right].
\end{split}\]
Question b
In the given general solution we now choose the constants \(c_1\) and \(c_2\) such that \(c_1=b\cdot i\) and \(c_2=-b \cdot i\), where \(b\) is a real number. Rewrite this solution to be expressed in terms of \(\cos(t)\) and \(\sin(t)\).
Hint
We could start by checking that
\[\begin{split}
\left[\begin{array}{c} f_1(t)\\ f_2(t)\end{array}\right] =b \cdot \left[\begin{array}{c} -(\mathrm e^{it}+\mathrm e^{-it})\\ 0\end{array}\right]+b \cdot \left[\begin{array}{c} 2i(\mathrm e^{it}-\mathrm e^{-it})\\ i(\mathrm e^{it}-\mathrm e^{-it})\end{array}\right].
\end{split}\]
Answer
\[\begin{split}
\left[\begin{array}{c} f_1(t)\\ f_2(t)\end{array}\right] = b \cdot \left[\begin{array}{c} -2\cos(t)\\ 0\end{array}\right]+b \cdot \left[\begin{array}{c} -4\sin(t)\\ -2\sin(t)\end{array}\right] =b \cdot \left[\begin{array}{c} -2\cos(t)-4 \sin(t)\\ -2\sin(t)\end{array}\right].
\end{split}\]
Exercise 3: Initial Conditions in a Real System of Linear Differential Equations
The system from Exercise 1 is considered once again, but this time as a real system of differential equations. From Exercise 2 (and Corollary 13.2.6) we conclude that the general, real solution to the system is:
\[\begin{split}
\left[\begin{array}{c} f_1(t)\\ f_2(t)\end{array}\right] = a \cdot \left[\begin{array}{c} 4\cos(t)-2 \sin(t)\\ 2\cos(t)\end{array}\right]+b\cdot \left[\begin{array}{c} -2\cos(t)-4 \sin(t)\\ -2\sin(t)\end{array}\right]\, , \quad \text{ where } a,b \in \mathbb{R}.
\end{split}\]
Determine the solution to the system that fulfills the initial conditions \(f_1(\pi)=1\) and \(f_2(\pi)=3\).
Hint
If \(t=\pi\) is inserted in the general solution and the initial conditions are applied then we achieve two linear equations, which the constants \(a\) and \(b\) must fulfill. Which two?
Hint
The two equations that \(a\) and \(b\) must fulfill are \(-4a+2b=1\) and \(-2a=3\).
Answer
\[\begin{split}
\left[\begin{array}{c} f_1(t)\\ f_2(t)\end{array}\right] = (-3/2) \cdot \left[\begin{array}{c} 4\cos(t)-2 \sin(t)\\ 2\cos(t)\end{array}\right]+(-5/2)\cdot \left[\begin{array}{c} -2\cos(t)-4 \sin(t)\\ -2\sin(t)\end{array}\right]=\left[\begin{array}{c} -\cos(t)+13 \sin(t)\\ -3\cos(t)+5 \sin(t)\end{array}\right].
\end{split}\]
Exercise 4: An Inhomogeneous, Real System of Linear Differential Equations
We are given the following real system of differential equations:
\[\begin{split}
\left\{
\begin{array}{rcl}
f_1'(t) & = & 2f_1(t)-5f_2(t)+\mathrm e^{2t}\\
f_2'(t) & = & f_1(t)-2f_2(t)-\mathrm e^{2t}
\end{array}.
\right.
\end{split}\]
Notice that the corresponding homogeneous system of differential equations is the system considered in Exercise 3.
Question a
We are informed that a particular solution to the given system exists of the form:
\[\begin{split}
\left[\begin{array}{c} f_1(t)\\ f_2(t)\end{array}\right] = \left[\begin{array}{c} d_1\cdot \mathrm e^{2t}\\ d_2\cdot \mathrm e^{2t}\end{array}\right]\, ,
\end{split}\]
where \(d_1\) and \(d_2\) are real numbers. Determine \(d_1\) and \(d_2\).
Hint
By inserting \(f_1(t)=d_1 \cdot \mathrm e^{2t}\) and \(f_1(t)=d_2 \cdot \mathrm e^{2t}\) into the system, we get \(2d_1 \mathrm e^{2t}= (2d_1-5d_2+1)\mathrm e^{2t}\) and \(2d_2 \mathrm e^{2t}= (d_1-2d_2-1)\mathrm e^{2t}\). Which equations must \(d_1\) and \(d_2\) thus fulfill?
Answer
The constants \(d_1\) and \(d_2\) must fulfill the equations \(2d_1=2d_1-5d_2+1\) and \(2d_2=d_1-2d_2-1.\) Solving these two equations will show that \(d_1=9/5 \, \) and \(d_2=1/5\).
Question b
Provide the general, real solution to the given inhomogeneous system of differential equations.
Answer
By adding the particular solution found in Question a to the general, real solution that was found in Exercise 3, we get:
\[\begin{split}
\left[\begin{array}{c} f_1(t)\\ f_2(t)\end{array}\right] = \left[\begin{array}{c} 9\mathrm e^{2t}/5\\ \mathrm e^{2t}/5\end{array}\right]+a \cdot \left[\begin{array}{c} 4\cos(t)-2 \sin(t)\\ 2\cos(t)\end{array}\right]+b\cdot \left[\begin{array}{c} -2\cos(t)-4 \sin(t)\\ -2\sin(t)\end{array}\right]\, , \quad \text{ where } a,b \in \mathbb{R}.
\end{split}\]
Exercise 5: A Coefficient Matrix that cannot be Diagonalized
We consider the following homogeneous, real system of differential equations:
\[\begin{split}
\left\{
\begin{array}{rcl}
f_1'(t) & = & 2f_1(t)+f_2(t)\\
f_2'(t) & = & -4f_1(t)-2f_2(t).
\end{array}
\right.
\end{split}\]
Question a
Determine the coefficient matrix \(\mathbf A\) of the system and investigate its eigenvalues and eigenvectors. Can the coefficient matrix be diagonalized?
Answer
We have:
\[\begin{split}{\mathbf A}= \left[\begin{array}{cc} 2 & 1\\ -4 & -2\end{array}\right].\end{split}\]
This matrix only has the eigenvalue \(0\), and we find that \(\mathrm{am}(0)=2\) and \(\mathrm{gm}(0)=1\). Of that reason \(\mathbf A\) cannot be diagonalised. A possible basis for the eigenspace \(E_0\) is:
\[\begin{split}\left\{ \left[\begin{array}{c} -1/2\\1\end{array}\right] \right\}.\end{split}\]
Another possible basis for \(E_0\) is
\[\begin{split}\left\{ \left[\begin{array}{c} 1\\-2\end{array}\right] \right\}.\end{split}\]
Question b
Now, first determine the general solution to the following homogeneous system of differential equations:
\[\begin{split}
\left[\begin{array}{c} g'_1(t)\\ g'_2(t)\end{array}\right] = {\mathbf J} \cdot \left[\begin{array}{c} g_1(t)\\ g_2(t)\end{array}\right],\end{split}\]
where
\[\begin{split}
{\mathbf J}= \left[\begin{array}{cc} 0 & 1\\0 & 0\end{array}\right].
\end{split}\]
Hint
If you write down which equations \(g_1(t)\) and \(g_2(t)\) must fulfill, you would achieve \(g_1'(t)=g_2(t)\) and \(g_2'(t)=0\). Which possibilities are there for \(g_2(t)\)?
Hint
You will find that \(g_2(t)\) must be a constant function, so you can write \(g_2(t)=c_1\). Now find all possibilities for \(g_1(t)\).
Answer
The general solution is
\[\begin{split}\left[\begin{array}{c} g_1(t)\\ g_2(t)\end{array}\right] = c_1 \cdot \left[\begin{array}{c} t\\ 1\end{array}\right]+c_2 \cdot \left[\begin{array}{c} 1\\ 0\end{array}\right]=\left[\begin{array}{c} c_1\cdot t+c_2\\ c_1\end{array}\right]\, , \quad \text{ where } c_1,c_2 \in \mathbb{R}.\end{split}\]
Question c
We are informed that \({\mathbf A}={\mathbf Q} \cdot {\mathbf J} \cdot {\mathbf Q}^{-1}\), where \(\mathbf J\) is as before and where
\[\begin{split}
{\mathbf Q}= \left[\begin{array}{cc} 1 & 0\\-2 & 1\end{array}\right].
\end{split}\]
It can be shown that this implies that
\[\begin{split}\left[\begin{array}{c} f_1(t)\\ f_2(t)\end{array}\right] \quad \text{ is a solution to the system } \left[\begin{array}{c} f'_1(t)\\ f'_2(t)\end{array}\right] = {\mathbf A} \cdot \left[\begin{array}{c} f_1(t)\\ f_2(t)\end{array}\right]\end{split}\]
if and only if
\[\begin{split}\left[\begin{array}{c} g_1(t)\\ g_2(t)\end{array}\right] = {\mathbf Q}^{-1} \cdot \left[\begin{array}{c} f_1(t)\\ f_2(t)\end{array}\right] \quad \text{ is a solution to the system } \left[\begin{array}{c} g'_1(t)\\ g'_2(t)\end{array}\right] = {\mathbf J} \cdot \left[\begin{array}{c} g_1(t)\\ g_2(t)\end{array}\right].\end{split}\]
Now find the general solution to the original homogeneous system of differential equations that had the coefficient matrix \(\mathbf A\).
Hint
Use the general solution from the previous Question and the given relation between solutions to the two mentioned systems of differential equations.
Answer
The general solution is
\[\begin{split}\left[\begin{array}{c} f_1(t)\\ f_2(t)\end{array}\right] = \left[\begin{array}{cc} 1 & 0\\-2 & 1\end{array}\right] \cdot \left[\begin{array}{c} c_1\cdot t+c_2\\ c_1\end{array}\right] = \left[\begin{array}{c} c_1\cdot t+c_2\\ -2 c_1\cdot t -2c_2 + c_1 \end{array}\right]
\, , \quad \text{ where } c_1,c_2 \in \mathbb{R}.\end{split}\]
