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
\[\begin{split}
E_i=\mathrm{span}_{\mathbb C}\left( \left[\begin{array}{c} 2+i\\ 1\end{array}\right] \right).\end{split}\]
Describe without further computations the eigenspace \(E_{-i}\).
Hint
Can complex conjugation be of use here?
Answer
Because the matrix \(\mathbf A\) is real (contains only real numbers) 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 one achieves by taking the complex conjugate of all elements 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 one from complex solutions can achieve real solutions. The general approach is described in Corollary 12.2.6 from 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 (3.7) and (3.8) 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 in terms of \(\cos(t)\) and \(\sin(t)\).
Hint
One could begin 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 12.2.6) 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 have 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 paricular solution to the given system exists in 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 some non-given real numbers. Determine \(d_1\) and \(d_2\).
Hint
If we insert \(f_1(t)=d_1 \cdot \mathrm e^{2t}\) and \(f_1(t)=d_2 \cdot \mathrm e^{2t}\) into the system, then we achieve that \(2d_1 \mathrm e^{2t}= (2d_1-5d_2+1)\mathrm e^{2t}\) 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 gives \(d_1=9/5 \, \) and \(d_2=1/5\).
Question b
Provide the general, real solution to the given inhomogeneous system of differential equations.
Answer
If the found particular solution from Question a is added to the found general, real solution as described in Exercise 3, then one gets:
\[\begin{split}
\left[\begin{array}{c} f_1(t)\\ f_2(t)\end{array}\right] = \left[\begin{array}{c} 9\mathrm e^{2t}/5\\ \mathrm e^{2t}/5\end{array}\right]+a \cdot \left[\begin{array}{c} 4\cos(t)-2 \sin(t)\\ 2\cos(t)\end{array}\right]+b\cdot \left[\begin{array}{c} -2\cos(t)-4 \sin(t)\\ -2\sin(t)\end{array}\right]\, , \quad \text{ where } a,b \in \mathbb{R}.
\end{split}\]
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 using SymPy. Can the coefficient matrix be diagonalized?
Answer
We have:
\[\begin{split}{\mathbf A}= \left[\begin{array}{cc} 2 & 1\\ -4 & -2\end{array}\right].\end{split}\]
SymPy can be used to show that \(0\) is the only eigenvalue of the coefficient matrix and 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
\[\begin{split}
{\mathbf Q}= \left[\begin{array}{cc} 1 & 0\\-2 & 1\end{array}\right].
\end{split}\]
You may check this equation using SymPy. 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
\[\begin{split}\left[\begin{array}{c} f_1(t)\\ f_2(t)\end{array}\right] = \left[\begin{array}{cc} 1 & 0\\-2 & 1\end{array}\right] \cdot \left[\begin{array}{c} c_1\cdot t+c_2\\ c_1\end{array}\right] = \left[\begin{array}{c} c_1\cdot t+c_2\\ -2 c_1\cdot t -2c_2 + c_1 \end{array}\right]
\, , \quad \text{ where } c_1,c_2 \in \mathbb{R}.\end{split}\]
