10.3.1: Basic Theory of Homogeneous Linear Systems (Exercises)
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( \newcommand{\kernel}{\mathrm{null}\,}\)
Q10.3.1
1. Prove: If y1, y2, …, yn are solutions of y′=A(t)y on (a,b), then any linear combination of y1, y2, …, yn is also a solution of y′=A(t)y on (a,b).
2. In Section 5.1 the Wronskian of two solutions y1 and y2 of the scalar second order equation
P0(x)y″+P1(x)y′+P2(x)y=0
was defined to be
W=|y1y2y′1y′2|.
- Rewrite (A) as a system of first order equations and show that W is the Wronskian (as defined in this section) of two solutions of this system.
- Apply Equation 10.3.6 to the system derived in (a), and show that W(x)=W(x0)exp{−∫xx0P1(s)P0(s)ds}, which is the form of Abel’s formula given in Theorem 9.1.3.
3. In Section 9.1 the Wronskian of n solutions y1, y2, …, yn of the n−th order equation
P0(x)y(n)+P1(x)y(n−1)+⋯+Pn(x)y=0
was defined to be
W=|y1y2⋯yny′1y′2⋯y′n⋮⋮⋱⋮y(n−1)1y(n−1)2⋯y(n−1)n|.
- Rewrite (A) as a system of first order equations and show that W is the Wronskian (as defined in this section) of n solutions of this system.
- Apply Equation 10.3.6 to the system derived in (a), and show that W(x)=W(x0)exp{−∫xx0P1(s)P0(s)ds}, which is the form of Abel’s formula given in Theorem 9.1.3.
4. Suppose
y1=[y11y21]andy2=[y12y22]
are solutions of the 2×2 system y′=Ay on (a,b), and let
Y=[y11y12y21y22]andW=|y11y12y21y22|
thus, W is the Wronskian of {y1,y2}.
- Deduce from the definition of determinant that W′=|y′11y′12y21y22|+|y11y12y′21y′22|.
- Use the equation Y′=A(t)Y and the definition of matrix multiplication to show that [y′11y′12]=a11[y11y12]+a12[y21y22] and [y′21y′22]=a21[y11y12]+a22[y21y22].
- Use properties of determinants to deduce from (a) and (a) that |y′11y′12y21y22|=a11Wand|y11y12y′21y′22|=a22W.
- Conclude from (c) that W′=(a11+a22)W, and use this to show that if a<t0<b then W(t)=W(t0)exp(∫tt0[a11(s)+a22(s)]ds)a<t<b.
5. Suppose the n×n matrix A=A(t) is continuous on (a,b). Let
Y=[y11y12⋯y1ny21y22⋯y2n⋮⋮⋱⋮yn1yn2⋯ynn],
where the columns of Y are solutions of y′=A(t)y. Let
ri=[yi1yi2…yin]
be the ith row of Y, and let W be the determinant of Y.
- Deduce from the definition of determinant that W′=W1+W2+⋯+Wn, where, for 1≤m≤n, the ith row of Wm is ri if i≠m, and r′m if i=m.
- Use the equation Y′=AY and the definition of matrix multiplication to show that r′m=am1r1+am2r2+⋯+amnrn.
- Use properties of determinants to deduce from (b) that det
- Conclude from (a) and (c) that W'=(a_{11}+a_{22}+\cdots+a_{nn})W,\nonumber and use this to show that if a<t_0<b then W(t)=W(t_0)\exp\left( \int^t_{t_0}\big[a_{11}(s)+a_{22}(s)+\cdots+a_{nn}(s)]\, ds\right), \quad a < t < b.\nonumber
6. Suppose the n\times n matrix A is continuous on (a,b) and t_0 is a point in (a,b). Let Y be a fundamental matrix for {\bf y}'=A(t){\bf y} on (a,b).
- Show that Y(t_0) is invertible.
- Show that if {\bf k} is an arbitrary n-vector then the solution of the initial value problem {\bf y}'=A(t){\bf y},\quad {\bf y}(t_0)={\bf k}\nonumber is {\bf y}=Y(t)Y^{-1}(t_0){\bf k}.\nonumber
7. Let
A=\left[\begin{array}{cc}{2}&{4}\\[4pt]{4}&{2}\end{array} \right], \quad {\bf y}_1=\left[\begin{array}{c} e^{6t} \\[4pt] e^{6t} \end{array}\right], \quad {\bf y}_2=\left[\begin{array}{r} e^{-2t} \\[4pt] -e^{-2t}\end{array}\right], \quad {\bf k}=\left[\begin{array}{r}-3 \\[4pt] 9\end{array}\right].\nonumber
- Verify that \{{\bf y}_1,{\bf y}_2\} is a fundamental set of solutions for {\bf y}'=A{\bf y}.
- Solve the initial value problem {\bf y}'=A{\bf y},\quad {\bf y}(0)={\bf k}. \tag{A}
- Use the result of Exercise 10.3.6 (b) to find a formula for the solution of (A) for an arbitrary initial vector {\bf k}.
8. Repeat Exercise 10.3.7 with
A=\left[\begin{array}{cc}{-2}&{-2}\\[4pt]{-5}&{1}\end{array} \right], \quad {\bf y}_1=\left[\begin{array}{r} e^{-4t} \\[4pt] e^{-4t}\end{array}\right], \quad {\bf y}_2=\left[ \begin{array}{r}-2e^{3t} \\[4pt] 5e^{3t}\end{array}\right], \quad {\bf k}=\left[\begin{array}{r} 10 \\[4pt]-4\end{array}\right].\nonumber
9. Repeat Exercise 10.3.7 with
A=\left[\begin{array}{cc}{-4}&{-10}\\[4pt]{3}&{7}\end{array} \right], \quad {\bf y}_1=\left[\begin{array}{r}-5e^{2t} \\[4pt] 3e^{2t} \end{array}\right], \quad {\bf y}_2=\left[\begin{array}{r} 2e^t \\[4pt]-e^t \end{array}\right], \quad {\bf k}=\left[\begin{array}{r}-19 \\[4pt] 11\end{array} \right ].\nonumber
10. Repeat Exercise 10.3.7 with
A=\left[\begin{array}{cc}{2}&{1}\\[4pt]{1}&{2}\end{array} \right], \quad {\bf y}_1=\left[\begin{array}{r} e^{3t} \\[4pt] e^{3t} \end{array}\right], \quad {\bf y}_2=\left[\begin{array}{r}e^t \\[4pt] -e^t\end{array}\right], \quad {\bf k}=\left[\begin{array}{r} 2 \\[4pt] 8 \end{array}\right].\nonumber
11. Let
\begin{aligned} A&= \left[\begin{array}{ccc}{3}&{-1}&{-1}\\[4pt]{-2}&{3}&{2}\\[4pt]{4}&{-1}&{-2}\end{array} \right] , \\[4pt] {\bf y}_1&=\left[\begin{array}{c} e^{2t} \\[4pt] 0 \\[4pt] e^{2t}\end{array} \right], \quad {\bf y}_2=\left[\begin{array}{c} e^{3t} \\[4pt]-e^{3t} \\[4pt] e^{3t}\end{array}\right], \quad {\bf y}_3=\left[\begin{array}{c} e^{-t} \\[4pt]-3e^{-t} \\[4pt] 7e^{-t} \end{array}\right], \quad {\bf k}=\left[\begin{array}{r} 2 \\[4pt]-7 \\[4pt] 20\end{array}\right].\end{aligned}\nonumber
- Verify that \{{\bf y}_1,{\bf y}_2,{\bf y}_3\} is a fundamental set of solutions for {\bf y}'=A{\bf y}.
- Solve the initial value problem {\bf y}'=A{\bf y}, \quad {\bf y}(0)={\bf k}. \tag{A}
- Use the result of Exercise 10.3.6 (b) to find a formula for the solution of (A) for an arbitrary initial vector {\bf k}.
12. Repeat Exercise 10.3.11 with
\begin{aligned} A&=\left[\begin{array}{ccc}{0}&{2}&{2}\\[4pt]{2}&{0}&{2}\\[4pt]{2}&{2}&{0}\end{array} \right], \\[4pt] {\bf y}_1&=\left[\begin{array}{c}-e^{-2t} \\[4pt] 0 \\[4pt] e^{-2t} \end{array}\right], \quad {\bf y}_2=\left[\begin{array}{c}-e^{-2t} \\[4pt] e^{-2t} \\[4pt] 0\end{array}\right], \quad {\bf y}_3=\left[\begin{array}{c} e^{4t} \\[4pt] e^{4t} \\[4pt] e^{4t}\end{array} \right], \quad {\bf k}=\left[\begin{array}{r} 0 \\[4pt]-9 \\[4pt] 12\end{array} \right].\end{aligned}\nonumber
13. Repeat Exercise 10.3.11 with
\begin{aligned} A&=\left[\begin{array}{ccc}{-1}&{2}&{3}\\[4pt]{0}&{1}&{6}\\[4pt]{0}&{0}&{-2}\end{array} \right], \\[4pt] {\bf y}_1&=\left[\begin{array}{c} e^t \\[4pt] e^t \\[4pt] 0\end{array}\right], \quad {\bf y}_2=\left[\begin{array}{c} e^{-t} \\[4pt] 0 \\[4pt] 0\end{array}\right], \quad {\bf y}_3=\left[\begin{array}{c} e^{-2t} \\[4pt]-2e^{-2t} \\[4pt] e^{-2t}\end{array}\right], \quad {\bf k}=\left[\begin{array}{r} 5 \\[4pt] 5 \\[4pt]-1 \end{array}\right].\end{aligned}\nonumber
14. Suppose Y and Z are fundamental matrices for the n\times n system {\bf y}'=A(t){\bf y}. Then some of the four matrices YZ^{-1}, Y^{-1}Z, Z^{-1}Y, Z Y^{-1} are necessarily constant. Identify them and prove that they are constant.
15. Suppose the columns of an n\times n matrix Y are solutions of the n\times n system {\bf y}'=A{\bf y} and C is an n \times n constant matrix.
- Show that the matrix Z=YC satisfies the differential equation Z'=AZ.
- Show that Z is a fundamental matrix for {\bf y}'=A(t){\bf y} if and only if C is invertible and Y is a fundamental matrix for {\bf y}'=A(t){\bf y}.
16. Suppose the n\times n matrix A=A(t) is continuous on (a,b) and t_0 is in (a,b). For i=1, 2, …, n, let {\bf y}_i be the solution of the initial value problem {\bf y}_i'=A(t){\bf y}_i,\; {\bf y}_i(t_0)={\bf e}_i, where
{\bf e}_1=\left[\begin{array}{c} 1\\[4pt]0\\[4pt] \vdots\\[4pt]0\end{array}\right],\quad {\bf e}_2=\left[\begin{array}{c} 0\\[4pt]1\\[4pt] \vdots\\[4pt]0\end{array}\right],\quad\cdots\quad {\bf e}_n=\left[\begin{array}{c} 0\\[4pt]0\\[4pt] \vdots\\[4pt]1\end{array}\right];\nonumber
that is, the jth component of {\bf e}_i is 1 if j=i, or 0 if j\ne i.
- Show that\{{\bf y}_1,{\bf y}_2,\dots,{\bf y}_n\} is a fundamental set of solutions of {\bf y}'=A(t){\bf y} on (a,b).
- Conclude from (a) and Exercise 10.3.15 that {\bf y}'= A(t){\bf y} has infinitely many fundamental sets of solutions on (a,b).
17. Show that Y is a fundamental matrix for the system {\bf y}'=A(t){\bf y} if and only if Y^{-1} is a fundamental matrix for {\bf y}'=- A^T(t){\bf y}, where A^T denotes the transpose of A. HINT: See Exercise 10.3.11.
18. Let Z be the fundamental matrix for the constant coefficient system {\bf y}'=A{\bf y} such that Z(0)=I.
- Show that Z(t)Z(s)=Z(t+s) for all s and t. HINT: For fixed s let \Gamma _{1}(t)=Z(t)Z(s) and \Gamma _{2}(t)=Z(t+s). Show that \Gamma _{1} and \Gamma_{2} are both solutions of the matrix initial value problem \Gamma '=A\Gamma , \:\Gamma (0)=Z(s). Then conclude from Theorem 10.2.1 that \Gamma _{1}=\Gamma _{2}.
- Show that (Z(t))^{-1}=Z(-t).
- The matrix Z defined above is sometimes denoted by e^{tA}. Discuss the motivation for this notation.