Math 309B/C Winter 2012, Homework 2


Due January 18

Also, I have a few examples of how to sketch solutions to some first order equations that should help you do some of the problems below.
  1. Suppose that $x^{(1)},...,x^{(n)}$ are solutions to a 1st order homogeneous system of $n$ equations $x'=Ax$ on the interval $(\alpha,\beta)$ (where $x(t)$ is an $n$-dimensional vector function and $A(t)$ is an $n\times n$ matrix function. Show that $W(t)=W(x^{(1)},...,x^{(n)})(t)\neq 0$ for all $t\in (\alpha,\beta)$ or $W(t)=0$ for all $t\in (\alpha,\beta)$. Hint: Remember that if $W(t_{0})=0$ for some $t_{0}$, then the vectors $x^{(1)}(t_{0}),...,x^{(n)}(t_{0})$ are linearly dependent, i.e., there is a linear combonation \[c_{1}x^{(1)}(t_{0})+\cdots + c_{n}x^{(n)}(t_{0})=0,\] where not all the $c_{j}$ are zero. Let's say it's $c_{1}$, so $c_{1}\neq 0$. Then $x^{(1)}(t_{0})$ can be written as a linear combination of $x^{(2)}(t_{0}),...,x^{(n)}(t_{0})$ (write down for yourself what this would look like, you just solve the above equation for $x^{(1)}(t_{0})$. Why will this same linear combination hold for all $t\in (\alpha,\beta)$ and not just $t_{0}$? (This uses a theorem we used in class last time). Why then does this show that $W(t)=0$ for all $t$?
    Answer: I'll give a slightly different proof than what I hinted at but it's pretty much the same. Suppose $W(t_0)=0$ for some $t_{0}$. Then $\det(x^{(1)}(t_0),...,x^{(n)}(t_0))=0$ (since this is what the Wronksian is). Hence, $x^{(1)}(t_0),...,x^{(n)}(t_0)$ are a linearly dependent set of vectors, so there are constants $c_1,...,c_n$ (not all zero) such that \[c_1 x^{(1)}(t_0)+...+c_n x^{(n)}(t_0)=0.\] Define \[ f(t)=c_1 x^{(1)}(t)+...+c_n x^{(n)}(t).\] Then by the above formula, $f(t_0)=0$ and $f$ solves $x'=Ax$ (since it is a linear combination of solutions). Hence, it solves the initial value problem \[\left\{ \begin{array}{c} x'=Ax \\ x(t_0)=0.\end{array}\right. .\] But $x=0$ also solves this same initial value problem, and by the uniqueness of solutions to initial value problems, these two functions $f$ and $0$ must be the same for all $t$, i.e. \[0=f(t)= c_1 x^{(1)}(t)+...+c_n x^{(n)}(t)\] for all $t$, so $x^{(1)}(t),...,x^{(n)}(t)$ are linearly dependent for all $t$, and hence $W(t)=0$ for all $t$.
    Now suppose $W(t_0)\neq 0$ for some $t_0$. Then $W(t)\neq 0$ for all $t$, since if $W(t_1)=0$ for some other number $t_1$, then by the previous work $W(t)=0$ for all $t$, which is impossible since $W(t_0)\neq 0$. Hence, $W(t)$ is either always zero or always nonzero.
  2. Let $x'=Ax$ be a 1st order homogeneous system of linear equations defined in $(\alpha,\beta)$ and $t_{0}\in (\alpha,\beta)$. Suppose $x^{(1)},...,x^{(n)}$ are solutions to the initial value problem $x^{(1)}(t_{0})=e_{1},...,x^{(n)}(t_{0})=e_{n}$, where $e_{1},...,e_{n}$ are the standard basis vectors. Show that these form a fundamental set of solutions. (Hint: remember what we concluded at the end of Wednesday's lecture was necessary for some functions to be a fundamental set of solutions). then the solutions.
    Answer: Recall that $x^{(1)},...,x^{(n)}$ are fundamental solutions if and only if their Wronskian is always nonzero. By a theorem (or really, by the previous problem), we only need to check that $W(t)\neq 0$ somewhere (note, you are allowed to use this result on future homeworks or exam problems, unless I'm asking you to prove it). So let's compute $W(t_0)$, \[W(t_0)=\det(x^{(1)}(t_0)|\cdots | x^{(n)}(t_0)) = \det(e_1|\cdots |e_n)=\det I=1\neq 0.\]
  3. Draw trajectories of solutions for the following equations:
    1. $x'=\left(\begin{array}{cc}3 & -2 \\ 2 & -2 \end{array}\right)x$.
      Answer: This matrix has eigenvectors $v_1=\left(\begin{array}{c} 1 \\ 2 \end{array}\right)$ and $v_2=\left(\begin{array}{c} 2 \\ 1 \end{array}\right)$ with eigenvalues $-1$ and $2$ respectively. Hence, the general solution is \[x(t)= c_1 \left(\begin{array}{c} 1 \\ 2 \end{array}\right) e^{-t} + c_2 \left(\begin{array}{c} 2 \\ 1 \end{array}\right)e^{2t}.\] To plot, notice that as $t\rightarrow\infty$, the first vector decays to zero and the function looks approximately like $c_2 \left(\begin{array}{c} 2 \\ 1 \end{array}\right)e^{2t}$, i.e. it's moving parallel to $v_2$, and as $t\rightarrow-\infty$, the opposite is happening: it is moving parallel with $v_1$. In either case, because the eigenvalues have opposite signs, the solution is unstable in both directions, i.e. the solutions move way from the origin as $t\rightarrow\pm\infty$, and so typical solutions should look like such (the lines parallel to $v_1$ and $v_2$ have been omitted):
    2. $x'=\left(\begin{array}{cc}1 & -2 \\ 3 & -4 \end{array}\right)x$.
    3. $x'=\left(\begin{array}{cc}1 & 1 \\ 4 & -2 \end{array}\right)x$
    4. $x'=\left(\begin{array}{cc}2 & -1 \\ 3 & -2 \end{array}\right)x$
    5. $x'=\left(\begin{array}{cc}-2 & 1 \\ 1 & -2 \end{array}\right)x$
      Answer:
      This matrix has eigenvectors $v_1=\left(\begin{array}{c} 1 \\ 1 \end{array}\right)$ and $v_2=\left(\begin{array}{c} 1 \\ -1 \end{array}\right)$ with eigenvalues $-1$ and $-3$ respectively. Hence, the general solution is \[x(t)= c_1 \left(\begin{array}{c} 1 \\ 1 \end{array}\right) e^{-t} + c_2 \left(\begin{array}{c} 1 \\ -1 \end{array}\right)e^{-3t}.\] Note that all solutions are decaying since both eigenvalues are negative, so all solutions go into the origin. Note that the first part of the general solution is much larger than the second for very large $t$, hence the solution gradually becomes more parallel with $v_1$ as $t\rightarrow\infty$. As $t\rightarrow-\infty$, the opposite happens: the solution becomes more parallel with $v_2$ since the second term in the general solution is much larger for these values of $t$. Hence, the solutions have the following sketches:
    6. $x'=\left(\begin{array}{cc}4 & -3 \\ 8 & -6 \end{array}\right)x$ This matrix has eigenvectors $v_1=\left(\begin{array}{c} 3 \\ 4 \end{array}\right)$ and $v_2=\left(\begin{array}{c} 1 \\ 2 \end{array}\right)$ with eigenvalues $0$ and $-1$ respectively. Hence, the general solution is \[x(t)= c_1 \left(\begin{array}{c} 3 \\ 4 \end{array}\right) + c_2 \left(\begin{array}{c} 1 \\ 2 \end{array}\right)e^{-t}.\] The sketches of solutions to this should look like lines parallel to $v_2$ descending straight onto the line through the origin parallel to $v_1$. The reason is that as $t\rightarrow\infty$, $x(t)\rightarrow c_1 \left(\begin{array}{c} 3 \\ 4 \end{array}\right)$. Hence, we know our solution converges to some point on the $v_1$ line. Note that $x'$ is always just a multiple of $\left(\begin{array}{c} 1 \\ 2 \end{array}\right)$, hence the direction $x$ is always parallel to this vector, and this is enough to conclude the description.
    7. $x'=\left(\begin{array}{cc}3 & -2 \\ 2 & -2 \end{array}\right)x$
    8. $x'=\left(\begin{array}{cc}3 & 6 \\ -1 & -2 \end{array}\right)x$
  4. Find the general solutions to the following systems. If there are any initial conditions, solve the initial value problem and describe the behavior of the solution as $t\rightarrow\infty$.
    1. $x'=\left(\begin{array}{ccc} 1 & 1 & 2 \\ 1 & 2 & 1 \\ 2 & 1 & 1 \end{array}\right)x$
      Answer: The general solution is \[x(t)=c_1 \left(\begin{array}{c} 1 \\ 1 \\ 1\end{array}\right)e^{4t} +c_2 \left(\begin{array}{c} 1 \\ -2 \\ 1\end{array}\right)e^{t} +c_3 \left(\begin{array}{c} 1 \\ 0 \\ -1\end{array}\right)e^{-t}.\] If a solution begins on the line parallel to $\left(\begin{array}{c} 1 \\ 0 \\ -1\end{array}\right)$, then it is stable and converges into the origin as $t\rightarrow \infty$ (since then $c_1=c_2=0$). Otherwise, the solutions will diverge to infinity in the direction \[c_1 \left(\begin{array}{c} 1 \\ 1 \\ 1\end{array}\right) +c_2 \left(\begin{array}{c} 1 \\ -2 \\ 1\end{array}\right),\] so notice that the direction it takes depends on the initial conditions (i.e. $c_1$ and $c_2$).
    2. $x'=\left(\begin{array}{ccc} 3 & 2 & 4 \\ 2 & 0 & 2 \\ 4 & 2 & 3 \end{array}\right)x$
    3. $x'=\left(\begin{array}{ccc} 1 & -1 & 4 \\ 3 & 2 & -1 \\ 2 & 1 & -1 \end{array}\right)x$
    4. $x'=\left(\begin{array}{ccc} 1 & 1 & 2 \\ 0 & 2 & 2 \\ -1 & 1 & 3 \end{array}\right)x$, $x(0)=\left(\begin{array}{c} 2 \\ 0 \\ 1 \end{array}\right)$. Answer: The general solution is \[x(t)=c_1\left(\begin{array}{c} 0 \\ -2 \\ 1\end{array}\right)e^{t} +c_2 \left(\begin{array}{c} 1 \\ 1 \\ 0\end{array}\right)e^{2t} +c_3\left(\begin{array}{c} 2 \\ 2 \\ 1\end{array}\right)e^{3t}.\] To solve the initial value problem, set \[\left(\begin{array}{c} 2 \\ 0 \\ 1 \end{array}\right)=x(0)=c_1\left(\begin{array}{c} 0 \\ -2 \\ 1\end{array}\right) +c_2 \left(\begin{array}{c} 1 \\ 1 \\ 0\end{array}\right) +c_3\left(\begin{array}{c} 2 \\ 2 \\ 1\end{array}\right) = \left(\begin{array}{c} 0 \\ -2 \\ 1\end{array} \begin{array}{c} 1 \\ 1 \\ 0\end{array} \begin{array}{c} 2 \\ 2 \\ 1\end{array}\right) \left(\begin{array}{c} c_1 \\ c_2 \\ c_3\end{array}\right),\] and now all you need to do is solve this matrix equation. The solution is \[\left(\begin{array}{c} c_1 \\ c_2 \\ c_3\end{array}\right)=\left(\begin{array}{c} 1 \\ 2 \\ 0\end{array}\right),\] and hence the solution to this initial value problem is \[ x(t)= \left(\begin{array}{c} 0 \\ -2 \\ 1\end{array}\right)e^{t} +2 \left(\begin{array}{c} 1 \\ 1 \\ 0\end{array}\right)e^{2t}.\]
    5. $x'=\left(\begin{array}{ccc} 0 & 0 & -1 \\ 2 & 0 & 0 \\ -1 & 2 & 4 \end{array}\right)x$, $x(0)=\left(\begin{array}{c} 7 \\ 5 \\ 5 \end{array}\right)$.