Math 309B/C Winter 2012, Homework 1

Due January 11

Write the following differential equations as first order linear equations. If it is an initial value problem, transform it into an initial value problem for a system of two or more first order equations.
1. $tu''+t^2u'+u=0$, $u(1)=0$, $u'(1)=1$
  - Answer: Since it's a two degree equation, we expect to write this as a system of two first order equations. Let $x_{1}=u,x_{2}=u'$. Then the first equation in our system is $x_{1}'=u'=x_{2}$, easy enough. For the second equation (i.e. the one for $x_{2}'$), we use the original equation to solve for $x_{2}'$ by substituting first $x_{2}$ in for $u'$: \[0=tu''+t^{2}u+u=t(u')'+t^{2}u'+u=tx_{2}'+t^{2}x_{2}+u\] and now substitute $x_{1}$ in for $u$, and we get \[0=tx_{2}'+t^{2}x_{2}+x_{1},\] and solving for $x_{2}'$ gives us the second equation in our system \[x_{2}'=t^{-1}x_{1}+tx_{2}.\] The initial conditions for this system are $x_{1}(1)=u(1)=0$ and $x_{2}(1)=u'(1)=1$.
2. $u'''+e^{2t}u''+e^{t}u'+u=1$, $u''(0)=1,u'(0)=2,u(0)=3$
3. $u^{(4)}-u=0$ (here, $u^{(4)}$ denotes the 4th derivative of $u$).
  - Answer: For this, we will need four equations. Set $x_{1}=u,x_{2}=u',x_{3}=u''$,and $x_{4}=u'''$. Then $x_{1}'=u'=x_{2},x_{2}'=u''=x_{3}$, and $x_{3}'=u''''=x_{4}$ are the first three equations in our system of equations. For the final one, we use the original equation to solve by substituting $x_{4}$ in for $u'''$ and $x_{1}$ in for $u$, \[0=u^{(4)}-u=(u''')'-u=x_{4}'-x_{1},\] and so the last equation in our system is $x_{4}'=x_{1}$.
Consider the linear homogeneous system \[\begin{array}{l} x'=p_{11}(t)x+p_{12}(t)y \\ y'=p_{21}(t)x+p_{22}(y)\end{array}\] Show that if $x=x_{1}(t),y=y_{1}(t)$ and $x=x_{2}(t),y=y_{2}(t)$ are two sets of solutions to this system, then so is $x=c_{1}x_{1}(t)+c_{2}x_{2}(t)$, $y=c_{1}y_{1}(t)+c_{2}y_{2}(t)$. This is called the principal of superposition, since we are taking two solutions and combining them to get another solution.
- Answer: First write this as a matrix equation, \[X'=PX,\] where $X=\left(\begin{array}{c} x \\ y \end{array} \right)$. (If you multiply this out, then you get $X'$ equals some vector, and so the first and second entries respectfully of these two vectors must equal, and those equations are exactly the system above). Thus, a pair of functions $x$ and $y$ solve this system if and only if the vector $X=\left(\begin{array}{c} x \\ y \end{array} \right)$ solves $X'=PX$. Suppose $x=x_{1}$ and $y=y_{1}$ solve the system and $x=x_{2},y=y_{2}$ solves the system (as in the problem). Let $X_{1}=\left(\begin{array}{c} x_{1} \\ y_{1} \end{array} \right)$ and $X_{2}'=PX_{2}$ if $X_{2}=\left(\begin{array}{c} x_{2} \\ y_{2} \end{array} \right)$. Then $X_{1}'=PX_{1}$ and $X_{2}'=PX_{2}$. Hence, if $X=c_{1}X_{1}+c_{2}X_{2}$, then \[X'=(c_{1}X_{1}+c_{2}X_{2})'=c_{1}X_{1}'+c_{2}X_{2}'=c_{1}PX_{1}+c_{2}PX_{2}=P(c_{1}X_{1}+c_{2}X_{2})=PX.\]
Let $A_{\theta}=\left(\begin{array}{cc} \cos\theta & \sin \theta \\ -\sin\theta & \cos\theta \end{array}\right).$ Compute the following:
1. $A_{\frac{\pi}{4}}A_{\frac{\pi}{4}}$
2. $A_{\frac{\pi}{3}}A_{-\frac{\pi}{3}}$
3. $A_{\frac{\pi}{6}}A_{\frac{\pi}{2}}$.
- Answer: This isn't really an answer, but this gives you a way of checking your work. Geometrically, the matrix $A_{\theta}$ as above is the rotation matrix: if you take $A_{\theta}x$, this is the vector $x$ rotated counter clockwise by angle $\theta$. Hence $A_{\theta}A_{\phi}x$ would be the vector we'd get if we rotated $x$ by $\phi$ and then by $\theta$, which is the same as just rotating by $\theta+\phi$, hence $A_{\theta}A_{\phi}=A_{\theta+\phi}$. So now you can compute what these matrices are and check your multipliciation.
Let $A=\left(\begin{array}{cc} 1+i & -1+2i \\ 3+2i & 2-i \end{array}\right)$ $B=\left(\begin{array}{cc} i & 3 \\ 2 & -2i\end{array}\right)$. Compute the following:
1. $A-2B$. Answer: $\left(\begin{array}{cc} 1-i & -7+2i \\ -1 +2i & =2 + 3i\end{array}\right)$.
2. $AB$.
  - Answer: $\left(\begin{array}{cc} -3 + 5i & 5 + 9i \\ 2 + i & 11 \end{array}\right)$.
  - $BA$
  - $A^{T}$
  - $\bar{A}$. Answer: $\left(\begin{array}{cc} 1-i & -1-2i \\ 3-2i & 2+i \end{array}\right)$.
  - $A^{*}$.Answer: $\left(\begin{array}{cc} 1-i & 3-2i \\ -1-2i & 2+i \end{array}\right)$.
If $A=\left(\begin{array}{ccc} 3i & 0 & 1+i \\ 0 & i & 1\end{array}\right)$, $x=\left(\begin{array}{c} 1 \\ 0 \\ i \end{array}\right)$, and $y=\left(\begin{array}{c} 1 \\ 2 \end{array}\right)$, compute
1. $\langle Ax,y\rangle$
2. $\langle x,A^{*}y\rangle$
Show that if $A$ is a 2x2 matrix with real interies (i.e. not complex), then for any two vectors $x$ and $y$, $\langle Ax,y\rangle=\langle x,A^{T}y\rangle$. Answer: This follows from the long equation in the answer to the previous problem. Otherwise, you can write $A=\left(\begin{array}{cc} a & b \\ c & d \end{array}\right)$, $x=\left(\begin{array}{c} x_{1} \\ x_{2}\end{array}\right)$, and $y = \left(\begin{array}{c} y_{1} \\ y_{2}\end{array}\right)$ and just multiply out $\langle Ax,\rangle$, you should get \[ax_{1}\bar{y}_{1}+bx_{2}\bar{y}_{1} + cx_{1}\bar{y}_{2} + dx_{2}\bar{y}_{2}\]
Compute the inverses of the following matrices, or show that they are singular:
1. $\left(\begin{array}{cc} 1 & 4 \\ -2 & 3 \end{array}\right)$. Answer $\frac{1}{11} \left(\begin{array}{cc} 3 & -4 \\ 2 & 1 \end{array}\right)$.
2. $\left(\begin{array}{ccc} 1 & 2 & 1 \\ -2 & 1 & 8 \\ 1 & -2 & -7 \end{array}\right)$. Answer: Not invertible.
3. $\left(\begin{array}{cccc} 1 & 0 & 0 & -1 \\ 0 & -1 & 1 & 0 \\ -1 & 0 & 1 & 0 \\ 0 & 1 & -1 & 1 \end{array}\right)$. Answer: $\left(\begin{array}{cccc} 1 & 1 & 0 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 \end{array}\right).$
4. $\left(\begin{array}{ccc} 1 & -1 & -1 \\ 2 & 1 & 0 \\ 3 & -2 & 1 \end{array}\right)$
5. $\left(\begin{array}{ccc} 1 & 2 & 1 \\ -2 & 1 & 9 \\ 1 & -2 & 1 \end{array}\right)$
Solve the following systems of linear equations, or show that no solution exists.
1. \[\begin{array}{r} x_{1}+2x_{2}-x_{3}=1 \\ 2x_{1}+x_{2}+x_{3}=1 \\ x_{1}-x_{2}+2x_{3}=1\end{array}\] Answer: No solution.
2. \[\begin{array}{r} x_{1}+2x_{2}-x_{3}=-2 \\-2x_{1} -4x_{2}+2x_{3}=4 \\ 2x_{1}+4x_{2}-2x_{3}=-4\end{array}\] Answer: Note that the second equation is -2 times the first, and the third is 2 times the first, so we need only find the solutions for the first equation $x_{1}+2x_{2}-x_{3}=-2$, so $x_{1}=-2x_{2}+x_{3}-2$. Hence any solution $x$ must be of the form \[x=\left(\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}\right) = \left(\begin{array}{c} -2x_{2}+x_{3}-2 \\ x_{2} \\ x_{3} \end{array}\right) = x_{2}\left(\begin{array}{c} -2 \\ 1 & 0 \end{array}\right) + x_{3}\left(\begin{array}{c} -1 \\ 0 \\ 1 \end{array}\right) + \left(\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\right).\]
3. \[\begin{array}{r} x_{1} - x_{3} = 0 \\ 3x_{1} + x_{2} + x_{3} = 1 \\ -x_{1}+x_{2}+2x_{3}=2 \end{array}.\]
Find the eigenvalues and eigenvectors for the following matrices.
1. $\left(\begin{array}{cc}5 & -1 \\ 3 & 1 \end{array}\right) $ Answer: For the characteristic polynomial, you should get $\lambda6{2}-6\lambda+ 8 = (\lambda -2)(\lambda-4)$, so the eigenvalues are $2$ and $4$. To find the eigenvector associated to $\lambda =2$, we solve $(A-2I)x=0$, where \[A-2I=\left(\begin{array}{cc} 3 & -1 \\ 3 & -1 \end{array}\right).\] Then one can solve and show that any multiple of $\left(\begin{array}{c} 1 3 \end{array}\right)$ is an eigenvalue. For $\lambda=4$, \[A-4I=\left(\begin{array}{cc} 1 & -1 \\ 3 & -3\end{array}\right),\] and $(A-4I)x=0$ whenever $x$ is a multiple of $\left(\begin{array}{c} 1 \\ 1 \end{array}\right)$, and so these are the eigenvectors for $\lambda=4$.
2. $\left(\begin{array}{cc} 1 & i \\ -i & 1 \end{array}\right) $
3. $\left(\begin{array}{ccc} 3 & 2 & 2 \\ 1 & 4 & 1 \\ -2 & -4 & -1 \end{array}\right)$. Answer: For the characteristic polynomial, you should get $-\lambda^{3}+6\lambda^{2}-11\lambda +6$. Now, in typical courses, if you have to solve a 3rd degree polynomial, then one of the roots should be something stupid, since no one expects you to actually know the formula to solve one of these. So start plugging in small integer numbers. Incidentally, one that works is $\lambda=1$. Now, divide this polynomial by $\lambda -1 $ and you will get $-\lambda^{2}+5\lambda -6=-(\lambda-2)(\lambda-3)$, so in total, the roots are $\lambda=1,2,3$. The associated eigenvectors are multiples of $\left(\begin{array}{c} 1 & 0 & -1 \end{array}\right)$, $\left(\begin{array}{c} 1 & -2 & 0 \end{array}\right)$, and $\left(\begin{array}{c} 0 & 1 & -1 \end{array}\right)$ respectively.
4. $\left(\begin{array}{ccc} 1 & 0 & 0 \\ 2 & 1 & -2 \\ 3 & 2 & 1 \end{array}\right)$.