Difference between revisions of "Tutorial Week 6"

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===Worksheet Questions===
 
===Worksheet Questions===
  
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#Salt water with concentration of 1 g/L flows into Tank A at a rate $a$ L/min. The mixed solution in Tank A flows into Tank B at a rate $b$ L/min. Another pipe takes the solution in Tank B back into Tank A at a rate $c$ L/min. Finally, solution drains out of Tank B at a rate $a$ L/min. '''Note''': the volume in each tank is the same and $a$+$c$=$b$ (to ensure that the volumes in the tanks are constant).
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You can print the [[Media:Tutorial7Worksheet.pdf|PDF]]. This worksheet relates to material from weeks 6 and 7.
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## Write down the system that gives the amount of salt in each tank at any given time.
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## Show that it is impossible for this system to have oscillations (damped or otherwise).
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#Plot the phase plane for the following systems of equations. Your phase plane should include eigen-directions for real eigenvalues and several solutions illustrating the general shapes of solutions in the phase plane.
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##\begin{eqnarray}
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−
x_1' &=& x_1 -8x_2 \\
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x_2' &=& 8x_1+x_2
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\end{eqnarray}
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##\begin{eqnarray}
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x_1' &=& x_2 \\
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x_2' &=& -3x_1-4x_2
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\end{eqnarray}
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===Solutions:===
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<ol>
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#('''6pts'''+'''2pts''')
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<li> Find the general solution to the following linear system of differential equations
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##\begin{eqnarray}
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\begin{equation}
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Q_1^\prime &=& (1)\cdot a -b\frac{Q_1}{V} + c\frac{Q_2}{V} \\
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\mathbf{x}'=
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Q_2^\prime &=& b\frac{Q_1}{V} -(a+c)\frac{Q_2}{V}  
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\left(
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\end{eqnarray} For each equation, '''1 pt''' for form (positive and negative terns in a DE), '''2 pts''' for the details of the terms (Q, V) so total '''6 pts'''.
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\begin{array}{cc}
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## Eigenvalues of the homogeneous equation are: $-b \pm \sqrt{bc}$ '''(1 pt)'''. Since b,c $\ge$ 0, the eigenvalues are real and the system can not have oscillations ('''1 pt''' - must state this explicitly).
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2&-1\\
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#('''3pts'''+'''6pts''')
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1&4
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##'''1 pt''' for eigenvalues (1+ 8i & 1-8i) , '''1 pt''' for growing spiral, '''1 pt''' for direction of rotation (must have some justification for choice)
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\end{array}
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##'''1 pt''' for eigenvalues (-1& -3), '''2 pts''' for eigenvectors (e.g. (-1,1) for -1, and (-1, 3) for -3), '''3 pts''' for drawing ('''1 pt''' for including a solution in each of the two eigen-directions with arrows correct, '''1 pt''' for shape of non-eigen solutions, '''1 pt''' for direction of arrows on non-eigen solutions).
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\right)\mathbf{x}
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Here are example answers to these two questions. The vector fields are not necessary but there should be arrows showing direction along the eigen-directions and the solution curves.
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\end{equation}</li>
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  https://www.desmos.com/calculator/oy3oslqfya
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  https://www.desmos.com/calculator/60djgxc0hn (Drag the blue point to see different solution curves.)
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−
[[File:Question 2b.jpeg]]
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<li>
 +
Consider the following system
 +
\begin{equation*}
 +
\mathbf{x}'=
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\left(
 +
\begin{array}{cc}
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1&-2\\
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2&-4
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\end{array}
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\right)\mathbf{x}+
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\mathbf{b},\quad
 +
 
 +
\text{where} \quad
 +
 
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\mathbf{b}=
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\left(
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\begin{array}{c}
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b_1\\
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b_2
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\end{array}
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\right)
 +
\end{equation*} </li>
 +
 
 +
<ol>
 +
<li>Determine the general solution when $b_1=1$ and $b_2=2$.</li>
 +
 
 +
<li>In part (a), $b_1$ and $b_2$ were specially chosen in that the row reduced form of the matrix equation had a full row of zeros (including the RHS) and therefore a solution. Write down an equation for $b_1$ and $b_2$ that ensures this will happen. </li>
 +
 
 +
<li>Now consider the case with $b_1=3$ and $b_2=3$. Because this $\mathbf{b}$ does not satisfy the
 +
equation from part (b), a different form for your particular guess is needed. By analogy with second order systems, we guess $\mathbf{x_p} = t \mathbf{v}+\mathbf{w}$. Now we take the following steps to find out the general solution for this case:
 +
 
 +
<ol>
 +
 
 +
<li>Plugging this $\mathbf{x_p} $
 +
into the system of ODEs, we find that we must have
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$\mathbf{v} = t \mathbf{A} \mathbf{v}+ \mathbf{A} \mathbf{w} + \mathbf{b}$
 +
for all $t$. This requires that $\mathbf{A} \mathbf{v}=0$ and $\mathbf{A} \mathbf{w}=\mathbf{v}-\mathbf{b};$ </li>
 +
 
 +
<li> $\mathbf{A} \mathbf{v}=0$ has a whole family of
 +
solutions. In fact , since $0$ is an eigenvalue of $\mathbf{A} $, $\mathbf{v}$ should be a corresponding eigenvector; </li>
 +
 
 +
<li> Next consider the equation $\mathbf{A} \mathbf{w}=\mathbf{v}-\mathbf{b}$ which we must solve for $\mathbf{w}$. Notice
 +
that for any vector $\mathbf{w}$, $\mathbf{A} \mathbf{w}$ will always have a second component equal to twice
 +
its first component. Thus, to be able to solve $\mathbf{A} \mathbf{w}=\mathbf{v}-\mathbf{b}$ , we must make sure
 +
that $\mathbf{v}-\mathbf{b}$ has second component equal to twice its first component. Find the vector $\mathbf{v}$ from the family of solutions to $\mathbf{A} \mathbf{v}=0$ that does this. Then find $\mathbf{w}$ and write down the general solution for this case.
 +
   
 +
  </li>
 +
</ol>
 +
 
 +
 
 +
  </li>
 +
</ol>
 +
</ol>
 +
 
 +
<!-- See history of "Tutorial Week 7" for solutions -->
 +
 
 +
[[Tutorial Week 6 Solutions]]

Latest revision as of 23:06, 30 December 2020

Worksheet Questions

You can print the PDF. This worksheet relates to material from weeks 6 and 7.

  1. Find the general solution to the following linear system of differential equations \begin{equation} \mathbf{x}'= \left( \begin{array}{cc} 2&-1\\ 1&4 \end{array} \right)\mathbf{x} \end{equation}


  2. Consider the following system \begin{equation*} \mathbf{x}'= \left( \begin{array}{cc} 1&-2\\ 2&-4 \end{array} \right)\mathbf{x}+ \mathbf{b},\quad \text{where} \quad \mathbf{b}= \left( \begin{array}{c} b_1\\ b_2 \end{array} \right) \end{equation*}
    1. Determine the general solution when $b_1=1$ and $b_2=2$.
    2. In part (a), $b_1$ and $b_2$ were specially chosen in that the row reduced form of the matrix equation had a full row of zeros (including the RHS) and therefore a solution. Write down an equation for $b_1$ and $b_2$ that ensures this will happen.
    3. Now consider the case with $b_1=3$ and $b_2=3$. Because this $\mathbf{b}$ does not satisfy the equation from part (b), a different form for your particular guess is needed. By analogy with second order systems, we guess $\mathbf{x_p} = t \mathbf{v}+\mathbf{w}$. Now we take the following steps to find out the general solution for this case:
      1. Plugging this $\mathbf{x_p} $ into the system of ODEs, we find that we must have $\mathbf{v} = t \mathbf{A} \mathbf{v}+ \mathbf{A} \mathbf{w} + \mathbf{b}$ for all $t$. This requires that $\mathbf{A} \mathbf{v}=0$ and $\mathbf{A} \mathbf{w}=\mathbf{v}-\mathbf{b};$
      2. $\mathbf{A} \mathbf{v}=0$ has a whole family of solutions. In fact , since $0$ is an eigenvalue of $\mathbf{A} $, $\mathbf{v}$ should be a corresponding eigenvector;
      3. Next consider the equation $\mathbf{A} \mathbf{w}=\mathbf{v}-\mathbf{b}$ which we must solve for $\mathbf{w}$. Notice that for any vector $\mathbf{w}$, $\mathbf{A} \mathbf{w}$ will always have a second component equal to twice its first component. Thus, to be able to solve $\mathbf{A} \mathbf{w}=\mathbf{v}-\mathbf{b}$ , we must make sure that $\mathbf{v}-\mathbf{b}$ has second component equal to twice its first component. Find the vector $\mathbf{v}$ from the family of solutions to $\mathbf{A} \mathbf{v}=0$ that does this. Then find $\mathbf{w}$ and write down the general solution for this case.



Tutorial Week 6 Solutions

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