MATH528

MATH528Due date: 08/29/18

Homework 06

1Exercises

Exercise. PS4.5.4

Solution. System $𝒚^{'} = 𝒇 (𝒚)$ has critical points at solutions of $𝒇 (𝒚) = 0$ ,

{\begin{cases} y_{1} (4 - y_{1}) = 0 \\ y_{2} = 0 \end{cases} ., (y_{1}, y_{2}) \in {(0, 0), (4, 0)} .

The Jacobian obtained by linearization is

𝐀 = \frac{\partial 𝒇}{\partial 𝒚} = (\begin{array}{ll} 4 - 2 y_{1} & 0 \\ 0 & 1 \end{array}) .

At critical point:

$(0, 0)$ , $𝐀$ has eigenvalues (4,1), unstable node
(4,0), $𝐀$ has eigenvalues (-4,1), saddle node

Exercise. PS4.5.5

Solution. System $𝒚^{'} = 𝒇 (𝒚)$ has critical points at solutions of $𝒇 (𝒚) = 0$ ,

{\begin{cases} y_{2} = 0 \\ y_{1} (- 1 + \frac{1}{2} y_{1}) = 0 \end{cases} ., (y_{1}, y_{2}) \in {(0, 0), (2, 0)} .

The Jacobian obtained by linearization is

𝐀 = \frac{\partial 𝒇}{\partial 𝒚} = (\begin{array}{ll} 0 & 1 \\ - 1 + y_{1} & 0 \end{array}) .

At critical point:

$(0, 0)$ , $𝐀$ has eigenvalues $(i, - i)$ , center
(2,0), $𝐀$ has eigenvalues (-1,1), saddle node

In[33]:=

Eigenvalues[{{0,1},{-1,0}}]

${i, - i}$

In[34]:=

Eigenvalues[{{0,1},{1,0}}]

${- 1, 1}$

In[35]:=

Exercise. PS4.5.6

Solution. System $𝒚^{'} = 𝒇 (𝒚)$ has critical points at solutions of $𝒇 (𝒚) = 0$ ,

{\begin{cases} y_{2} = 0 \\ y_{1} (1 + y_{1}) = 0 \end{cases} ., (y_{1}, y_{2}) \in {(0, 0), (- 1, 0)} .

The Jacobian obtained by linearization is

𝐀 = \frac{\partial 𝒇}{\partial 𝒚} = (\begin{array}{ll} 0 & 1 \\ - 1 - y_{1} & 0 \end{array}) .

At critical point:

$(0, 0)$ , $𝐀$ has eigenvalues $(i, - i)$ , center
(-1,0), $𝐀$ has eigenvalues $(i \sqrt{2}, - i \sqrt{2})$ , center

In[33]:=

Eigenvalues[{{0,1},{-1,0}}]

${i, - i}$

In[35]:=

Eigenvalues[{{0,-2},{1,0}}]

${i \sqrt{2}, - i \sqrt{2}}$

In[36]:=

Exercise. PS4.5.7

Solution. System $𝒚^{'} = 𝒇 (𝒚)$ has critical points at solutions of $𝒇 (𝒚) = 0$ ,

{\begin{cases} - y_{1} + y_{2} - y_{2}^{2} = y_{2} (2 - y_{2}) = 0 \\ - y_{1} - y_{2} = 0 \end{cases} ., (y_{1}, y_{2}) \in {(0, 0), (- 2, 2)} .

The Jacobian obtained by linearization is

𝐀 = \frac{\partial 𝒇}{\partial 𝒚} = (\begin{array}{ll} - 1 & 1 - 2 y_{2} \\ - 1 & - 1 \end{array}) .

At critical point:

$(0, 0)$ , $𝐀$ has eigenvalues $(- 1 + i, - 1 - i)$ , decaying spiral
(-2,2), $𝐀$ has eigenvalues $(- 1 - \sqrt{3}, \sqrt{3} - 1)$ , sadle point

In[36]:=

Eigenvalues[{{-1,1},{-1,-1}}]

${- 1 + i, - 1 - i}$

In[37]:=

Eigenvalues[{{-1,-3},{-1,-1}}]

${- 1 - \sqrt{3}, \sqrt{3} - 1}$

In[38]:=

Exercise. PS4.6.2

Solution. System is $𝒚^{'} = 𝐀 𝒚 + 𝒓 (t)$ ,

(\begin{array}{c} y_{1}^{'} \\ y_{2}^{'} \end{array}) = (\begin{array}{cc} 1 & 1 \\ 3 & - 1 \end{array}) (\begin{array}{c} y_{1} \\ y_{2} \end{array}) + 10 (\begin{array}{c} \cos t \\ - \sin t \end{array})

Solve eigenproblem $𝐀 𝐗 = 𝐗 𝚲$ to find eigendecomposition

𝐗 = (\begin{array}{cc} - 1 & 1 \\ 3 & 1 \end{array}), 𝚲 = (\begin{array}{cc} - 2 & 0 \\ 0 & 2 \end{array}) .

Solution of homogeneous system

𝒚_{h} = c_{1} (\begin{array}{c} - 1 \\ 3 \end{array}) e^{- 2 t} + c_{2} (\begin{array}{c} 1 \\ 1 \end{array}) e^{2 t} .

Seek particular solution by undetermined coefficients

𝒚_{p} = (\begin{array}{c} b_{11} \cos t + b_{12} \sin t \\ b_{21} \cos t + b_{22} \sin t \end{array}) = (\begin{array}{cc} b_{11} & b_{12} \\ b_{21} & b_{22} \end{array}) = 𝐁 (\begin{array}{c} \cos t \\ \sin t \end{array}) .

Compute

𝒚_{p}^{'} = 𝐀 𝒚_{p} + 𝒓 (t) \Rightarrow 𝐁 (\begin{array}{c} - \sin t \\ \cos t \end{array}) = 𝐀 (\begin{array}{c} \cos t \\ \sin t \end{array}) + 10 (\begin{array}{c} \cos t \\ - \sin t \end{array}) \Rightarrow

\begin{array}{l} - b_{11} \sin t + b_{12} \cos t = - 2 \cos t + 10 \cos t \\ - b_{21} \sin t + b_{22} \cos t = 2 \sin t - 10 \sin t \end{array} \Rightarrow 𝐁 = (\begin{array}{cc} 0 & 8 \\ 8 & 0 \end{array}) .

General solution is

𝒚 = c_{1} (\begin{array}{c} - 1 \\ 3 \end{array}) e^{- 2 t} + c_{2} (\begin{array}{c} 1 \\ 1 \end{array}) e^{2 t} + 8 (\begin{array}{c} \sin t \\ \cos t \end{array})

In[44]:=

A={{1,1},{3,-1}}; L=DiagonalMatrix[Eigenvalues[A]]

$(\begin{array}{cc} - 2 & 0 \\ 0 & 2 \end{array})$

In[42]:=

X=Transpose[Eigenvectors[A]]

$(\begin{array}{cc} - 1 & 1 \\ 3 & 1 \end{array})$

In[43]:=

Inverse[X]

$(\begin{array}{cc} - \frac{1}{4} & \frac{1}{4} \\ \frac{3}{4} & \frac{1}{4} \end{array})$

In[45]:=

y={ -c1 Exp[-2t] + c2 Exp[2t] + 8 Sin[t], 3c1 Exp[-2t] + c2 Exp[2t] + 8 Cos[t]}

${c1 (- e^{- 2 t}) + c2 e^{2 t} + 8 \sin (t), 3 c1 e^{- 2 t} + c2 e^{2 t} + 8 \cos (t)}$

In[47]:=

Simplify[D[y,t] == A.y + 10 {Cos[t],-Sin[t]}]

${- 8 \sin (t) - 10 \cos (t), 8 \cos (t) - 22 \sin (t)} = {0, 0}$

In[48]:=

Exercise. PS4.6.3

Solution.

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Exercise. PS4.6.4

Solution.

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Exercise. PS4.6.5

Solution.

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2Problems

Problem. PS1.1.16

Solution.

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Problem. PS1.1.17

Solution.

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Problem. PS1.1.18

Solution.

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Problem. PS1.3.22

Solution.

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1Exercises

2Problems

3Projects

3.1PS2.1.16