Differential Equations

Have a go first. Then open the working one line at a time.

1. Eigenvalues

   \(\det\begin{pmatrix}5-\lambda&1\\2&4-\lambda\end{pmatrix}=0\), \(\lambda^2-9\lambda+18=0\), \(\lambda=3\) or \(\lambda=6\)

   Both are positive.

2. Eigenvectors

   \(\lambda=3\): \(2x+y=0\), \(\mathbf p_1=\begin{pmatrix}1\\-2\end{pmatrix}\); \(\lambda=6\): \(-x+y=0\), \(\mathbf p_2=\begin{pmatrix}1\\1\end{pmatrix}\)

   Use the first row of \(A-\lambda I\) each time.

3. General solution

   \(\begin{pmatrix}x\\y\end{pmatrix}=Ae^{3t}\begin{pmatrix}1\\-2\end{pmatrix}+Be^{6t}\begin{pmatrix}1\\1\end{pmatrix}\)

4. Particular solution

   \(A+B=7\) and \(-2A+B=1\) give \(A=2,\ B=5\), so \(x=2e^{3t}+5e^{6t}\), \(y=-4e^{3t}+5e^{6t}\)

   Put \(t=0\) in the general solution to match the starting numbers.

5. The phase portrait

   Both eigenvalues positive: unstable node

   All curves leave the origin. Near the origin they follow the slower direction \(\mathbf p_1\); far away they follow the faster direction \(\mathbf p_2\). The dashed lines are the eigenvector directions.

6. Is the model suitable?

   \(x\) and \(y\) grow without limit as \(t\to\infty\)

   A culture cannot grow for ever: food and space run out. The model works for a short time only.

Have a go first. Then open the working one line at a time.

1. Eigenvalues and eigenvectors

   \(\lambda^2-3\lambda-10=0\), \(\lambda=-2\) or \(\lambda=5\); \(\lambda=5\): \(\mathbf p=\begin{pmatrix}1\\2\end{pmatrix}\); \(\lambda=-2\): \(\mathbf p=\begin{pmatrix}1\\1\end{pmatrix}\)

   One positive and one negative eigenvalue: a saddle point.

2. Particular solution

   \(A+B=9\) and \(2A+B=13\) with \(A\) for \(\lambda=5\): \(A=4,\ B=5\). \(x=4e^{5t}+5e^{-2t}\), \(y=8e^{5t}+5e^{-2t}\)

   Check \(t=0\): \(x=9\), \(y=13\).

3. The solution curve

   Start at \((9,13)\). Draw the lines \(y=x\) and \(y=2x\) for the eigenvectors

   As t grows the \(e^{-2t}\) terms vanish and the curve bends towards the line \(y=2x\). It leaves the window when y reaches 50.

4. The long-term ratio

   As \(t\to\infty\): \(x\approx4e^{5t}\), \(y\approx8e^{5t}\), so \(y\div x\to2\)

   The long-term ratio comes from the eigenvector of the larger eigenvalue \(\begin{pmatrix}1\\2\end{pmatrix}\): 2 perch for each trout.

Have a go first. Then open the working one line at a time.

1. Follow the curve

   The curve starts at \((12,14)\), goes slightly to the right, then turns down and left

   x increases for a short time before it decreases. The turning point is the maximum of x.

2. Maximum of x

   Highest point on the curve: \(x\approx14.7\)

   Read the largest x-value along the curve, where it turns back. An estimate close to this is fine.

3. Long-term state

   All the arrows point towards \((10,5)\)

   The curve ends there, so \(x\to10\) and \(y\to5\). That is the long-term state.

4. Statement (i)

   Above \(y=5\) every arrow points downwards: true

   Check the whole region \(y>5\), not just the curve.

5. Statement (ii)

   At \((11,20)\) the arrow points to the right: x increases there. False

   One arrow that breaks the statement is enough. To the right of \(x=10\) but high up, arrows can still point right.

Have a go first. Then open the working one line at a time.

1. Set up Euler's method

   \(x_{n+1}=x_n+0.1(-3x_n-y_n)\), \(y_{n+1}=y_n+0.1(5x_n-y_n)\), \(t_n=0.1n\)

   Use the same spreadsheet as in the last lesson, filled down to \(t=8\).

2. x and y against t

   From the calculator: scatter plots of \(x_n\) and of \(y_n\) against \(t_n\)

   Both graphs oscillate and the swings shrink. They cross zero at different times: y is a quarter of a turn ahead or behind x.

3. The solution curve in the x-y plane

   From the calculator: scatter plot of \(y_n\) against \(x_n\)

   A spiral going in towards the origin, anticlockwise.

4. Describe

   x and y oscillate about 0 with decreasing amplitude, and both tend to 0

   Check with the eigenvalues: \(\lambda^2+4\lambda+8=0\), \(\lambda=-2\pm2i\). The real part is negative, so the origin is a stable focus.