mathematics-class-12-unit-09-chapter-03-differential-equations-3-8-vmgki92kzm

### <Success style="font-size:25px"> Differential Equations L-3 </Success>
### <primary style="font-size:24px"> Predator-prey model again geometric view point </primary>
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- We now take up the pair of Volterra-Lotka differential equations for the two species ecology consisting of a predator and a prey.

- $\frac{d x}{d t}=-a x+b x y, \quad \frac{d y}{d t}=k y-c x y$

- Each solution is a pair of functions $(x(t), y(t))$ and so can be thought of as a parametrized curve $C$ in the plane. We now try to determine the cartesian equation of this curve $C$. Using the chain rule (and the dots denoting the derivative with respect to time $t$ )

- $\frac{d x}{d y}=\frac{\dot{x}}{\dot{y}}=\frac{-a x+b x y}{k y-c x y}=\frac{x(b y-a)}{y(k-c x)}$

- This is a variable separable equation !

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<footer style = "font-size: 18px"> $\rightarrow$ Differential equation lecture-3 $\rightarrow$ <span style = "color:blue"> Predator-prey model again geometric view point </span> $\rightarrow$ Predator-prey model again Geometric view point contd $\rightarrow$ Predator-prey model again geometric view point </footer>

### <Success style="font-size:25px"> Differential Equations L-3 </Success>
### <primary style="font-size:24px"> Predator-prey model again Geometric view point contd </primary>
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- Since $x$ and $y$ denote populations, they are assumed to be positive. Also we assume that by $-a \neq 0$ and $k-c x \neq 0$. Proceeding as usual,

- $\frac{k-c x}{x} \frac{d x}{d y}=\frac{b y-a}{y}$

- Integrating with respect to $y$ we get

- $k \log x-c x+a \log y-b y=E$

- where $E$ is the constant of integration. Note that $(1.15)$ is the cartesian equation of the parametrized curve $(x(t), y(t))$ - called a phase curve.
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<footer style = "font-size: 18px">Differential equation lecture-3 $\rightarrow$ Predator-prey model again geometric view point $\rightarrow$ <span style = "color:blue"> Predator-prey model again Geometric view point contd </span> $\rightarrow$ Predator-prey model again geometric view point $\rightarrow$ Phase curves of the Volterra-Lotka system </footer>

### <Success style="font-size:25px"> Differential Equations L-3 </Success>
### <primary style="font-size:24px"> Predator-prey model again geometric view point </primary>
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- Note that (1.15) is a very interesting equation. Although the individual $x(t)$ and $y(t)$ keep varying with time, the combination

- $k \log x(t)-c x(t)+a \log y(t)-b y(t)$

- is always constant in time. This should be compared with the familiar situation in physics with conservative mechanical systems namely the law of conservation of energy. Thus equation (1.15) can be termed as the "law of conservation of ecological energy"
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<footer style = "font-size: 18px">Predator-prey model again geometric view point $\rightarrow$ Predator-prey model again Geometric view point contd $\rightarrow$ <span style = "color:blue"> Predator-prey model again geometric view point </span> $\rightarrow$ Phase curves of the Volterra-Lotka system $\rightarrow$ Example on phase diagram arising from physics </footer>

### <Success style="font-size:25px"> Differential Equations L-3 </Success>
### <primary style="font-size:24px"> Phase curves of the Volterra-Lotka system </primary>
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- It is natural to ask for a picture of the phase curves represented by (1.15). Unfortunately the curve sketching would involve calculus of several variables and so is beyond the scope of the present lecture series. I would only mention that the curves are closed curves in the plane.

- The student whose curiosity has been aroused may consult: P. Glendinning: Stability, instability and chaos, Cambridge University Press, 1999, pp 130-132.

- On page 3 of the following article in the website you can find a beautiful sketch of the phase curves for the Volterra-Lotka model:

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<footer style = "font-size: 18px">Predator-prey model again Geometric view point contd $\rightarrow$ Predator-prey model again geometric view point $\rightarrow$ <span style = "color:blue"> Phase curves of the Volterra-Lotka system </span> $\rightarrow$ Example on phase diagram arising from physics $\rightarrow$ An exercise on sketching phase curves </footer>

### <Success style="font-size:25px"> Differential Equations L-3 </Success>
### <primary style="font-size:24px"> Example on phase diagram arising from physics </primary>
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- Let us discuss the phase diagram of some systems of differential equations:

- $\frac{d x}{d t}=y, \quad \frac{d y}{d t}=-x$

- This is really the equation of SHM rewritten as a system. We see that

- $\frac{d y}{d x}=\frac{-x}{y}$

- and integrating this last equation gives

- $x^2+y^2=c$

- The phase curves are concentric circles. Note that equation (1.17) carries LESS information than (1.16). Solving (1.16) amounts to finding $x$ and $y$ INDIVIDUALLY as functions of time. Whereas ALL that we get out of (1.17) is a relationship (1.18) between $x(t)$ and $y(t)$.
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<footer style = "font-size: 18px">Predator-prey model again geometric view point $\rightarrow$ Phase curves of the Volterra-Lotka system $\rightarrow$ <span style = "color:blue"> Example on phase diagram arising from physics </span> $\rightarrow$ An exercise on sketching phase curves $\rightarrow$ An exercise on sketching phase curves </footer>

### <Success style="font-size:25px"> Differential Equations L-3 </Success>
### <primary style="font-size:24px"> An exercise on sketching phase curves </primary>
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- Consider the pair of differential equations

- $\frac{d y}{d t}=2 x y, \quad \frac{d x}{d t}=1+x^2$

- Work in the first quadrant $x>0, y>0$ and obtain the first order differential equations for the phase curves

- $\frac{d y}{d x}=f(x, y)$

- Find the solutions of (1.20) and so determine the phase curves for (1.19). Sketch these curves. Is it possible to find the solutions $x(t), y(t)$ of the original system (1.19)?
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<footer style = "font-size: 18px">Phase curves of the Volterra-Lotka system $\rightarrow$ Example on phase diagram arising from physics $\rightarrow$ <span style = "color:blue"> An exercise on sketching phase curves </span> $\rightarrow$ An exercise on sketching phase curves $\rightarrow$ Phase curve </footer>

### <Success style="font-size:25px"> Differential Equations L-3 </Success>
### <primary style="font-size:24px"> An exercise on sketching phase curves </primary>
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- Again consider the pair of differential equations

- $\frac{d x}{d t}=\frac{1}{1+y^2}, \quad \frac{d y}{d t}=\frac{1}{1+x^2}$

- Find the phase curves of (1.21) by obtaining a differential equation of the form (1.20) and solving it. Sketch the phase curves. Here one can solve the two equations in (1.21) independently. However, you would only get $x$ and $y$ implicitly as functions of $t$.
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<footer style = "font-size: 18px">Example on phase diagram arising from physics $\rightarrow$ An exercise on sketching phase curves $\rightarrow$ <span style = "color:blue"> An exercise on sketching phase curves </span> $\rightarrow$ Phase curve $\rightarrow$ Exercises </footer>

### <Success style="font-size:25px"> Differential Equations L-3 </Success>
### <primary style="font-size:24px"> Phase curve </primary>
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- $\frac{d y}{d x}=\frac{1+y^2}{1+x^2}$

- The solutions are

- $\tan ^{-1} x=\tan ^{-1} y+C$

- $x=\tan \left(\tan ^{-1} y+C\right)=\frac{y+\tan C}{1-y \tan C}$

- Sketch these curves. These are a family of rectangular hyperbolas In the next exercise we shall see a whole family of related equations ALL of them having the SAME phase curves!
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<footer style = "font-size: 18px">An exercise on sketching phase curves $\rightarrow$ An exercise on sketching phase curves $\rightarrow$ <span style = "color:blue"> Phase curve </span> $\rightarrow$ Exercises $\rightarrow$ Exercises </footer>

### <Success style="font-size:25px"> Differential Equations L-3 </Success>
### <primary style="font-size:24px"> Exercises </primary>
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- Let us now take $\phi(x, y)=x^2$ (work in the first quadrant $x>0, y>0$ ) what is the solution of $(1.22)$ with $x(0)=1 / \sqrt{2}$ and $y(0)=1 / \sqrt{2}$ ? Hint: Since the point $(x(t), y(t))$ traces a circle, try $x(t)=c \cos \psi(t)$ and $y(t)=c \sin \psi(t)$. Then $c=1$ (why?). Obtain a differential equation for $\psi(t)$. The exercise shows that when we change the function $\phi(x, y)$ the phase curves do not change but the parametrization changes.
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<footer style = "font-size: 18px">Phase curve $\rightarrow$ Exercises $\rightarrow$ <span style = "color:blue"> Exercises </span> $\rightarrow$ Remark-1 $\rightarrow$ Remark-2 </footer>

### <Success style="font-size:25px"> Differential Equations L-3 </Success>
### <primary style="font-size:24px"> Remark-1 </primary>
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- These specific systems (1.16), (1.19) and (1.22) are RARE instances where the pair of differential equations can be solved explicitly for $x(t)$ and $y(t)$ as for instance in the case of SHM

- $\frac{d x}{d t}=y, \quad \frac{d y}{d t}=-x$

- we have the explicit solution

- $x(t)=c \sin t, \quad y(t)=c \cos t $

- (and similarly for the system (1.19)).
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<footer style = "font-size: 18px">Exercises $\rightarrow$ Exercises $\rightarrow$ <span style = "color:blue"> Remark-1 </span> $\rightarrow$ Remark-2 $\rightarrow$ The pendulum revisited the geometric viewpoint </footer>

### <Success style="font-size:25px"> Differential Equations L-3 </Success>
### <primary style="font-size:24px"> Remark-2 </primary>
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- However in the case of the Volterra-Lotka (reproduced here for convenience)

- $\frac{d x}{d t}=-a x+b x y, \frac{d y}{d t}=k y-c x y$

- It is NOT possible to find $x(t)$ and $y(t)$ individually and we have to be content with the equation

- $k \log x-c x+a \log y-b y=E$

- spelling out the law of conservation of ecological energy! But equation (1.25) is adequate for most practical purposes.
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<footer style = "font-size: 18px">Exercises $\rightarrow$ Remark-1 $\rightarrow$ <span style = "color:blue"> Remark-2 </span> $\rightarrow$ The pendulum revisited the geometric viewpoint $\rightarrow$ The pendulum revisited The geometric viewpoint contd </footer>

### <Success style="font-size:25px"> Differential Equations L-3 </Success>
### <primary style="font-size:24px"> The pendulum revisited the geometric viewpoint </primary>
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- Recall that the equation of the pendulum is

- $\frac{d^2 y}{d t^2}+\frac{g}{l} \sin y=0$

- Let us introduce the (angular) velocity variable $z(t)=\frac{d y}{d t}$ so that $\frac{d z}{d t}=\frac{d^2 y}{d t^2}=-\frac{g}{1} \sin y$ and finally,

- $\frac{d y}{d t}=z(t), \quad \frac{d z}{d t}=-\frac{g}{l} \sin y$

- for the parametrized curves $(y(t), z(t))$.
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<footer style = "font-size: 18px">Remark-1 $\rightarrow$ Remark-2 $\rightarrow$ <span style = "color:blue"> The pendulum revisited the geometric viewpoint </span> $\rightarrow$ The pendulum revisited The geometric viewpoint contd $\rightarrow$ A comment on notation </footer>

### <Success style="font-size:25px"> Differential Equations L-3 </Success>
### <primary style="font-size:24px"> The pendulum revisited The geometric viewpoint contd </primary>
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- Obtain a combination of $z(t)$ and $y(t)$ that is constant in time and interpret it physically. Have you seen the corresponding phase diagram somewhere? If not, can you use your physical intuition to sketch these curves?

- Again finding the individual functions $y(t)$ and $z(t)$ is not easy - they involve elliptic functions! All we get easily is a relationship between $y(t)$ and $z(t)$.
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<footer style = "font-size: 18px">Remark-2 $\rightarrow$ The pendulum revisited the geometric viewpoint $\rightarrow$ <span style = "color:blue"> The pendulum revisited The geometric viewpoint contd </span> $\rightarrow$ A comment on notation $\rightarrow$ A comment on notation contd </footer>

### <Success style="font-size:25px"> Differential Equations L-3 </Success>
### <primary style="font-size:24px"> A comment on notation </primary>
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- In books you often see a differential equation being written as

- $M(x, y) d x+N(x, y) d y=0$

- What is the meaning of this? Although the expression $M d x+N d y$ can be defined with mathematical precision, we shall not do so. So what do we do with (1.28)? We proceed to clarify this now.
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<footer style = "font-size: 18px">The pendulum revisited the geometric viewpoint $\rightarrow$ The pendulum revisited The geometric viewpoint contd $\rightarrow$ <span style = "color:blue"> A comment on notation </span> $\rightarrow$ A comment on notation contd $\rightarrow$ M dx + N dy = 0 </footer>

### <Success style="font-size:25px"> Differential Equations L-3 </Success>
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- We have seen that differential equations arising naturally in science and technology are pairs (or triples etc., but we shall restrict to pairs) of simultaneous ordinary differential equations such as (1.16), (1.19) or (1.22) and they are all of the form

- $\frac{d x}{d t}=N(x, y), \quad \frac{d y}{d t}=-M(x, y)$

- with $t$ representing time. For a mechanical system, the pair (1.29) spells out the components $N(x, y)$ and $-M(x, y)$ of the instantaneous velocity of a moving point and the problem of solving the pair differential equations amounts to finding the two functions $x(t)$ and $y(t)$.
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<footer style = "font-size: 18px">The pendulum revisited The geometric viewpoint contd $\rightarrow$ A comment on notation $\rightarrow$ <span style = "color:blue"> A comment on notation contd </span> $\rightarrow$ M dx + N dy = 0 $\rightarrow$ M dx + N dy = 0 </footer>

### <Success style="font-size:25px"> Differential Equations L-3 </Success>
### <primary style="font-size:24px"> M dx + N dy = 0 </primary>
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- However nothing prevents us from proceeding as

- $\frac{d x}{d y}=\frac{\dot{x}}{\dot{y}}=-\frac{N(x, y)}{M(x, y)}$

- In view of the equal roles played by $x$ and $y$ in the system, it is convenient think of the equation

- $M(x, y) d x+N(x, y) d y=0$

- as representing either (1.30) or (1.31).

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<footer style = "font-size: 18px">A comment on notation contd $\rightarrow$ M dx + N dy = 0 $\rightarrow$ <span style = "color:blue"> M dx + N dy = 0 </span> $\rightarrow$ M dx + N dy = 0 $\rightarrow$ M dx + N dy = 0 </footer>

### <Success style="font-size:25px"> Differential Equations L-3 </Success>
### <primary style="font-size:24px"> M dx + N dy = 0 </primary>
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- So from a practical point of view interesting first order differential equations (1.28) actually arise out of a study of pairs of equations (1.29) and the purpose of solving (1.28) (that is to say $(1.30)$ or (1.31)) is to determine the phase curves for (1.29).

- Stated differently, the pair of differential equations

- $\frac{d x}{d t}=N(x, y), \quad \frac{d y}{d t}=-M(x, y)$

- is to be regarded as the MORE fundamental object of study and the equation

- $M(x, y) d x+N(x, y) d y=0$

- as arising out of (1.29). The interest lies in the family of curves represented by $(1.28)$
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<footer style = "font-size: 18px">M dx + N dy = 0 $\rightarrow$ M dx + N dy = 0 $\rightarrow$ <span style = "color:blue"> M dx + N dy = 0 </span> $\rightarrow$ M dx + N dy = 0 $\rightarrow$ Differential equations of a one parameter family of curves </footer>

### <Success style="font-size:25px"> Differential Equations L-3 </Success>
### <primary style="font-size:24px"> M dx + N dy = 0 </primary>
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- It should be evident that for a non-vanishing function $\mu(x, y)$, the pair

- $\frac{d x}{d t}=\mu(x, y) N(x, y), \quad \frac{d y}{d t}=-\mu(x, y) M(x, y)$

- has the SAME phase curves as the pair (1.29). The solutions $x((t), y(t))$ of (1.29) and (1.32) differ in parametrizations.

- We have seen many examples of phase curves already and an instance of reparametrization mentioned above.

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<footer style = "font-size: 18px">M dx + N dy = 0 $\rightarrow$ M dx + N dy = 0 $\rightarrow$ <span style = "color:blue"> M dx + N dy = 0 </span> $\rightarrow$ Differential equations of a one parameter family of curves $\rightarrow$ One parameter family of curves </footer>

### <Success style="font-size:25px"> Differential Equations L-3 </Success>
### <primary style="font-size:24px"> Differential equations of a one parameter family of curves </primary>
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- We have seen that a differential equation

- $M(x, y) d x+N(x, y) d y=0$

- gives a one parameter family of curves in the plane (we have seen many examples). Conversely, given a one parameter family of curves can we construct a differential equation of the type (1.28)? One parameter families of curves are VERY pretty objects. They appear naturally in Electrostatics as equipotential curves for distribution of charges (you could start with a three dimensional configuration where these would be surfaces in $\mathbb{R}^3$ and slice the picture along the $x-y$ plane to get curves). As a starter, look at the pictures on page 635 in Halliday-Resnick's book (cited earlier and the same edition!).
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<footer style = "font-size: 18px">M dx + N dy = 0 $\rightarrow$ M dx + N dy = 0 $\rightarrow$ <span style = "color:blue"> Differential equations of a one parameter family of curves </span> $\rightarrow$ One parameter family of curves $\rightarrow$ Families of curves contd </footer>

### <Success style="font-size:25px"> Differential Equations L-3 </Success>
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- Let us now look at some interesting examples of such families. Among the simplest examples are the family of concentric circles

- $x^2+y^2=c$

- Differentiating and clearing the factor of 2 :

- $x+y \frac{d y}{d x}=0$

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<footer style = "font-size: 18px">M dx + N dy = 0 $\rightarrow$ Differential equations of a one parameter family of curves $\rightarrow$ <span style = "color:blue"> One parameter family of curves </span> $\rightarrow$ Families of curves contd $\rightarrow$ Families of curves contd </footer>

### <Success style="font-size:25px"> Differential Equations L-3 </Success>
### <primary style="font-size:24px"> An exercise on families of curves </primary>
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- (10) We have regarded $y$ as an implicit function of $x$. Is this tenable near the points $(0,0)$ and $(2 c, 0)$ ? Near these points can we instead regard $x$ as a function of $y$ and then differentiate equation (1.34) with respect to $y$ ? Sketch relevant figures and try to argue geometrically. Do you end up with the same equation (1.36)? Are there merits in writing equations (1.30)-(1.31) in the more symmetrical form (1.28)?

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<footer style = "font-size: 18px">Families of curves contd $\rightarrow$ Families of curves contd $\rightarrow$ <span style = "color:blue"> An exercise on families of curves </span> $\rightarrow$ Families of curves contd $\rightarrow$ Thank you </footer>