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Math Help - Phase Lines

  1. #1
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    Phase Lines

    -How can I sketch the phase lines for these differential equations? What does it mean and how can I identify the equilibrium points as sinks, sources, or nodes? Letís say the equations:
    dy/dt = (y-1) sin y and dy/dt = 3 y^3 Ė 12y^2
    I have no idea how to start this one, how can I label each equilibrium point as I sketch these two differential equation. Iím guessing but since I have dx/dt= f(x,y) and dy/dt= g(x,y). At each point, (x,y), then dy/dx= g(x,y)/f(x,y). Should I choose a number of points in the plane, and calculate dy/dx for each of them. Because that will give me the slope of the line through that point that is tangent to the solution curve through that point.

    -Now, what about if a differential equation and various initial conditions are specified. How can I sketch the graphs of the solution satisfying these initial conditions? Suppose we use the equations above, to make it more simple and understandable,
    dy/dt = (y-1) sin y , and letís use y(0) = 1, y(-2) = -1, y(0) = Ĺ, y(0) = 2
    same for dy/dt = 3 y^3 Ė 12y^2 but letís use y(0) = 1, y(1) = 0, y(0) = -7, y(0) = 5
    After computing how can I put all the graphs on one pair of axes.
    This one Iím completely blank
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  2. #2
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    So the first question
    For the equilibrium points I assumed that they are at equilibrium when the equation is equal to 0
    So dy/dt=(y-1)sin y
    they are at 0 when y is equal 0,1,+-pi,+-2pi
    but how do I know which one is a sink,source or node?
    The same as for the second
    the equilibrium for dy/dt = 3 y^3 – 12y^2 is when equation is equal to 0
    so eq. points are 0 and 4 because dy/dt = 3 y^3 – 12y^2=(3y^2(y-4))
    again how do I know which one is a sink, source or node



    The second question I know that y(0)=1/2 is the only one that satisfy dy/dt = (y-1) sin y...why? Someone told me its because its between 0 less than y(t) less than kpi?

    The same for the second one, I know y(0)=1 because its between 0 and 4, why? How can I derive the answer? I'm confused, especially when it comes to graphing
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  3. #3
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    I was also just reading, but the book doesn't make anything clear. It says y=y1 is called sink because the neighboring solutions are attracted to it as t goes to infinity. Equilibria that repel neighboring solutions, like y=u2, are known as sources; all other equilibria are called nodes, y=y3. Sources and nodes are unstable equilibria. That doesn't help at all
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  4. #4
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    Quote Originally Posted by hansbahia View Post
    -How can I sketch the phase lines for these differential equations? What does it mean and how can I identify the equilibrium points as sinks, sources, or nodes? Let’s say the equations:
    dy/dt = (y-1) sin y and dy/dt = 3 y^3 – 12y^2
    I strongly recommend you go back and read the problem again. Are these two separate problems? You talk, later,
    about points in the plane and you would need two equations but they must be in x and y, not both in y.

    I have no idea how to start this one, how can I label each equilibrium point as I sketch these two differential equation. I’m guessing but since I have dx/dt= f(x,y) and dy/dt= g(x,y).

    Well, that's the problem. In what you wrote, you do NOT have "dx/dt= f(x,y) and dy/dt= g(x,y)"!

    At each point, (x,y), then dy/dx= g(x,y)/f(x,y). Should I choose a number of points in the plane, and calculate dy/dx for each of them. Because that will give me the slope of the line through that point that is tangent to the solution curve through that point.

    -Now, what about if a differential equation and various initial conditions are specified. How can I sketch the graphs of the solution satisfying these initial conditions? Suppose we use the equations above, to make it more simple and understandable,
    dy/dt = (y-1) sin y , and let’s use y(0) = 1, y(-2) = -1, y(0) = Ĺ, y(0) = 2
    same for dy/dt = 3 y^3 – 12y^2 but let’s use y(0) = 1, y(1) = 0, y(0) = -7, y(0) = 5
    After computing how can I put all the graphs on one pair of axes.
    This one I’m completely blank
    Once again, go back and check what the equation really are.
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  5. #5
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    Yea you are right, there is only y
    sketch the phase lines for the given differential equation. Identify the equilibrium points as sinks, sources, or nodes.
    (i) dy/dt = (y-1) sin y (ii) dy/dt = 3 y^3 – 12y^2
    b) In this problem, a differential equation and various initial conditions are specified. Sketch the graphs of the solution satisfying these initial conditions. For each problem, put all your graphs on one pair of axes.
    (i) Equation from a) part (i): y(0) = 1, y(-2) = -1, y(0) = Ĺ, y(0) = 2
    (ii) Equation from a) part (ii): y(0) = 1, y(1) = 0, y(0) = -7, y(0) = 5

    These are the questions

    But I can't understand how to indentify what they are asking. My book only says if y=y1 is called sink because the neighboring solutions are attracted to it as t goes to infinity. Equilibria that repel neighboring solutions, like y=u2, are known as sources; all other equilibria are called nodes, y=y3. Sources and nodes are unstable equilibria. That doesn't help at all. I tried to look online for phase lines, but there is nothing, not even examples
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  6. #6
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    I also know that
    If both arrows point toward the critical point, it is stable (a sink): nearby solutions will converge asymptotically to the critical point, and the solution is stable under small perturbations, meaning that if the solution is disturbed, it will return to (converge to) the solution.

    If both arrows point away from the critical point, it is unstable (a source): nearby solutions will diverge from the critical point, and the solution is unstable under small perturbations, meaning that if the solution is disturbed, it will not return to the solution.

    Otherwise – if one arrow points towards the critical point, and one points away – it is semi-stable (a node): it is stable in one direction (where the arrow points towards the point), and unstable in the other direction (where the arrow points away from the point).

    But how do I know where to point the arrows? And in letter "b)" how can I sketch
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  7. #7
    Behold, the power of SARDINES!
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    Quote Originally Posted by hansbahia View Post
    Yea you are right, there is only y
    sketch the phase lines for the given differential equation. Identify the equilibrium points as sinks, sources, or nodes.
    (i) dy/dt = (y-1) sin y (ii) dy/dt = 3 y^3 Ė 12y^2
    b) In this problem, a differential equation and various initial conditions are specified. Sketch the graphs of the solution satisfying these initial conditions. For each problem, put all your graphs on one pair of axes.
    (i) Equation from a) part (i): y(0) = 1, y(-2) = -1, y(0) = Ĺ, y(0) = 2
    (ii) Equation from a) part (ii): y(0) = 1, y(1) = 0, y(0) = -7, y(0) = 5

    These are the questions

    But I can't understand how to indentify what they are asking. My book only says if y=y1 is called sink because the neighboring solutions are attracted to it as t goes to infinity. Equilibria that repel neighboring solutions, like y=u2, are known as sources; all other equilibria are called nodes, y=y3. Sources and nodes are unstable equilibria. That doesn't help at all. I tried to look online for phase lines, but there is nothing, not even examples
    First you need to graph a direction field! The ODE gives you the slope of the function at every ordered pair (x,y)

    After you have made this plot imagine the it represents a river the is flowing. If you drop a cork at the initial condition where will the flow take it.

    Here is an example

    Phase Lines-phase.jpeg

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  8. #8
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    Now I understand!
    However instead of drawing direction field, I used a different method that I found on this website--- http://www.math.utah.edu/~gustafso/2250phaseline.pdf
    The way I did was...
    first I found the zeros which for (i) are 0,1,+-pi,2+-pi and (ii) are 0 and 4
    since dy/dt=f(t)... then the derivative of f(t) will give me the slope. Then I plug each equilibrium point that I found into y. Then If the result ends up "+", its a source, if its "-" its a sink, and 0 its a node.
    so for (i)0-sink 1-source, pi-sink, -pi-source, 2pi-source, -2pi-sink and for (ii) 0-node, 4-source.

    Now can you help me for part b) I'm stuck to where to start
    Draw was pretty easy too for part 1.
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  9. #9
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    I will have to ask again- are these two separate problems? TheEmptySet is assuming that one of the equations is for x, not y, and so writing the problem in the phase plane. But your "so for (i)0-sink 1-source, pi-sink, -pi-source, 2pi-source, -2pi-sink and for (ii) 0-node, 4-source" seems to indicate that you are thinking of these as points on a phase line for the first equation only.
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