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Euler methond and the improved Euler method

  1. Nov 22, 2009 #1
    I am confused by both of these. I get what it does, and how it does it, but when using it, somehow I ultimately screw up.

    Could someone just post a quick description of what to do in conversational English, please? I am pretty sure that would solve this random mental block I have.

    Thanks,
    Nkk
     
  2. jcsd
  3. Nov 23, 2009 #2
  4. Nov 23, 2009 #3

    HallsofIvy

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    Euler method: The differential equation, dy/dx= f(x,y) tells you how to find the slope of the tangent line to y= y(x) at each point. Starting at, say, [itex](x_0,y_0)[/itex], calculate dy/dx and draw the tangent line, [itex]y= f'(x_0,y_0)(x- x_0)+ y_0[/itex].

    That will approximate the graph of y= y(x) for a short distance. So follow that tangent from [itex](x_0,y_0)[/itex] a short distance to [itex](x_0+ h, f'(x_0,y_0)h+ y_0)[/itex]. Now, using that point as "[itex](x_0,y_0)[/itex]" do it again.

    Improved Euler's method: The Euler's method isn't exact because the tangent line is only approximate to the actual curve. And there is a "double" problem. Following the tangent line not only gives an point that is a little off, you will be calculating the next slope at the wrong place! Suppose [math]f'(x_0,y_0)[/math] is positive but decreasing. Following the tangent line not only gives an point that is a little too high (you didn't follow the downard curve of the graph), you will be calculating the next slope at the wrong place! One way to allow for that is to
    1) Find [itex]f(x_0,y_0)[/itex] to find the slope and follow the tangent line to [itex](x_1,y_1)[/itex]
    2) Find [itex]f(x_1,y_1)[/itex] to approximate the slope of the tangent line at the new point.
    3) Use the average of those two slopes as the slope of your tangent line rather than the original.
     
  5. Nov 27, 2009 #4
    Euler method is a Taylor expansion of first order (linear approximation) of the solution at every point.
    You can try using the chain rule to make a second order approximation to the function:

    [tex]y^{'}=f(x,y)[/tex]
    [tex]
    y^{''}=f_{x}(x,y)+f_{y}(x,y)y'=f_{x}(x,y)+f_{y}(x,y)f(x,y)[/tex]

    So at point [tex](x_{0},y_{0})[/tex] the next point [tex](x_{0}+h,y)[/tex] will satisfy
    [tex]y=y_{0}+f(x_{0},y_{0})h +[f_{x}(x_{0},y_{0})+f_{y}(x_{0},y_{0})f(x_{0},y_{0})]\frac{h^{2}}{2}[/tex]
     
  6. Dec 2, 2009 #5
    Just saw in one textbook. Improved Euler method is also known as Heun method. It is a kind of predictor-corrector method.

    Beware! There is another method called Modified Euler Method.
     
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