Picard's Iteration

**JonathanEyoon** · February 13th 2009, 06:53 PM

Guys I really need help with understanding how to use this method for my O.D.E. class. The main part I don't understand is, I don't get what f(t,Yn(t)) is supposed to be. I also do not understand how to do the next iteration after the first and the one after that and so on. I know their is a pattern but could someone show me how to do all this stuff to help me prepare for a Test? For instance

dy/dx = (x^2)(y)
y(0) = 1

This is from my notes, Why is f(t,Yn(t)) = to [t^2 - 1]? What procedure is there for figuring this out? Also if possible, could you tell it to me in a way for which it will work for any Picard Iteration problem the professor gives us? I just need to know how f(t,Yn(t)) becomes what it is.

**Opalg** · February 14th 2009, 03:04 AM

Here's a procedure that should always work. Start with the equation $\tfrac{dy}{dx} = x^2y$ . For convenience, replace the variable x by t, so that y is a function of t: $\tfrac{dy}{dt} = t^2y(t)$ . Integrate both sides from 0 to x: $y(x)-y(0) = \int_0^xt^2y(t)\,dt$ , or $y(x) = 1 + \int_0^xt^2y(t)\,dt$ .

The idea now is to start with an approximate solution $y_0(t)$ which you substitute into the integral, getting a new function $y_1(x)$ which you then substitute into the integral to get another function $y_2(x)$ , and so on.

The first approximation $y_0(t)$ is just the constant function 1 (because this is given as the value of y at x=0). Then
$y_1(x) = 1 + \int_0^xt^2y_0(t)\,dt = 1 + \int_0^xt^2\,dt = 1+x^3/3$ . So $y_1(t) = 1+t^3/3$ . Next,
$y_2(x) = 1 + \int_0^xt^2y_1(t)\,dt = 1 + \int_0^xt^2(1+t^3/3)dt = \ldots$ , and so on.

**HallsofIvy** · February 14th 2009, 05:37 AM

Originally Posted by JonathanEyoon

Guys I really need help with understanding how to use this method for my O.D.E. class. The main part I don't understand is, I don't get what f(t,Yn(t)) is supposed to be. I also do not understand how to do the next iteration after the first and the one after that and so on. I know their is a pattern but could someone show me how to do all this stuff to help me prepare for a Test? For instance

dy/dx = (x^2)(y)
y(0) = 1

This is from my notes, Why is f(t,Yn(t)) = to [t^2 - 1]? What procedure is there for figuring this out? Also if possible, could you tell it to me in a way for which it will work for any Picard Iteration problem the professor gives us? I just need to know how f(t,Yn(t)) becomes what it is.

It isn't equal to $t^2- 1$ . Where did you get that? You must be looking at an example that has nothing to do with this given problem. f(t, Yn(t)) is simply f applied to t and whatever you have just calculated Yn(t) to be!

In this case $dy/dx= x^2y$ . Since we are given that y(0)= 1, a reasonable starting approximation for y is the constant y= 1; Then we have $dy/dx= x^2(1)$ and, integrating $y= (1/3)x^3+ C$ . In order to have y(0)= 1, we must have y(0)= (1/3)0+ C= 1 so C= 1 and $y= (1/3)x^3+ 1$ . Replacing the y in the differential equation with that, $dy/dx= x^2((1/3)x^3+ 1)= (1/3)x^5+ x^2$ and integrating, $y= (1/18)x^6+ (1/3)x^3+ C$ . Again, y(0)= 1 gives C= 1. Take $dy/dx= f(x,y(x))= x^2((1/18)x^3+ (1/3)x^3+ 1)$ and integrate again.

**JonathanEyoon** · February 14th 2009, 09:42 AM

I just figured out that the example the professor showed us in class was wrong. You are right, it should most definitely not be what I have above. Thanks guys for your input~! It really helped me get this.

Originally Posted by HallsofIvy

It isn't equal to $t^2- 1$ . Where did you get that? You must be looking at an example that has nothing to do with this given problem. f(t, Yn(t)) is simply f applied to t and whatever you have just calculated Yn(t) to be!

In this case $dy/dx= x^2y$ . Since we are given that y(0)= 1, a reasonable starting approximation for y is the constant y= 1; Then we have $dy/dx= x^2(1)$ and, integrating $y= (1/3)x^3+ C$ . In order to have y(0)= 1, we must have y(0)= (1/3)0+ C= 1 so C= 1 and $y= (1/3)x^3+ 1$ . Replacing the y in the differential equation with that, $dy/dx= x^2((1/3)x^3+ 1)= (1/3)x^5+ x^2$ and integrating, $y= (1/18)x^6+ (1/3)x^3+ C$ . Again, y(0)= 1 gives C= 1. Take $dy/dx= f(x,y(x))= x^2((1/18)x^3+ (1/3)x^3+ 1)$ and integrate again.

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