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Mathematica Solutions to the ISSAC '97 Systems
Challenge
Wolfram Research, Inc.
Problem 10
Consider the following initial value problem.
,
Find the smallest positive number r such that the solution
has a derivative singularity at x = r. Calculate r
to 13 significant digits. Is y(r) infinite or finite? If
y(r)
is finite then compute it to 13 significant digits.
Result
r=1.6443766903388...
y(r)=0.93193876511028...
Method 1: Solve the ODE for and ![[Graphics:ISSACChallengegr188.gif]](ISSACChallengegr188.gif)
Let's start with the straightforward numerical solution.
![[Graphics:ISSACChallengegr7.gif]](ISSACChallengegr7.gif) ![[Graphics:ISSACChallengegr189.gif]](ISSACChallengegr189.gif)
![[Graphics:ISSACChallengegr7.gif]](ISSACChallengegr7.gif) ![[Graphics:ISSACChallengegr190.gif]](ISSACChallengegr190.gif)
Let's look at the graphs of the function and its derivative. There
seems to be a singularity of on the interval (0,2) at the
point .
![[Graphics:ISSACChallengegr7.gif]](ISSACChallengegr7.gif) ![[Graphics:ISSACChallengegr193.gif]](ISSACChallengegr193.gif)
![[Graphics:ISSACChallengegr7.gif]](ISSACChallengegr7.gif)
If has a singularity at ,
then near this singularity the differential equation reduces to .
![[Graphics:ISSACChallengegr7.gif]](ISSACChallengegr7.gif) ![[Graphics:ISSACChallengegr198.gif]](ISSACChallengegr198.gif)
![[Graphics:ISSACChallengegr7.gif]](ISSACChallengegr7.gif) ![[Graphics:ISSACChallengegr199.gif]](ISSACChallengegr199.gif)
The solution of this differential equation has the form .
We conclude that is finite at
and that there is a square root singularity at .
We first solve for up to some point to the left
of the singularity and then switch to the differential equation for .
We can then solve across the singularity for ;
it is not a singularity for . We set the appropriate
options in NDSolve to achieve the required precision goal.
![[Graphics:ISSACChallengegr7.gif]](ISSACChallengegr7.gif) ![[Graphics:ISSACChallengegr208.gif]](ISSACChallengegr208.gif)
Now we change from y[x] to x[y] and continue solving.
![[Graphics:ISSACChallengegr7.gif]](ISSACChallengegr7.gif) ![[Graphics:ISSACChallengegr209.gif]](ISSACChallengegr209.gif)
![[Graphics:ISSACChallengegr7.gif]](ISSACChallengegr7.gif) ![[Graphics:ISSACChallengegr210.gif]](ISSACChallengegr210.gif)
To get an approximation to we use the calculated
solution.
![[Graphics:ISSACChallengegr7.gif]](ISSACChallengegr7.gif) ![[Graphics:ISSACChallengegr212.gif]](ISSACChallengegr212.gif)
![[Graphics:ISSACChallengegr7.gif]](ISSACChallengegr7.gif) ![[Graphics:ISSACChallengegr213.gif]](ISSACChallengegr213.gif)
We use the resulting interpolating function to find the minimum of
, which is the point .
![[Graphics:ISSACChallengegr7.gif]](ISSACChallengegr7.gif) ![[Graphics:ISSACChallengegr216.gif]](ISSACChallengegr216.gif)
![[Graphics:ISSACChallengegr7.gif]](ISSACChallengegr7.gif) ![[Graphics:ISSACChallengegr217.gif]](ISSACChallengegr217.gif)
Here is a plot of the real part of near the square root branch
point. The violet line is the path of integration.
![[Graphics:ISSACChallengegr7.gif]](ISSACChallengegr7.gif)
Method 2: Change to the polar coordinate system centered at
{0,1}.
We change variables from ,
to , phi.
![[Graphics:ISSACChallengegr7.gif]](ISSACChallengegr7.gif) ![[Graphics:ISSACChallengegr223.gif]](ISSACChallengegr223.gif)
![[Graphics:ISSACChallengegr7.gif]](ISSACChallengegr7.gif) ![[Graphics:ISSACChallengegr224.gif]](ISSACChallengegr224.gif)
This is the new differential equation in .
![[Graphics:ISSACChallengegr7.gif]](ISSACChallengegr7.gif) ![[Graphics:ISSACChallengegr226.gif]](ISSACChallengegr226.gif)
![[Graphics:ISSACChallengegr7.gif]](ISSACChallengegr7.gif) ![[Graphics:ISSACChallengegr227.gif]](ISSACChallengegr227.gif)
![[Graphics:ISSACChallengegr7.gif]](ISSACChallengegr7.gif) ![[Graphics:ISSACChallengegr228.gif]](ISSACChallengegr228.gif)
Here is a plot of the solution.
![[Graphics:ISSACChallengegr7.gif]](ISSACChallengegr7.gif) ![[Graphics:ISSACChallengegr229.gif]](ISSACChallengegr229.gif)
![[Graphics:ISSACChallengegr7.gif]](ISSACChallengegr7.gif)
Now the solution always stays in the order 1, so there is not an
accuracy problem.
The maximum of determines .
![[Graphics:ISSACChallengegr7.gif]](ISSACChallengegr7.gif) ![[Graphics:ISSACChallengegr233.gif]](ISSACChallengegr233.gif)
![[Graphics:ISSACChallengegr7.gif]](ISSACChallengegr7.gif) ![[Graphics:ISSACChallengegr234.gif]](ISSACChallengegr234.gif)
Substituting back into the solution of the differential equation
gives .
![[Graphics:ISSACChallengegr7.gif]](ISSACChallengegr7.gif) ![[Graphics:ISSACChallengegr236.gif]](ISSACChallengegr236.gif)
![[Graphics:ISSACChallengegr7.gif]](ISSACChallengegr7.gif) ![[Graphics:ISSACChallengegr237.gif]](ISSACChallengegr237.gif)
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