Company
Mathematica Solutions to the ISSAC '97 Systems Challenge

Wolfram Research, Inc.


Problem 10

Consider the following initial value problem.

[Graphics:ISSACChallengegr178.gif]
[Graphics:ISSACChallengegr179.gif], [Graphics:ISSACChallengegr180.gif]

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 [Graphics:ISSACChallengegr187.gif] and [Graphics:ISSACChallengegr188.gif]

Let's start with the straightforward numerical solution.

[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr189.gif]
[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr190.gif]

Let's look at the graphs of the function and its derivative. There seems to be a singularity of [Graphics:ISSACChallengegr191.gif] on the interval (0,2) at the point [Graphics:ISSACChallengegr192.gif].

[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr193.gif]
[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr194.gif]

If [Graphics:ISSACChallengegr195.gif] has a singularity at [Graphics:ISSACChallengegr196.gif], then near this singularity the differential equation reduces to [Graphics:ISSACChallengegr197.gif].

[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr198.gif]
[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr199.gif]

The solution of this differential equation has the form [Graphics:ISSACChallengegr200.gif]. We conclude that [Graphics:ISSACChallengegr201.gif] is finite at [Graphics:ISSACChallengegr202.gif] and that there is a square root singularity at [Graphics:ISSACChallengegr203.gif].

We first solve for [Graphics:ISSACChallengegr204.gif] up to some point to the left of the singularity and then switch to the differential equation for [Graphics:ISSACChallengegr205.gif]. We can then solve across the singularity for[Graphics:ISSACChallengegr206.gif]; it is not a singularity for [Graphics:ISSACChallengegr207.gif]. We set the appropriate options in NDSolve to achieve the required precision goal.

[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr208.gif]

Now we change from y[x] to x[y] and continue solving.

[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr209.gif]

[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr210.gif]

To get an approximation to [Graphics:ISSACChallengegr211.gif] we use the calculated solution.

[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr212.gif]
[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr213.gif]

We use the resulting interpolating function to find the minimum of [Graphics:ISSACChallengegr214.gif], which is the point [Graphics:ISSACChallengegr215.gif].

[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr216.gif]
[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr217.gif]

Here is a plot of the real part of [Graphics:ISSACChallengegr218.gif] near the square root branch point. The violet line is the path of integration.

[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr219.gif]


Method 2: Change to the polar coordinate system centered at {0,1}.

We change variables from [Graphics:ISSACChallengegr220.gif], [Graphics:ISSACChallengegr221.gif] to [Graphics:ISSACChallengegr222.gif], phi.

[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr223.gif]

[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr224.gif]

This is the new differential equation in [Graphics:ISSACChallengegr225.gif].

[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr226.gif]
[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr227.gif]

[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr228.gif]

Here is a plot of the solution.

[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr229.gif]
[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr230.gif]

Now the solution always stays in the order 1, so there is not an accuracy problem.

The maximum of [Graphics:ISSACChallengegr231.gif] determines [Graphics:ISSACChallengegr232.gif].

[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr233.gif]
[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr234.gif]

Substituting back into the solution of the differential equation gives [Graphics:ISSACChallengegr235.gif].

[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr236.gif]
[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr237.gif]