### Moral: When you write an exam, check your problems.

Mathematics: Differential Equations

I almost had some egg on my face. In differential equations class, we had just covered the basic existance and uniqueness theorems for linear first order differential equations. That is, if one is given y' = f(t, y) where f is continuous at (t0, y0) and continuous on some open rectangle containing (to, yo) then the differential equation y'=f(t,y), y(t0) = y0 has at least one solution. Furthermore, if the partial derivative of f(t,y) with respect to y is similarly continuous at (t0, y0), then the solution is unique.

Example: y' = t*exp(y), y(t0) = y0 has a unique solution for any initial condition,

y' = t*y^(2/3) has a soluton of all initial conditions, but fails to have a unique solution for any initial condition y(t0) = 0, as the partial derivative with respect to y is t*(2/3)*y^(-1/3) which fails to be continuous when y = 0.

So, on the test I gave y' = y^(2/3)*ln(t)*tan(y) and asked if there was a guarantee for a unique solution at y(2) = 0.

Answer: not that obvious! taking the partial derivative with respect to y we get:

ln(t) (y^(2/3) * sec^2(y) + (2/3)^y(-1/3)*tan(y)). Pluggin in t = 2 causes no difficulty, but what about y = 0? The first term gives 0 (no problems) but what about the second?

Actually, if you go through the proof (of the uniqueness theorem), it is the fact that we have some boundedness condition for the partial with respect to y (or a Lipshitz condition, at least).

Note what happens if we apply L'Hopital's rule to the second term:

lim as y goes to zero of (2/3)*(tan(y)/y^(1/3)) = (2/3)(sec^2(y)/ (1/3)y^(-2/3)) = (2/3)(3)sec^2(y)y^(2/3) = 2*1*0 = 0 !!!

In other words, technically speaking, the partical derivative may not have been continuous, but the singularity is removable and hence the uniqueness condition holds.

I didn't anticipate this at all and was lucky that I caught this prior to passing out the exams. Moral: test your problems ahead of time.

I almost had some egg on my face. In differential equations class, we had just covered the basic existance and uniqueness theorems for linear first order differential equations. That is, if one is given y' = f(t, y) where f is continuous at (t0, y0) and continuous on some open rectangle containing (to, yo) then the differential equation y'=f(t,y), y(t0) = y0 has at least one solution. Furthermore, if the partial derivative of f(t,y) with respect to y is similarly continuous at (t0, y0), then the solution is unique.

Example: y' = t*exp(y), y(t0) = y0 has a unique solution for any initial condition,

y' = t*y^(2/3) has a soluton of all initial conditions, but fails to have a unique solution for any initial condition y(t0) = 0, as the partial derivative with respect to y is t*(2/3)*y^(-1/3) which fails to be continuous when y = 0.

So, on the test I gave y' = y^(2/3)*ln(t)*tan(y) and asked if there was a guarantee for a unique solution at y(2) = 0.

Answer: not that obvious! taking the partial derivative with respect to y we get:

ln(t) (y^(2/3) * sec^2(y) + (2/3)^y(-1/3)*tan(y)). Pluggin in t = 2 causes no difficulty, but what about y = 0? The first term gives 0 (no problems) but what about the second?

Actually, if you go through the proof (of the uniqueness theorem), it is the fact that we have some boundedness condition for the partial with respect to y (or a Lipshitz condition, at least).

Note what happens if we apply L'Hopital's rule to the second term:

lim as y goes to zero of (2/3)*(tan(y)/y^(1/3)) = (2/3)(sec^2(y)/ (1/3)y^(-2/3)) = (2/3)(3)sec^2(y)y^(2/3) = 2*1*0 = 0 !!!

In other words, technically speaking, the partical derivative may not have been continuous, but the singularity is removable and hence the uniqueness condition holds.

I didn't anticipate this at all and was lucky that I caught this prior to passing out the exams. Moral: test your problems ahead of time.