數學證明題~>

F(x, y2) - F(x,y1) <= L(y2 - y1)

So, F(x,y) is lipschitz.

y2' - y1' <= L(y2 - y1)

Separating variables,

S d(y2 - y1) / (y2 - y1) <= S L dx

ln│y2 - y1│ <= e^(Lx) + C

│y2 - y1│ <= Ae^(Lx), where A is a constant

For f(x) and g(x) are the solutions to the differential equation,

│f(x) - g(x)│ <= Ae^(Lx) ... (1)

Put x = a,

│f(a) - g(a)│ <= Ae^(La)

A <= e^(-La)│f(a) - g(a)│ ... (2)

Putting back (2) into (1):

│f(x) - g(x)│ <= e^L(x - a)│f(a) - g(a)│ for x > a