Let f(x) and g(x) be two solutions of the differential equation y'=F(x,y) in a domain where F satisfies the condition:
y1<y2 implies F'(x,y2)-F(x,y1)<等於L(y2-y1) show that /f(x)-g(x)/<e^L(x-a)/f(a)-g(a)/ if x>a
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
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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