Showing uniqueness of complex ODE

[dochalladay32]

Homework Statement

We are given (1-z)*f'(z)-3*f(z) = 0, f(0) = 2 valid on the open disk centered at 0 with radius 1 and told to prove there is a unique solution to the differential equation. The hint he gave was to find a factor that makes the left side the derivative of a product, but that didn't really help me.

Homework Equations

The found factor was (1-z)^2, so you now have (1-z)^3*f'(z)-3(1-z)^2*f(z) = 0.

The Attempt at a Solution

I integrated both sides as I thought you would do after making the LHS a derivative and you can solve it, but that doesn't tell me that it is unique or not... although it has been a long time since I've done existence/uniqueness so maybe that is all I have to do. I also tried looking at it as f'(z)=3/(1-z)*f(z) and I figured it had something to do with the analyticity of 3/(1-z), but I could never get anywhere with that idea.

It was just an extra credit problem on the last homework so we haven't gone into ODE's in the complex plane, but with Spring Break this week, won't get to see the solution for awhile so was just curious if anyone knew how to prove it, or could tell me if I was using the hint correctly.

 

[tiny-tim]

hidochalladay32!

(try using the X² button just above the Reply box )

(1-z)*f'(z) - 3*f(z) = 0

(1-z)³*f'(z) - 3(1-z)²*f(z) = 0​

if the second equation is true, then either the first equation is true or z = 1

 

[dochalladay32]

Well yes, that is true, but what I'm not able to figure out is how to show uniqueness, whether with the original equation or the new one.

 

[tiny-tim]

well, the new equation is the same as

d/dt (1-z)³*f(z) = 0 …​

surely that means (1-z)³*f(z) is a constant, and sooo … ?

 

[dochalladay32]

God... I feel stupid hahaha. I completely missed that when I wrote that down before.