# Homework Help: Solution to complex valued ODE

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1. Feb 15, 2017

### MxwllsPersuasns

1. The problem statement, all variables and given/known data
Let f : I → C be a smooth complex valued function and t0 ∈ I fixed.

(i) Show that the initial value problem z'(t) = f(t)z(t) z(t0) = z0 ∈ C has the unique solution z(t) = z0exp(∫f(s)ds) (where the integral runs from t0 to t. Hint : for uniqueness let w(t) be another solution of the same initial value problem and contemplate the expression w(t)/z(t).

(ii) Show that if f : I → iR is imaginary valued, the length of the solution z(t) is constant, i.e. |z(t)| = |z0|. Does the converse also hold, i.e. if you know that the length of the solution to the ODE z' = fz is preserved, then f necessarily has to be imaginary valued.

(iii) Show that if z1(t), z2(t) solve the ODE z'(t) = f(t)z(t), then z1(t) = cz2(t) for some c ∈ C (this is another way to state the uniqueness property).

2. Relevant equations

Not entirely sure what equations I should put in here but I suppose I could put z = x + iy

3. The attempt at a solution
I'm currently working on part i) but will continue to post here as I work further.

So I almost did separation of variables here but then realized I could just directly integrate (rather than get dz/z on the LHS and f(t)dt on the RHS) but doing the direct integration I see that if I could just integrate out Z(s) to get Z(t0) in the front but that's obviously not possible as the integral of two functions requires the use of integration by parts.

I integrated by parts using u = z(t) and v = f(t) to find ∫f(t)z(t)dt = z(t)f(t) - ∫f(s)z'(s)ds and so I tried continuing on but am not quite sure it's going to be fruitful. I tried a few things but none really seem right. Can anyone tell me where I'm going wrong?

2. Feb 16, 2017

### Orodruin

Staff Emeritus
Why are you trying to solve the differential equation when you are handed the solution? You just need to check that it is a solution and that the solution is unique.

If you do want to solve the equation, you should do so with an integrating factor.

3. Feb 16, 2017

### MxwllsPersuasns

Oh wow.. I'm stupid. I guess I thought based on the wording of the problem I had to derive the solution from the equation in order to show that.. but I can just as easily differentiate z(t) and just see that I get z'(t) = f(t)z(t) -- excellent!

I shall work more on this problem when I get out of my research lab later this evening and I expect I'll probably post here on this thread once or twice more tonight, feel free to check back in if you like, you've always been a big help. Thank you Orodruin.

4. Feb 16, 2017

### Orodruin

Staff Emeritus
My PhD student says I have a similar effect on him. I am pretty sure he is smarter than me anyway.
I am older and wiser and have already made all (ehrrm, well, debatable) the mistakes though ...

5. Feb 16, 2017

### MxwllsPersuasns

Well I'll gladly feel like a dummy if it means I get to move past my hangups and draw a clearer picture of whats going on. So I showed that z'(t) = f(t)z(t) but now I'm confused on how to show the solution is unique. I remember talking about uniqueness of solutions back in ODEs but that was 3 and a half years ago now so It's pretty fuzzy.

6. Feb 16, 2017

### MxwllsPersuasns

Actually I'm thinking about it and I could just change out the constant in front of the rest of the expression for z(t)? Like for instance if I'm talking about the function w(t) then w0 could be the constant and then when the question asks to contemplate z(t)/w(t) the ratio of these two solutions would only be off by a constant factor (z0/w0, is this correct for uniqueness?

7. Feb 16, 2017

### Orodruin

Staff Emeritus
What happens if you differentiate z/w?

8. Feb 16, 2017

### MxwllsPersuasns

Oh interesting! So you get zero... so if you take the derivative of the ratio of two unique solutions do you get zero, in general? Is this some sort of Uniqueness theorem or test for uniqueness or something? That's pretty nifty.

9. Feb 16, 2017

### MxwllsPersuasns

Also, I'm curious, for part ii) I showed that when f is imaginary valued then |z(t)| = |z0| using the notion that the absolute value of a complex number is itself times its complex conjugate and then the 'i' got negated and the exponents canceled but I'm having more trouble showing the converse. My intuition says that it doesn't hold.. meaning that some function could have |z(t)| = |z0| but not necessarily be imaginary valued. I'm wondering what your thoughts are on this? If you wouldn't mind. Or what you can help guide me with I should say

10. Feb 16, 2017

### Orodruin

Staff Emeritus
What is $|e^{x+iy}|$? If $f(t)$ has a real part, what happens to the solution?

11. Feb 16, 2017

### MxwllsPersuasns

Oh so the real part would remain i.e., |ex+iy| = |exeiy| = sqrt{e2xeiye-iy} = sqrt{e2x} = ex right? Or did I miss something? In this case then the absolute value would not be constant since the real part would remain and the real part involves x's which are parameterized by t. Yes?

12. Feb 16, 2017

### MxwllsPersuasns

Oh as an aside as well I'm curious.. I want to show that z→az with a ∈ ℂ and a = |a|e corresponds to x → |a|ex = |a|R(α)x with x = (x1, x2) and where R(α) is the standard 2d rotation matrix. I showed that e is responsible for the rotation component by way of Eulers formula equation the exponential to polar form but I'm unsure how to show that e is related to the 2d rotation matrix. I mean I guess other than the fact that you'd need a 2x2 matrix like that to rotate the 2 components of x if you put them in vector form. I'm just not seeing it I guess.