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I've been looking through old questions for some time and I found an interesting topic (Complex differential equation). I decided to try solving a complex differential equation with a similar premise. I looked at the equation $$z' = \overline{\mathbb{z}} +it$$ I followed a similar strategy to the post linked, giving $z'' = \overline{\mathbb{z'}} + i$ and then taking the conjugate of the original equation, where $\overline{\mathbb{z'}} = z - it$

Finally, I plugged this back into the equation for $z''$ to get $$z'' = z + i -it$$ $$z'' - z = i - it$$

This gives a non-homogeneous differential equation without any conjugates, which I am more familiar with solving. I decided to attempt solving it through the use of power series, where the solution, $z(t)$ is of the form $$z(t) = \displaystyle\sum_{n=0}^{\infty} C_nt^n$$

I then took the second derivative of this and plugged it into our derived formula to get $$\displaystyle\sum_{n=0}^{\infty} C_n(n)(n-1)t^{n-2} - \displaystyle\sum_{n=0}^{\infty} C_nt^n = i - it$$

Fixing the power and index of term 1 then yields $$\displaystyle\sum_{n=0}^{\infty} t^n[C_{n+2}(n+2)(n+1) - C_n] = i - it $$

Because the right hand side is non-zero, we obviously can't assume that the terms in square brackets sum to zero for all n. From here I decided to evaluate the summation at the first few values of n and compare it to the right hand side to build a recursive relation.

$n =0$ $$t^0(C_2(2)(1) - C_0) = t^0(i)$$ $$C_2(2)(1) - C_0 = i$$ $$C_2 = \frac{C_0 + i}{2}$$

$n =1$ $$t^1(C_3(3)(2) - C_1) = t^1(-i)$$ $$C_3 = \frac{C_1-i}{(3)(2)}$$

$n=2$ $$t^2(C_4(4)(3) - C_2) = t^2(0)$$ $$C_4 = \frac{C_0+i}{(4)(3)(2)}$$

$n=3$ $$t^3(C_5(5)(4) - c_3) = t^3(0)$$ $$C_5 = \frac{C_1 - i}{(5)(4)(3)(2)}$$

This pattern continues for later values of n, giving us the relations

$$ C_n = \frac{C_0+i}{n!} , n \in \mathbb{2Z} , n\geq2$$ $$ C_n = \frac{C_1-i}{n!} , n \in \mathbb{2Z+1} , n>2$$

With these relations, I was able to then build the final solution of $z(t)$ $$z(t) = \displaystyle\sum_{n=0}^{\infty} C_nt^n$$ $$z(t) = C_0+C_1t+(C_0+i)\displaystyle\sum_{n=1}^{\infty} \frac{t^{2n}}{(2n)!}+(C_1-i)\displaystyle\sum_{n=1}^{\infty} \frac{t^{2n+1}}{(2n+1)!}$$

We can then use the hyperbolic sine and cosine taylor series to simplify our answer, where $$\cosh(t) = \displaystyle\sum_{n=0}^{\infty} \frac{t^{2n}}{(2n)!} = \displaystyle\sum_{n=1}^{\infty} \frac{t^{2n}}{(2n)!}+1$$ $$\sinh(t) = \displaystyle\sum_{n=0}^{\infty} \frac{t^{2n+1}}{(2n+1)!} = \displaystyle\sum_{n=1}^{\infty} \frac{t^{2n+1}}{(2n+1)!}+t$$

This gave me my final solution for $z(t)$ $$z(t) = (C_0+i)\cosh(t) + (C_1-i)\sinh(t)-i-it$$

Finally, my problem lies in actually plugging the solution back into the original differential equation. I won't write it out now because I've definitely written too much already, but it's clear that both sides would not equal each other. I was hoping for someone to know where I went wrong, or if there's a better way to solve this type of problem. Sorry for any weird formatting or notation. This is my first time using stack exchange and latex. I've also only learned math up to Calc 3 and ODEs, so there might be some concepts I'm missing that screwed my answer. Thanks!

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  • $\begingroup$ I have to commend you for your effort! If this was your first time using TeX and solving any sort of equation like this, it is very impressive! $\endgroup$ – Ninad Munshi Oct 21 '19 at 20:36
  • $\begingroup$ Your solution is in fact correct. You can absorb the $i$'s into the arbitrary constants. The constants have to be complex numbers so when you plug in to the differential equation you have to take their conjugate. $\endgroup$ – Ninad Munshi Oct 21 '19 at 20:38
  • $\begingroup$ I found your mistake. When you reindexed to fix powers, you should've had to pull out terms from the summation that look like $C_0 + C_1 t$ but I don't see that in your work. Reindexing is a tricky business! Also, don't forget that the second derivative summation starts at $2$, not $0$. $\endgroup$ – Ninad Munshi Oct 21 '19 at 20:53
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Here's an alternative solution that proves your answer is mostly correct. Split up the first differential equation into real and imaginary components denoting $z(t) = x(t) + iy(t)$:

$$\begin{cases} x' = x \\ y' = -y + t \\ \end{cases}$$

$$\implies x(t) = C_1 e^t \hspace{20 pt} y(t) = C_2e^{-t} + t-1$$

Putting these solutions together, we have

$z(t) = C_1 e^t + C_2 e^{-t} + it - i$

Which is almost the same as your solution because $\cosh$ and $\sinh$ are linear combinations of those exponentials, the only part that went awry is the inhomogeneous term. But the hard part has no mess ups.

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  • $\begingroup$ Thank you for all your help! my question is for the C1 and C2 terms, when taking the conjugate of them, would they simply become -C1 and -C2? If so, wouldn't plugging into the original equation still lead to different signs on each side? this form does fix the problem with the non-homogenous terms! $\endgroup$ – alienboy3735 Oct 21 '19 at 21:00
  • $\begingroup$ @alienboy3735 they're not purely imaginary so saying $C_1^* = -C_1$ is incorrect. They're just conjugate. $\endgroup$ – Ninad Munshi Oct 21 '19 at 21:53
  • $\begingroup$ Understood. Thank you for your assistance. :) $\endgroup$ – alienboy3735 Oct 21 '19 at 22:49

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