Linear differential equation of first order in EXP(C)

[mathmari]

Hey!

I want to check if there is a solution of a linear differential equation of first order in the ring of exponential sums $\text{EXP}(\mathbb{C})$. I have done the following: The general linear differential equation of first order is $$ax'(z)+bx(z)=y(z) \tag{*}$$
where $x,y \in \text{EXP}(\mathbb{C})$.

We have that $$y(z)=\sum_{i=1}^N \alpha_i e^{\mu_i z}, \ \ x(z)=\sum_{i=1}^N \beta_i e^{k_i z} \ \ \text{ so } \ \ x'(z)=\sum_{i=1}^N \beta_i k_i e^{k_i z}$$

$$(*) \Rightarrow a\sum_{i=1}^N \beta_i k_i e^{k_i z}+b\sum_{i=1}^N \beta_i e^{k_i z}= \sum_{i=1}^N \alpha_i e^{\mu_i z} \Rightarrow \sum_{i=1}^N \left [a\beta_i k_i +b\beta_i \right ]e^{k_i z}=\sum_{i=1}^N \alpha_i e^{\mu_i z}$$ For $k_i=\mu_i$ we have the following:

$$\sum_{i=1}^N \left [\beta_i (a\mu_i +b) \right ]e^{\mu_i z}=\sum_{i=1}^N \alpha_i e^{\mu_i z} \Rightarrow \sum_{i=1}^N \left [(a \mu_i +b)\beta_i -\alpha_i\right ]e^{\mu_i z}=0$$ 1. If for at least one $i$ with $\alpha_i\ne0$ we have $a\mu_i+b=0$, there is no solution.

2. If for at least one $i$ with $\alpha_i = 0$ we have $a\mu_i+b \neq 0$, we have that $\beta_i=0$.

3. If for all $i$ it holds that $\alpha_i = 0 \land a\mu_i+b = 0$, then thee are infinitely many solutions.

4. If for all $i$ it holds that $\alpha_i\ne0 \land a\mu_i+b\ne0$, then there is exactly one solution, with $\beta_i=\frac{\alpha_i}{a\mu_i+b}$.
Is everything correct? Have I forgotten a case? (Wondering)

 

[mathmari]

I did it also as follows by taking cases:

• $\textbf{Case 1.}$
  $a=0, b \neq 0$
  
  Then we have $bx(z)=y(z)$.
  So, the solution is $$x(z)=\frac{1}{b}y(z) \in \text{EXP}(\mathbb{C})$$
• $\textbf{Case 2.}$
  $a\neq 0, b=0$
  
  Then we have $ax'(z)=y(z)$.
  
  We have that $\displaystyle{x'(z)=\frac{1}{a}y(z)}$, where $\displaystyle{y(z)=\sum_{i=1}^N \alpha_i e^{\mu_i z}}$, so $\displaystyle{x'(z)=\frac{1}{a}\sum_{i=1}^N \alpha_i e^{\mu_i z}}$.
  
  • If $\exists j$ such that $\mu_j =0 \land \alpha_j \neq 0$ then $\displaystyle{x'(z)=\frac{1}{a}\sum_{i=1}^{j-1} \alpha_i e^{\mu_i z}+\frac{\alpha_j}{a}+\frac{1}{a}\sum_{i=j+1}^N \alpha_i e^{\mu_i z}}$ then $\displaystyle{x(z)=\frac{1}{a}\sum_{i=1}^{j-1} \frac{\alpha_i}{\mu_i} e^{\mu_i z}+\frac{\alpha_j}{a}z+\frac{1}{a}\sum_{i=j+1}^N \frac{\alpha_i}{\mu_i} e^{\mu_i z}} \notin \textbf{EXP}{(\mathbb{C})}$. So, there is no solution.
    
  • If $\mu_i \neq 0$ for $i$ for which $\alpha_i \neq 0$ then the solution is $\displaystyle{x(z)=\frac{1}{a}\sum_{i=1}^N \frac{\alpha_i}{\mu_i} e^{\mu_i z}}\in \text{EXP}(\mathbb{C})$.
• $\textbf{Case 3.}$
  $a=0, b=0$
  
  Then we have $0=y(z)$.
  
  • If $y(z)=0$ then we have infinitely many solutions.
  • If $y(z) \neq 0$ then we have no solution.
• $\textbf{Case 4.}$
  $a \neq 0, b \neq 0$
  
  Then we have $$ax'(z)+bx(z)=y(z) \Rightarrow x'(z)+\frac{b}{a} x(z)=\frac{1}{a}y(z) \ \ \ \ \ (*)$$
  
  We have that $$y(z)=\sum_{i=0}^N \alpha_i e^{\mu_i z}, x(z)=\sum_{i=0}^N \beta_i e^{k_i z}=\beta_0 +\sum_{i=1}^N \beta_i e^{k_i z}, also x'(z)=\sum_{i=1}^N \beta_i k_i e^{k_i z} $$
  
  $$(*) \Rightarrow \sum_{i=1}^N \beta_i k_i e^{k_i z}+\frac{b}{a} \left [\beta_0 +\sum_{i=1}^N \beta_i e^{k_i z}\right ]=\frac{1}{a} \left [\sum_{i=0}^N \alpha_i e^{\mu_i z}\right ] \\
  \Rightarrow \frac{b}{a}\beta_0 +\sum_{i=1}^N \left [\beta_i k_i +\frac{b}{a}\beta_i \right ]e^{k_i z}=\frac{1}{a}\alpha_0+\sum_{i=1}^N \alpha_i e^{\mu_i z} $$
  
  So, the following must hold:
  
  $$\left\{\begin{matrix}
  \frac{b}{a}\beta_0=\frac{1}{a}\alpha_0\\
  \beta_i k_i+\frac{b}{a}\beta_i=\alpha_i, \ \ i=1, \dots , N\\
  k_i=\mu_i, \ \ i=1, \dots , N
  \end{matrix}\right. \Rightarrow \left\{\begin{matrix}
  \beta_0=\frac{\alpha_0}{b}\\
  \beta_i \mu_i+\frac{b}{a}\beta_i=\alpha_i, \ \ i=1, \dots , N\\
  k_i=\mu_i, \ \ i=1, \dots , N
  \end{matrix}\right. \Rightarrow \left\{\begin{matrix}
  \beta_0=\frac{\alpha_0}{b}\\
  \beta_i \left (\mu_i+\frac{b}{a}\right )=\alpha_i, \ \ i=1, \dots , N\\
  k_i=\mu_i, \ \ i=1, \dots , N
  \end{matrix}\right. $$
  
  • If $\displaystyle{\mu_i \neq -\frac{b}{a}}$, then the solution is $\displaystyle{x(z)=\frac{\alpha_0}{b}+\sum_{i=1}^N \frac{\alpha_i}{\mu_i+\frac{b}{a}}e^{\mu_i z}}$.
    
  • If $\displaystyle{\mu_i=-\frac{b}{a}}$ then if $\alpha_i=0$ then there are infinitely many solutions, but if $\alpha_i \neq 0$ then there is no solution.

Is this correct? Have I taken into consideration all the possible cases? (Wondering)

 

[I like Serena]

Hey mathmari! (Wave)

It seems you assume that $\{ e^{\mu_i z} \}$ forms an independent basis as long as the $\mu_i$ are distinct.
However, that is not the case.
For instance $1\cdot e^{i\pi/2 z} = -1 \cdot e^{-i\pi/2 z}$.
(Worried)

 

[mathmari]

It seems you assume that $\{ e^{\mu_i z} \}$ forms an independent basis as long as the $\mu_i$ are distinct.
However, that is not the case.
For instance $1\cdot e^{i\pi/2} = -1 \cdot e^{-i\pi/2}$.
And $2e^{1+i} + 3e^{2+i} = (2e+3e^2)e^i$.
(Worried)

Ahaa... Ok... (Thinking)
And what could I change? (Wondering)

 

[I like Serena]

Hey mathmari! (Wave)

It seems you assume that $\{ e^{\mu_i z} \}$ forms an independent basis as long as the $\mu_i$ are distinct.
However, that is not the case.
For instance $1\cdot e^{i\pi/2 z} = -1 \cdot e^{-i\pi/2 z}$.
(Worried)

Ahaa... Ok... (Thinking)
And what could I change? (Wondering)

Nevermind, I think $\{ e^{\mu_i z} :i=0..N, \text{ distinct } \mu_i\in \mathbb C \}$ does form a basis.
So I think your reasoning is correct, although you should probably mention that. (Thinking)

 

[mathmari]

Nevermind, I think $\{ e^{\mu_i z} :i=0..N, \text{ distinct } \mu_i\in \mathbb C \}$ does form a basis.
So I think your reasoning is correct, although you should probably mention that. (Thinking)

Do you mean that I should mention that $\{ e^{\mu_i z} :i=0..N, \text{ distinct } \mu_i\in \mathbb C \}$ forms a basis? So, in the ring $\text{EXP}(\mathbb{C})$ the linear differential equation of first order has always a solution except at the following cases:

• $a \neq 0 \land b =0 \land \mu_j =0 \land \alpha_j \neq 0$
  
• $a=0 \land b=0 \land y \neq 0$
  
• $a \neq 0 \land b \neq 0 \land \mu_i =-\frac{b}{a} \land \alpha_i \neq 0$

Is this correct? (Wondering)