MHB Linear differential equations with constants coefficients

[mathmari]

Hey!

Each element of the ring $\mathbb{C}[z, e^{\lambda z} \mid \lambda \in \mathbb{C}]$ is of the form $\displaystyle{\sum_{k=1}^n α_kz^{d_k}e^{β_kz}}$.

A differential equation in this ring is of the form $$Ly = \sum_{k=0}^m \alpha_k y^{(k)}(z)=\sum_{l=1}^n C_lz^{d_l} e^{\beta_l z} , \ \ \alpha_k , \beta_l \in \mathbb{C} \ \ \ \ (*)$$

At the linear differential equations we can apply the superposition principle. That means that we can split the problem $(*)$ into the subproblems $$Ly(z)=0, \ \ Ly(z)=C_lz^{d_l} e^{\beta_l z}, \ \ l=1, 2, \dots , n$$ so into an homogeneous and $n$ non-homogeneous equations. We solve these equations and then we add the solution of the homogeneous $y_{H}(z)$ and the solutions $y_{p_i}(z)$of the $n$ non-homogeneous equations.
So we get the general solution of the equation $(*)$, which is $$y(z)=y_{H}(z)+\sum_{l=1}^n y_{p_i}(z)$$

To solve the homogeneous equation $$\sum_{k=0}^m \alpha_k y^{(k)}(z)=0$$ we find the characteristic equation and its eigenvalues $\lambda_1, \dots , \lambda_m$.

• If $\lambda_1, \dots , \lambda_m$ are eigenvalues of multiplicity $1$, then the solution of $Ly(z)=0$ is $$y_{H}(z)=\sum_{i=1}^m c_i e^{\lambda_i z}.$$
• If $\lambda_i$ is an eigenvalues of multiplicity $M>1$, then the $$e^{\lambda_i z}, ze^{\lambda_i z}, z^2e^{\lambda_i z}, \dots , z^{M-1}e^{\lambda_i z}$$ are $M$ linear independent solutions of $Ly(z)=0$.

To solve the equation $$\sum_{k=0}^m \alpha_k y^{(k)}(z)=C_lz^{d_l} e^{\beta_l z} \ \ \ \ \ (**)$$ we do the following:

• If $\beta_l$ is not one of the eigenvalues:
  
  We suppose that the solution is in the ring $\mathbb{C}[z, e^{\lambda z} \mid \lambda \in \mathbb{C}]$, so it is of the form $$y(z)=e^{\beta_l z} x(z)$$
  
  So $$y^{(k)}(z)=\sum_{j=0}^k \binom{k}{j}\beta_l^j e^{\beta_l z}x^{(k-j)}(z)$$
  
  Substituting in $(**)$ we get the following equation, the order of which is the same as the order of $(**)$, $$\sum_{k=0}^m \beta_k x^{(k)}(z)=C_lz^{d_l}$$
  
  The solution of the above equation will be polynomials.
  
  The first non-zero term $\beta_k$ determines the degree of $x$.
  
  For example if $\beta_o \neq 0$ then the solution will have degree at most $l$, and will be of the form $$x(z)=\gamma_l z^l +\dots +\gamma_0$$
  Then $$x'(z)=l\gamma_l z^{l-1}+\dots +\gamma_1 \\ \dots \\ x^{(l)}(z)=l!\gamma_l$$
  
  Then $$\sum_{k=0}^m \beta_k x^{(k)}(z)=C_lz^{d_l} \Rightarrow \beta_o\gamma_kz^{d_l}+(\beta_0\gamma_{l-1}+l\gamma_la_1)z^{d_l-1}+ \dots =C_lz^{d_l} $$
  
  So it must stand $$\beta_0\gamma_l=C_l \Rightarrow \gamma_l=\frac{C_l}{\beta_0} \\ \beta_0\gamma_{l-1}+l\gamma_la_1=0 \\ \dots $$
  
  So to solve the differential equation we have to solve the above system.
  
  We write this in the form of matrix. Can we be sure that we will find a unique solution? (Wondering)
  
  In this ring the number of solutions is equal to the number of the order, right? How can we justify that? (Wondering)
• If $\beta_l$ is one of the eigenvalues $\lambda_i$, and the multiplicity of $\lambda_i$ is $L$:
  
  We suppose that the solution is in the ring $\mathbb{C}[z, e^{\lambda z} \mid \lambda \in \mathbb{C}]$, so it is of the form $$y(z)=z^L e^{\beta_l z} x(z)=e^{\beta_l z}\tilde{x}(z)$$ and we continue as in the previous case.

Is everything correct?

Could I improve something at the formulation? (Wondering)

 

[Nikolas7]

At first glance, all is ok, but need to read more carefully