Exploring Homogeneous Equations and Unique Solutions

[mathmari]

Hey!

I saw in my notes the part that to show the uniqueness we have to prove that $Lx=0$ has only trivial solution.
($L$ is the differential operator)

To solve the homogeneous equation $$\sum_{k=0}^m \alpha_k x^{(k)}(z)=0$$ 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 $Lx(z)=0$ is $$x_{H}(z)=\sum_{i=1}^m c_i e^{\lambda_i z}.$$
- 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 $Lx(z)=0$. So aren't there also solutions other than $x=0$ ? Does this mean that the solution is not unique? Or have I understood it wrong? (Wondering)

 

[Sudharaka]

Hey!

I saw in my notes the part that to show the uniqueness we have to prove that $Lx=0$ has only trivial solution.
($L$ is the differential operator)

To solve the homogeneous equation $$\sum_{k=0}^m \alpha_k x^{(k)}(z)=0$$ 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 $Lx(z)=0$ is $$x_{H}(z)=\sum_{i=1}^m c_i e^{\lambda_i z}.$$
- 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 $Lx(z)=0$. So aren't there also solutions other than $x=0$ ? Does this mean that the solution is not unique? Or have I understood it wrong? (Wondering)

Hi mathmari, :)

I am not getting what you are trying to prove here. A homogeneous linear differential equation of the $n-th$ order would have a characteristic equation of order $n$. This would give it $n$ linearly independent solutions. Thus the general solution will be the sum of all the linearly independent solutions. It is not true that a homogeneous linear differential equation has only the trivial solution.