MHB Are Unique Solutions Possible for Homogeneous Equations?

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To establish uniqueness for homogeneous equations, it is essential to demonstrate that the equation Lx=0 has only the trivial solution. The characteristic equation derived from the homogeneous differential equation reveals eigenvalues, which determine the nature of solutions. If the eigenvalues have multiplicity greater than one, multiple linearly independent solutions exist, suggesting non-uniqueness. Consequently, a homogeneous linear differential equation generally has more than just the trivial solution. This indicates that the solution is not unique, contradicting the initial assumption.
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Hey! :o

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)
 
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mathmari said:
Hey! :o

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.
 

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