Are Unique Solutions Possible for Homogeneous Equations?

  • Context: MHB 
  • Thread starter Thread starter mathmari
  • Start date Start date
  • Tags Tags
    Homogeneous
Click For Summary
SUMMARY

The discussion centers on the uniqueness of solutions for homogeneous linear differential equations, specifically those represented by the operator $L$. It is established that the equation $Lx=0$ can have multiple solutions depending on the multiplicity of the eigenvalues derived from the characteristic equation. If eigenvalues $\lambda_1, \dots , \lambda_m$ have multiplicity greater than one, the solutions include $M$ linearly independent functions, indicating that non-trivial solutions exist. Therefore, the assertion that the solution is unique is incorrect; multiple solutions can arise from eigenvalues of higher multiplicity.

PREREQUISITES
  • Understanding of homogeneous linear differential equations
  • Familiarity with characteristic equations and eigenvalues
  • Knowledge of linear independence in the context of differential equations
  • Basic concepts of differential operators
NEXT STEPS
  • Study the theory of linear differential equations and their solutions
  • Learn about the implications of eigenvalue multiplicity in differential equations
  • Explore the concept of linear independence and its role in solution sets
  • Investigate the application of differential operators in various mathematical contexts
USEFUL FOR

Mathematicians, students of differential equations, and anyone interested in the theory of linear operators and their solutions.

mathmari
Gold Member
MHB
Messages
4,984
Reaction score
7
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)
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
Replies
0
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 36 ·
2
Replies
36
Views
5K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K