MHB Are Unique Solutions Possible for Homogeneous Equations?

  • Thread starter Thread starter mathmari
  • Start date Start date
  • Tags Tags
    Homogeneous
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.
 
There is the following linear Volterra equation of the second kind $$ y(x)+\int_{0}^{x} K(x-s) y(s)\,{\rm d}s = 1 $$ with kernel $$ K(x-s) = 1 - 4 \sum_{n=1}^{\infty} \dfrac{1}{\lambda_n^2} e^{-\beta \lambda_n^2 (x-s)} $$ where $y(0)=1$, $\beta>0$ and $\lambda_n$ is the $n$-th positive root of the equation $J_0(x)=0$ (here $n$ is a natural number that numbers these positive roots in the order of increasing their values), $J_0(x)$ is the Bessel function of the first kind of zero order. I...
Are there any good visualization tutorials, written or video, that show graphically how separation of variables works? I particularly have the time-independent Schrodinger Equation in mind. There are hundreds of demonstrations out there which essentially distill to copies of one another. However I am trying to visualize in my mind how this process looks graphically - for example plotting t on one axis and x on the other for f(x,t). I have seen other good visual representations of...
Back
Top