# Uniqueness of Solutions for 2nd Order Linear Homogeneous ODEs

• MHB
• ognik
In summary, the author suggests that if $y(x)$ is a solution to $y^{\prime\prime}+ay^{\prime}+by=0$ with $y(0)=y_0$, then $y(x)$ is unique and besides $y(x)$ is analytic. However, they are not able to prove the uniqueness with the given information.
ognik
Hi, please review my answer, I suspect I am missing some fine points...

y(x) is a solution to a 2nd order, linear, homogeneous ODE. Also y(x0)=y0 and dy/dz=y'0
Show that y(x) is unique, in that no other solution passes through (x0, y0) with a slope of y'0.

Expanding y(x) in a Taylor series, $y(x) = \sum_{n=0}^{\infty} {y}^{n}({x}_{0}) \frac{{ (x -{x}_{0}) }^{n}}{n!} = y({x}_{0}) + {y}^{'}({x}_{0}) + \sum_{n=2}^{\infty} {y}^{n}({x}_{0}) \frac{{ (x -{x}_{0}) }^{n}}{n!}$

So I assume another solution is h(x) - but I am told that h(x0) = y0 and that the slope of h(x) is dy/dx, so when I expand h(x) in a Taylor series, I will get a series identical to y(x) - and because the 1st derivatives are equal, all higher derivatives will also be equal.

Therefore the function y(x) is the only unique solution.

ognik said:
I will get a series identical to y(x) - and because the 1st derivatives are equal, all higher derivatives will also be equal.
You can't assure it. The question is that if $y(x)$ is a solution of $y^{\prime\prime}+ay^{\prime}+by=0$ with $y(0)=y_0$ and $y^{\prime}(0)=y^{\prime}_0,$ by a well-known theorem, $y(x)$ is unique and besides $y(x)$ is analytic. In the proof of the theorem we don't use series expansions.

You can verify the uniqueness, with the hypothesis of being $y(x)=\displaystyle\sum_{n\ge 0}a_n(x-x_0)^n,$ substituting this expansion in the differential equation and using the initial conditions. You'll obtain necessary conditions for the coefficients $a_n$ (i.e. uniqueness).

The book suggested the approach I used ("Hint. Assume a second solution satisfying these conditions and compare
the Taylor series expansions.") but you seem to have a good point so I looked at it as well.

But I'm not sure the problem is to expand about point x0, I think y(x0)=y0 is just an initial condition?

Anyway, this is in a section on Frobenius's method, so I should use the substitution $y=\sum_{\lambda = 0}^{\infty}{a}_{\lambda}{x}^{k+\lambda}$, differentiate and substitute back into a standard ODE (y'' + Py' + Qy = 0). I get the expected indicial eqtn, with k=0 or 1. Following Frobenius from there, I don't see how to show the coefficiants I get mean only 1 coefficient, in fact I managed to get 2 coefficiants :-( ?

## 1. What is the "Check uniqueness theorem"?

The "Check uniqueness theorem" is a mathematical concept that states that if a function is continuous and has a unique value at every point, then it is unique for all points within a given interval.

## 2. How does the "Check uniqueness theorem" differ from the "Existence theorem"?

The "Check uniqueness theorem" and the "Existence theorem" are closely related and are often used together. The main difference is that the "Existence theorem" guarantees the existence of a solution to a problem, while the "Check uniqueness theorem" guarantees that the solution is unique.

## 3. What are the applications of the "Check uniqueness theorem"?

The "Check uniqueness theorem" is widely used in various fields of mathematics, including calculus, differential equations, and optimization. It is also applicable in physics and engineering for solving real-world problems that require unique solutions.

## 4. Can the "Check uniqueness theorem" be extended to multivariable functions?

Yes, the "Check uniqueness theorem" can be extended to multivariable functions, also known as vector-valued functions. In this case, the function must have a unique value at every point in the given domain.

## 5. Is the "Check uniqueness theorem" a sufficient condition for the uniqueness of a solution?

No, the "Check uniqueness theorem" is not a sufficient condition for the uniqueness of a solution. It only guarantees the uniqueness of a solution within a given interval or domain. Other conditions, such as smoothness and boundary conditions, may also be necessary for a unique solution.

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