Vibration Problems -- Why must there be two solutions to the DE?

[Trying2Learn]

TL;DR Summary

Why must there be two solutions

Consider the second order linear differential equation that describes vibration:
Mass, damping, stifffness and free vibration

If the roots of the characteristic equation are repeated, we have one solution and we must search for a second. We assume it by taking the first solution and multiplying it by time.

Why?

I mean: I get that it works, but all the textbooks say that we MUST search for a second solution.

Now, I know we can do that because we have two initial conditions and a second order equation.

But can someone explain why the books say "we MUST search for another solution. For it seems to me, right now, a case of a hammer in search of a nail. How does one justify a search for a second solution?

 

[BvU]

How does one justify a search for a second solution?

Is there another way to get room for a second initial condition ?

 

[Trying2Learn]

Is there another way to get room for a second initial condition ?

Ah, but that is my point. You seem to be saying (and I understand and almost agree), that just because we have two initial conditions, we must have two equations.

That is what confuses me. Why does "two initial conditions" DEMAND that there be two solutions?

Can you formalize this connection between # of I.C. and # of solutions?

 

[fresh_42]

Ah, but that is my point. You seem to be saying (and I understand and almost agree), that just because we have two initial conditions, we must have two equations.

That is what confuses me. Why does "two initial conditions" DEMAND that there be two solutions?

It does not in general, only if they are taken at the same point. Consider the solutions as trajectories through a vector field constructed by attaching the direction of the trajectory at each point. Then a solution is a flow through this vector field.

A second initial condition can either be one of the same trajectory at a different point, or one which directs in a different direction at the same point, and thus a second trajectory. And even if taken at a different point, we cannot directly tell to which trajectory aka solution it belongs.

 

[Trying2Learn]

O

It does not in general, only if they are taken at the same point. Consider the solutions as trajectories through a vector field constructed by attaching the direction of the trajectory at each point. Then a solution is a flow through this vector field.

A second initial condition can either be one of the same trajectory at a different point, or one which directs in a different direction at the same point, and thus a second trajectory. And even if taken at a different point, we cannot directly tell to which trajectory aka solution it belongs.

So let me stick with the case where we resolve the solutoin at the same point (displacment and velocity I.C.)

You said "Then a solution is a flow through this vector field."

Those are nice words (please pardon an ostensible scarcasm -- I do not intend to be truly sarcastic, but only insofar as to manifest my ignorance). Why do those words demand the need for a second solution?

Later, you said: "or one which directs in a different direction at the same point, and thus a second trajectory."

OK, so I like the first clause, above, but I do not know why even that DEMANDS a second solution when the roots are repeated.

I feel there is something here that is important to understand about critical damping but I do not see it yet.

 

[fresh_42]

I neither know the equations you are talking about, nor the initial conditions and even less the vector field they define. Hence I can only give a general answer.

We have a real world problem, so tiny wobbles of the initial conditions can result in two completely different trajectories. In the image below e.g. a stable orbit or an ejection along the axis. Other possibilities occur, if the two solutions are independent. Then we have such a thing as two different vector fields within the same frame. Then chance decides which way will be realized in the real world.

E.g. we can have an equilibrium situation at a certain point, placing a pen on its peak. It will almost certainly skip, but it may also stay at rest until disturbed. Both are solutions of the same system. They have the different initial conditions at the same point of release: undisturbed or with some rest energy which causes the pen to skip, which is a completely different solution than the one with the pen standing up.

Equilibria can be classified by looking at the signs of the eigenvalues of the linearization of the equations about the equilibria. That is to say, by evaluating the Jacobian matrix at each of the equilibrium points of the system, and then finding the resulting eigenvalues, the equilibria can be categorized. Then the behavior of the system in the neighborhood of each equilibrium point can be qualitatively determined, (or even quantitatively determined, in some instances), by finding the eigenvector(s) associated with each eigenvalue.

An equilibrium point is hyperbolic if none of the eigenvalues have zero real part. If all eigenvalues have negative real part, the equilibrium is a stable equation. If at least one has a positive real part, the equilibrium is an unstable node. If at least one eigenvalue has negative real part and at least one has positive real part, the equilibrium is a saddle point.

https://en.wikipedia.org/wiki/Equilibrium_point

 

[BvU]

Ah, but that is my point. You seem to be saying (and I understand and almost agree), that just because we have two initial conditions, we must have two equations.

That is what confuses me. Why does "two initial conditions" DEMAND that there be two solutions?

Can you formalize this connection between # of I.C. and # of solutions?

Yes. That is to say between the order of the equation and # of solutions. The characteristic equation of a second order differential equation has two identical roots in case of critical damping. How to deal with them is shown here

 

[Trying2Learn]

Thank you, again, to all. This has been great.

 

[Trying2Learn]

Yes. That is to say between the order of the equation and # of solutions. The characteristic equation of a second order differential equation has two identical roots in case of critical damping. How to deal with them is shown here

Yes, I know how to get the second equation.

However, you wrote: "That is to say between the order of the equation and # of solutions." -- but why? That is my question.

Yes, we learn the two roots are identical and books immediately begin the process of attempting a second solution, without saying 'why' (not how -- how is not my issue), why we "must."

 

[martinbn]

Summary:: Why must there be two solutions

Consider the second order linear differential equation that describes vibration:
Mass, damping, stifffness and free vibration

If the roots of the characteristic equation are repeated, we have one solution and we must search for a second. We assume it by taking the first solution and multiplying it by time.

Why?

I mean: I get that it works, but all the textbooks say that we MUST search for a second solution.

Now, I know we can do that because we have two initial conditions and a second order equation.

But can someone explain why the books say "we MUST search for another solution. For it seems to me, right now, a case of a hammer in search of a nail. How does one justify a search for a second solution?

You need two linearly independent solutions for the following reason. Consider the space of all solutions, it is a linear space because the equation is linear. Map it to $\mathbb R^2$ by sending the solution $y(t)$ to $(y_0,y'_0)$. This is a linear map and by the existence and uniqueness theorem it is bijective. So the space of all solutions is two dimensional and to write a generic solution you need two linearly independent ones.

 

[Trying2Learn]

You need two linearly independent solutions for the following reason. Consider the space of all solutions, it is a linear space because the equation is linear. Map it to $\mathbb R^2$ by sending the solution $y(t)$ to $(y_0,y'_0)$. This is a linear map and by the existence and uniqueness theorem it is bijective. So the space of all solutions is two dimensional and to write a generic solution you need two linearly independent ones.

OK; this is really very close to what I am looking for.

I understand if you do not have the time...

But could you repeat that, with a bit less sophistication?

 

[martinbn]

Are you familiar with vector spaces and bases?

 

[Trying2Learn]

Are you familiar with vector spaces and bases?

Are you familiar with vector spaces and bases?

Yes, I have a strong understanding of vectors, the basis, a bit on covariatn and covectors (but not strong),
and how we form a basis for a vector space... and how the modern definition of vector supercedes the arrow with a tail, notion.

 

[Delta2]

It is a theorem in mathematics that the set of all functions that are solutions of a linear differential equation is a vector space , sub space of the vector space of all functions (of a real variable).

The vector space of all functions has infinite dimension (that is its basis consists of infinitely many functions).

However it is also a theorem that the vector subspace of the functions that are solutions to a specific linear differential equation has finite dimension and this dimension is equal to the order of the linear differential equation let's say n. The exact n vectors(functions) that consist the basis of this vector subspace depend on which exactly is the linear differential equation.

So in your example the order of the linear differential equation is 2, this means that the vector subspace $V$ that contains all the solutions to that linear ODE has dimension 2, which means that its basis contains two linear independent functions which belong to $V$ and hence satisfy the linear ODE. Once we know the basis, let's say it consists from the functions $f_1(x),f_2(x)$ , an element of $V$ can be written as linear combination of the vectors of the basis that is for any function $f$ that belongs to $V$ (and hence satisfies the linear ODE) we can write it as $$f=\lambda_1f_1+\lambda_2f_2 (1)$$ $$\lambda_1,\lambda_2 \in \mathbb{R}$$. So (1) gives the general solution to the linear ode of order 2. In your example you had found $f_1$ but you had to go find $f_2$ so that you find out the basis of $V$ and hence your can write the general solution as (1).

 

[Trying2Learn]

It is a theorem in mathematics that the set of all functions that are solutions of a linear differential equation is a vector space , sub space of the vector space of all functions (of a real variable).

The vector space of all functions has infinite dimension (that is its basis consists of infinitely many functions).

However it is also a theorem that the vector subspace of the functions that are solutions to a specific linear differential equation has finite dimension and this dimension is equal to the order of the linear differential equation let's say n. The exact n vectors(functions) that consist the basis of this vector subspace depend on which exactly is the linear differential equation.

So in your example the order of the linear differential equation is 2, this means that the vector subspace $V$ that contains all the solutions to that linear ODE has dimension 2, which means that its basis contains two linear independent functions which belong to $V$ and hence satisfy the linear ODE. Once we know the basis, let's say it consists from the functions $f_1(x),f_2(x)$ , an element of $V$ can be written as linear combination of the vectors of the basis that is for any function $f$ that belongs to $V$ (and hence satisfies the linear ODE) we can write it as $$f=\lambda_1f_1+\lambda_2f_2 (1)$$ $$\lambda_1,\lambda_2 \in \mathbb{R}$$. So (1) gives the general solution to the linear ode of order 2. In your example you had found $f_1$ but you had to go find $f_2$ so that you find out the basis of $V$ and hence your can write the general solution as (1).

This was exactly what I was looking for!

Thank you!

The theorem is, obviously, beyond me. But once I accept it, I can understand. Thank you.

May I risk one more question?

I have also been having some difficulty trying to understand how, in the Euler-Lagrange Equations from Hamilton's Principle and cal.variations... why is it that we can vary the position and velocity independently.

I believe this answer you gave here, satisfies that question, too.

Am I correct?

 

[Delta2]

I have also been having some difficulty trying to understand how, in the Euler-Lagrange Equations from Hamilton's Principle and cal.variations... why is it that we can vary the position and velocity independently.

That's an interesting question but I don't see how it relates to this talk about vector spaces and their dimension. Can you tell me how do you think it relates?

 

[Trying2Learn]

That's an interesting question but I don't see how it relates to this talk about vector spaces and their dimension. Can you tell me how do you think it relates?

I don't know : -)

So I will guess.

We have a second order DiffEq of motion. We need to obtain it from the physics and free body diagrams (as per Newton).

However, Hamilton suggested the variational approach to get the Euler Lagrange equations.

And we get the SAME equation. So we know it is correct.

Now in the variational method, we vary position and velocity independently.

So is this not the same as saying that in solving a diff.Eq. at which the conditions are stated, NOT at different points (boudnary value problem), but at the SAME point (Initial value problem), that the two vectors (one, basiclly for position and one for velocity), must be an independent basis from each other?

 

[Trying2Learn]

That's an interesting question but I don't see how it relates to this talk about vector spaces and their dimension. Can you tell me how do you think it relates?

I mean, sure, in the case of an Initial value problem, we specify the position and velocity AT THE SAME POINT (unlike a boundary value problem where the points are in a different place).

So while there are TWO solutions for the position, we resolve the associated constant coefficients with the TWO initial conditions (one of which is velocity).

So, while we ultimately solved for position, we indirectly solved for velocity.

So, in specifying two solutions, we are indirectly also solving for velocity, independently.

Yes?

I am trying to learn this, so forgive me if I mess up.

 

[Delta2]

Sorry I can't understand exactly what you trying to say at the last two posts, but I think you mixing up the concept of linearly independent vectors with the concept of independent variables of a function.

 

[Trying2Learn]

Sorry I can't understand exactly what you trying to say at the last two posts, but I think you mixing up the concept of linearly independent vectors with the concept of independent variables of a function.

Yes, in hindsight, with the simplicity (NOT in the pejorative) of your response, I see how much my struggle brought in issues similar in appearance, only. I think, now, the Lagrangian issue is not related.

Thank you so much for the rest!