Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Can anybody cast some physical insight into this? I've been studying differential equations on my own and don't understand how you can have a whole host of general solutions. It seems like a rather curious situation which we don't come across in other areas of mathematics. Is there anything more to the discussion that I'm missing?

share|improve this question
    
To study differential equations, you should look at a numerical book, to learn to simulate them. The analytic methods are generally primitive and misleading outside of one and two dimensional phase spaces, because of chaotic behavior. –  Ron Maimon Jul 18 '12 at 17:09

3 Answers 3

The solutions of a homogeneous linear ordinary differential equation form a vector space, ie $0$ is always a valid solution and arbitrary linear combinations of solutions will yield another valid solution. Finding the general solution of such an equation means finding a basis for that vector space.

In contrast, solutions to an inhomogeneous linear ordinary differential equation form an affine space over the space of homogeneous solutions. This means that $0$ is in general not a solution, and the difference between two solutions of the inhomogeneous equation will be a solution of the homogeneous one.

Thus, you only need to find a single inhomogeneous solution: The general solution of the inhomogeneous equation can be expressed of the sum of that particular solution and the general solution of the homogeneous one.

In case of non-linear differential equations, the integration constants (or equivalently, the choice of initial conditions) still parametrize the space of general solutions, and choosing values for these parameters will get you a particular one. However, the simple relation between the general solution of an inhomogeneous equation and the general solution of the corresponding homogeneous one does not hold for arbitrary differential equations.

share|improve this answer
    
This answer is giving an answer specific to linear ODEs, while the phenomenon is general. But it's a fine answer for the linear case. –  Ron Maimon Jul 19 '12 at 1:55
    
@Ron: added a small note about the non-linear case –  Christoph Jul 19 '12 at 8:03
    
I don't agree it is a small note--- the reason I wrote a separate answer is because the linear case is misleading, the real reason for the arbitrary constants is not a vector space argument, it's an initial conditions argument, and it works regardless of the linearity of the equation. Neither are nonlinear equations "more complicated", they are just nonlinear. Sometimes they are simpler. –  Ron Maimon Jul 19 '12 at 16:27
    
@Ron: I slightly reworded the last part –  Christoph Jul 19 '12 at 16:49
    
thanks, I appreciate it, +1. –  Ron Maimon Jul 19 '12 at 17:02

The differential equation needs initial conditions to simulate on a computer (which you should do once, if you want to understand what they are). A differential equation is a rule on a time-grid of size $\epsilon$ for $x(t+\epsilon)$, given $x(t)$. For example;

$$ {dx\over dt}= x + x^2 + x^3 $$

means

$$ x(t+\epsilon) = x(t) + \epsilon ( x(t) + x(t)^2 + x(t)^3) $$

in the limit $\epsilon\rightarrow 0$.

You can use the equation above to step forward in time, but you need to give the value of $x(0)$. Then from this value, you calculate $x(\epsilon)$, then $x(2\epsilon)$, then $x(3\epsilon)$ and so on into the future. A special solution is one for a given initial value, while the general solution is expressed in terms of $x(0)$ or some equivalent arbitrary constant that can be used to find $x(0)$.

The theorem you want is the uniqueness theorem for the initial value problem of ordinary differential equations. The mathematical proof tends to use a less useful approximation technique than lattice stepping, but people who do computer simulations or computer games always use some sort of lattice for ODE's.

share|improve this answer

To answer your remark about the host of solutions for an ode: try to solve for (x,y) the equation: x+y=1. There is an infinity of solutions, (x,y)=(x, 1-x). But you can write them as follow: (x,y)=(0,1)+x*(1,-1)

(0,1) is a peculiar solution of the full equation.

(1,-1) is solution of the homogenous equation: x+y=0 and x in the expression x*(1,-1) plays the role of the constant C you introduce in the solution of an ode.

For linear ode, the situation is quite similar. If you want a physical meaning, one can say that the homogeneous equation (without rhs) is used to study instabilities, if any of its solutions are an increasing function of time, then the system is instable. Any perturbation will blow out with time. Once you know the system is stable, it is worth studying the full equation with a rhs that describes the driving external force. Mathematically the full solution is the sum of the solutions of the homogeneous equation (general solution) plus the solution of the non homogeneous one (peculiar solution) exactly as in the exemple above. Why do we need the general solution if for time long enough they become negligeable in the stable case? Mathematicians were assigned the task to solve the ode, not to discuss their interest, but even in physics, sometimes the transiants (homogenous solutions) are important, especialy when you have to deal with the boundary conditions. For instance, if you study the flow in a pipe, you must impose a zero velocity at the wall. This can be achieved with a proper choice of the constants which appears in the solution of the homogeneous equation.

share|improve this answer
    
This answer is giving an answer specific to linear ODEs, while the phenomenon is general. But it's a fine answer for the linear case. –  Ron Maimon Jul 19 '12 at 1:54

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.