fisico30 said:
Thanks both Mapes and arildno.
That is very clear.
It seems that a differential equation trying to describe a physical phenomenon via a dependent variable can really be of different types depending on the domain of existence and on the independent variables.
A PDE like Navier-Stokes is a prime example: for v(x,y,x,t) there is a \frac{dv}{dt} term which is linear and the \{v}\frac{dv}{dx} which is nonlinear...
Also,
1) For linear PDEs, there is a distinction between parabolic, hyperbolic, and elliptic and depending on B^2-4AC. Is there a similar distinction for nonlinear equations?
Not to my knowledge.
2) can you tell from just analyzing a nonlinear term in an equation what type of behavior the equation/phenomenon has?
Very occasionally. Most often, no.
For example, one nasty feature with non-linear equations is that the may exhibit singularities dependent upon initial conditions. This is not the case with linear diff.eqs, whose singularities (if any!) will always coincide with the singularities in the coefficient functions.
3)The word nonlinearity, to me, suggests "mixing". Is that a correct interpretation? How can I see that in a simple nonlinear equation? Does the the nonlinear equation has to be a vector equation with coupling between components to see that? What if it is just a scalar nonlinear ODE. What type of mixing, if any, occurs then?
with gratitude,
fisico30
Very good question!
In order to answer this, we need the concept of an "differential operator":
We can regard an operator as "doing something" to our unkown function:
Suppose we have a first-order diff eq like:
\frac{dy}{dx}+A(x)y=B(x)
Now, we can think of
L=\frac{d}{dx}+A(x) as something which is "done to" (or sort of "multiplied with") our function "Y", so that we may rewrite our diff.eq as:
L(y)=B(x)
(Note the similarity of L-notation with "normal" function notation!)
We can then state the condition for "linearity":
A diff.eq is "linear" if and only if its associated operator L, for constants a, b, and functions y(x), z(x) follws the principle:
L(ay+bz)=aL(y)+bL(z)
That is, the output of an operator applied to a linear combination of two functions should be the same linear combination of the "operated" output of the two functions involved.
To see that the given example IS linear, we have:
L(ay+bz)=\frac{d}{dx}(ay(x)+bz(x))+A(x)*(ay(x)+b(z(x))=a(\frac{dy}{dx}+A(x)y)+b(\frac{dz}{dx}+A(x)z)=aL(y)+bL(z)
Thus, linearity of operator is shown!
Let us not take another diff.eq:
\frac{dy}{dx}y=B(x)
Here, let us regard the operator L as applied to a sum of two functions y(x) and z(x).
L(y+z)=(\frac{d}{dx}(y+z))(y+z)=(\frac{dy}{dx}+\frac{dz}{dx})(y+z)=\frac{dy}{dx}y+\frac{dz}{dx}z+\frac{dy}{dx}z+\frac{dz}{dx}y\neq{L}(y)+L(z)= \frac{dy}{dx}y+\frac{dz}{dx}z
Thus, L in this case is non-linear.
Thus, your idea that "non-linearities" sort of mix together solutions or at least functions is spot on; the associated differential operator is unable to keep the two component functions separate as a linear combination, cross-terms of mixture appears.
