linear and non-linear ODE

Thread Starter

PG1995

Joined Apr 15, 2011
816
Hi

Please have a look on the attachment. You can find my questions there. Please help me with them. Thanks a lot.

The equation dy/dx = y/(y+x) is non-linear because it cannot be written in the form shown in green highlight in the attachment.

ODE: An ordinary differential equation (or ODE) is a relation that contains functions of only one independent variable, and one or more of their derivatives with respect to that variable. This means there could be more than one dependent variables such as y, z, w, etc, but they will all depend on one independent variable, say, x.

Regards
PG
 

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Papabravo

Joined Feb 24, 2006
12,511
There are two big ideas that are associated with things that are linear. The first big idea is that if you add two things together and perform some operation, that is the same as performing the same operation on each of the things individually and then adding them together. The second big idea is that if you multiply a thing by a constant and perform some operation on that thing, that is the same as performing the operation and then multiplying the result by the constant.

For example - take differentiation. Is it a linear operation? It is, because the derivative of (f(x) + g(x)) is equal to the f'(x) + g'(x)

AND

the derivative of (a*f(x)) is equal to a * f'(x)

You can make exactly the same argument for integration and indeed many other operations.

Now taking the sine of something is not a linear operation because the sin(x + y) is not equal to the sin(x) + sin(y)

AND

the sin(ax) is not equal to a*sin(x)
 

steveb

Joined Jul 3, 2008
2,436
On question 1, don't think of linear as implying constant slope. The above post provides the correct definition of linear.

To see why a general line is not linear consider the line function\(y(t)=m\;x(t)+b\), where m is slope and b is offset. Now let \(x(t)=a_1\;f(t)+a_2\;g(t)\), where \(a_1\) and \(a_2\) are constants . The output y is as follows

\(y(t)=m(a_1\;f(t)+a_2\; g(t))+b\)

But linearity implies the following

\(y(t)=a_1(m\; f(t)+b)+a_2(m\; g(t)+b)\)

Clearly these two things are not equal unless b=0. Hence, offsets create nonlinearity.
 
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Thread Starter

PG1995

Joined Apr 15, 2011
816
Thank you, Papabravo, Steve.

Okay. It's easy to understand when it's said "it's a linear equation" which simply means an equation of a line and as a line is straight it intuitively implies that it has a constant slope. Perhaps, it's more about semantics than mathematics. If a*y = f(a*x) then f(x) is said to be linear; it doesn't matter if the degree of x is one or not. Why didn't they choose some other names for such types of functions other than linear? If possible, please let me know it. Thank you.

Best wishes
PG
 

steveb

Joined Jul 3, 2008
2,436
Perhaps, it's more about semantics than mathematics. If a*y = f(a*x) then f(x) is said to be linear; it doesn't matter if the degree of x is one or not. Why didn't they choose some other names for such types of functions other than linear? If possible, please let me know it. Thank you.
Note that the degree of x in terms of powers does need to be one, but offsets are not allowed. Above, I probably mispoke when I said don't think of constant slope. Constant slope is ok, if there is no offset. Higher degree derivatives and integrals are allowed because these are linear operations. Things like squares, transcendental functions and multiplication of variables are nonlinear. The definition of linearity is really the important things and provides a simple test when you are in doubt.

I agree that the choice of the term "linear" is not the best because it does cause some confusion. The biggest confusion is that an offset to an otherwise linear function or system induces nonlinearity, while an offset to a line still leaves you with a line. The term "superposition" is perhaps a better one because it does not conflict with common language. Linear systems obey the principle of superposition which is ... you guessed it ... the ability to scale and add solutions and still have a valid solution (i.e. the definition of linearity).
 

someonesdad

Joined Jul 7, 2009
1,583
You've run into what programmer's call overloading -- something can mean different things in different contexts. Check out the wikipedia page on linear for some of the details: http://en.wikipedia.org/wiki/Linearity.

As Papabravo intimated, the big idea with linearity in the context of differential equations is that if f(x) and g(x) are solutions of a linear differential equation, then an arbitrary linear combination of them such as \(af(x) + bg(x)\) is also a solution where a and b are constants. This idea will also come up in the context of linear vector spaces, of which you'll no doubt study later. It's a deep and important property and is why much of the analysis of the physical world is done by trying to linearize things: because we can solve the resulting equations.

The classic example studied in a basic physics class is the equation of motion of a pendulum. Alas, in its generality it's a nonlinear differential equation whose solution involves Jacobian elliptic functions, something you don't teach to a beginner. To avoid this problem, the equation is linearized by restricting the problem to small angles of oscillation, which turns the sine term into a linear term -- because sin(x) is about x when x is small and in radians. Then you wind up with the familiar \(\ddot x + x = 0\) equation (in suitable units). You can verify that both sin(t) and cos(t) solve this equation; therefore, \(a sin(t) + b cos(t)\) (a and b constants again) is also a solution because the equation is a linear differential equation.
 
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