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Homogenous and linear differential equations

  • #1

Homework Statement


I am currently in quantum chemistry, and in class one day my professor spent some time talking about Maxwell's equations. I am looking at my notes, trying to piece together Maxwell's equations, differential equations, and the principle of superposition, since this is not in my book. I haven't taken dif eq either, so that might be part of the problem.

I was told (if I understand my notes) that the solutions to Maxwell's equations are differential equations that have the quality of being homogenous and linear, therefore the superposition principle will work. 1. What does it mean for a function to be homogenous and linear?

Also in my notes, I wrote: E1 = a solution to Maxwell's eq
E2 = a solution to Maxwell's equations

For any constants, λ1 and λ2 the linear combinations λ1E1+λ2E2 will be a solution to Maxwell's equations. 2. This may be outside the realm of the pure math people, but why does this combination describe a light field?

Homework Equations


I don't think there are any.

The Attempt at a Solution


1. Well I was told that a homogenous linear equation takes this form:

Σfi(x)⋅dig(x)/dxi=0
g(x) is the probe function
The fact the it is equal makes it homogenous - but can't you make any equation equal to zero by subtracting it from both sides?
There must be no powers in the differential, what would that look like? Must g(x) just be a linear function, then?
Also, the power of the coefficient doesn't matter.

So I know these things, but I do not know what the mean, really, I don't know how to recognize a linear homogenous (LH) function, and what the significance of a LH function is.

2. I don't know if E stands for energy or electric field - I think electric field, since in class he spoke about how E2 is proportional to intensity. Either way, I don't see a light field.
 

Answers and Replies

  • #2
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586
the solutions to Maxwell's equations are differential equations that have the quality of being homogenous and linear, therefore the superposition principle will work
This sentence is correct if you remove "solutions to". But with the correction that only two of the Maxwell's equations are homogeneous.(Meaning that they have terms not containing one of the fields E or B.)

And the below sentence is the definition of a linear DE.
E1 = a solution to Maxwell's eqs
E2 = a solution to Maxwell's eqs
For any constants, λ1 and λ2 the linear combinations λ1E1+λ2E2 will be a solution to Maxwell's equations.
 
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  • #3
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We are dealing with differential equations, moving any term to one side does not make it homogeneous. It is homogeneous if there are only terms with differentials and y which equate to zero, for example ay'' + by' + cy = 0. If we have a lone function of x in the equation such that ay'' + by' + cy = f(x) OR ay'' + by' + cy - f(x) = 0, then the equation is not homogeneous.

Linear refers to the type of equation and is a result of the linear property of derivatives. Namely, d/dx[af(x) + bg(x)] = ad/dx[f(x)] + bd/dx[g(x)]. If the derivative and function terms, y'', y', y appear in the numerator, then the solution will be linear. That is, if you find two solutions y1(x) and y2(x), then the general solution is a linear combination of y1 and y2
 
  • #4
We are dealing with differential equations, moving any term to one side does not make it homogeneous. It is homogeneous if there are only terms with differentials and y which equate to zero, for example ay'' + by' + cy = 0. If we have a lone function of x in the equation such that ay'' + by' + cy = f(x) OR ay'' + by' + cy - f(x) = 0, then the equation is not homogeneous.

Linear refers to the type of equation and is a result of the linear property of derivatives. Namely, d/dx[af(x) + bg(x)] = ad/dx[f(x)] + bd/dx[g(x)]. If the derivative and function terms, y'', y', y appear in the numerator, then the solution will be linear. That is, if you find two solutions y1(x) and y2(x), then the general solution is a linear combination of y1 and y2
So in your example - are y'', y' functions or differentials in the form dy2/dx and dy/dx? I guess I do not know if the y's in this: ay'' + by' + cy = 0 are just y's or a function... if it's a function, and it's in terms of x and you have ay'' + by' + cy = f(x), then why can't you subtract f(x) to make it homogeneous? You said it's because there are only terms with differentials, but is "y" a differential? Isn't either just a variable or a function in terms of a variable?
 
  • #5
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Yes, y is a function of x, y(x). The difference is y(x) is in an undetermined form, whereas f(x) is in a closed form (determined). Thus, our equation is homogeneous if one one side we have the undetermined forms, y''(x), y'(x), y(x), all equal to zero. If there is a closed form function into the equation which stands as its own term, it no longer becomes homogeneous.

ay''(x) + by'(x) + cy(x) = 0 ---> homogeneous, note that a, b, c could be scalars or functions of x.
ay''(x) + by'(x) + cy(x) + f(x) = 0 ---> not homogeneous
The goal here is to solve for y(x). If we have just the homogeneous form, it is much simpler than if the extra term is added into the equation, because it affects our equation explicitly in its closed form.

For example,
2y''(x) + 3y'(x) + ln(x)y(x) = 0 is homogeneous, the only closed form function of x is attached to a term containing the solution y(x).
whereas
3y''(x) + 4y'(x) + 2y(x) - ln(x) = 0 is inhomogeneous, because the closed form function of x is its own term, namely, - ln(x).
 
  • #6
Yes, y is a function of x, y(x). The difference is y(x) is in an undetermined form, whereas f(x) is in a closed form (determined). Thus, our equation is homogeneous if one one side we have the undetermined forms, y''(x), y'(x), y(x), all equal to zero. If there is a closed form function into the equation which stands as its own term, it no longer becomes homogeneous.

ay''(x) + by'(x) + cy(x) = 0 ---> homogeneous, note that a, b, c could be scalars or functions of x.
ay''(x) + by'(x) + cy(x) + f(x) = 0 ---> not homogeneous
The goal here is to solve for y(x). If we have just the homogeneous form, it is much simpler than if the extra term is added into the equation, because it affects our equation explicitly in its closed form.

For example,
2y''(x) + 3y'(x) + ln(x)y(x) = 0 is homogeneous, the only closed form function of x is attached to a term containing the solution y(x).
whereas
3y''(x) + 4y'(x) + 2y(x) - ln(x) = 0 is inhomogeneous, because the closed form function of x is its own term, namely, - ln(x).
Okay I think I have almost got it, but in your last example, how do we know that -ln(x) isn't y(x)? Does that make sense? How can you tell if you have f(x) or y(x)? I don't know what a closed form function is.
 
  • #7
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So, - ln(x) could be y(x). The difference is that the latter is explicitly stated while y(x) is at the moment unknown, keep in mind the ultimate goal is to find y(x). The term homogeneous is a way for us to classify the equation based on the structure of the equation with respect to y(x). If we have terms with the unknown function y(x) and no extraneous functions in seperate terms, it is homogeneous.

Closed form means explicitly stated, like sin(x) is in closed form.

So consider one more example:
ay''(x) + by'(x) + cy(x) + sin(x) = 0 this equation is in-homogeneous.
Whether or not y(x) = sin(x), this equation is not homogeneous because as the equation stands, we do not know y(x) AND there is an extra term to deal with in the equation, that is not necessarily the same as y(x). The homogeneous/inhomogeneous label is just a way to classify the equation and affects how we will go about solving the problem.

In practice, an inhomogeneous equation is solved first by finding the solution to the homogeneous equation y(x) and then adding on an extra term to our solution which would account for the inhomogeneous term. You can think of the inhomogeneous equation as a more complicated version of the homogeneous equation. The homogeneous equation relates the sum of the derivatives of some unknown function and the unknown function itself to zero. The inhomogeneous equation relates the sum of the derivatives of some unknown function and the unknown function itself to some other functions, the inhomogeneous term.

ay''(x) + by'(x) + cy(x) = 0 is much simpler to solve than ay''(x) + by'(x) + cy(x) = f(x). y(x) is unknown, f(x) is known.
 
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