Partial differential equations represented as "operators"


by kingwinner
Tags: differential, equations, operators, partial, represented
kingwinner
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#1
Sep9-09, 10:18 PM
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1. The problem statement, all variables and given/known data
Partial differential equations (PDEs) can be represented in the form Lu=f(x,y) where L is an operator.
Example:
Input: u(x,y)
Operator: L=∂xy + cos(x) + (∂y)2
=> Output: Lu = uxy+cos(x) u + (uy)2

2. Relevant equations
N/A


3. The attempt at a solution
I don't understand the parts in red.
What does (∂y)2 mean? Shouldn't it mean (∂y)(∂y) which (I believe) means taking the second partial derivative with respect to y. But the above example seems to imply that (∂y)2 means take the partial derivative with respect to y and then square the result. But I just can't possibly understand how (∂y)2u would be equal to (∂yu)2


I don't have a lot of experience working with operators, so I am feeling really confused. Could someone please kindly explain?
Thank you very much!
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Dick
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#2
Sep9-09, 10:25 PM
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I think you are right. It should be u_yy, the second derivative. Not (u_y)^2. It's a linear operator. Is it a typo?
kingwinner
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#3
Sep9-09, 10:31 PM
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Quote Quote by Dick View Post
I think you are right. It should be u_yy, the second derivative. Not (u_y)^2. It's a linear operator. Is it a typo?
But if that is the case, why don't they write ∂yy (instead of (∂y)2? (I am not sure whether it's a typo, but it seems to me that the context of it highly suggests it isn't)

Also, if they really meant (uy)2, how could it be expressed as an "operator"?

Thanks!

Dick
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#4
Sep9-09, 10:41 PM
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Partial differential equations represented as "operators"


This is getting a little overly semantic. del_yy=(del_y)^2. If they mean O(u)=(u_y)^2, that is an operator. But it's not a linear operator. And I wouldn't write it as del_yy or (del_y)^2.
kingwinner
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#5
Sep9-09, 11:15 PM
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For example, if we have the PDE:
uxy+cos(x) u + (uy)2 = x

How can we express it in "operator" form Lu=f(x,y)?

OK, in this case, f(x,y)=x, but what would L be?
L=∂xy + cos(x) + ?????
Dick
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#6
Sep10-09, 06:42 AM
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Quote Quote by kingwinner View Post
For example, if we have the PDE:
uxy+cos(x) u + (uy)2 = x

How can we express it in "operator" form Lu=f(x,y)?

OK, in this case, f(x,y)=x, but what would L be?
L=∂xy + cos(x) + ?????
I told you, that's not a linear operator. Why do you think you should be able to express it using the notation of linear operators?
kingwinner
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#7
Sep10-09, 02:38 PM
P: 1,270
I think I may be missing something...so whenever we write L= (...) + (...) + (...), is L necessarily always a LINEAR operator? Can non-linear operators be expressed this way?


Also, there is something in general that I don't understand about operators...(I'm just guessing that I can do things like the following)

If L=∂xy + cos(x) + (∂y)2, to find Lu, should we just imagine multiplying L with u (just like multiplying real numbers?), so
Lu = [∂xy + cos(x) + (∂y)2] u
=∂xy u + cos(x) u + (∂y)2 u (by distributive law of multiplication) ?
Dick
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#8
Sep10-09, 02:48 PM
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If L is just a sum of differential operators and functions as you are writing then every term in Lu will only have one u in it. So you are going to have a hard time getting (u_y)^2. And yes, you just distribute the operator, so in your example you get Lu=u_xy+cos(x)u+u_yy. Or u_yx+cos(x)u+u_yy depending on how you define ∂xy.
kingwinner
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#9
Sep10-09, 02:53 PM
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Quote Quote by Dick View Post
If L is just a sum of differential operators and functions as you are writing then every term in Lu will only have one u in it. So you are going to have a hard time getting (u_y)^2. And yes, you just distribute the operator, so in your example you get Lu=u_xy+cos(x)u+u_yy. Or u_yx+cos(x)u+u_yy depending on how you define ∂xy.
Can we distribute it that way for any operator in general, or just for linear operators?
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#10
Sep10-09, 03:05 PM
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Quote Quote by kingwinner View Post
Can we distribute it that way for any operator in general, or just for linear operators?
If you write L=(L1+L2+L3) then, yes, Lu=L1u+L2u+L3u. That's what addition of operators means, any operators.
kingwinner
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#11
Sep11-09, 12:59 AM
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Quote Quote by Dick View Post
If you write L=(L1+L2+L3) then, yes, Lu=L1u+L2u+L3u. That's what addition of operators means, any operators.
Are "operators" the same as "functions"?
Dick
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#12
Sep11-09, 08:23 AM
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Quote Quote by kingwinner View Post
Are "operators" the same as "functions"?
What do YOU think? Can't you look up the definitions?


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