Partial differential equations represented as "operators"by kingwinner Tags: differential, equations, operators, partial, represented 

#1
Sep909, 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 = u_{xy}+cos(x) u + (u_{y})^{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})^{2}u would be equal to (∂_{y}u)^{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! 



#2
Sep909, 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?




#3
Sep909, 10:31 PM

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Also, if they really meant (u_{y})^{2}, how could it be expressed as an "operator"? Thanks! 



#4
Sep909, 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.




#5
Sep909, 11:15 PM

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For example, if we have the PDE:
u_{xy}+cos(x) u + (u_{y})^{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) + ????? 



#6
Sep1009, 06:42 AM

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#7
Sep1009, 02:38 PM

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I think I may be missing something...so whenever we write L= (...) + (...) + (...), is L necessarily always a LINEAR operator? Can nonlinear 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) ? 



#8
Sep1009, 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.




#9
Sep1009, 02:53 PM

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#10
Sep1009, 03:05 PM

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#11
Sep1109, 12:59 AM

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#12
Sep1109, 08:23 AM

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