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Subject:
partial differential equations
Category: Science > Math Asked by: madukar-ga List Price: $2.00 |
Posted:
15 Jan 2003 14:06 PST
Expires: 14 Feb 2003 14:06 PST Question ID: 143177 |
show that the solutions of the differential equation u'''- 3u''+4u = 0 form a vector space . find a basis for it. |
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Subject:
Re: partial differential equations
Answered By: calebu2-ga on 15 Jan 2003 14:25 PST Rated: |
madukar-ga, First of all, let me give you the definition of a vector space from mathworld.com (http://mathworld.wolfram.com/VectorSpace.html) In terms of our question we can simplify the requirements as follows (because functions in general already satisfy most of the rules) : 1) 0 must be a solution to the differential equation (Additive Identity) 2) If u=f(x) and u=g(x) are solutions to the differential equation then a*f(x)+b*g(x) must also be a solution for all constants a and b. So first, is u=0 a solution If u = 0, then u' = 0, u'' = 0 and u''' = 0. So u''' - 3u'' + 4u = 0 - 3*0 + 4*0 = 0. Hence u = 0 is a solution. Secondly, let f and g be solutions. ie. f''' - 3f'' + 4f = 0 and g''' - 3g'' + 4g = 0 Consider h(x) = af + bg. h' = af' + bg' h'' = af'' + bg'' h''' = af''' + bg''' So h''' - 3h'' + 4h = af''' + bg''' - 3(af'' + bg'') + 4(af + bg) = a(f''' - 3f'' + 4f) + b(g''' - 3g'' + 4g) = a*0 + b*0 = 0 for all a,b. So both requirements 1 and 2 are satisfied. The other requirements follow from the vector properties of functions in general. Hence the solutions to u''' - 3u'' + 4u = 0 form a vector space. Hope this helps. calebu2-ga Search strategy : "Vector space" within mathworld.com | |
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madukar-ga rated this answer: |
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Subject:
Re: partial differential equations
From: kennyh-ga on 18 Jan 2003 23:28 PST |
It seems that the answer is not complete. Since, it did not show the basis and independency. Given u'''- 3u''+4u = 0, by solving its characteristic equation say m^3 - 3 m^2 + 4 = 0 or (m + 1) (m - 2)^2 = 0. So m = -1, -2 -2 and we obtain 3 functions e^(-x), e^(2x) & xe^(2x) which form a basis of a 3-dim vector space. Let's call them f1,f2, & f3. Testing {f1,f2,f3} are indep. : Consider af1 + bf2 +cf3= 0, That is: a e^(-x) + be^(2x) + cxe^(2x) = 0...(1) Set x = 0, we have a + b = 0 ..(1) [ Since e^(-x) <> 0 for all x, cancel it(by mult. e^x)] Multiply e^x to (1): a + b e^(3x) + c x e^(3x) = 0..(2) take dev. wrt x of (2) 3b e^(3x) + c e^(3x) + 3cx e^(3x) = 0 Cancel e^(3x): 3b + c + 3cx = 0 Set x = 0, 3b + c = 0 ..(3) Set x = 1, 3b + 4c = 0...(4) (3)&(4) imply b=c=0 goto(1) a = 0 This shows {f1, f2, f3} are linearly indep and therefore they form a basis of the solution space of the given diff. eq. (aasume we know that the sol. space of a k-degree linear diff eq. is of dim k) Kenny |
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