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# Meth MATH 556

UM

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This 3 page Class Notes was uploaded by Mrs. Preston Lehner on Thursday October 29, 2015. The Class Notes belongs to MATH 556 at University of Michigan taught by Staff in Fall. Since its upload, it has received 12 views. For similar materials see /class/231507/math-556-university-of-michigan in Mathematics (M) at University of Michigan.

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Date Created: 10/29/15

l NORMALIZATION OF SOLUTIONS 11 Initial conditions The general solution of any homogeneous lin ear second order ODE 1 i ptx qtz 0 has the form 01x1 ngg where 01 and 02 are constants The solutions 12 are often called basic7 but this is a poorly chosen name since it is important to understand that there is absolutely nothing special about the solutions 12 in this formula beyond the fact that neither is a multiple of the other For example the ODE i 0 has general solution at b We can take 1 t and 2 1 as basic solutions and have a tendency to do this or else list them in the reverse order so 1 1 and x2 t But equally well we could take a pretty randomly chosen pair of polynomials of degree at most one such as 1 4 t and 2 3 7 2t as basic solutions This is because for any choice of a and b we can solve for 01 and 02 in at b 01x1 02x2 The only requirement is that neither solution is a multiple ofthe other This condition is expressed by saying that the pair zlx2 is linearly independent Given a basic pair of solutions 12 there is a solution of the initial value problem with zt0 ax t0 b of the form x 01x1 02x2 The constants 01 and 02 satisfy a 011 022t0 b Clil Cgig For instance the ODE i 7 z 0 has exponential solutions e and e which we can take as 12 The initial conditions 0 2i0 4 then lead to the solution x clet Cge t as long as 0102 satisfy 7t 7 2 0 ole0 62670 01 02 ole0 027e 0 cl 7 02 This pair of linear equations has the solution 01 302 71 so z 3e 7 equot 12 Normalized solutions Very often you will have to solve the same differential equation subject to several different initial conditions It turns out that one can solve for just two well chosen initial conditions and then the solution to any other lVP is instantly available Here7s how 1 2 De nition 121 A pair of solutions 172 of 1 is normalized at to if 9M 1 962 0 z1 07 i2 1 By existence and uniqueness of solutions with given initial condi tions7 there is always exactly one pair of solutions which is normalized at to For example7 the solutions of i 0 which are normalized at 0 are 1 172 It To normalize at to 17 we must nd solutions7 polynomials of the form at b7with the right values and derivatives at t1 These are 1 17z2 t7 1 to to to to For another example7 the harmonic oscillator77 5539 mix 0 has basic solutions coswnt and sinwnt They are normalized at 0 only if run 17 since asinhunt run coswnt has value run at t 07 rather than value 1 We can x this as long as run 31 0 by dividing by u so 2 coswnt wgl sinwnt is the pair of solutions to i wiz 0 which is normalized at to 0 Here is another example The equation i 7 z 0 has linearly inde pendent solutions die 7 but these are not normalized at any to for example because neither is ever zero To nd 1 in a pair of solutions normalized at to 07 we take zl aet bequot and nd 11 such that z10 1 and 210 0 Since 21 aet 7 be 7 this leads to the pair of equations a b 17a 7 b 07 with solution a b 12 To nd 12 aet be t z20 07z 20 1 imply 1 b 07a7b 1 or a 12b 712 Thus our normalized solutions 1 and 2 are the hyperbolic sine and cosine functions t 7t t m it cosht6 6 sinht These functions are important precisely because they occur as nor malized solutions of i 7 z 0 Normalized solutions are always linearly independent 1 cant be a multiple of 2 because z1t0 31 0 while z2t0 07 and 2 cant be a multiple of zl because 922 7E 0 while z1t0 0 Now suppose we wish to solve 1 with the general initial conditions lf 1 and 2 are a pair of solutions normalized at to then the solution x with 050 a iltt0gt b is az1b2 The constants of integration are the initial conditions If I want x such that i z 0 and 0 3 2 for example we have x 3cost 2 sin t Or for an other example the solution of i 7 z 0 for which 0 2 and 4 is z 2 cosht 4sinht You can check that this is the same as the solution given above Exercise 122 Check the identity coshzt 7 sinth 1 13 ZSR and ZIR There is an interesting way to decompose the solution of a linear initial value problem which is appropriate to the inhomogeneous case and which arises in the systemsignal approach Two distinguishable bits of data determine the choice of solution the initial condition and the input signal Suppose we are studying the initial value problem 3 i 100 qtz ft7 020 07 020 i0 There are two related initial value problems to consider ZSR The same ODE but with rest initial conditions or zero state plttgtz qlttgtz N we 0 we 0 Its solution is called the Zero State Response or ZSR It depends entirely on the input signal and assumes zero initial conditions We7ll write xf for it using the notation for the input signal as subscript ZIR The associated homogeneous ODE with the given initial condi tions i 1002 909 07 t0 07 020 i0 Its solution is called the the Zero Input Response or ZIR lt de pends entirely on the initial conditions and assumes null input signal We7ll write x for it where h indicates homogeneous7 By the superposition principle the solution to 3 is precisely z xf zh The solution to the initial value problem 3 is the sum of a ZSR and a ZlR in exactly one way

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