Lecture 10 Phys 3750 D M Riffe -1- 2/1/2013 General Solution using Normal Modes Overview and Motivation: Last time we solved the initial-value problem (IVP) for the 1D wave equation on a finite domain with "closed-closed" bc's using the general form of the solution () ( )()ctxgctxftxq−++=, . Today we solve the same problem using the normal mode solutions for this system. Key Mathematics: We utilize some integrations involving harmonic functions. I. The Problem Defined We are looking for the general solution to the wave equation () ()22222,,xtxqcttxq∂∂=∂∂ (1) on the finite domain Lx ≤≤0 subject to the initial conditions () ()xaxq =0, and () ()xbxtq=∂∂0, (2a), (2b) and the boundary conditions ()0,0 =tq and ()0, =tLq . (3a), (3b) This time we write the solution ()txq , as linear combination1 of the normal-mode solutions ()txqn, ()()()∑∞==1,,nntxqtxq , (4) where the normal modes can be expressed as ()() ( )()tintinnnnneBeAxktxqωω−+= sin,. (5) Here Lnknπ= is the wave vector and nnck=ω is the angular frequency. As before, let's make the normal-mode solutions explicitly real by setting *nnAB = . Then Eq. (5) can be written as 1 As written, Eq. (4) looks like a simple sum, not a linear combination, but as Eq. (5) shows, we have kept undetermined amplitudes as part of our normal modes, so Eq. (4) may be justifiably thought of as a linear combination of normal modes.Lecture 10 Phys 3750 D M Riffe -2- 2/1/2013 ()() ( ) () ()()()[]tataxktxqnnnnnnωωsinImcosResin,−= , (6) where we have defined a new amplitude 2nnAa=. Using Eq. (6) we may thus express the general solution [via Eq. (4)] as ( ) ( ) () ( ) () ( )[]∑∞=−=1sinImcosResin,nnnnnntataxktxqωω, (7) As we shall shortly see, the amplitudes na are determined by the initial conditions. Recall, however, that normal modes already satisfy the bc's, and so the general solution as expressed by Eq. (7) automatically satisfies those bc's. We thus need not consider the bc's any further. II. The Initial Value Problem (Yet Again!) Let's now apply the initial conditions and see what we get. From Eq. (7) we obtain for the initial displacement ()()()∑∞==1ResinnnLnaxxaπ, (8) and for the initial velocity, we obtain after differentiating Eq. (7) ()()()∑∞=−=1ImsinnnLnnaxxbπω. (9) In Eqs. (8) and (9) we have used Lnknπ=. So what do we have here? Well, perhaps not surprisingly, we have two equations for the amplitudes na in terms of the initial conditions. However, unlike the (finite) N-oscillator case, the rhs's of Eqs. (8) and (9) have an infinite number of amplitudes because the wave equation has an infinite number of normal modes! Aside: The N -oscillator problem This looks pretty grim, but perhaps a look back at the N-oscillator case will give us some insight into the current problem. For the N oscillator problem the equation equivalent to Eq. (8) is the extension of Eq. (19) from the Lecture (6) notes to N oscillators, which we can write asLecture 10 Phys 3750 D M Riffe -3- 2/1/2013 ()()()()()()()()()∑=++++=NnnNnNnNnNnNaNqqqq11111321Resin3sin2sin1sin0000ππππMM, (10) where we have again made the assignment 2nnAa=. Now recall what we did there. To find any particular amplitude ()maRe (which we label by m ) we take the m th eigenvector expressed as a row vector and multiply Eq. (10) by that vector. We can write this multiplication as () ()() ()()()()()()()()()()()=∑=++++++++NnnNnNnNnNnNNmNmNmNmaNqqqqN111113211111Resin3sin2sin1sin0000sin3sin2sin1sinππππππππMMK , (11) Now recall what happens in that case: when the rhs is multiplied by the m th eigenvector, the only term in the sum that survives in the sum is the on with the same eigenvector. That is, only the mn= term survives, which transforms Eq. (11) into () ()() ()()()()()()()()()()()=++++++++mNmNmNmNmNNmNmNmNmaNqqqqN Resin3sin2sin1sin0000sin3sin2sin1sin11113211111ππππππππMMK, (12) Notice that this equation only has one coefficient ()maRe on the rhs and so it can now be solved for that coefficient,Lecture 10 Phys 3750 D M Riffe -4- 2/1/2013 ()() ()() ()()()()()()() ()() ()()()()()()=++++++++++++NNqqqqNaNmNmNmNmNmNmNmNmNNmNmNmNmm111111113211111sin3sin2sin1sinsin3sin2sin1sin0000sin3sin2sin1sinReππππππππππππMKMK. (13) The key point here is that multiplying Eq. (10) by the m th eigenvector allows us to find ()maRe in terms of the initial condition on the displacement of the system. Notice that Eq. (10) is an equation for the initial displacement of the system in terms of the set of coefficients ()naRe , while Eq. (13) is an equation for any coefficient ()maRe (labeled by m) in terms of the initial displacement of the system. We can thus think of Eq. (13) as the inversion of Eq. (10). Now the notation used in Eqs. (10) – (13) is rather cumbersome. Fortunately there is a more succinct way to express these equations. Looking at Eq. (10) we first note that the j th element of that equation can be written as ()()()∑−+=NnnNnjajq11Resin0π. (14) We now notice that if we multiply this equation by ()jNm1sin+π and them sum the equation on j , ()()() ()()nNjNnNnNmNjjNmajjqj Resinsin0sin111111∑∑∑==++=+=πππ, (15) then this is indeed multiplication of Eq. (10) by the transpose of the m th eigenvector! That is, Eq. (15) is the same equation as Eq. (11), only in a much more succinct form. We now switch the order of the sums on the rhs ()() ( )()()∑∑∑==++=+=NnNjNnNmnNjjNmjjaqj111111sinsinRe0sinπππ (16)Lecture 10 Phys 3750 D M Riffe -5- 2/1/2013 and notice that the sum on j on the rhs is simply the product of the (transpose of) m th eigenvector with the n th eigenvector, which is nonzero only if they are the same eigenvector. That is, the sum on j on the rhs is nonzero only if mn = . Thus Eq. (16) simplifies to ()() ( )()∑∑=+=+=NjNmmNjjNmjaqj11211sinRe0sinππ. (17) This is equivalent to Eq.