# MIT 18 303 - The 1-D Wave Equation (24 pages)

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## The 1-D Wave Equation

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## The 1-D Wave Equation

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Pages:
24
School:
Massachusetts Institute of Technology
Course:
18 303 - Linear Partial Differential Equations
##### Linear Partial Differential Equations Documents
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The 1 D Wave Equation 18 303 Linear Partial Di erential Equations Matthew J Hancock Fall 2006 1 1 D Wave Equation Physical derivation Reference Guenther Lee 1 2 Myint U Debnath 2 1 2 4 Oct 3 2006 We consider a string of length l with ends xed and rest state coinciding with x axis The string is plucked into oscillation Let u x t be the position of the string at time t Assumptions 1 Small oscillations i e the displacement u x t is small compared to the length l a Points move vertically In general we don t know that points on the string move vertically By assuming the oscillations are small we assume the points move vertically b Slope of tangent to the string is small everywhere i e ux x t 1 so stretching of the string is negligible Rlp c arc length t 0 1 u2x dx l 2 String is perfectly exible it bends This implies the tension is in the tan gent direction and the horizontal tension is constant or else there would be a preferred direction of motion for the string Consider an element of the string between x and x x Let T x t be tension and x t be the angle wrt the horizontal x axis Note that tan x t slope of tangent at x t in ux plane 1 u x t x 1 Newton s Second Law F ma states that F x 2u t2 2 where is the linear density of the string M L 1 and x is the length of the segment The force comes from the tension in the string only we ignore any external forces such as gravity The horizontal tension is constant and hence it is the vertical tension that moves the string vertically obvious Balancing the forces in the horizontal direction gives T x x t cos x x t T x t cos x t const 3 where is the constant horizontal tension Balancing the forces in the vertical direc tion yields F T x x t sin x x t T x t sin x t T x x t cos x x t tan x x t T x t cos x t tan x t Substituting 3 and 1 yields F tan x x t tan x t u u x x t x t x x 4 Substituting F from 2 into Eq 4 and dividing by x gives 2u t t2 u x x x t x u x x t for x x x Letting x 0 gives the 1 D Wave Equation 2 2u 2 u c t2 x2 Note that c has units c 1 1 h Force Density i1 2 c2 0 5 LT 1 of speed Boundary conditions Ref Guenther Lee 4 2 p 94 Myint U Debnath 4 4 In order to guarantee that Eq 5 has a unique solution we need initial and boundary conditions on the displacement u x t There are now 2 initial conditions and 2 boundary conditions 2 E g The string is xed at both ends u 0 t u l t 0 t 0 homogeneous type I BCs E g The string is connected to frictionless cylinders of mass m that move vertically on tracks at x 0 l Performing a force balance at either x 0 or x 1 gives T sin mg 6 In other words the vertical tension in the string balances the mass of the cylinder But T cos const and tan ux so that 6 becomes ux T cos tan mg Rearranging yields mg x 0 1 These are Type II BCs If the string is really tight and the cylinders are very light then mg 1 and we approximate ux 0 at x 0 1 and the BCs become Type II homogeneous BCs ux 1 2 Initial conditions Ref Guenther Lee 4 2 p 94 We need to specify both the initial position of the string and the initial velocity u x 0 f x and ut x 0 g x 0 x l The idea is the same as nding the trajectory of a falling body we need to know both the initial position and initial velocity of the particle to compute its trajectory We can also see this mathematically The Taylor series of u x t about t 0 is t3 t2 3 u 3 x 0 2 t 3 From the initial conditions u x 0 f x ut x 0 g x and the PDE gives u x t u x 0 ut x 0 t utt x 0 utt x 0 c2 uxx x 0 c2 f x 3u x 0 c2 utxx x 0 c2 g x 3 t Higher order terms can be found similarly Therefore the two initial conditions for u x 0 and ut x 0 are su cient to determine u x t near t 0 To summarize the dimensional basic 1 D Wave Problem with Type I BCs xed ends is PDE BC IC utt c2 uxx 0 x l u 0 t 0 u l t u x 0 f x 7 t 0 ut x 0 g x 3 8 0 x l 9 1 3 Non dimensionalization We now scale the basic 1 D Wave Problem The characteristic quantities are length L and time T Common sense suggests choosing L l the length of the string We introducing the non dimensional variables x x L t t T From the chain rule u x t u x t L f x f x L g x T g x L u u t L u L t T t t t u u x u L x x x x and similarly for higher derivatives Substituting the dimensionless variables into 1 D Wave Equation 7 gives T 2 c2 u t t 2 u x x L This suggests choosing T L c l c so that u t t u x x t 0 0 x 1 10 The BCs 8 become The ICs 9 become u 0 t 0 u 1 t u x 0 f x 1 3 1 t 0 u t x 0 g x 11 0 x 1 12 Dimensionless 1 D Wave Problem with xed ends Dropping hats the dimensionless 1 D Wave Problem is from 10 12 PDE BC IC 2 utt uxx 0 x 1 u 0 t 0 u 1 t u x 0 f x 13 t 0 ut x 0 g x 14 0 x 1 15 Separation of variables solution Ref Guenther Lee 4 2 Myint U Debnath 6 2 and 7 1 7 3 Substituting u x t X x T t into the PDE 13 and dividing by X x T t gives X x T t 16 X x T t 4 where is a constant The negative sign is for convention The BCs 14 become u 0 t X 0 T t 0 u 1 t X 1 T t 0 which implies X 0 X 1 0 17 Thus the …

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