1D Wave Equation General Solution / Gaussian Function
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1 D Wave Euation General Solution / Gaussian Function Phys 375 Overview and Motivation: Last time we derived the partial differential euation known as the (one dimensional wave euation Today we look at the general solution to that euation As a specific eample of a localized function that can be useful when studying waves, we introduce the Gaussian function Key Mathematics: We reacuaint ourselves with the chain rule (for taking derivatives and look at the Gaussian function and the integral of the Gaussian function, which is known as the error function I Solutions to the Wave Euation A General Form of the Solution Last time we derived the wave euation (, t (, t c ( from the long wave length limit of the coupled oscillator problem Recall that c is a (constant parameter that depends upon the underlying physics of whatever system is being described by the wave euation Now it may surprise you, but the solution to E ( can, uite generally, be written in very succinct form as (, t f ( + ct + g( ct, ( where f and g are any "well-behaved" functions We won't worry about the details of what well-behaved means, but certainly we certainly want their second derivatives to eist As we previously discussed, f ( + ct travels in the direction at the speed c and g( ct travels in the + direction at the same speed To see that E ( is a solution to E ( let's calculate the second and t derivatives of f ( + ct and g( ct To do this we need the chain rule, which can be written for the case at hand as [ k( t ] h, dh dk k k h (3 Applying this rule to f ( + ct, for eample, we have D M Riffe -- /4/3
2 f ( + ct f + ct Phys 375 ( + ct ( f ( + ct, (4 where f ( + ct is the derivative of f ( + ct with respect to its argument Applying the chain rule again to calculate the second derivative of f ( + ct with respect to give us ( + ct f ( + ct f f + ct ( + ct ( f ( + ct (5 Similarly, we have for the t derivatives f ( + ct f + ct ( + ct ( c f ( + ct (6 and ( + ct f ( + ct ( + ct f c cf ( + ct c From Es (5 and (7 we see that f ( + ct (7 ( + ct f ( + ct f c (8 and so f ( + ct indeed solves the wave euation Proof that g( ct also satisfies E ( follows from an essentially identical calculation B More Specific Solutions So the (totally unknown functions f ( + ct and g( ct are solutions, but only in a very general sense As we shall see, any solution has the form of E (, but how do we know what the solution will be in any given situation? Well, as with earlier problems that we have looked at in this class, the situation can be specified by initial conditions and the boundary conditions Recall, for the coupled oscillator problem the initial conditions were specified by values for j ( and & j ( Now, however, the spatial variable is not the discrete inde j but the continuous variable The corresponding initial conditions can thus be written as (, a( (9a D M Riffe -- /4/3
3 and Phys 375 (, b(, (9b where a ( and b ( are assumed to be known functions We could have just stuck with (, and (,, but in the long run we save a bit of notational cumbersomeness by using a ( and b ( instead For the coupled-oscillator system the boundary conditions are ( t and N + ( t There will be similar instances for the wave euation where we will be interested in waves on a finite sized system In such cases we will need to specify the condition on (, t at the system boundaries Indeed, you have already seen an eample of this in Eercise 74 from the last lecture notes II Initial Value Problem (IVP for an Infinite System Here we write the most general solution to the wave euation, given the initial conditions a ( and b ( To keep things simple at this point we will assume that the ends of the (D system where these waves eist are at and + In this way we do not have to deal with any boundary conditions (We will deal with bc's in the net lecture! From E ( (, t f ( + ct + g( ct, ( we have for t ( f ( g( ( a + That is simple enough What about the other initial condition Well, taking the t derivative of E ( and setting it eual to b ( at t gives us ( c[ f ( g ( ] ( b Now E ( can be formally integrated, which gives us c b ( d f ( g( ( D M Riffe -3- /4/3
4 Phys 375 where can have any (constant value If we now take the sum and difference of Es ( and ( we obtain f c ( a( + b( d (3a and g c ( a( b( d (3b Using E (3 in E ( finally gives us the general solution to the initial-value problem + c ct (, t a( + ct + a( ct + b( d (4 ct Notice that the undetermined constant has disappeared E (4 is remarkably simple II The Gaussian Function and Two Initial-Value-Problem Eamples A Gaussian Function A very useful function in physics is the Gaussian, which is defined as G ( e (5 As shown in the picture on the top of the following page, the Gaussian is peaked at and has a width that is proportional to the parameter In fact, the full width at half maimum (FWHM, which is the width of the peak at half its maimum height, is eual to ln( 665 B IVP Solution with Gaussian Initial Position Let's see what the solution (, t looks like with the initial conditions a( AG (, ( A is just some arbitrary amplitude and b ( (Physically, how would you describe this set of initial conditions? Using E (4 we rather trivially obtain D M Riffe -4- /4/3
5 Phys 375 A (, t [ G ( + ct + G ( ct ], (6 or more eplicitly, GAUSSIAN FUNCTIONS sigma sigma 3 sigma 5 G( 5 A [ ] 5 5 ( ( + ct, ( ct t e + e, (7 So the solution consists of two Gaussian functions, one moving in the direction and one in the + direction, both at the speed c The amplitude of each function is ½ the amplitude of the initial Gaussian displacement The following picture illustrates this solution as a function of for several times t For simplicity we have set A,, and c SOLUTION FOR INITIAL GAUSS DISPLACEMENT t t t 3 t 6 (,t D M Riffe -5- /4/3
6 Phys 375 B IVP Solution with Gaussian Initial Velocity / Error Function Let's look at another eample using the Gaussian function This time let's have the initial displacement of the system be zero so that a (, but let's have a Gaussian initial-velocity function so that b( BG (, where B is some arbitrary velocity amplitude In this case we get from E (4 B c + ct ( (, t e d (8 ct So what is the integral of a Gaussian function? Well, it is known as the error function Specifically, the error function is defined as erf ( e d (9 π The following figure shows a plot of erf ( vs As the graph indicates, ( defined such that erf (, erf (, and erf ( erf is ERROR FUNCTION erf( All this is fine and well, but what do we do about the in E (8, which does not appear in E (9? We must do a little math (! and change variables in E (8 Let's define a new integration variable y, dy d Then E (8 becomes B c ( + ct y (, t e dy ( ( ct 5 5 D M Riffe -6- /4/3
7 We can now use E (9, the definition of the error function, to write Phys 375 π B 4c (, t { erf [( + ct ] erf [( ct ]} ( Notice that again we have the sum of two functions, each traveling in opposite directions at the speed c The following picture plots the solution vs for several values of t (with B,, and c all set to SOLUTION FOR INITIAL GAUSSIAN VELOCITY t t t 3 t 6 (,t Eercises *8 The chain rule Let h (, y + y + y (a Directly calculate h and h y (b Now define two new independent variables u ( + y and u ( y (c Rewrite h (, y in terms of u and v That is, find h ( u, v (d Now starting with h ( u, v and thinking of it as h ( u(, t, v(, t, calculate h h h u h v using the chain rule That is, calculate this derivative using + Does u v this agree with your answer in (a? (e Follow a procedure analogous to that in (d to calculate h y Again, does this agree with your answer in (a? Technically speaking, we should give the function h ( u, v another name [ ( u v physicists, we are rather lazy and typically still call the new function h h,, say ], but being D M Riffe -7- /4/3
8 *8 Verify for t that E (4 and its time derivative reduce to the initial conditions (, a( and (, b(, respectively Phys 375 *83 If the initial displacement is zero, then E (4, the solution to the initial value c + ct problem, can be written as (, t b( d By calculating second derivatives in ct and t, directly demonstrate that this function solves E (, the wave euation **84 The error function (a Using an appropriate change of integration variable in the euation erf ( e d, show that ( ( erf e π d π (b Using a computer mathematic package, plot erf ( over an appropriate range of for, 3, and 5 **85 The initial value problem Consider E (4, the general solution to the initial value problem (a Eplain why E (4 in not a function of the variable (This is a basic feature of the definite integral Consult a calculus book if necessary (b What are c, a (, and b ( in E (4? 4 (c Consider the specific case where ( b c + Using a a and ( ( c, plot ( computer mathematics package and letting b over an appropriate range of (d Using the initial conditions given in (c, solve E (4 (Do not set c to zero! The integral can be done either with a change of variable, a computer mathematics package, or can be looked up in a table of integrals (such as found in the CRC Handbook of Chemistry and Physics (e Show that your solution can be written in the form (, t f ( + ct + g( ct Thus identify f ( and g ( (f Again, using a computer mathematics package and letting c, plot (, t as a function of for t,,, and 3 Be careful to let your graph include all interesting parts of the solution! D M Riffe -8- /4/3
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