College of Engineering and Computer Science Mechanical Engineering Department Engineering Analysis Notes Larry Caretto February 28 2009 Solutions to the Wave Equation The wave equation The one dimensional wave equation shown below describes the propagation of a disturbance u over space and time For example u might be the amplitude of a vibrating string which varies with space and time 2 2u 2 u c 2t x 2 1 The D Alambert solution and its proof The D Alambert solution to equation 1 is written in terms of coordinates and defined as follows x ct and x ct 2 The D Alambert solution to the wave equation is written in terms of two arbitrary functions F and G This gives the following solution u F G F x ct G x ct 3 To show that this is a solution we have to rewrite the wave equation in terms of these two functions This means that we have to transform the coordinates in the wave equation from x t to To start the coordinate transformations we look at the first derivative terms If we have u as a function of and we can get the time derivatives with respect to t by the following equation u u u t t t 4 The D Alambert solution gives simple expressions for the and derivatives since F is a function of only and G is a function of only u dF F G F d 5 Here we use the notation F for the first derivative of F with respect to we will subsequently define a second derivative in a similar manner F dF d F d 2 F d 2 6 In a similar fashion we can write Jacaranda Engineering Room 3333 Email lcaretto csun edu Mail Code 8348 Phone 818 677 6448 Fax 818 677 7062 u F G dG G d 7 Finally from the definitions of x ct and x ct we can write the following partial derivatives c t c t 8 Combining this equation with equations 4 5 and 7 gives the following result u cF cG c F G t 9 We find the second time derivative by taking the time derivative of this first derivative We can simplify the result using the equations above for coordinate transformation the D Alambert solution that u F G and the definitions of F and G as ordinary second derivatives 2u u u u c c F G c c F G c 2 F G 2 t t t t t t t 10 We can repeat this process for the x derivatives the main difference is in the partial derivatives of the new coordinates with respect to x 1 x 1 x 11 With these relationships we can obtain the first derivative with respect to x as follows u u u F G F G 1 1 F G 12 x x x The second derivative is obtained by taking the derivative of the first derivative 2u u u u 1 F G 1 F G F G 2 x x x x x x x 13 We can now show that the two expressions for the second derivatives in equations 10 and 13 satisfy the original two dimensional wave equation in 1 2 2u 2 u c 2t x 2 2u c 2 F G 2 t c2 2u c 2 F G 2 x 14 Thus the D Alambert solution u F x ct G x ct where F and G are arbitrary functions satisfies the differential equation We now have to show how we can use this solution to satisfy the differential equation and the initial conditions Jacaranda Engineering Room 3333 Email lcaretto csun edu Mail Code 8348 Phone 818 677 6448 Fax 818 677 7062 The D Alambert solution with initial conditions We want to be able to satisfy arbitrary initial conditions on the displacement u and its first derivative the velocity at t 0 These arbitrary initial conditions are written as follows u 0 x f x and u t g x 15 x 0 In this section we show that the solution to the wave equation 1 with the boundary conditions in equation 15 can be written as follows 1 1 u t x f x ct f x ct 2 2c x ct g d 16 x ct In this equation we have used the D Alambert solution for the first two terms where the arbitrary functions F and G in the D Alambert solution are both the initial condition function f We say that u t 0 x f x and u t x has a contribution f x ct and f x ct This means that we use the same functional form but the argument to the function f at later times is computed as x ct and x ct For example if f x 1 when x 0 then f x ct would equal 1 when x ct 0 This would occur at all points where x ct The final term in equation 16 is the integral of the initial derivative condition that is integrated over the dummy variable The first two terms have just been shown to satisfy the differential equation Recall that any function of x ct or x ct satisfied the differential equation In order to show that the integral term in equation 16 satisfies the differential equation we need to use the general formula for differentiation under the integral sign y b y b x dx y b a y a a y 17 We can apply this general result to compute the space and time derivatives for the wave equation First find the time derivatives Using the general rule above we find the following result t x ct g d x ct x ct x ct g x ct g x ct t t 18 cg x ct c g x ct c g x ct g x ct The second derivative of the integral is just the derivative of the first derivative which becomes 2 t 2 x ct g d x ct x ct g d cg x ct cg x ct t t x ct t 19 x ct g x ct x ct g x ct c c 2 g x ct g x ct x ct t x ct t We have used the usual notation g to indicate an ordinary first derivative Jacaranda Engineering Room 3333 Email lcaretto csun edu Mail Code 8348 Phone 818 677 6448 Fax 818 677 7062 g dg d 20 In a similar fashion we can take the second order space derivative of the integral by starting with the first derivative x ct x ct x ct g g d g x ct g x ct x x ct x x x g 1 g x ct 1 g x ct g x ct g x ct x 21 We then take the second derivative from the result above x ct x ct 2 g d g d g x ct g x ct 2 x x ct x x x ct x x ct g x ct x ct …