College of Engineering and Computer ScienceMechanical Engineering DepartmentEngineering Analysis NotesLarry Caretto February 28, 2009Solutions to the Wave EquationThe wave equationThe 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.22222xuctu[1]The D’Alambert solution and its proofThe D’Alambert solution to equation [1] is written in terms of coordinates ξ and η, defined as follows:ctxandctx [2]The D’Alambert solution to the wave equation is written in terms of two arbitrary functions, F(ξ) and G(η). This gives the following solution.)()()()( ctxGctxFGFu [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 (ξ,η). 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. ututtu[4]The D’Alambert solution gives simple expressions for the ξ and η derivatives since F is a function of ξ only and G is a function of η only. )(')()()( FddFGFu[5]Here we use the notation F’(ξ) for the first derivative of F with respect to ; we will subsequently define a second derivative in a similar manner.22)()('')()('dFdFddFF [6]In a similar fashion we can writeJacaranda (Engineering) Room 3333 Mail Code Phone: 818.677.6448Email: [email protected] 8348 Fax: 818.677.7062 )(')()()( GddGGFu[7]Finally, from the definitions of ξ = x + ct and η = x - ct, we can write the following partial derivatives.ctct[8]Combining this equation with equations [4], [5], and [7] gives the following result.)''('' GFccGcFtu[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.   )''''()''()''(222GFcGFccGFcctuttuttuttu[10]We can repeat this process for the x-derivatives; the main difference is in the partial derivatives ofthe new coordinates with respect to x.11 xx[11]With these relationships, we can obtain the first derivative with respect to x as follows.)(')(')]()([)1()]()([)1(GFGFGFuxuxxu[12]The second derivative is obtained by taking the derivative of the first derivative.   )''''()''()1()''()1(22GFGFGFxuxxuxxuxxu[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].   ''''''''222222222222GFcxucGFctuxuctu[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 Mail Code Phone: 818.677.6448Email: [email protected] 8348 Fax: 818.677.7062The D’Alambert solution with initial conditionsWe 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.)()(),0(0xgtuandxfxux[15]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. ctxctxdgcctxfctxfxtu)(21)()(21),([16]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.)()()()()(ayabybdxxyybya[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. )()()()()()()()()()(ctxgctxgcctxgcctxcgctxgtctxctxgtctxdgtctxctx[18]The second derivative of the integral is just the derivative of the first derivative, which becomes  )(')(')()()()()()()()()()(222ctxgctxgcctxctxgtctxctxctxgtctxcctxcgctxcgtdgttdgtctxctxctxctx[19]We have used the usual notation, g’, to indicate an ordinary first derivative.Jacaranda (Engineering) Room 3333 Mail Code Phone: 818.677.6448Email: [email protected] 8348 Fax: 818.677.7062  ddgg '[20]In a similar fashion we can take the second-order space derivative of the integral by starting with the first