Partial Differential Equations
7.4 Wave Equation
The wave equation models the propagation of waves, such as sound waves, light waves, or water waves, through a medium. It captures how these waves travel and change over time and space. The wave equation for the initial boundary value problem for the displacement (deflection) of a vibrating string whose endpoints are held fixed is
[asciimath](del^2u)/(delt^2)(x,t)=alpha^2(del^2u)/(delx^2)(x,t) \ ,[/asciimath] [asciimath]0ltxltL,[/asciimath] [asciimath]tgt0[/asciimath] (7.4.1)
[asciimath]u(0,t)=u(L,t)=0\ ,[/asciimath] [asciimath]t>0[/asciimath]
[asciimath]u(x,0)=f(x)\ ,[/asciimath] [asciimath](delu)/(delt)(x,0)=g(x)\ ,[/asciimath] [asciimath]0<=x<=L[/asciimath]
Using the method of Separation of Variables, we can find the formal solution to this initial boundary value problem:
[asciimath]u(x,t)=sum_(n=1)^oo[A_n cos((npialphat)/L)+(B_nL)/(npialpha)sin((npialphat)/L )][/asciimath] [asciimath]sin((npix)/L)[/asciimath] (7.4.2)
where
[asciimath]f(x)=sum_(n=1)^ooA_n sin((npix)/L)[/asciimath] and [asciimath]g(x)=sum_(n=1)^ooB_n sin((npix)/L)[/asciimath]
are the Fourier sine series of [asciimath]f(x)[/asciimath] and [asciimath]g(x)[/asciimath] on [asciimath][0,L][/asciimath] and
[asciimath]A_n=2/Lint_0^Lf(x)sin((npix)/L) dx[/asciimath] and [asciimath]B_n=2/Lint_0^Lg(x)sin((npix)/L) dx[/asciimath]
Find the solution to the vibrating string problem
[asciimath](del^2u)/(delt^2)(x,t)=4(del^2u)/(delx^2)(x,t) \ ,[/asciimath] [asciimath]0ltxltpi[/asciimath], [asciimath]tgt0[/asciimath]
[asciimath]u(0,t)=u(pi,t)=0\ ,[/asciimath] [asciimath]t>0[/asciimath]
[asciimath]u(x,0)=sin(3x)-1/2sin(5x) ,[/asciimath] [asciimath](delu)/(delt)(x,0)=sin(4x)+2sin(6x)\ ,[/asciimath] [asciimath]0<=x<=pi[/asciimath]
Show/Hide Solution
Comparing the equation with Equation 7.4.1, we see that [asciimath]alpha=2[/asciimath], [asciimath]L=pi[/asciimath] , [asciimath]f(x)=sin(3x)-1/2sin(5x)[/asciimath], and [asciimath]g(x)=sin(4x)+2sin(6x)[/asciimath]. Since [asciimath]f[/asciimath] and [asciimath]g[/asciimath] are in terms of sine functions, we can determine the values of the coefficients [asciimath]A_n[/asciimath] and [asciimath]B_n[/asciimath] by equating [asciimath]f[/asciimath] and [asciimath]g[/asciimath] to [asciimath]u(x,0)[/asciimath] and [asciimath]u_t(x,0)[/asciimath], respectively.
Substituting [asciimath]t=0[/asciimath] into Equation 7.4.2, we obtain
[asciimath]u(x,0)=[/asciimath] [asciimath]sum_(n=1)^oo (A_n cos(0)+B_n/(2n)sin(0) )sin(nx)[/asciimath] [asciimath]=sum_(n=1)^oo A_nsin(nx)[/asciimath]
From initial boundary values, we have
[asciimath]u(x,0)=sin(3x)-1/2sin(5x)[/asciimath]
Thus
[asciimath]sum_(n=1)^oo A_nsin(nx)[/asciimath] [asciimath]=sin(3x)-1/2sin(5x)[/asciimath]
Equating the coefficients of like terms, we see that
[asciimath]A_3=1,[/asciimath] and [asciimath]A_5=-1/2[/asciimath]
with the remaining coefficients being zero. Similarly, by partially differentiating Equation 7.4.2 with respect to [asciimath]t[/asciimath] and substituting [asciimath]t=0[/asciimath], we obtain
[asciimath](delu)/(delt)(x,t)[/asciimath] [asciimath]=[/asciimath][asciimath]sum_(n=1)^oo (-2nA_n sin(2nt)+B_ncos(2nt) )sin(nx)[/asciimath]
[asciimath](delu)/(delt)(x,0)[/asciimath] [asciimath]=[/asciimath][asciimath]sum_(n=1)^oo (-2nA_n sin(0) +B_ncos(0) )sin(nx)[/asciimath] [asciimath]=sum_(n=1)^oo B_nsin(nx)[/asciimath]
From initial boundary values, we have
[asciimath](delu)/(delt)(x,0)=sin(4x)+2sin(6x)[/asciimath]
Thus
[asciimath]sum_(n=1)^oo B_nsin(nx)[/asciimath] [asciimath]=sin(4x)+2sin(6x)[/asciimath]
Equating the coefficients of like terms, we see that
[asciimath]B_4=1,[/asciimath] and [asciimath]B_6=2[/asciimath]
with the remaining coefficients being zero.
The solution to the problem is
[asciimath]u(x,t)=sum_(n=1)^oo (A_n cos(2nt)+B_n/(2n)sin(2nt))sin(nx)[/asciimath]
[asciimath]u(x,t)=cos(6t)sin(3x)+1/8sin(8t)sin(4x)[/asciimath] [asciimath]-1/2cos(10t)sin(5x)+1/6sin(12t)sin(6x)[/asciimath]
The figure below shows the sketch of [asciimath]u(x,t)[/asciimath].
Try an Example
Section 7.4 Exercises
- Find the solution to the initial boundary value wave problem
[asciimath](del^2u)/(delt^2)(x,t) =9 (del^2u)/(delx^2)(x,t) ,[/asciimath] [asciimath]0ltxltpi,\ tgt0[/asciimath]
[asciimath]u(0,t)= u(pi,t)=0,[/asciimath] [asciimath]t>0[/asciimath]
[asciimath]u(x,0)=-sin(x)+3sin(7x),[/asciimath] [asciimath](delu)/(delt)(x,0)=-2sin(4x)+sin(10x)[/asciimath] [asciimath]0 le x le pi[/asciimath]
Show/Hide Answer
[asciimath]u(x,t)=-cos(3t)sin(x)-1/6sin(12t)sin(4x)+[/asciimath][asciimath]3cos(21t)sin(7x)+1/30sin(30t)sin(10x)[/asciimath]
- Find the solution to the initial boundary value wave problem
[asciimath](del^2u)/(delt^2)(x,t) =4 (del^2u)/(delx^2)(x,t) ,[/asciimath] [asciimath]0ltxltpi,\ tgt0[/asciimath]
[asciimath]u(0,t)= u(pi,t)=0,[/asciimath] [asciimath]t>0[/asciimath]
[asciimath]u(x,0)=-sin(x)+3sin(3x),[/asciimath] [asciimath](delu)/(delt)(x,0)=-4sin(2x)+3sin(6x)[/asciimath] [asciimath]0 le x le pi[/asciimath]
Show/Hide Answer
[asciimath]u(x,t)=-cos(2t)sin(x)-sin(4t)sin(2x)+[/asciimath] [asciimath]3cos(6t)sin(3x)+1/4sin(12t)sin(6x)[/asciimath]
