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Section 6.8 The wave equation

Objective:

1. The solution set of a wave equation (vibration of a string) with different initial conditions

The wave equation problem is utt(x,t)=a2uxx(x,t),t>0u(0,t)=0,u(L,t)=0,t>0u(x,0)=f(x),ut(x,0)=g(x),0xL  where  f(x) is the initial position function and  g(x) is the initial velocity.  We assume u(x,t)=X(x)T(t), then translate the problem into solving X(x)+λX(x)=0,X(0)=X(L)=0T(t)+a2λT(t)=0. The only non-trivial solutions of  X(x)+λX(x)=0,X(0)=X(L)=0 are  λn=(nπL)2 and  Xn=cnsin(nπLx). Hence we need to solve  T(t)+a2(nπL)2T(t)=0. 

Case (1):  g(x)=0.} Then  ut(x,0)=g(x)=0=X(x)T(0) implies  T(0)=0. The solution of  T(t)+a2(nπL)2T(t)=0 with  T(0)=0 will be  Tn(t)=kncos(anπLt). We let un(x,t)=cos(anπLt)sin(nπLx). and write u(x,t)=n=1dnun(x,t)=n=1dncos(anπLt)sin(nπLx). We use the initial condition  u(x,0)=f(x) to solve for  dnu(x,0)=f(x)=n=1dnun(x,0)=n=1dnsin(nπLx). We see that  f(x) is a sine Fourier series and hence  dn=2L0Lf(x)sin(nπLx)dx. 

 

 

 

Example 1: Suppose the displacement  u(x,t) of a piece of flexible string is given by the initial-boundary value problem utt(x,t)=9uxx(x,t),t>0u(0,t)=0,u(π,t)=0,t>0u(x,0)=10sin(2x)20sin(4x),ut(x,0)=0,0xπ.

(a) What is the physical meaning of its boundary conditions?

(b) What are the initial displacement and initial velocity of the string at the mid- point,  x=π2. 

(c) State the general form of its solution.

 

 

 

Exercise 1: Suppose the displacement  u(x,t) of a piece of flexible string is given by the initial-boundary value

problem utt(x,t)=16uxx(x,t),t>0u(0,t)=0,u(6,t)=0,t>0u(x,0)=10sin(πx)30sin(2πx),ut(x,0)=0,0x6.

(a) What is the physical meaning of its boundary conditions?

(b) What are the initial displacement and initial velocity of the string at the mid- point,  x=3. 

(c) State the general form of its solution.

 

 

 

Case (2):  f(x)=0.} Then  u(x,0)=f(x)=0=X(x)T(0) implies  T(0)=0. The solution of  T(t)+a2(nπL)2T(t)=0 with  T(0)=0 will be  Tn(t)=knsin(anπLt). We let  un(x,t)=sin(anπLt)sin(nπLx) and write  u(x,t)=n=1dnun(x,t)=n=1dnsin(anπLt)sin(nπLx). We use the initial condition  ut(x,0)=g(x) to solve for  dnut(x,0)=g(x)=n=1dnnπaLsin(nπLx). We see that  g(x) is a sine Fourier series and hence  dnnπaL=2L0Lg(x)sin(nπLx)dx, i.e.  dn=2nπa0Lg(x)sin(nπLx)dx. 

 

 

 

Example 2: Suppose the displacement  u(x,t) of a piece of flexible string is given by the initial-boundary value problem utt(x,t)=49uxx(x,t),t>0u(0,t)=0,u(2,t)=0,t>0u(x,0)=0,ut(x,0)=20x2,0x2.

(a) What is the physical meaning of its boundary conditions?

(b) What are the initial displacement and initial velocity of the string at the mid- point,  x=1. 

(c) State the general form of its solution.

 

 

 

Exercise 2: Suppose the displacement  u(x,t) of a piece of flexible string is given by the initial-boundary value

problem utt(x,t)=9uxx(x,t),t>0u(0,t)=0,u(4,t)=0,t>0u(x,0)=0,ut(x,0)=10+2x,0x4.

(a) What is the physical meaning of its boundary conditions?

(b) What are the initial displacement and initial velocity of the string at the mid- point,  x=2. 

(c) State the general form of its solution.

 

 

 

Case (3):  f(x)0 and  g(x)0.} We let  v(x,t) and  w(x,t) be the functions satisfy the boundary problems vtt(x,t)=a2vxx(x,t),t>0v(0,t)=0,v(L,t)=0,t>0v(x,0)=f(x),vt(x,0)=0,0xL with the solution, v(x,t)=n=1dncos(anπLt)sin(nπLx), and dn=2L0Lf(x)sin(nπLx)dx. wtt(x,t)=a2wxx(x,t),t>0w(0,t)=0,w(L,t)=0,t>0w(x,0)=0,wt(x,0)=g(x),0xL with the solution, w(x,t)=n=1ensin(anπLt)sin(nπLx), and en=2nπa0Lg(x)sin(nπLx)dx. Then  u(x,t)=v(x,t)+w(x,t) satisfy utt(x,t)=a2uxx(x,t),t>0u(0,t)=0,u(L,t)=0,t>0u(x,0)=f(x),ut(x,0)=g(x),0xL with solution u(x,t)=v(x,t)+w(x,t)=n=1dncos(anπLt)sin(nπLx)+n=1ensin(anπLt)sin(nπLx) and  dn=2L0Lf(x)sin(nπLx)dxen=2nπa0Lg(x)sin(nπLx)dx.

 

 

 

Example 3: Suppose the displacement  u(x,t) of a piece of flexible string is given by the initial-boundary value problem utt(x,t)=16uxx(x,t),t>0u(0,t)=0,u(2π,t)=0,t>0u(x,0)=20cos(2x),ut(x,0)=20x2,0x2π.

(a) What is the physical meaning of its boundary conditions?

(b) What are the initial displacement and initial velocity of the string at the mid- point,  x=π. 

(c) State the general form of its solution.

 

 

 

Exercise 3: Suppose the displacement  u(x,t) of a piece of flexible string is given by the initial-boundary value problem utt(x,t)=4uxx(x,t),t>0u(0,t)=0,u(10,t)=0,t>0u(x,0)=20+10cos(πx)30cos(2πx),ut(x,0)=10+2x,0x10.

(a) What is the physical meaning of its boundary conditions?

(b) What are the initial displacement and initial velocity of the string at the mid- point,  x=5. 

(c) State the general form of its solution.

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