7 Homogeneous Wave PDE in 2D

 7.1 Rectangular membrane
 7.2 circular membrane

7.1 Rectangular membrane

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Using Cartesian coordinates. Wave displacement is u ≡ u(x,y,t)  (out of page).

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Boundary conditions on x

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And boundary conditions on y

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Initial conditions

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Let u = X (x)Y (y)T (t)  . Plug in the PDE

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Hence

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Hence time ODE becomes

 ′′  2
T + c λT = 0

And the space ODE is

X-′′  Y′′
 X  + Y  = − λ

Separating this again

X ′′       Y′′
--- = − λ −---
 X         Y

Using new separation variable μ  , we obtain two new ODE’s

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Or

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Solving for X  ODE, and knowing that μ > 0  from earlier, we obtain

X = A cos(√ μx)+ B sin (√ μx)

Applying B.C. at x = 0

0 = A

Hence            (√ --)
X (x) = B sin   μx . Applying B.C. at x = L

         √--
0 = Bsin( μL )

Hence

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Therefore the X  solution is

              ((nπ ) )
Xn (x) = Bn sin  -L- x     n = 1,2,3,⋅⋅⋅

Solving the Y  ODE

      (    (   ) )
Y′′ + Y λ−   nπ-2  = 0
             L

The solution is

         (∘ ---(---)2-)        (∘ ----(--)2-)
Y = A cos    λ −  nπ- y  + B sin    λ −  nπ- y
                 L                     L

Applying first B.C.

0 = A

Hence

         (∘ ---(-nπ)2-)
Y = B sin    λ −  --- y
                 L

Applying second B.C.

        (∘ ----(---)2 )
0 = B sin  λ −  nπ-  H
                 L

Hence

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Hence the Y  solution is

             (    )
Ynm = Bnm sin  m-πy      n = 1,2,3,⋅⋅⋅,m = 1,2,3,⋅⋅⋅
               H

Now we solve the time T  ode

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Combining all solution , and merging all constants into two, we find

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We now use initial conditions to find Anm, Bnm  . At t = 0

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Applying first initial condition to (1)  gives

            (                 )
         ∞∑    ∞∑         (m π )     ((nπ ) )
f (x,y) =         Anm sin  -H-y   sin   -L- x
         n=1 m=1

Applying 2D orthogonality gives

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Taking time derivative of (1)

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AT t = 0  the above becomes

        ∑∞ ∑∞   ∘ ----      ((nπ-) )   ( m-π )
g(x,y) =       c  λnmBnm sin    L  x  sin  H  y
        n=1m=1

Applying 2D orthogonality gives

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Summary of solution

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Here are few animations, all using L = 1,H = 2,c = 0.1  . For different modes. The mode n,m  is given below each animation. These run for 100  seconds. Initial conditions are u(x,y,0) = xcos(y)  and -∂u(x,y,0) = 0
∂t  . Boundary conditions zero on all 4 edges.

  

  

  

7.2 circular membrane

7.2.1 not circularly symmetric

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Using polar coordinates. Disk has radius a  . Wave displacement is u ≡ u (r,θ,t)  (out of page).

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Boundary conditions on r

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And boundary conditions on θ

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Initial conditions

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Let u = T (t)R (r)Θ (θ)  . Plug in the PDE

-1T ′′RΘ = R′′T Θ + 1R ′TΘ + -1Θ ′′RT
c2                r       r2

Dividing by RT Θ

1 T ′′   R′′   1R ′  1 Θ ′′
c2-T-=  R--+ rR- + r2-Θ-

Hence

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The time ODE is

T′′+ c2λT = 0

Now we separate again the space ODE’s (remember to move the λ  with the R  and not the Θ  )

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Let separation constant be μ  , therefore

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With periodic boundary conditions and

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Now it is in SL form, where            μ
p = r,q = − r,σ = r  . This is singular SL. Can be written as

     1     (    μ)
R ′′ +-R ′ + λ− -2  R = 0
     r         r

Before we solve the above R  ODE, we solve the Θ′′ + μΘ = 0  to find μ  Eigenvalues. The solution is

         √ --        √ --
Θ = A cos( μθ)+ B sin ( μθ)

With B.C Θ (− π) = Θ(π)  and Θ′(− π) = Θ′(π)  . From first B.C. we obtain

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Looking at second B.C. Θ ′(− π) = Θ ′(π )

          √ --   √--   √ --    √ --
Θ′(θ) = − A μsin ( μθ)+   μB cos( μθ)

Hence

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From (1,2), we see that both are satisfied if

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Hence

Θn = An cos(nθ)+ Bnsin(nθ)

There is another solution for μ = 0  which is constant (that is why one of the sums below starts from n = 0  ). We can combine the zero eigenvalue with the above and write

Θn  = Ancos(nθ)+ Bn sin (nθ)    n = 0,1,2,3,...

Since at n = 0  the above reduces to constant A0  .

Now that we know       2
μn = n  , from solving the θ  part, we go and solve the r  ODE. For each n  , the solution to the r  (Bessel) ode

          (     2)
R′′ + 1R′ + λ− n-  R = 0
     r         r2

The solution turns out to be  

           ( ∘---- )
Rnm (r) = Jn  λnmr      m = 1,2,3,⋅⋅⋅

Where λnm  is found from roots of      (√ ----)
0 = Jn  λnma giving the eigenvalues. Now the time ODE is solved

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Hence the solution is

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We now break this sum as follows

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Or

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Then we break the above into 4 sums

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Finally, we merge constants in the above as follows

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Hence the final solution now becomes

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Now initial conditions u(r,θ,0) = f (r,θ)  is used to find Anm,Cnm  using orthogonality. At t = 0  the solution simplifies to (all terms with    ( √----)
sin  c λnmt vanish giving

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Hence

         ∞   ∞                          ∞  ∞
f (r,θ) = ∑  ∑  A   J  (∘ λ--r)cos(nθ)+ ∑   ∑  C   J (∘ λ---r)sin (n θ)
         n=0 m=1  nm n     nm           n=1m=1  nm  n    nm
(4)

When iterating over m  index, the terms cos(nθ)  and sin(nθ)  will be constant. So for each n  , we have ∑ ∞  AnmJn (√ λnmr)
  m=1 and ∑ ∞   CnmJn (√λnmr )
  m=1 . So orthogonality is carried out on the m  index on the Bessel functions, as in ∫a
 0  J (√ λ--r)J  (√λ---′r)σdr
 n    nm   n    nm  . But σ = r  in this problem (from the Bessel ODE above). Multiplying (4) by    (√----′)
Jn   λnm r and integrating

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Or

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Replacing m ′ back with m  , the above becomes

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We now apply orthogonality on m  using the cos  . Hence

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Or

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The ∫     ′       ′
  sin (n θ)cos(nθ)σdθ  term goes to zero, and we are left with (also, changed   ′
n back to n  since we are doing with summation)

∫ π ∫ a        (∘ ---- )                   ∫ π ∫ a 2(∘ ---- )   2
       f (r,θ)Jn   λnmr  cos(nθ)r dθdr = Anm        Jn   λnmr  cos (nθ)rdθdr
 − π 0                                      −π  0

Hence

       ∫  ∫          (√ ----)
       -π−π-0af (r,θ)Jn--λnmr-cos(nθ)r dθdr
Anm  =   ∫ π ∫aJ2 (√λnmr )cos2 (nθ)r dθdr
          −π  0 n

We now need to apply orthogonality on n  using the sin  . to find Cnm  . Using (5)

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Hence

∫ π (∫ a        ( ∘---- )   )                ∫ π ∫ a 2(∘ ---- )  2
 −π   0 f (r,θ)Jn   λnmr  rdr  sin (nθ)dθ = Cnm  −π  0 Jn   λnmr  sin  (nθ)r dθdr

Therefore

       ∫π ∫ a        (√ ----)
       -−π∫-0-f∫ (r,θ)(Jn--λn)mr-sin-(nθ)r dθdr
Cnm  =    −ππ a0 J2n √λnmr  sin2(nθ)r dθdr

Now we will look at the second initial conditions ∂∂ut (r,θ,0) = g(r,θ).  Taking derivative w.r.t. time t  of the solution in (3) gives

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At time t = 0  the above becomes (all terms with    (√ ----)
sin c  λnmt vanish).

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Now orthogonality is used. At t = 0  the above becomes

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Similarly to the above we now find Bnm  and Dnm  . The only difference, is that now we have extra  √ ----
c  λnm  terms that show up. The final result will be

       ∫π ∫ a        (√---- )
B   = -√−π-0∫g(r,∫θ)Jn(√λnmr-)cos(nθ)r dθdr
 nm   c  λnm −ππ a0 J2n  λnmr  cos2 (n θ) r dθdr

And

       ∫π  ∫a        (√ ----)
      --−π-0-g∫(r,∫θ)Jn(--λnmr)sin(nθ)r dθdr
Dnm = c√ λnm  π−π a0 J2n √λnmr  sin2 (nθ)r dθdr

Summary of solution

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       ∫π ∫ a        (√ ----)
Anm  = -−π∫-0π-f∫ (ar,θ)(J√n--λn)mr-cos(nθ)r dθdr
          −π  0 J2n  λnmr  cos2 (nθ)r dθdr

       ∫π ∫ a        (√ ----)
Cnm  = -−π∫-0π-f∫ (ar,θ)(J√n--λn)mr-sin-(nθ)r dθdr
          − π 0 J2n  λnmr  sin2(nθ)r dθdr

       ∫π ∫ a        (√---- )
B   = -√−π-0∫g(r,∫θ)Jn(√λnmr-)cos(nθ)r dθdr
 nm   c  λnm −ππ a0 J2n  λnmr  cos2 (n θ) r dθdr

       ∫π  ∫a        (√ ----)
D   = -√−π-0-g∫(r,∫θ)Jn(√-λnmr)sin(nθ)r dθdr
 nm   c  λnm  π−π a0 J2n  λnmr  sin2 (nθ)r dθdr

With λnm  being the solutions for       (√ ----)
0 = Jn   λnma . For each n  , we find λn,1,λn,2,λn,3,⋅⋅⋅ , which are the zeros of the Bessel Jn (x)  function. So for each n  , we have infinite number of zeros. This generates all the needed λnm  . Hence √----
 λnma  = BesselJZero (n,m)  , therefore √----
 λnm = BesselJaZero(n,m-)

The following animations run for 80 seconds. They are for different n,m  modes. All have zero for initial velocity, which means g(r,θ) = 0  . Radius used is a = 1  and c = 0.2  , initial position used is f (r,θ) = rθ  .

Cases for n = 0

  
  

Cases for n = 1

  
  

Cases for n = 2

  
  

Cases for n = 3

  
  

7.2.2 case circularly symmetric

In this case, there is no θ  dependency in boundary conditions or in initial conditions. Using polar coordinates. Disk has radius a  . Wave displacement is u ≡ u(r,t)  (out of page).

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Boundary conditions on r

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Initial conditions

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Let u = T (t)R (r)  . Plug in the PDE

 1  ′′     ′′   1  ′
c2T  R = R T + rR T

Dividing by RT

   ′′    ′′     ′
1-T--= R--+  1R-
c2T     R    rR

Hence

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The time ODE is

 ′′  2
T + c λT = 0

And the r  ODE is (Sturm-Liouville)

rR ′′ + R ′ +λrR = 0

Where p = r,q = 0,σ = r  . This is singular SL.  The solution turns out to be  

             (∘ ---)
Rn (r) = AnJ0   λnr     n = 1,2,3,⋅⋅⋅

Where λn  is found from roots of       (√---)
0 = Jn  λna giving the eigenvalues. Now the time ODE is solved

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Hence the solution is

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Now initial conditions u(r,0) = f (r)  is used to find An  using orthogonality. At t = 0  the solution simplifies to

         ∞
u (r,0) = ∑  A J (∘ λ-r)
        n=1  n 0    n

Hence

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Now we will look at the second initial conditions ∂u (r,0) = g (r).
∂t  Taking derivative w.r.t. time t  of the solution in (1) gives

∂u       ∑∞    ∘---     ( ∘ ---)  ( ∘---)      ∘ ---  ( ∘ ---)   (∘ ---)
∂t-(r,t) =   − c λnAn sin c  λnt J0   λnr  + Bnc  λn cos  c λnt  J0   λnr
         n=1

At time t = 0  the above becomes

      ∑∞    ∘ ---  (∘ ---)
g(r) =   Bnc  λnJ0    λnr
      n=1

Now orthogonality is used. The above becomes

      ∫ ag(r)J0(√λnr )rdr
Bn  = -0√---∫a-2-(√----)----
      c λn  0 J0  λnr rdr

Summary of solution

        ∑∞       ( ∘ --)   (∘ ---)        ( ∘ ---)  ( ∘ ---)
u(r,t) =   An cos c  λnt J0   λnr  + Bnsin c  λnt J0    λnr
        n=1

     ∫a       (√ ---)
A  = -0∫f (r)J(0√--λn)r-rdr
 n      a0 J20   λnr rdr

      ∫ a      (√--- )
Bn  = -0√-g(r∫)J0-(√λnr-)rdr-
      c λn  a0 J20  λnr rdr

With λn  being the solutions for       (√---)
0 = J0  λna . We have infinite number of zeros. This generates all the needed λn  . Hence √ ---
  λna = BesselJZero (0,n)  , therefore √ ---
  λn = BesselJaZero(0,n)-

This animation runs for 40 seconds. Using radius a = 1  and zero initial velocity. Initial position is u (r,0) = f (r) = r.  And c = 0.2  . Number of terms in the sum used was 30  . Which means                    ( √---)   (√ ---)
u (r,t) = ∑30n=1An cos c λnt J0  λnr . The Bn  terms are all zero since initial velocity g (r)  was zero.