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Analytical solution to specific Stockes first problem PDE

Nasser M. Abbasi

June 12, 2017 compiled on — Monday June 12, 2017 at 01:18 AM

Solve

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Initial conditions

u(0,x ) = 0

Boundary conditions

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Let

u = v +uE
(2)

where uE(x,t)  is steady state solution that only needs to satisfy boundary conditions and v(x,t)  satisfies the PDE itself but with homogenous B.C.  At steady state, the PDE becomes

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The solution is u  (t) = (L−x)sin (t)
 E        L  .  Hence (2) becomes

               ( L − x)
u(x,t) = v(x,t) +  --L--  sin(t)

Substituting the above in (1) gives

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With boundary conditions u0(0,t) = 0,u (L, t) = 0  . This is now in standard form and separation of variables can be used to solve it.

         (     )
Q (x,t) =  x-−-L  cos(t)
            L

Now acts as a source term. The eigenfunctions are known to be            (√---)
Φn (x) = sin  λnx where     (nπ-)2
λn =  L  . Hence by eigenfunction expansion, the solution to (3) is

        ∑∞
v (x,t) =   Bn (t)Φn(x)
        n=1
(3A)

Substituting this into (3) gives

∞∑                  ∞∑
   dBn-(t)Φn (x) = k   Bn (t)Φ′n′(x )+ Q(x,t)
n=1  dt            n=1
(4)

Expanding Q (x,t)  using same basis (eigenfunctions) gives

         ∞
Q (x,t) = ∑ q (t)Φ (x)
         n=1 n    n

Applying orthogonality

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But      ∫                  ∫
∑ ∞n=1 0LΦn (x)Φm (x)dx = 0LΦ2m(x)dx = L2-  since           (    )
Φn (x) = sin nπL x and the above simplifies to

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But Q (x,t) = (x−-L)cos(t)
          L  , hence

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Therefore         ∑ ∞               ∑ ∞   −2        (nπ )
Q (x,t) =  n=1 qn (t)Φn (x) =  n=1 nπ cos(t)sin-L x and (4) becomes

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This is an ODE in Bn (t)  whose solution is

             (    )     2(  2 2        2   )
B  (t) = C e−k n2Lπ22t − 2L--kn-π-cost+-L--sint--
  n      n                nπ(L4 + k2n4π4)

From (3A) v(x,t)  now becomes

         ∞       ( 22)    (    )     2(  2 2       2    )   (    )
v (x,t) = ∑ Cne −k nπL2-tsin  nπ-x − 2L---kn-π-cost+-L--sint-sin nπ-x
        n=1                 L         nπ (L4 + k2n4π4)        L
(5)

To find Cn  , from initial conditions, at t = 0  the above becomes

                          (     )
   ∑∞       (nπ- )   --2L2-kn2π2----  ( nπ-)
0 =    Cnsin  L x −  nπ(L4 +k2n4π4) sin  L x
   n=1

Hence

           (     )
     --2L2--kn2π2---
Cn = n π(L4 + k2n4π4)

Therefore (5) becomes

           (      (     )      (   )       (                 ))
        ∑∞   --2L2-kn2π2----− k nL2π22-t  2L2--kn2π2cost+-L2sint-    ( nπ-)
v(x,t) =     nπ (L4 + k2n4π4)e        −     nπ (L4 + k2n4π4)     sin  L x
        n=1

And since u = v+ uE  then the solution is

        (    (      (     )     (   )       (                 ))          )  (     )
         ∑∞   --2L2--kn2π2--- −k n2Lπ22-t   2L2-kn2π2cost+-L2-sint--    (nπ- )    L-−-x
u(x,t) =      nπ (L4 + k2n4π4)e       −     n π(L4 + k2n4π4)     sin  L x   +    L    sin (t)
         n=1

To simulate

Here is the animation from the above

  

Here is the numerical solution to compare with

Here is the animation from the above

  

Reference: stokes second problem question and answer