1 Summary table

 1.1 Using Mathematica to obtain the eigenvalues and eigenfunctions for heat PDE in 1D
  1.1.1 u(0) = 0,u(L ) = 0
  1.1.2  ′       ′
u(0) = 0,u (L ) = 0
  1.1.3  ′
u(0) = 0,u(L) = 0
  1.1.4          ′
u(0) = 0,u(L) = 0
  1.1.5                ′
u(0) = 0,u(L)+ u (L ) = 0
  1.1.6        ′       ′
u(0)+ u(0) = 0,u (L) = 0
  1.1.7 u(− L) = 0,u (L) = 0

Heat PDE ∂u   ∂2u
∂t = k∂x2   in 1D  (in a rod)






Left side Right side initial condition u(x,0)  λ = 0  λ > 0










u (0) = 0  u(L) = 0  {  x  0 < x < L-
  L− x L< x < 2L
       2  No λn = (nLπ)2,(n√ =-1,2),3,⋅⋅⋅
Xn = Bn s∑in   λnx   (√---)
u(x,t) =  ∞n=1Bn sin   λnx  e− kλnt





u (0) = 0  u(L) = 0  100  No      (  )
λn =  nLπ 2,(n√ =-1,2),3,⋅⋅⋅
Xn = Bn sin   λnx   (√---)
u(x,t) = ∑ ∞n=1Bn sin  λnx  e− kλnt





u (0) = T0  u(L) = 0  x  No λn = (nπ)2,n = 1,2,3,⋅⋅⋅
Xn = BLn sin(√ λnx)
u(x,t) = T0 − T0x + ∑ ∞ Bn sin(√ λnx)e−kλnt
             L      n=1





∂u(0)= 0
 ∂x  ∂u(L)-= 0
 ∂x  x  λ0 = 0
X0 = A0       (nπ)2
λn =  L   ,n( =√-1,2),3,⋅⋅⋅
Xn = An cos  λ∑nx∞       (√ ---) −kλnt
u(x,t) = A0 +  n=1 An cos   λnx e





∂u(0)
 ∂x = 0  u(L) = T0  0  No      (nπ)2
λn =  2L-  ,n( =√-1,3),5,⋅⋅⋅
Xn = An cos ∑λn∞x           (√ ---) −kλ t
u(x,t) = T0 + n=1,3,5,⋅⋅⋅An cos   λnx e   n





∂u(∂0x)= 0  u(L) = 0  f (x)  No      (  )
λn =  n2Lπ 2,n( =√-1,3),5,⋅⋅⋅
Xn = An c∑os  λnx       (√---)
u(x,t) =  ∞n=1,3,5⋅⋅⋅An cos  λnx  e− kλnt





u (0) = 0  ∂u(L)-= 0
 ∂x  f (x)  No λn = (nπ)2,n = 1,3,5,⋅⋅⋅
Xn = B2Ln sin(√ λnx)
u(x,t) = ∑ ∞     Bn sin(√ λnx)e−kλnt
          n=1,3,5⋅⋅⋅





u (0) = 0  u(L)+  ∂u(L)= 0
        ∂x  f (x)  No tan(√λ--L)+ √ λ-= 0
X  = B nsin(√ λ-nx)
un(x,t)n = ∑ ∞  Bn sin (√λ-x) e− kλnt
          n=1  n      n





u (0) + ∂u∂(0x)= 0  u(0) = 0  0 λ  = 0
X0 = A
  0   0     (√ ---)  √ ---
tan   λnL −   λn = 0





u (− 1) = 0  u(1) = 0  f (x)  No √λ--= nπ    n = 1,2,3,⋅⋅⋅
X  n= A 2cos(√λ-x) ,sin(√ λ-x)
un(x,t)n = ∑ ∞   n A  cos(√λnx )e−λnt + ∑ ∞    B  sin(√ λ-x)e−λnt
          n=1,3,⋅⋅⋅  n      n            n=2,4,⋅⋅⋅ n       n





Heat PDE        2
∂∂ut = α ∂∂ux2-− βu  in 1D  (in a rod) with α,β > 0  for 0 < x < π







Left side Right side initial condition λ = 0  λ > 0  analytical solution           u (x,t)












∂u(0,t)
 ∂x  = 0  ∂u(π,t)
 ∂x   = 0  u (x,0) = x  λ0 = 0
X0 = A0  λn = n2,n = 1,2,3,⋅⋅⋅
X (x) = A0 + ∑ ∞n=1 Ancos(nx)  π    ( −βt   )  -2∑ ∞  ((−1)n−1)        −(n2α+β)t
2 + c0 e  − 1 + π   n=1   n2   cos(nx)e






(TO DO) Heat PDE for periodic conditions u (− L) = u(L)  and ∂u(−L)-  ∂u(L)-
 ∂x   =  ∂x

     (n π)2
λn =  -L-  ,n = 1,2,3,⋅⋅⋅

              ◜--------------------λ>◞0◟---------------------◝
        λ◜=◞0◟◝  ∞∑        (∘ ---) −kλnt  ∑∞       (∘ ---) − kλnt
u(x,t) = a0 +    An cos   λnx e     +    Bn sin    λnx e
              n=1                     n=1

1.1 Using Mathematica to obtain the eigenvalues and eigenfunctions for heat PDE in 1D

  1.1.1 u(0) = 0,u(L ) = 0
  1.1.2  ′       ′
u(0) = 0,u (L ) = 0
  1.1.3  ′
u(0) = 0,u(L) = 0
  1.1.4          ′
u(0) = 0,u(L) = 0
  1.1.5                ′
u(0) = 0,u(L)+ u (L ) = 0
  1.1.6        ′       ′
u(0)+ u(0) = 0,u (L) = 0
  1.1.7 u(− L) = 0,u (L) = 0

1.1.1 u(0) = 0,u (L ) = 0

For eigenvalue

{  2   2   2    2     2}
 π-2, 4π2-, 9π2 , 16π2-, 25π2
 L   L   L    L    L

π2n2
-L2--

For eigenfunctions

{   (   )    (    )    (    )     (    )    (    )}
  sin  πx-,sin  2πx- ,sin  3πx- ,sin  4πx- ,sin  5πx-
      L         L         L         L         L

  ( πnx)
sin   L

1.1.2 u′(0) = 0,u′(L ) = 0

For eigenvalue

{    2   2    2    2}
  0, π-, 4π-, 9π-, 16π
    L2  L2  L2   L2

π2(n − 1)2
----2----
   L

For eigenfunctions

{     (   )    (    )     (    )    (    )}
 1,cos πx- ,cos  2πx- ,cos  3πx- ,cos  4πx-
        L         L         L          L

   (π(n − 1)x )
cos ---------
        L

1.1.3  ′
u(0) = 0,u(L) = 0

For eigenvalue

{ π2  9π2 25π2 49π2 81π2  121π2}
 4L2-,4L2,4L2-,-4L2 ,-4L2 ,-4L2

π2(1−-2n)2
   4L2

For eigenfunctions

{cos(πx-),cos( 3πx-) ,cos( 5πx) ,cos(7πx-) ,cos( 9πx) }
     2L        2L        2L         2L        2L

   ( π(1− 2n)x)
cos  ----------
        2L

1.1.4          ′
u(0) = 0,u (L) = 0

Eigenvalue are the same as above

{ π2  9π2 25π2 49π2 81π2  121π2}
 4L2-,4L2,4L2-,-4L2 ,-4L2 ,-4L2

π2(1−-2n)2
   4L2

For eigenfunctions

{    (πx)    ( 3πx)     (5πx )    ( 7πx)    ( 9πx)    ( 11πx) }
  sin  --- ,sin  ---- ,sin  ---- ,sin  ---- ,sin  ---- ,sin  -----
      2L       2L         2L        2L        2L         2L

     (          )
− sin  π(1-−-2n)x-
          2L

1.1.5 u(0) = 0,u (L)+ u′(L ) = 0

For eigenvalue

For eigenfunctions

{sin(2.02876x),sin(4.91318x),sin(7.97867x),sin(11.0855x),sin(14.2074x),sin(17.3364x)}

1.1.6 u(0)+ u′(0) = 0,u′(L ) = 0

For eigenvalue

For eigenfunctions, mathematica only gives plots, so not shown.

1.1.7 u(− L) = 0,u (L ) = 0

For eigenvalue

{                         }
  π2--π2-9π2- 4π2 25π2 9π2
  4L2 ,L2,4L2 ,L2 ,4L2 , L2

 2 2
π-n--
4L2

For eigenfunctions

{   (        )    (        )    (          )    (         )    (         ) }
     π(L-+-x)       π(L+-x)-      3π(L-+-x)       2π-(L-+-x)-      5π(L+-x)-
 sin     2L     ,sin     L     ,sin     2L     ,sin      L      ,sin     2L

  ((       )n   (    )n)
1i  − ie− iπ2xL-  −  ieiπ2Lx
2