IX.3 Finite Fourier Transformvps/ME505/IEM/09 03.pdf · IX.3 Finite Fourier Transform . IX.3.1 Introduction - Finite Integral Transform 756 . Table “Finite Fourier Transform”

Chapter IX The Integral Transform Methods IX.3 Finite Fourier Transform November 26, 2018 755

IX.3 Finite Fourier Transform

IX.3.1 Introduction - Finite Integral Transform 756

Table “Finite Fourier Transform” 760

IX.3.2 Heat Equation in the Finite Layer 762 IX.3.3 Conduction and Advection 768

IX.3.4 Heat Equation in the Sphere 774

IX.3.5 Examples – plug flow over heating ribs 781


IX.3.1 INRODUCTION The integral transforms we studied before (Fourier and Laplace) can be symbolically denoted as a pair of operators

:T u u→ 1 :T − u u→

defined with the help of the integral transform kernel ( ),K xλ

( ) ( ) ( ),D

u u x K x dxλ λ= ∫

( ) ( ) ( )D

u x u K ,x dλ λ λ= ∫

These transforms can be applied for solution of IBVP of classical PDE’s. In general, the procedure consisted of the following steps: 1) application of the appropriate integral transform eliminates the

corresponding derivative from the equation; 2) the transformed equation is solved for the transformed unknown

function; 3) the solution of IBVP is obtained by the inverse integral transform.

Integral Transform over Finite Interval The integral transforms were applied in the infinite and semi-infinite

domains. Now, we want to discover how to solve IBVP for PDE’s in the finite domain [ ]21 L,L∈ξ with the help of the integral transform. To do this we want integral transforms to posses the same property – the ability to eliminate the corresponding derivative from the PDE. Let us investigate this problem. The general form of an integral transform in the Cartesian coordinate system over the finite domain [ ]1 2,x L L∈ can be written as

( ) ( ) ( )2

1

L

L

u u x K ,x dxλ λ= ∫

where ( ),K xλ is the kernel of the integral transform. Apply this transform to the second derivative with respect to x and see how to choose ( ),K xλ (apply integration by parts twice):

( ) ( )2

1

2

2 ,L

L

u xK x dx

xλ

∂

∂ ∫ ( ) ( )2

1

L

L

u xK ,x d

xλ

∂ = ∂

∫

( ) ( ) ( ) ( )2 2

11

, ,L L

LL

u x u xK x K x dx

x x xλ λ

∂ ∂ ∂= − ∂ ∂ ∂

∫

( ) ( ) ( )2 2

11

, ,L L

LL

u xK x K x du

x xλ λ

∂ ∂= − ∂ ∂

∫

( ) ( ) ( ) ( ) ( )2 2 2

1 11

2

2, , ,L L L

L LL

u xK x K x u x u K x dx

x x xλ λ λ

∂ ∂ ∂ = − + ∂ ∂ ∂ ∫

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2

1

L 22 1

2 1 2 2 1 1 2L

u L u LK ,L K ,L K ,L u L K ,L u L u K ,x dx

x x x x xλ λ λ λ λ

∂ ∂ ∂ ∂ ∂ = − − + + ∂ ∂ ∂ ∂ ∂ ∫ ( )◊

x


Obviously, the integral term turns into the transformed function if, in general, the kernel of transform has the property

( ) ( ) ( )2

2 , ,K x c K xx

λ λ λ∂=

∂

This condition reminds the eigenvalue problem.

Auxiliary Sturm-Liouville Problem Let us consider the corresponding auxiliary Sturm-Liouville Problem (SLP) with the parameter λ in the Cartesian coordinate system: ( ) ( )2 0X x X xλ′′ + = ( ) ( ) 0LXhLXk 1111 =+′− (SLP) ( ) ( )2 2 2 2 k X L h X L 0′ + = which according to the Sturm-Liouville Theorem possesses infinitely many values of the parameter nλ , ,...2,1n = (eigenvalues) for which SLP has non-trivial solutions ( )nX x (eigenfunctions). Therefore,

eigenfunctions ( )nX x satisfy boundary conditions: ( ) ( ) 0LXhLXk 1n11n1 =+′− ( ) ( )2 n 2 2 n 2 k X L h X L 0 ′ + =

for all ,...2,1n = . In particular, the general solution of the given equation of SLP is given by

Solution of the SLP ( ) 1 2cos sinnX c x c xξ λ λ= +

In addition, the set of all eigenfunctions ( ){ }nX x , ,...2,1n = is a complete set of orthogonal functions, for which we can introduce the norm by

( )2

1

1 2

2 L

n nL

X X x dx

= ∫

Let the kernel of the integral transform be

Kernel ( ) ( ) ( ), n

n nn

X xK x K x

Xλ= = ( )∗

Differention of the kernel ( )∗ with respect to ξ twice yields

( )nK x′′ ( )n

n

X xX′′

= ( )

n

nn

XX ξλ ′′−

=2

( )2n nK xλ= −

The property which is wanted. From this follows that the last term in the equation ( )◊ becomes

( ) ( ) ( )2 2

1 1

22 2 2

2 , ,L L

n n n n nL L

u K x dx uK x dx u ux

λ λ λ λ λ λ∂= − = − = −

∂∫ ∫

Now, consider homogeneous boundary conditions, which are obviously satisfied by the kernel ( )nK x because it is constructed from the solution of SLP: if 0k,k 21 ≠ then

( ) ( ) 01111 =+′− LKhLKk nn ⇒ ( ) ( )11

11 LK

khLK nn =′

( ) ( ) 02222 =+′ LKhLKk nn ⇒ ( ) ( )22

22 LK

khLK nn

−=′


Substitute the kernel and the derivative of the kernel into the first terms of equation ( )◊ , then

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2

1

L 22 1

2 1 2 2 1 1 2L

u L u LK ,L K ,L K ,L u L K ,L u L u K ,x dx

x x x x xλ λ λ λ λ

∂ ∂ ∂ ∂ ∂ − − + + ∂ ∂ ∂ ∂ ∂ ∫

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 1 22 1 2 2 1 1n n n n n n

u L u LK L K L K L u L K L u L u

x x x xλ

∂ ∂ ∂ ∂= − − + −

∂ ∂ ∂ ∂

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 1 22 12 1 2 2 1 1

2 1n n n n n n

u L u L h hK L K L K L u L K L u L ux x k k

λ∂ ∂ −

= − − + −∂ ∂

0k,k 21 ≠ ( ) ( ) ( ) ( ) ( ) ( )1 2 21 2n 1 1 n 2 2 n n

1 2

u L u Lh hK L u L K L u L ux k x k

λ∂ ∂

= − + + + − ∂ ∂

Therefore, if we choose eigenfunctions ( )nX x (and the corresponding

kernel of the integral transform ( ) ( )nn

n

X xK x

X= ) such that they satisfy

the same homogeneous boundary conditions as of the function ( )u x in the BVP for PDE’s, then both the first two terms in the previous expression become zero. So, we can make the conclusion that in this case, application of the integral transform to the second derivative yields

( )2

1

L 22

n n n2L

u K x dx ux

λ∂= −

∂∫

if the kernel ( )nK x is constructed from the solution of the corresponding SLP with the same homogeneous boundary conditions as for the function

( )u x in BVP for PDE. If conditions for ( )u x are not homogeneous and both are of Robin type, then

( ) ( )1 1 11

1 1

u L h fu Lx k k

∂− + =

∂

( ) ( )2 2 22

2 2

u L h fu L

x k k∂

+ =∂

and the derivative is transformed to

( ) ( ) ( )2

1

L 221 2

n n 1 n 2 n n21 2L

f fu K x du K L K L uk kx

λ∂= + −

∂∫

If 0h,h 21 ≠ then

( ) ( ) 01111 =+′− LKhLKk nn ⇒ ( ) ( )1n1

11n LK

hk

LK ′=

( ) ( ) 02222 =+′ LKhLKk nn ⇒ ( ) ( )2n2

22n LK

hkLK ′−

=

( ) ( ) 21 2n 1 n 2 n n

1 2

f f... K L K L uk k

λ= + − ( ) ( ) ( ) ( ) ( ) ( )1 2 21 2n 1 1 n 2 2 n n

1 2

u L u Lk kK L u L K L u L uh x h x

λ∂ ∂

′ ′= − + − + − ∂ ∂


Finite Fourier Transform over [ ]1 2,L L The integral transform with the desired operational property

nu ( ) ( )2

1

L

nL

u x K x dx= ∫ ( ) ( )2

1

Ln

nL

X xu x dx

X= ∫

( ) ( )2

1

1 L

nn L

u x X x dxX

= ∫

The orthogonal set of eigenfunctions ( ){ }nX x , ,...2,1n = can be applied

for the generalized Fourier series expansion of the function ( )u x

( ) ( )1

n nn

u x a X x∞

=

= ∑

where the Fourier coefficients are

( ) ( )2

1

2

1 L

n nLn

a u x X x dxX

= ∫

Then expansion can be written as

( )u x ( )1

n nn

a X x∞

=

= ∑ ( ) ( ) ( )2

1

21

1 L

n nn Ln

u x X x dx X xX

∞

=

=

∑ ∫

( ) ( ) ( )2

11

Ln n

n n nL

X x X xu x dx

X X

∞

=

=

∑ ∫

( ) ( ) ( )2

11

L

n nn L

u x K x dx K x∞

=

=

∑ ∫

( )1

n nn

u K x∞

=

= ∑

which can be treated as the inverse transform in the form of the series with

the same kernel ( ) ( )nn

n

X xK x

X= .

Therefore, we have the following integral transform pairs:

Finite Fourier Transform over [ ]1 2,L L nu ( ) ( )2

1

L

nL

u K dξ ξ ξ= ∫ integral transform

( )u x ( )n nn 1

u K x∞

=

= ∑ inverse transform

Finite Fourier Transform over [ ]0,L nu ( ) ( )L

n0

u K dξ ξ ξ= ∫ integral transform

( )u x ( )n nn 1

u K x∞

=


Application of the Finite Fourier Transform over [ ]0,L to 2nd derivative:

( ) ( )2

20

L

n

u xK x dx

x ∂ ∂

∫ ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )L 2

n n n n n20

u L u 0K L K 0 K L u L K 0 u 0 u K x dx

x x x x x∂ ∂ ∂ ∂ ∂ = − − + + ∂ ∂ ∂ ∂ ∂

∫

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2n n n n n n

u L u 0K L K 0 K L u L K 0 u 0 u

x x x xλ

∂ ∂ ∂ ∂ = − − + − ∂ ∂ ∂ ∂ ( )◊◊


Finite Fourier Transform nu ( ) ( )L

n0

u x K x dx= ∫ integral transform

( )u x ( )n nn 1

u K x∞

=


Boundary Conditions

[ ]1 1 0x 0k u h u f=

′− + = 11

1

hHk

=

[ ]2 2 Lx Lk u h u f=

′ + = 22

2

hHk

=

Kernel

( ) ( )nn

n

X xK x

X=

( ) ( )nn

n

X xK x

X′

′ =

Operational Property

( )L 2

n20

u K x dxx

∂ ∂

∫

D [ ] 0x 0u f=

=

D [ ] Lx Lu f=

=

nnLπλ = n 1,2,...=

nnX sin xLπ =

n2 nK sin xL L

π =

nn 2 nK cos xL L Lπ π ′ =

( ) ( ) 2

0 n L n n nf K 0 f K L uλ′ ′− −

( )n 1 20 L n n

2 n f 1 f uL L

π λ+ + − −

N [ ] 0x 0u f=

′ =

D [ ] Lx Lu f=

=

n1n2 L

πλ = +

n 0,1,2,...=

n1X cos n x2 L

π = +

n2 1K cos n xL 2 L

π = +

n1 2 1K n sin n x2 L L 2 L

π π ′ = − + +

( ) ( ) 2

0 n L n n nf K 0 f K L uλ′− − −

D [ ] 0x 0u f=

=

N [ ] Lx Lu f=

′ =

n1n2 L

πλ = +

n 0,1,2,...=

n1X sin n x2 L

π = +

n2 1K sin n xL 2 L

π = +

n1 2 1K n cos n x2 L L 2 L

π π ′ = + +

( ) ( ) 2

0 n L n n nf K 0 f K L uλ′ + −

N [ ] 0x 0u f=

′ =

N [ ] Lx Lu f=

′ =

nnLπλ = n 0,1,2,...=

0X 1= , nnX cos xLπ =

01KL

= , n2 nK cos xL L

π =

n 1,2,...=

( ) ( ) 2

0 n L n n nf K 0 f K L uλ− + −

( )

[ ][ ]

n

2

D,N or R

x=0D,N or R

x L

Where X x are the eigenfunctions of the corresponding Sturm-Liouville Problem:X + X 0

X 0

X 0

λ

=

′′ =

=

=


D [ ] 0x 0u f=

=

R [ ]2 2 Lx Lk u h u f=

′ + =

nλ are positive roots of the equation:

( ) ( )2cos L H sin L 0λ λ λ+ =

( )n nX sin xλ=

( )( )n

nn

n

sin xK

sin 2 LL2 4

λ

λλ

=

−

( )( )

n nn

n

n

cos xK

sin 2 LL2 4

λ λ

λλ

′ =

−

( ) ( ) 2L0 n n n n

2

ff K 0 K L uk

λ′ + −

N [ ] 0x 0u f=

′ =

R [ ]2 2 Lx Lk u h u f=

′ + =


( ) ( )2sin L H cos L 0λ λ λ− =

( )n nK cos xλ=

( )( )n

nn

n

cos xK

sin 2 LL2 4

λ

λλ

=

+

( ) ( ) 2L0 n n n n

2

ff K 0 K L uk

λ− + −

R [ ]1 1 0x 0k u h u f=

′− + =

D [ ] Lx Lu f=

=


( ) ( )1cos L H sin L 0λ λ λ+ =

( )n nX sin x Lλ= −

( )( )

nn

n

n

sin x LK

sin 2 LL2 4

λ

λλ

− =

−

( ) ( )nn

n

sin LK 0

Xλ−

=

( )( )

n nn

n

n

cos x LK

sin 2 LL2 4

λ λ

λλ

− ′ =

−

( ) nn

n

K LXλ′ =

( ) ( ) 20n L n n n

1

fK 0 f K L u

kλ′− −

or

( ) ( ) 20n L n n n

1

fK 0 f K L u

kλ′ ′− −

R [ ]1 1 0x 0k u h u f=

′− + =

N [ ] Lx Lu f=

′ =


1sin L H cos L 0λ λ λ− =

( )n nX cos x Lλ= −

( )( )

nn

n

n

cos x LK

sin 2 LL2 4

λ

λλ

− =

+

( ) ( ) 2

0 n L n n nf K 0 f K L uλ+ −

R [ ]1 1 0x 0k u h u f=

′− + =

R [ ]2 2 Lx Lk u h u f=

′ + =


( ) ( ) ( ) ( )21 2 1 2H H sin L H H cos L 0λ λ λ λ− + + =

( ) ( )n n n 1 nX cos x H sin xλ λ λ= +

( ) ( )( )

n n 1 nn

2 2n 1 2 1

2 2n 2

cos x H sin xK

H H HL2 2H

λ λ λ

λ

λ

+=

+ + + +

( ) ( ) 20 Ln n n n

1 2

f fK 0 K L uk k

λ+ −


IX.3.2 Heat Equation in the Finite Layer Consider heat conduction in the 1-dimensional slab with heat generation

Equation: ( ),2

2

1 u u S x tt xα

∂ ∂= +

∂ ∂ [ ]L,0x ∈ 0t >

Initial condition: ( ) ( )xu0,xu 0=

Boundary conditions: ( )1 1 1 1,x 0

uk h u h u tx ∞

=

∂ − + = ∂ ( )0f t= Robin

[ ] ( )Lx Lu u t=

= ( )Lf t= Dirichlet 1) Integral transform According to the table FFT, the kernel of the integral transform

corresponding to Robin-Dirichlet boundary conditions is:

( ) ( )n

nn X

LxsinxK −=

λ , where ( )n

nn 4

L2sin2LX

λλ

−=

and eigenvalues nλ are the positive roots of the equation 0LsinHLcos =+ λλλ

2) Transformed equation According to the table FFT (R-D), the second derivative of ( )t,xu w.r.t. x is transformed to

( )L 2

n20

u K x dxx

∂ ∂

∫( ) ( ) ( ) ( )0 2

n L n n n1

f tK 0 f t K u t

kλ′= − −

( ) ( ) ( ) ( )0 2nn L n n

1 n

f tK 0 f t u t

k Xλ

λ= − −

Transform the source function and initial condition:

( )[ ] ( ) ( )tSdxxKt,xS nn

L

0

=∫

( )[ ] ( ) n,0n

L

00 udxxKxu =∫

Then the transformed equation has the form ( ) ( ) ( ) ( ) ( ) ( )n 0 2n

n L n n n1 n

u t f t1 K 0 f t u t S tt k X

λ λα

∂= − − + ∂

( ) ( ) ( )n 2n n n

u tu t Q t

tαλ

∂= − +

∂

left end is exposed to aconvective enviromentwith time dependenttemperature

( )L

right end is following the prescribed time-dependent temperature f t

( )0u x

( )u x,t

0 Lx

[ ] ( )Lx Lu u t=

=( )1 1 1 1,

x 0

uk h u h u tx ∞

=

∂ − + = ∂


( ) ( ) ( )n 2n n n

u tu t Q t

tαλ

∂+ =

∂

where ( ) ( ) ( ) ( ) ( )0 nn n L n

1 n

f tQ t K 0 f t S t

k Xλα α

= − +

The transformed solution is determined by variation of parameter:

( ) ( ),

2 2 2n n n

tt t

n 0 n n0

u t u e e e Q dαλ αλ αλ τ τ τ− −= + ∫

4) If ( ) nn QtQ ≡ does not depend on time, then the solution reduces to

( ) 2 2na tn n

n 0 ,n2 2 2 2n n

Q Qu t u e

a aλ

λ λ−

= + −

If in addition, the initial condition is zero, ( ) ( ), 0u x 0 u x 0= = , then

( ) ( )2n tn

n 2n

Qu t 1 e αλ

αλ−= −

3) Solution of IBVP The solution can be obtained by the inverse transform

( ) ( ) ( )∑∞

=

=1n

nn xKtut,xu

( ) ( ) ( )2 2 2n n n

tt t

0 ,n n nn 1 0

u x,t u e e e Q d K xαλ αλ αλ τ τ τ∞

− −

=

= +

∑ ∫

4) Steady state solution In the case of constant boundary conditions and with a uniform volumetric heat generation, ( ),S x t q k= , the steady state solution is [Jonathan Stoddard, Section ]: FIT-04.mws

Without heat generation:

( ) ( ) L11

0L1s fLx

Lhkffhxu +−

+−

=

5) Example (FIT-01.mws) ( )S x,t 0=

0u 0=

0f const= Lf const=

( ) 0 nn n L

1 n

fQ K 0 fk X

λα

= −


( )u x,t ( ) ( )2nn t

n2n 1 n

K x1 1 e Qαλ

α λ

∞−

=

= −∑

( ) ( ) 2n t

n nn n2 2

n 1 n 1n n

K x K x e1 1Q Qαλ

α λ α λ

−∞ ∞

= =

= −∑ ∑

Obviously, the steady state solution ( ) ( )s tu x limu x,t

→∞ =

exists only

when the function nQ is time independent. Then, provided that t → ∞ , from the solution for transformed function in 3), we have

( ) ns ,n 2 2

n

Qu t

a λ=

And then by inverse transform

( ) ( )ns n2

n 1 n

K x1u x Qα λ

∞

=

= ∑

This is the Fourier series representation of the steady state solution (4). FIT-01 .mws Heat Equation in the finite layer, Robin - Dirichlet boundary conditions

> restart;with(plots): > Digits:=50;

:= Digits 50

> L:=3;k1:=2;h1:=3;H1:=h1/k1;a:=1; := L 3

:= k1 2

:= h1 3

:= H1 32

:= a 1

> f0:=1;fL:=2; := f0 1

:= fL 2

> u0:=0; := u0 0

> S:=0; := S 0

characteristic equation: > w(x):=x*cos(x*L)+H1*sin(x*L);

:= ( )w x + x ( )cos 3 x 32 ( )sin 3 x

> plot(w(x),x=0..6);

1λ

steady state transient

solution solution


Eigenvalues: > lambda:=array(1..200);

:= λ ( )array , .. 1 200 [ ]

> n:=1: for m from 1 to 50 do y:=fsolve(w(x)=0,x=m/2..(m+1)/2): if type(y,float) then lambda[n]:=y: n:=n+1 fi od: > for i to 3 do lambda[i] od;

0.87171750182652512021379384729076722823460402426027

1.8021775720308982288404308033660197498395520355731

2.7827873192762392106009190946661202624265539349707

> N:=n-1; := N 24

> n:='n':i:='i':m:='m':y:='y':x:='x': Eigenfunctions: > X[n]:=sin(lambda[n]*(x-L));

:= Xn ( )sin λn ( ) − x 3

> NX[n]:=sqrt(L/2-sin(2*lambda[n]*L)/4/lambda[n]);

:= NXn12 − 6

( )sin 6 λn

λn

Kernel: > K[n](x):=X[n]/NX[n];

:= ( )Kn x2 ( )sin λn ( ) − x 3

− 6( )sin 6 λn

λn

> K[n](0):=subs(x=0,K[n](x));

:= ( )Kn 02 ( )sin −3 λn

− 6( )sin 6 λn

λn

Transformed Initial Condition and Source function: > u0t[n]:=int(u0*K[n](x),x=0..L): > St[n]:=int(S*K[n](x),x=0..L): > Q[n]:=K[n](0)*f0/k1-lambda[n]*fL/NX[n]+St[n]: > QQ[n](tau):=subs(t=tau,Q[n]): > ut[n]:=u0t[n]*exp(-a^2*lambda[n]^2*t)+exp(-a^2*lambda[n]^2*t)*int(exp(a^2*lambda[n]^2*tau)*QQ[n](tau),tau=0..t): Solution - Inverse Transform: > u(x,t):=sum(ut[n]*K[n](x),n=1..N): > plot3d(u(x,t),x=0..L,t=0..10,axes=boxed);


Steady State Solution: > us:=(h1*fL-f0)*(x-L)/(k1+h1*L)+fL;

:= us + 5 x11

711

Temperature profiles for different moments of time: > z1:=subs(t=.01,u(x,t)):z2:=subs(t=.2,u(x,t)):z3:=subs(t=1,u(x,t)): > z4:=subs(t=3.0,u(x,t)):z5:=subs(t=10.0,u(x,t)): > plot({us,z1,z2,z3,z4,z5},x=0..L,color=black);

t

x

( )u x,t

( )u x,t

xt 0.01=

t 0.2=

t 1.0=

t 10.0=

t 3.0=

steady state solution


6) Time-dependent boundary condition ( )1 1 1 1,x 0

uk h u h u tx ∞

=

∂ − + = ∂ ( )1 0h sin t f t= = Robin

[ ] Lx Lu u=

= Lf 0= = Dirichlet

( ) ( ),

2 2 2n n n

tt t

n 0 n n0

u t u e e e Q dαλ αλ αλ τ τ τ− −= + ∫

Assume S=0, u0=0 ( ) ( ) ( )sin1n n

1

hQ t K 0 tk

α= ( ) ( )sin1 nH K 0 tα=

( ) ( ) ( )sin2 2n n

tt

n 1 n0

u t e e H K 0 dαλ αλ τα τ τ−= ∫

( ) ( ) ( )sin2 2n n

tt

n 1 n0

u t H K 0 e e dαλ αλ τα τ τ−= ∫

( ) ( ) cos sin2 2n n2

n

t t2t n

n 1 n 2 4n

e t e t 1u t H K 0 e1

αλ αλαλ αλα

α λ−

− + +=

+

( ) ( ) cos sin2n t1 n 2

n n2 4n

H K 0u t t t e

1αλα

αλα λ

− = − + + +

( ) ( ) ( )cos2n t1 n 2 4

n n2 4n

H K 0u t 1 t e

1αλα

α λ ω δα λ

− = + − + +

( ) ( ) ( )∑∞

=

=1n

nn xKtut,xu


IX.3.3. Conduction and Advection Consider the steady state heat transfer in the fluid flowing through the rectangular duct with a constant velocity v. Energy is transferred jointly by conduction and a bulk motion (advection). Fluid enters the duct with initial temperature eT . The walls of the duct are thermostated with the temperature wT . There is a uniformly distributed energy source of strength g. Heat Equation:

kg

xTvaT 22 −=

∂∂

−∇ p2 c 1ak

ρα

= =

kg

xTva

zT

yT

xT 2

2

2

2

2

2

2

−=∂∂

−∂∂

+∂∂

+∂∂ ( )x 0,∈ ∞ ( )y 0,M∈ ( )z 0,L∈

By the change of dependent variable ( ) ( ) wTz,y,xTz,y,xu −= the problem is reduced to the following BVP:

kg

xuva

zu

yu

xu 2

2

2

2

2

2

2

−=∂∂

−∂∂

+∂∂

+∂∂

with boundary conditions:

0x = ( )M,0y ∈ ( )L,0z ∈ 0we uTTu =−= Dirichlet 0y = 0x > ( )L,0z ∈ 0u = Dirichlet My = 0x > ( )L,0z ∈ 0u = Dirichlet 0z = 0x > ( )M,0y ∈ 0u = Dirichlet Lz = 0x > ( )M,0y ∈ 0u = Dirichlet

Integral Transforms: 1) Integral transforms All three space variables are involved in the problem. We need elimination of the partial derivatives with respect to two variables to reduce the equation from partial to ordinary. Two of the variables, y and z, have the finite domain. According to table SLP, the kernel of integral transform corresponding to Dirichlet-Dirichlet boundary conditions is:

Transform in z-variable: The first transform for variable z in the interval [ ]L,0z ∈ :

Eigenvalues L

nn

πλ = Eigenfunctions zsinX nn λ=

Kernel ( ) zsinL2

XXzK n

n

nn λ==

Integral transform pair:

( ) ( ) ( )dzzKz,y,xuy,xu n

L

0n ∫= ( ) ( ) ( )∑

∞

=

=1n

nn zKy,xuz,y,xu

Transform in y-variable: The second transform for variable y in the interval [ ]M,0y ∈ :

Eigenvalues Mm

mπλ = Eigenfunctions ysinX mm µ=

Kernel of integral transform ( ) ysinM2

XX

zK mm

mm µ==


( ) ( ) ( )dyyKy,xuxu m

M

0nm,n ∫= ( ) ( ) ( )∑

∞

=

=1m

mm,nn yKxuy,xu

Reconstruction of the twice transformed function is given by

( ) ( ) ( ) ( )yKyKxuz,y,xu n1m 1n

mm,n∑∑∞

=

∞

=

= where

L

M


( ) ( ) ( ) ( ) ( ) ( ) ( ), , , , ,M L L M

n m n m m n0 0 0 0

u x u x y z K z dz K y dy u x y z K y dy K z dz

= ==

∫ ∫ ∫ ∫

Transformed source and initial conditions:

( )xg m,n ( ) ( ) ( )∫ ∫=M

0nm

L

0

dzdyzKyKk

z,y,xg

( ) ( )∫ ∫=M

0

L

0nm dzdyzKyK

kg

( ) ( )n 1 m 12

g 2 LM 1 1 1 1k nmπ

+ + = + − + − (assume

g cons t= )

m,n,0u ( ) ( )∫ ∫=M

0nm

L

00 dzdyzKyKu

( ) ( ) ( )n 1 m 1e w 2

2 LMT T 1 1 1 1nmπ

+ + = − + − + −

Transformed equation: 2) Transformed equation All boundary conditions in the transformed

domain are homogeneous, therefore, the equation is transformed to

( )2

n,m n,m2 2 2n m n,m n,m2

u ua v u g

xxλ µ

∂ ∂− − + = −

∂∂ m,n,00xm,n uu =

=

General solution of the equation consists of the general solution of the homogeneous part plus a particular solution of the non-homogeneous equation:

( )( ) ( ) ( ) ( )2 22 2 2 2 2 2 2 2

n n n n1 1a v a v 4 x a v a v 4 x2 2 n,m

n,m 1 2 2 2n m

gu x c e c e

λ µ λ µ

λ µ

− + + + + +

= + ++

For the solution to be bounded we have to choose 0c2 = , then with notation

( ) ( )22 2 2 2n,m n n

1 a v a v 42

β λ µ = − + +

The solution becomes

( ) n ,m x n,mn,m 1 2 2

n m

gu x c eβ

λ µ= +

+

The coefficient 1c can be found from the boundary condition, that yields a solution of the transformed equation

solution of transformed equation ( ) 2m

2n

m,nx2m

2n

m,nm,n,0m,n

ge

guxu m,n

µλµλβ

++

+

−=

3) Solution of BVP now can found by the double inverse transform

( )z,y,xu ( ) ( ) ( )n,m m nm 1 n 1

u x K y K z∞ ∞

= =

= ∑ ∑

( ) ( )n ,m xn,m n,m0,n,m m n2 2 2 2

m 1 n 1 n m n m

g gu e K y K zβ

λ µ λ µ

∞ ∞

= =

= − +

+ + ∑ ∑

Then change of the variable yields

Solution: ( ) ( )z,y,xuTz,y,xT w +=

( ) ( )zKyKg

eg

uT n1m 1n

m2m

2n

m,nx2m

2n

m,nm,n,0w

m,n∑∑∞

=

∞

=

++

+−+=

µλµλβ


duct-03.mws Heat Transfer in the 3-D rectangular duct with advection

> restart;with(plots): > Tw:=40;Te:=20;g:=0;k:=0.1;M:=1;L:=2;v:=1;a:=5;dT:=Te-Tw;

:= Tw 40

:= Te 20

:= g 0

:= k 0.1

:= M 1 := L 2

:= v 1

:= a 5

:= dT -20 Eigenvalues: > lambda[n]:=n*Pi/L;mu[m]:=m*Pi/M;

:= λnn π2

:= µm m π

Kernel: > K[m](y):=sin(mu[m]*y)*sqrt(2/M);K[n](z):=sin(lambda[n]*z)*sqrt(2/L);

:= ( )Km y ( )sin m π y 2

:= ( )Kn z

sin n π z

2

> u0[n,m]:=dT*int(int(K[m](y),y=0..M)*K[n](z),z=0..L);

:= u0 ,n m −40 2 ( ) − ( )cos m π 1 ( ) − ( )cos n π 1

m π2 n

> u0[n,m]:=subs({cos(m*Pi)=(-1)^m,cos(n*Pi)=(-1)^n},u0[n,m]);

:= u0 ,n m −40 2 ( ) − ( )-1 m 1 ( ) − ( )-1 n 1

m π2 n

> g[n,m]:=g/k*int(int(K[n](y),y=0..L)*K[m](z),z=0..M); := g ,n m 0.

> beta[n,m]:=(a^2*v-sqrt((a^2*v)^2+4*(lambda[n]^2+mu[m]^2)))/2;

:= β ,n m − 252

+ + 625 n2 π2 4 m2 π2

2

> u[n,m]:=((u0[n,m]-g[n,m]/k/(lambda[n]^2+mu[m]^2))*exp(beta[n,m]*x) +g[n,m]/k/(lambda[n]^2+mu[m]^2))*K[m](y)*K[n](z);

:= u ,n m −80 ( ) − ( )-1 m 1 ( ) − ( )-1 n 1 e

− /25 2

+ + 625 n2 π2 4 m2 π2

2x

( )sin m π y

sin n π z

2m π2 n


SOLUTION:

> T(x,y,z):=Tw+sum(sum(u[n,m],n=1..50),m=1..50): > ux(x):=subs({z=L/2,y=M/2},T(x,y,z)): > plot(ux(x),x=0..10);

> uxy(x,y):=subs({z=L/5},T(x,y,z)): > plot3d(uxy(x,y),x=0..5,y=0..M,axes=boxed);

> uyz0(y,z):=subs({x=0},T(x,y,z)): > uyz1(y,z):=subs({x=1},T(x,y,z)): > uyz2(y,z):=subs({x=2},T(x,y,z)): > plot3d({uyz0(y,z),uyz1(y,z)+15,uyz2(y,z)+30},y=0..M,z=0..L);

> animate3d(T(x,y,z),y=0..M,z=0..L,x=0..4,frames=200);

temperature profile along the central line

eT

eT

eT

eT

temperature profile in the plane z L 5=

wT

wT

x 2=

x 1=

x 0=

for animationsee web site

temperature distributionwith the change of x

temperature distributionwith the change of x




IX.3.4. Heat Equation in the Sphere Consider heat conduction in the solid sphere with angular symmetry. The spatially, non-stationary temperature field ( )t,ru depends only on the radial variable r. The Heat Equation:

( )2

2 22 2

1 ur u atr r

∂ ∂=

∂∂ [ ]1r,0r ∈ 0t >

with initial condition: ( ) ( )ru0,ru 0= and convective boundary condition:

( )1

1

r rr r

uk h u ur ∞ =

=

∂= −

∂

where ∞u is a temperature of the surroundings (generally, a function of time). We can rewrite the boundary condition in the standard form

1r r

huu h ur k k

∞

=

∂ + = ∂

1) Introduce the new dependent variable ( ) ( )t,rrut,rU = , then the HE becomes

t

UarU 22

2

∂∂

=∂∂ [ ]1r,0r ∈ 0t >

which now formally is the 1-d Heat Equation in r in the finite interval [ ]1r,0r ∈ , which requires two boundary conditions. The first condition at 0r = is obtained directly from the equation used for a change of

variable: 0ruU

0r0r==

== Dirichlet

Consider the second boundary condition at 1rr = :

k

huukh

ru

1rr

∞

=

=

+

∂∂

k

hur

Ukh

rU

r1rr

∞

=

=

+∂∂

k

hur

Ukh

rU

rU

r1

1rr2

∞

=

=

+−

∂∂

k

huUr1

kh

r1

rU

r1

1rr

∞

=

=

−+

∂∂

k

rhuUr1

kh

rU 1

rr11

∞

=

=

−+

∂∂

1

h 1Hk r

= − , ( )1

1r

hu rf tk∞=

( )tfHUrU

11

rrr

=

+

∂∂

=

Robin

2) Integral transform Consider the finite Fourier transform corresponding to the case of Dirichlet-Robin boundary conditions:

Eigenvalues nλ are positive roots of the equation: 0rsinHrcos 11 =+ λλλ

Eigenfunctions rsinX nn λ= , ( )n

1n1n 4

r2sin2rX

λλ

−=

The kernel ( )( )

n

1n1

n

n

nn

4r2sin

2r

rsinXXrK

λλ

λ

−==

( ) ( )

reduction to1-d Cartesianformulation by change of variableU r,t ru r,t=

( )U r,t

r 0U 0=

= ( )1

1

rr r

U HU f tr =

∂ + = ∂

1r r

huu h ur k k

∞

=

∂ + = ∂ h,u∞

conveciveenviroment

boundary condition

two boundary conditions



( ) ( ) ( )drrKt,rUtU n

r

0n

1

∫= ( ) ( ) ( )∑∞

=

=1n

nn rKtUt,rU

According to the table FIT, the derivative is transformed to

( )drrKxU

n

r

02

21

∫

∂∂ ( ) ( ) ( ) ( )tUtfrKtf

X n2nr1n0

n

n1

λλ−+=

( ) ( ) ( )tUtfrK n2nr1n 1

λ−=

Transformation of the initial condition:

( ) ( ) ( ) ( ) ( )drrKrrudrrK0,rru0U n

r

00n

r

0n

11

∫∫ ==

3) Transformed equation Applying the integral transform one gets

( ) ( ) ( )t

UatUtfrK n2n

2nr1n 1 ∂

∂=− λ

( ) ( ) ( )2

r1nn2

2nn

atfrK

tUat

U 1=+∂

∂ λ ( )0UU n0tn ==

Solution by variation of parameter:

( )tUn ( ) ( ) ( )∫

−−

+=t

02

r1nat

at

an d

atfrK

eee0U 12

2n

2

2n

2

2n

ττλλλ

( ) ( ) ( )2 2 2n n n

2 2 2

1

tt tn 1a a a

n r20

K rU 0 e e e f d

a

λ λ λτ

τ τ− −

= + ∫

4) Case 1rf const= if

1rf is a constant, then

( )tUn ( ) ( )∫

−−

+=t

0

at

a2

r1nta

n deea

frKe0U 2

2n

2

2n

12

2n

ττλλλ

( ) ( )t

0

at

a2n

2

2r1nt

an

2

2n

2

2n

12

2n

eeaa

frKe0U

+=

−− τλλλ

λ

( ) ( )

−+=

−−

1eefrK

e0Ut

at

a2n

r1nta

n2

2n

2

2n

12

2n λλλ

λ

( ) ( )

−+=

− ta

2n

r1nta

n2

2n

12

2n

e1frK

e0Uλλ

λ

ttanconsf1r

= ( ) ( ) ( ) t

a2n

r1nn2

n

r1n 2

2n

11 efrK

0UfrK λ

λλ

−

−+=

Solution of IBVP can be found by the inverse transform in the form of an

infinite series 1r

f1r

f

( )t,ru ( )r

t,rU= ( ) ( )∑

∞

=

=1n

nn rKtUr1

( ) ( ) ( ) ( )∑∞

=

−

−+=

1nn

ta

2n

r1nn2

n

r1n rKefrK

0UfrK

r1 2

2n

11

λ

λλ


5) Example (turkey-2.mws) Roasting of a turkey (turkey-3.mws smooth solution)

The turkey is assumed to be a sphere of radius 1r 0.1 m= with the uniform initial temperature o

0u 20 C= . It is exposed to the convective environment at ou 150 C∞ = with the convective

coefficient 2

Wh 5 m K

= . The turkey is considered to be done when its minimum temperature reaches

odoneu 75 C= . Thermophysical properties of turkey meat used for calculation are from the table

(Section VIII.1.15 p.580). > restart;with(plots): > r1:=0.1;k:=0.6;h:=5;u0:=20;uinf:=150;a:=2774;

:= r1 0.1

:= k 0.6

:= h 5

:= u0 20

:= uinf 150

:= a 2774

> H[2]:=h/k-1/r1; := H2 -1.666666667

> fL:=h*uinf*r1/k; := fL 125.0000000

Characteristic equation: > w(x):=x*cos(x*r1)+H[2]*sin(x*r1);

:= ( )w x − x ( )cos 0.1 x 1.666666667 ( )sin 0.1 x

> plot(w(x),x=0..50);

Eigenvalues: > lambda:=array(1..200);

:= λ ( )array , .. 1 200 [ ]

> n:=1: for m from 1 to 50 do y:=fsolve(w(x)=0,x=10*m..10*(m+1)): if type(y,float) then lambda[n]:=y: n:=n+1 fi od: > for i to 4 do lambda[i] od;

14.56892774

46.76766900

78.32706546

> N:=n-1; := N 16

> n:='n':i:='i':m:='m':y:='y':x:='x': Eigenfunctions and kernel (See table (5) Dirichlet-Robin): > X[n]:=sin(lambda[n]*r);

:= Xn ( )sin λn r

1λ


> XN[n]:=sqrt(r1/2-sin(2*lambda[n]*r1)/4/lambda[n]);

:= XNn − 0.05000000000 14

( )sin 0.2 λn

λn

> K[n](r):=X[n]/XN[n];

:= ( )Kn r( )sin λn r

− 0.05000000000 14

( )sin 0.2 λn

λn

> K[n](L):=subs(r=r1,K[n](r)): > U0[n]:=u0*int(r*K[n](r),r=0..r1);

:= U0n −200. ( )− + 1. ( )sin 0.1 λn 0.1 λn ( )cos 0.1 λn

−1. ( )− + 5. λn 25. ( )sin 0.2000000000 λn

λnλn

2

> U[n]:=fL*K[n](L)/lambda[n]^2+(U0[n]-fL*K[n](L)/lambda[n]^2)*exp(-lambda[n]^2*t/a^2): Solution: > u(r,t):=sum(U[n]*K[n](r),n=1..N)/r: > u2(r,t):=subs(r=-r,u(r,t)): > t1:=1*60*10:t2:=1*60*60:t2:=3*60*60:t3:=5*60*60:t4:=7*60*60: z1:=subs(t=t1,u2(r,t)):z2:=subs(t=t2,u2(r,t)):z3:=subs(t=t3,u2(r,t)):z4:=subs(t=t4,u2(r,t)): > plot({z1,z2,z3,z4,z5},r=-r1..r1,color=black,axes=boxed);

> with(plots): > animate(u(r,t),r=0..r1,t=0..36000,frames=150);

t 10 min=

t 1 hour=

t 7 hours=

t 5 hours=

( )u r,t




Examples 1. Consider heat conduction in the 2-dimensional cross-section of the long column:

Equation: 2 2

2 2

1 u u ut x yα

∂ ∂ ∂= +

∂ ∂ ∂ ( ) ( ) ( ), , ),x y 0 L M∈ × 0t >

Initial condition: ( ) ( ), 0u x 0 u x 0= =

2. Point heat source – moving or stationary. Impulse point source. 06-03 not moving source - Copy.mws

the temperature of the left side is periodic

right side is insulated( )u x,y,t

0 Lx

x L

u 0x =

∂ = ∂ ( ) ( )u 0,y,t sin t=

( )y M

u x,y,t 1=

= −

( )y M

u x,y,t 1=

=

M


IX.3.5 3. [Adam] Conduction and advection (plug flow)

Equation: 2 2

p n2 2

c vu u u 0x k x y

ρ∂ ∂ ∂− + =

∂ ∂ ∂ ( ) ( ) ( ), , ,x y 0 M∈ −∞ ∞ ×

Boundary conditions: ( ) ( ) ( ),N

s 2k 2k 1k 0

u x 0 q H x x H x xy +

=

∂= ⋅ − − − ∂ ∑

( ),u x M 0y

∂=

∂

Notation p nc va 0

kρ

= >

plug flow

y M is a line of symmetry (insulated)=

( )u x,y 0→

0 x

y M

u 0y =

∂= ∂

nv

L

M

sy 0

u qy =

∂= ∂

y 0

u 0y =

∂= ∂

2L

y 0

u 0y =

∂= ∂

0L 1L 2L 3L 4L 5L

x → −∞

( )u x,y < ∞

x → ∞

x

y 0

u 0y =

∂= ∂


1) 1 step

Equation: 2 2

p n2 2

c vu u u 0x k x y

ρ∂ ∂ ∂− + =

∂ ∂ ∂ ( ) ( ) ( ), , ,x y 0 M∈ −∞ ∞ ×

Boundary conditions: ( ) ( ) ( ), su x 0 q H x H x Ly

∂= − ⋅ − − ∂

(Attention: sq 0− < for heating)

( ),u x M 0y

∂=

∂

1) Finite Fourier Transform in y : ( )nu x ( ) ( )M

n0

u x,y K y dy= ∫ integral transform

( )u x,y ( ) ( )n nn 1

u x K y∞

=


N [ ] 0 sy 0u f q=

′ = = − heating

N [ ] My Mu f 0=

′ = = insulation

N [ ] 0y 0u f=

′ =

N [ ] My Mu f=

′ =

nnMπλ = n 0,1,2,...=

0Y 1= , nnY cos yMπ =

01KM

= , n2 nK cos yM M

π =

n 1,2,...=

( ) ( ) 2

0 n M n n nf K 0 f K M uλ− + −

Transformed equation:

( ) ( )2

p n 2n nn n 0 n2

c vu u u f x K 0 0x k x

ρλ∂ ∂

− − − =∂ ∂

plug flow


( )u x,y 0→

0 x

y M

u 0y =

∂= ∂

nv

L

M

sy 0

u qy =

∂= ∂

y 0

u 0y =

∂= ∂

0L 1L 2L 3L 4L 5L

x → −∞

( )u x,y < ∞

x → ∞

x

y 0

u 0y =

∂= ∂


Consider conjugate problems:

I x 0< 2

p n 2n nn n2

c vu u u 0x k x

ρλ∂ ∂

− − =∂ ∂

I IIn nx 0 x 0

u u= =

=

I II

n n

x 0 x 0

u ux x

= =

∂ ∂=

∂ ∂

II 0 x L< < ( )2

p n 2n nn n s n2

c vu u u q K 0 0x k x

ρλ∂ ∂

− − + =∂ ∂

III x L> 2

p n 2n nn n2

c vu u u 0x k x

ρλ∂ ∂

− − =∂ ∂

II IIIn nx L x L

u u= =

=

II IIIn n

x L x L

u ux x

= =

∂ ∂=

∂ ∂

Plot of Mu x,10


General Case Arbitrary number N of heating ribs

For better performance x-variable has to be normalized xXL

= or ,2 n

xXm L

=⋅

?

( )2k 1L g k x g+ = ∆ + ⋅ ∆ + ∆ , ,...,k 0 1 N 1= −

2k 2k 1L L x−= + ∆ ,...,k 1 N=

2 N 1 2 NL L g L+ = + ∆ ≡

Equation: 2 2

p n2 2

c vu u u 0x k x y

ρ∂ ∂ ∂− + =

∂ ∂ ∂ ( ) ( ) ( ), , ,x y 0 M∈ −∞ ∞ ×

Boundary conditions: I-II-III ( ),u x M 0y

∂=

∂ x−∞ < < ∞

I ( ),u x 0 0y

∂=

∂ x 0−∞ < <

II ( ),u x 0y

∂∂

( ) ( ) ( )

( )F x

N

0 s 2k 1 2kk 1

f x q H x x H x x−=

= = − ⋅ − − − ∑

0 x L< <

III ( ),u x M 0y

∂=

∂ L x< < ∞

plug flow


( )u x,y 0→

0x

y M

u 0y =

∂= ∂

nv

M

sy 0

u qy =

∂= ∂

y 0

u 0y =

∂= ∂

0L 1L 2L N 1L −

x → −∞

( )u x,y < ∞x → ∞

x

y 0

u 0y =

∂= ∂

III

y 0

u 0y =

∂= ∂

III

...NL N 1L L+ =

y 0

u 0y =

∂= ∂ y 0

u 0y =

∂= ∂ s

y 0

u qy =

∂= ∂


1) Finite Fourier Transform in y : ( )nu x ( ) ( )M

n0

u x,y K y dy= ∫ integral transform

( )0u x ( ) ( )M

00

u x, y K y dy= ∫

( )u x,y ( ) ( )n nn 0

u x K y∞

=


N [ ] 0y 0u f=

′ =

N [ ] My Mu f=

′ =

0 0λ = nnMπλ = n 1,2,...=

0Y 1= , nnY cos yMπ =

01KM

= , n2 nK cos yM M

π =

n 1,2,...=

Operational property:

2

2

uFy

∂=

∂

( ) ( ) 20 n M n n nf K 0 f K M uλ= − + −

Transformed equation:

( ) ( )( )n

2p n 2n n

n n 0 n2

g x

c vu uu f x K 0

k xxρ

λ∂ ∂

− − =∂∂

( ) ( ) ( )0 s 0g x q K 0 F x= − ⋅ ⋅

( ) ( ) ( )n s ng x q K 0 F x= − ⋅ ⋅

Consider conjugate problems: n 0,1,2,...=

I x 0< 2

p n 2n nn n2

c vu u u 0x k x

ρλ∂ ∂

− − =∂ ∂

I IIn nx 0 x 0

u u= =

=

I II

n n

x 0 x 0

u ux x

= =

∂ ∂=

∂ ∂

II 0 x L< < ( ) ( )2

p n 2n nn n 0 n2

c vu uu f x K 0 0

k xxρ

λ∂ ∂

− − + =∂∂

III x L> 2

p n 2n nn n2

c vu u u 0x k x

ρλ∂ ∂

− − =∂ ∂

II IIIn nx L x L

u u= =

=

II III

n n

x L x L

u ux x

= =

∂ ∂=

∂ ∂


,

22

1 n na am 02 2

λ = − + <

, ,

22

2 n na am 02 2

λ = + + >

Conjugate conditions:

In x

u 0→−∞

=

I IIn nx 0 x 0

u u= =

=

I II

n n

x 0 x 0

u ux x

= =

∂ ∂=

∂ ∂

I x 0< n 0= 2

0 02

u ua 0x x

∂ ∂− =

∂ ∂

,

I0 1 0u c= ,

ax2 0c e+

,I ax

0 2 0u c e=

( ) ,I ax

0 2 0u ac e′ =

n 0> 2

2n nn n2

u ua u 0x x

λ∂ ∂− − =

∂ ∂

,

In 1 nu c= ,

1 2m x m x2 ne c e+

,2m xI

n 2 nu c e=

( ) ,2m xI

n 2 n 2u c m e′ =

II III

n n

x L x L

u ux x

= =

∂ ∂=

∂ ∂

II IIIn nx L x L

u u= =

=

II 0 x L< < n 0= ( )2

0 002

u ua g x

xx∂ ∂

− =∂∂

( ) ( ), , , ,

II ax ax0 3 0 4 0 1 0 2 0u c 1 c e s x 1 s x e= ⋅ + + ⋅ +

( ) ( ) ( ) ( ), , , ,II ax ax ax

0 4 0 1 0 2 0 2 0u c ae s x 1 s x e s x ae′ ′ ′= + ⋅ + +

ax

axax

1 eW ae

0 ae= =

( ) ( )

axax

1 0ax0

0 eW g x e

g x ae= = − ( )01 g xW

W a−

=

( ) ( )00

1 0W g x

0 g x= = ( )02 g xW e ax

W a= −

( ) ( ) ( ) ( ) ( ) ( ), ,

x x0 s s

1 0 0 0 0 10 0

g x q qs x dx K 0 F d K 0 F x

a a aτ τ

−= = + = + ⋅∫ ∫

( ) ( ) ( ) ( ) ( ) ( ), ,

x x0 ax as s

2 0 0 0 0 20 0

g x q qs x e dx K 0 F e d K 0 F x

a a aττ τ− −= = − = − ⋅∫ ∫


n 0> ( )2

2n nn n s2

u ua u g x

xxλ

∂ ∂− − =

∂∂

( ) ( ), , , ,

, , , ,1 n 2 n 1 n 2 nm x m x m x m xII

n 3 n 4 n 3 n 4 nu c e c e s x e s x e= ⋅ + + ⋅ + ⋅

( ) ( ) ( ) ( ) ( ), , , , , ,, , , , , , , , , ,

1 n 2 n 1 n 2 n 1 n 2 nm x m x m x m x m x m xIIn 3 n 1 n 4 n 2 n 3 n 4 n 3 n 1 n 4 n 2 nu c m e c m e s x e s x e s x m e s x m e′ ′ ′= ⋅ + + ⋅ + ⋅ + ⋅ + ⋅

( ), ,

, ,

, , , ,, ,

1 n 2 n

1 n 2 n

1 n 2 n

m x m xm x m x

2 n 1 nm x m x1 n 2 n

e eW m m e e

m e m e= = − ⋅ ⋅

( )

( ),

,

,,

2 n

2 n

2 n

m xm x

nm xn 2 n

0 eW g x e

g x m e= = − ⋅ ( )

,

, ,

1 nm xn1

2 n 1 n

g xW eW m m

−= − ⋅−

( )

( ),

,

,,

1 n

1 n

1 n

m xm x

nm x1 n n

e 0W g x e

m e g x= = ⋅ ( )

,

, ,

2 nm xn2

2 n 1 n

g xW eW m m

−= + ⋅−

( ) ( ) ( ) ( ) ( ) ( ), ,, ,

, , , , , ,

1 n 1 n

x xm x mn s s

1 n n n n 12 n 1 n 2 n 1 n 2 n 1 n0 0


m m m m m mττ τ− −−

= ⋅ = + = +− − −∫ ∫

( ) ( ) ( ) ( ) ( ) ( ), ,, ,

, , , , , ,

2 n 2 n

x xm x mn s s

2 n n n n 22 n 1 n 2 n 1 n 2 n 1 n0 0


m m m m m mττ τ− −= ⋅ = − = −

− − −∫ ∫

( ),1 ns 0 0=

( ) ( ) ( ), ,, ,

s1 n n n 1

2 n 1 n

qs L K 0 F L

m m= +

−0=

( ) ( ) ( ) ,,

, ,

1 nm xs1 n n

2 n 1 n

qs x K 0 F x e

m m−′ = +

− ( ),1 ns 0 0′ = ( ),1 ns L 0′ =

( ) ( ) ( ) ,,

, ,

2 nm xs2 n n

2 n 1 n

qs x K 0 F x e

m m−′ = −

− ( ),2 0s 0 0′ = ( ),1 ns L 0′ =

III L x< < ∞ n 0= 2

0 02

u ua 0x x

∂ ∂− =

∂ ∂

, ,III ax

0 5 0 6 0u c c e= +

( ) ,III

0 5 0u c′ =

2

2n nn n2

u ua u 0x x

λ∂ ∂− − =

∂ ∂

, ,1 2m x m xIII

n 5 n 6 nu c e c e= +

( ) ,1m xIII

n 5 n 1u c m e′ =


System of equations for coefficients:

I II0 0x 0 x 0

u u= =

= ,2 0c , ,3 0 4 0c c= +

I II

0 0

x 0 x 0

u ux x

= =

∂ ∂=

∂ ∂ ,2 0c a ,4 0c a=

II III

0 0x L x Lu u

= == ( ) ( ), , , , ,

aL aL3 0 4 0 1 0 2 0 5 0c 1 c e s L 1 s L e c⋅ + + ⋅ + =

II III

0 0

x L x L

u ux x

= =

∂ ∂=

∂ ∂ ( ), ,4 0 2 0c s L= −

I IIn nx 0 x 0

u u= =

= , , ,2 n 3 n 4 nc c c= +

I II

n n

x 0 x 0

u ux x

= =

∂ ∂=

∂ ∂ , , , , , ,2 n 2 n 3 n 1 n 4 n 2 nc m c m c m= +

II III

n nx L x Lu u

= == ( ) ( ), , , ,

, , , , ,1 n 2 n 1 n 2 n 1m L m L m L m L m L

3 n 4 n 3 n 4 n 5 nc e c e s L e s L e c e⋅ + + ⋅ + ⋅ =

II III

n n

x L x L

u ux x

= =

∂ ∂=

∂ ∂ ( ) ( ), , , ,

, , , , , , , , , ,1 n 2 n 1 n 2 n 1m L m L m L m L m L

3 n 1 n 4 n 2 n 3 n 1 n 4 n 2 n 5 n 1 nc m e c m e s L m e s L m e c m e⋅ + + ⋅ + ⋅ =

Coefficients: n 0= n 0>

,2 0c ( ),1 0s L= ,2 nc ( ),2 ns L= −

,3 0c 0= ,3 nc 0=

,4 0c ( ),2 0s L= − ,4 nc ( ),2 ns L= −

,5 0c ( ),1 0s L= ,5 nc ( ),1 ns L=


Solution of the transformed equation:

I x 0< n 0= ,I ax

0 2 0u c e=

n 0> ,

2m xIn 2 nu c e=

II 0 x L< < n 0= ( ) ( ), , , ,II ax ax

0 3 0 4 0 1 0 2 0u c 1 c e s x 1 s x e= ⋅ + + ⋅ +

n 0> ( ) ( ), , , ,, , , ,

1 n 2 n 1 n 2 nm x m x m x m xIIn 3 n 4 n 3 n 4 nu c e c e s x e s x e= ⋅ + + ⋅ + ⋅

III L x< < ∞ n 0= ,III

0 5 0u c=

n 0> ,

1m xIIIn 5 nu c e=

Solution:

I x 0< ( )Iu x, y ( ) ( )In n

n 0u x K y

∞

=

= ∑

II 0 x L< < ( )IIu x, y ( ) ( )IIn n

n 0u x K y

∞

=

= ∑

III L x< < ∞ ( )IIIu x, y ( ) ( )IIIn n

n 0u x K y

∞

=

= ∑


Maple code: ADAM-09 I-II-III zones N ribs.mw



Homework:

Equation: 2 2

p n2 2

c vu u u 0x k x y

ρ∂ ∂ ∂− + =

∂ ∂ ∂ ( ) ( ) ( ), , ,x y 0 M∈ −∞ ∞ ×

Boundary conditions: ( ) ( ) ( ) or

,1 2

s 2k 2k 1k 0

u x 0 q H x x H x xy +

=

∂= ⋅ − − − ∂ ∑

( ),u x M 0y

∂=

∂

Transformed equation: ( ) ( )2

p n 2n nn n 0 n2

c vu u u f x K 0 0x k x

ρλ∂ ∂

− − − =∂ ∂

plug flow


( )u x,y 0→

0 x

y M

u 0y =

∂= ∂

nv

L

M

sy 0

u qy =

∂= ∂

y 0

u 0y =

∂= ∂

2L

y 0

u 0y =

∂= ∂

0L 1L 2L 3L 4L 5L

x → −∞

( )u x,y < ∞

x → ∞

x

y 0

u 0y =

∂= ∂


22

1 na am 02 2

λ = − + <

, 2

22 n

a am 02 2

λ = + + >

In x

u 0→−∞

=

I x 0< 2

0 02

u ua 0x x

∂ ∂− =

∂ ∂ I II

n nx 0 x 0u u

= ==

2

2n nn n2

u ua u 0x x

λ∂ ∂− − =

∂ ∂

I IIn n

x 0 x 0

u ux x

= =

∂ ∂=

∂ ∂

,

I0 1 0u c= ,

ax2 0c e+

,I

n 1 nu c= ,1 2m x m x

2 ne c e+

,

I ax0 2 0u c e=

,2m xI

n 2 nu c e=

( ) ,I ax

0 2 0u ac e′ =

( ) ,2m xI

n 2 n 2u c m e′ =

II 0 10 L x L= < < ( )2

0 0s 02

u ua q K 0 0x x

∂ ∂− + =

∂ ∂

1 1

II IIIn nx L x L

u u= =

=

( )2

2n nn n s n2

u ua u q K 0 0x x

λ∂ ∂− − + =

∂ ∂

1 1

II IIIn n

x L x L

u ux x

= =

∂ ∂=

∂ ∂

( ), ,

s 0II ax0 3 0 4 0

q K 0u c c e x

a= + +

( ), ,

1 2m x m x s nIIn 3 n 4 n 2

n

q K 0u c e c e

λ= + +

( ) ( ),

s 0II ax0 4 0

q K 0u ac e

a′ = +

( ) , ,1 2m x m xII

n 3 n 1 4 n 2u c m e c m e′ = +

III 1 2L x L< < 2

0 02

u ua 0x x

∂ ∂− =

∂ ∂

2 2

III IVn nx L x L

u u= =

=

22n nn n2

u ua u 0x x

λ∂ ∂− − =

∂ ∂

2 2

III IVn n

x L x L

u ux x

= =

∂ ∂=

∂ ∂

, ,III ax

0 5 0 6 0u c c e= +

, ,1 2m x m xIII

n 5 n 6 nu c e c e= +


( ) ,III ax

0 6 0u c ae′ =

( ) , ,1 2m x m xIII

n 5 n 1 6 n 2u c m e c m e′ = +

IV 2 3L x L< < ( )2

0 0s 02

u ua q K 0 0x x

∂ ∂− + =

∂ ∂

3 3

IV Vn nx L x L

u u= =

=

( )2

2n nn n s n2

u ua u q K 0 0x x

λ∂ ∂− − + =

∂ ∂

3 3

IV Vn n

x L x L

u ux x

= =

∂ ∂=

∂ ∂

( ), ,

s 0IV ax0 7 0 8 0

q K 0u c c e x

a= + +

( ), ,

1 2m x m x s nIVn 7 n 8 n 2

n

q K 0u c e c e

λ= + +

( ) ( ),

s 0IV ax0 8 0

q K 0u c ae

a′ = +

( ) , ,1 2m x m xIV

n 7 n 1 8 n 2u c m e c m e′ = +

V x L> 2

0 02

u ua 0x x

∂ ∂− =

∂ ∂

4 4

V VIn nx L x L

u u= =

=

22n nn n2

u ua u 0x x

λ∂ ∂− − =

∂ ∂

4 4

V VIn n

x L x L

u ux x

= =

∂ ∂=

∂ ∂

, ,V ax0 9 0 10 0u c c e= +

, ,1 2m x m xV

n 9 n 10 nu c e c e= +

( ) ,V ax0 10 0u c ae′ =

( ) , ,1 2m x m xV

n 9 n 1 10 n 2u c m e c m e′ = +

VI 0 10 L x L= < < ( )2

0 0s 02

u ua q K 0 0x x

∂ ∂− + =

∂ ∂

5 5

VI VIIn nx L x L

u u= =

=

( )2

2n nn n s n2

u ua u q K 0 0x x

λ∂ ∂− − + =

∂ ∂

5 5

VI VIIn n

x L x L

u ux x

= =

∂ ∂=

∂ ∂

( ), ,

s 0VI ax0 11 0 12 0

q K 0u c c e x

a= + +

( ), ,

1 2m x m x s nVIn 11 n 12 n 2

n

q K 0u c e c e

λ= + +

( ) ( ),

s 0VI ax0 12 0

q K 0u c ae

a′ = +

( ) , ,1 2m x m xVI

n 11 n 1 12 n 2u c m e c m e′ = +


VII x L> 2

0 02

u ua 0x x

∂ ∂− =

∂ ∂

5

VIIn x L

u>

< ∞

2

2n nn n2

u ua u 0x x

λ∂ ∂− − =

∂ ∂

, ,

VII ax0 13 0 14 0u c c e= +

, ,1 2m x m xVII

n 13 n 14 nu c e c e= +

, ,

VII0 13 0 14 0u c c= + axe

, ,1m xVII

n 13 n 14 nu c e c= + 2m xe

,

VII0 13 0u c=

,1m xVII

n 13 nu c e=

( )VII0u 0′ =

( ) ,1m xVII

n 13 n 1u c m e′ =

n=0

L0 ,I ax

0 2 0u c e= ( ), ,

s 0II ax0 3 0 4 0

q K 0u c c e x

a= + +

( ) ,I ax

0 2 0u ac e′ = ( ) ( ),

s 0II ax0 4 0

q K 0u ac e

a′ = +

L1 ( ), ,

s 0II ax0 3 0 4 0

q K 0u c c e x

a= + + , ,

III ax0 5 0 6 0u c c e= +

( ) ( ),

s 0II ax0 4 0

q K 0u ac e

a′ = + ( ) ,

III ax0 6 0u c ae′ =

L2 , ,III ax

0 5 0 6 0u c c e= + ( ), ,

s 0IV ax0 7 0 8 0

q K 0u c c e x

a= + +

( ) ,III ax

0 6 0u c ae′ = ( ) ( ),

s 0IV ax0 8 0

q K 0u c ae

a′ = +

L3 ( ), ,

s 0IV ax0 7 0 8 0

q K 0u c c e x

a= + + , ,

V ax0 9 0 10 0u c c e= +

( ) ( ),

s 0IV ax0 8 0

q K 0u c ae

a′ = + ( ) ,

V ax0 10 0u c ae′ =

L4 , ,V ax0 9 0 10 0u c c e= + ( )

, ,s 0VI ax

0 11 0 12 0

q K 0u c c e x

a= + +

( ) ,V ax0 10 0u c ae′ = ( ) ( )

,s 0VI ax

0 12 0

q K 0u c ae

a′ = +

L5 ( ), ,

s 0VI ax0 11 0 12 0

q K 0u c c e x

a= + + ,

VII0 13 0u c=


( ) ( ),

s 0VI ax0 12 0

q K 0u c ae

a′ = + ( )VII

0u 0′ =

n=0 L0 ,2 0c , ,3 0 4 0c c= +

,2 0ac ( ),

s 04 0

q K 0ac

a= +

L1 ( ), ,

1aL s 03 0 4 0 1

q K 0c c e L

a+ + , ,

1aL5 0 6 0c c e= +

( ),

1aL s 04 0

q K 0ac e

a+ ,

1aL6 0c ae=

L2 , ,2aL

5 0 6 0c c e+ ( ), ,

2aL s 07 0 8 0 2

q K 0c c e L

a= + +

,2aL

6 0c ae ( ),

2aL s 08 0

q K 0c ae

a= +

L3 ( ), ,

3aL s 07 0 8 0 3

q K 0c c e L

a+ + , ,

3aL9 0 10 0c c e= +

( ),

3aL s 08 0

q K 0c ae

a+ ,

3aL10 0c ae=

L4 , ,4aL

9 0 10 0c c e+ ( ), ,

4aL s 011 0 12 0 4

q K 0c c e L

a= + +

,4aL

10 0c ae ( ),

4aL s 012 0

q K 0c ae

a= +

L5 ( ), ,

5aL s 011 0 12 0 5

q K 0c c e L

a+ + ,13 0c=

( ),

5aL s 012 0

q K 0c ae

a+ 0=


n>0

L0 ,2m xI

n 2 nu c e= ( ), ,


n

q K 0u c e c e

λ= + +

( ) ,2m xI

n 2 n 2u c m e′ =

L1 ( ), ,


n

q K 0u c e c e

λ= + + , ,

1 2m x m xIIIn 5 n 6 nu c e c e= +

( ) , ,1 2m x m xII

n 1 n 1 2 n 2u c m e c m e′ = + ( ) , ,1 2m x m xIII

n 5 n 1 6 n 2u c m e c m e′ = +

L2 , ,1 2m x m xIII

n 5 n 6 nu c e c e= + ( ), ,


n

q K 0u c e c e

λ= + +

( ) , ,1 2m x m xIII

n 5 n 1 6 n 2u c m e c m e′ = + ( ) , ,1 2m x m xIV

n 7 n 1 8 n 2u c m e c m e′ = +

L3 ( ), ,


n

q K 0u c e c e

λ= + + , ,

1 2m x m xVn 9 n 10 nu c e c e= +

( ) , ,1 2m x m xIV

n 7 n 1 8 n 2u c m e c m e′ = + ( ) , ,1 2m x m xV

n 9 n 1 10 n 2u c m e c m e′ = +

L4 , ,1 2m x m xV

n 9 n 10 nu c e c e= + ( ), ,

1 2m x m x s nVIn 11 n 12 n 2

n

q K 0u c e c e

λ= + +

( ) , ,1 2m x m xV

n 9 n 1 10 n 2u c m e c m e′ = + ( ) , ,1 2m x m xVI

n 11 n 1 12 n 2u c m e c m e′ = +

L5 ( ), ,

1 2m x m x s nVIn 11 n 12 n 2

n

q K 0u c e c e

λ= + + ,

1m xVIIn 13 nu c e=

( ) , ,1 2m x m xVI

n 11 n 1 12 n 2u c m e c m e′ = + ( ) ,1m xVII

n 13 n 1u c m e′ =


n>0

L0 ,2 nc ( ), ,

s n3 n 4 n 2

n

q K 0c c

λ= + +

,2 n 2c m , ,3 n 1 4 n 2c m c m= +

L1 ( ), ,

1 1 2 1m L m L s n3 n 4 n 2

n

q K 0c e c e

λ+ + , ,

1 1 2 1m L m L5 n 6 nc e c e= +

, ,1 1 2 1m L m L

3 n 1 4 n 2c m e c m e+ , ,1 1 2 1m L m L

5 n 1 6 n 2c m e c m e= +

L2 , ,1 2 2 2m L m LIII

n 5 n 6 nu c e c e= + ( ), ,

1 2 2 2m L m L s n7 n 8 n 2

n

q K 0c e c e

λ= + +

, ,1 2 2 2m L m L

5 n 1 6 n 2c m e c m e+ , ,1 2 2 2m L m L

7 n 1 8 n 2c m e c m e= +

L3 ( ), ,

1 3 2 3m L m L s n7 n 8 n 2

n

q K 0c e c e

λ+ + , ,

1 3 2 3m L m L9 n 10 nc e c e= +

, ,1 3 2 3m L m L

7 n 1 8 n 2c m e c m e+ , ,1 3 2 3m L m L

9 n 1 10 n 2c m e c m e= +

L4 , ,1 4 2 4m L m L

9 n 10 nc e c e+ ( ), ,

1 4 2 4m L m L s n11 n 12 n 2

n

q K 0c e c e

λ= + +

, ,

1 4 2 4m L m L9 n 1 10 n 2c m e c m e+ , ,

1 4 2 4m L m L11 n 1 12 n 2c m e c m e= +

L5 ( ), ,

1 5 2 5m L m L s nVIn 11 n 12 n 2

n

q K 0u c e c e

λ= + + ,

1 5m L13 nc e=

, ,1 5 2 5m L m L

11 n 1 12 n 2c m e c m e+ ,1 5m L

13 n 1c m e=


22

1 na am 02 2

λ = − + <

, 2

22 n

a am 02 2

λ = + + >

In x

u 0→−∞

=

I x 0< 2

0 02

u ua 0x x

∂ ∂− =

∂ ∂ I II

n nx 0 x 0u u

= ==

2

2n nn n2

u ua u 0x x

λ∂ ∂− − =

∂ ∂

I IIn n

x 0 x 0

u ux x

= =

∂ ∂=

∂ ∂

,

I0 1 0u c= ,

ax2 0c e+

,I

n 1 nu c= ,1 2m x m x

2 ne c e+

,

I ax0 2 0u c e=

,2m xI

n 2 nu c e=

( ) ,I ax

0 2 0u ac e′ =

( ) ,2m xI

n 2 n 2u c m e′ =

II 0 10 L x L= < < ( )2

0 0s 02

u ua q K 0 0x x

∂ ∂− + =

∂ ∂

1 1

II IIIn nx L x L

u u= =

=

( )2

2n nn n s n2

u ua u q K 0 0x x

λ∂ ∂− − + =

∂ ∂

1 1

II IIIn n

x L x L

u ux x

= =

∂ ∂=

∂ ∂

( ), ,

s 0II ax0 3 0 4 0

q K 0u c c e x

a= + +

( ), ,


n

q K 0u c e c e

λ= + +

( ) ( ),

s 0II ax0 4 0

q K 0u ac e

a′ = +

( ) , ,1 2m x m xII

n 3 n 1 4 n 2u c m e c m e′ = +

III 1 2L x L< < 2

0 02

u ua 0x x

∂ ∂− =

∂ ∂

2 2

III IVn nx L x L

u u= =

=

22n nn n2

u ua u 0x x

λ∂ ∂− − =

∂ ∂

2 2

III IVn n

x L x L

u ux x

= =

∂ ∂=

∂ ∂

, ,III ax

0 5 0 6 0u c c e= +

, ,1 2m x m xIII

n 5 n 6 nu c e c e= +


( ) ,III ax

0 6 0u c ae′ =

( ) , ,1 2m x m xIII

n 5 n 1 6 n 2u c m e c m e′ = +

IV 2 3L x L< < ( )2

0 0s 02

u ua q K 0 0x x

∂ ∂− + =

∂ ∂

3 3

IV Vn nx L x L

u u= =

=

( )2

2n nn n s n2

u ua u q K 0 0x x

λ∂ ∂− − + =

∂ ∂

3 3

IV Vn n

x L x L

u ux x

= =

∂ ∂=

∂ ∂

( ), ,

s 0IV ax0 7 0 8 0

q K 0u c c e x

a= + +

( ), ,


n

q K 0u c e c e

λ= + +

( ) ( ),

s 0IV ax0 8 0

q K 0u c ae

a′ = +

( ) , ,1 2m x m xIV

n 7 n 1 8 n 2u c m e c m e′ = +

V x L> 2

0 02

u ua 0x x

∂ ∂− =

∂ ∂

4 4

V VIn nx L x L

u u= =

=

22n nn n2

u ua u 0x x

λ∂ ∂− − =

∂ ∂

4 4

V VIn n

x L x L

u ux x

= =

∂ ∂=

∂ ∂

, ,V ax0 9 0 10 0u c c e= +

, ,1 2m x m xV

n 9 n 10 nu c e c e= +

( ) ,V ax0 10 0u c ae′ =

( ) , ,1 2m x m xV

n 9 n 1 10 n 2u c m e c m e′ = +

VI 0 10 L x L= < < ( )2

0 0s 02

u ua q K 0 0x x

∂ ∂− + =

∂ ∂

5 5

VI VIIn nx L x L

u u= =

=

( )2

2n nn n s n2

u ua u q K 0 0x x

λ∂ ∂− − + =

∂ ∂

5 5

VI VIIn n

x L x L

u ux x

= =

∂ ∂=

∂ ∂

( ), ,

s 0VI ax0 11 0 12 0

q K 0u c c e x

a= + +

( ), ,

1 2m x m x s nVIn 11 n 12 n 2

n

q K 0u c e c e

λ= + +

( ) ( ),

s 0VI ax0 12 0

q K 0u c ae

a′ = +

( ) , ,1 2m x m xVI

n 11 n 1 12 n 2u c m e c m e′ = +


VII x L> 2

0 02

u ua 0x x

∂ ∂− =

∂ ∂

5

VIIn x L

u>

< ∞

2

2n nn n2

u ua u 0x x

λ∂ ∂− − =

∂ ∂

, ,

VII ax0 13 0 14 0u c c e= +

, ,1 2m x m xVII

n 13 n 14 nu c e c e= +

, ,VII0 13 0 14 0u c c= + axe

, ,1m xVII

n 13 n 14 nu c e c= + 2m xe

,

VII0 13 0u c=

,1m xVII

n 13 nu c e=

( )VII0u 0′ =

( ) ,1m xVII

n 13 n 1u c m e′ =

n=0

L0 ,I ax

0 2 0u c e= ( ), ,

s 0II ax0 3 0 4 0

q K 0u c c e x

a= + +

( ) ,I ax

0 2 0u ac e′ = ( ) ( ),

s 0II ax0 4 0

q K 0u ac e

a′ = +

L1 ( ), ,

s 0II ax0 3 0 4 0

q K 0u c c e x

a= + + , ,

III ax0 5 0 6 0u c c e= +

( ) ( ),

s 0II ax0 4 0

q K 0u ac e

a′ = + ( ) ,

III ax0 6 0u c ae′ =

L2 , ,III ax

0 5 0 6 0u c c e= + ( ), ,

s 0IV ax0 7 0 8 0

q K 0u c c e x

a= + +

( ) ,III ax

0 6 0u c ae′ = ( ) ( ),

s 0IV ax0 8 0

q K 0u c ae

a′ = +

L3 ( ), ,

s 0IV ax0 7 0 8 0

q K 0u c c e x

a= + + , ,

V ax0 9 0 10 0u c c e= +

( ) ( ),

s 0IV ax0 8 0

q K 0u c ae

a′ = + ( ) ,

V ax0 10 0u c ae′ =

L4 , ,V ax0 9 0 10 0u c c e= + ( )

, ,s 0VI ax

0 11 0 12 0

q K 0u c c e x

a= + +

( ) ,V ax0 10 0u c ae′ = ( ) ( )

,s 0VI ax

0 12 0

q K 0u c ae

a′ = +

L5 ( ), ,

s 0VI ax0 11 0 12 0

q K 0u c c e x

a= + + ,

VII0 13 0u c=

( ) ( ),

s 0VI ax0 12 0

q K 0u c ae

a′ = + ( )VII

0u 0′ =


Documents

IX.3 Finite Fourier Transformvps/ME505/IEM/09 03.pdf · IX.3 Finite Fourier Transform . IX.3.1 Introduction - Finite Integral Transform 756 . Table “Finite Fourier Transform”