Lecture 9

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Lecture 9

Partial Differential Equation

So far, we have learned how to deal with an ordinary differential equation or a system of ordinary differential equations that consist of only one independent variable.

Now, what if we have a differential equation that consists of multiple independent variables instead?

Such equation is called partial differential equation, and, like in partial derivative, the symbol  is used to indicate wherever there is differentiation operation instead of the symbol d. Notation such as u_xy are also used to short for



²

u  x  y

.

Many existing engineering phenomena can be described using partial differential equation. Often time, they involve an independent variable of time

t

and other independent variables of distance

x

, y, and/or

z

. For instance,

 Wave equation in three dimensions is given as

 

 







 



 



 



2 2 2 2 2 2 2 2 2

z u y

u x

c u t

u

, (1)

where

u  u  x , y , z , t 

.

 Heat equation in three dimensions is given as

 

 







 



 



 



2 2 2 2 2 2 2

z u y

u x

c u t

u

(2)

where

u  u  x , y , z , t 

 Laplace equation in three dimensions is given as

2

0

2 2 2 2

2





 



 



z u y

u x

u

, (3)

where

u  u  x , y , z 

 Poisson equation in three dimensions is given as

 x y z 

z f u y

u x

u

₂

, ,

2 2 2 2 2

 

 



 



(4)

(2)

where

u  u  x , y , z 

Solutions of partial differential equation are also linear. That is, if

u

₁ and

u

₂ are the solutions of a partial differential equation, then

u  c

₁

u

₁

 c

₂

u

₂ is also the solution of the partial differential equation.

Occasionally, partial differential equation can be solved directly by using one of the methods similar to those used for solving ordinary differential equation. For instance,

 Solve

u

xy

 u

x.

Let u_x  p, then p_y  p. So,

p

y

/ p  1  ln p  y  c ~   x  p  c   x e

^y. Consequently,

 

^x ^dx ^d

 

^y

c e

u  ^y





Notice that whenever we performed integration,

o if we integrated with respect to

x

, then a function of y would be added in place of the constant of integration;

o if we integrated with respect to y, then a function of

x

would be added in place of the constant of integration.

Method of Separation of Variables

This is one of most important methods which are generally used to solve partial differential equation that consists of two independent variables.

It assumes that the solution is a product of two separate functions – one function is dependent of one of the independent variables, and the other function is dependent of the other independent variables.

That is, if the solution is

u   x , y

, then

  x y F     x G y

u , 

. (5)

Suppose the differential equation is

y u x u



 



. (6)

We find that

   

G y x

x F x u







 and

   

y y x G y F

u



 



. (7a,b)

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Then,

       

t y x G F y x G

x F



 



 , (8a)

   

t y G y G x

x F x

F 

 



 1

1

. (8b)

As we can see, the left hand side is only dependent of

x

while the right hand side is only dependent of y. The only way that this can happen is when both the left and the right hand sides are equal the same constant.

So, we can set

   

x k x F x

F 



1

and

   

t k y G y

G 



1

, (9a,b)

and solve each equation individually.

This is the essence of the method of separation of variables. To understand this method, let’s consider some of the equations mentioned at the beginning.

One-Dimensional Wave Equation

Let’s look at one-dimensional wave equation

 

 







 



2 2 2 2 2

x c u t

u

, (10)

where

u  u   x , t

. The simplest example of this type of equation is the vibration of a string.

Since this is a partial differential equation with two independent variables, we can apply the method of separation of variable. Let

  x t F     x G t

u , 

. (11)

Then,

    G x x

x F x

u

2 2 2 2



 



and

   

2 2 2

2

t t x G t F

u



 



. (12a,b)

Substituting (12) into (10) yields

        G t x

x c F

t t x G

F

₂

2 2 2 2



 



. (13)

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Consequently,

   

2 2 2

2 2

1 1

x x F x F t

t G t G

c 

 



. (14)

Set each side of (14) to be equal to a constant k. Then, we find that

        0

1

2 2 2

2



 

 

 

 kF x

x x k F

x x F x

F

^, ^(15a)

        0

1

2

2 2 2

2

 



 

 

 c kG t

t t k G

t t G t G

c

^. ^(15b)

If k is positive, i.e.

k  p

², by solving (15) we get

 

x Ae^px Be ^px

F   ^ , (16a)

 

t Ce^cpt De ^cpt

G   ^ . (16b)

However, if k is negative, i.e.

k   p

², by solving (15) we get

  x A   px B   px

F  cos  sin

, (17a)

  t C   cpt D   cpt

G  cos  sin

. (17b)

If k is zero, by solving (15) we get

  x Ax B

F  

(18a)

  t Ct D

G  

(18b)

To find out whether k is positive or negative, and to find constants A,B,C,D, we need two types of conditions:

 Boundary conditions – conditions on distance, e.g. values of

u   0 , t

and

u   L , t

. For the wave equation, we need boundary conditions because it involves /x. In addition, we need 2 boundary conditions because the derivative is second order.

 Initial conditions – conditions on time, values of

u   x , 0

and

 u   x , 0 /  t

. For the wave equation, we need indition conditions because it involves /t. In addition, we need 2 initial conditions because the derivative is second order.



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So for this case, let the boundary conditions be

  0 , t  0

u

and

u   L , t  0

. (19a,b)

Let the initial conditions be

  x f   x

u , 0 

and

 u   x , 0 /  t  g   x

. (20a,b)

Consequently, from (19),

  0  0

F

and

F   L  0

. (21a,b)

Of course, one could argue that there is always the case that

G   t  0

which would also satisfy (19).

However, such case would lead to trivial solution (u0) which is not what we are looking for. Thus, we shall reject this conclusion.

We shall see that

 if k is positive, (16a) cannot satisfy (21) unless both A and B are zero;

 if k is zero, (18a) can also not satisfy (21) unless both A and B are zero;

 However, such problem would not arise if k is negative.

Thus,

F   x

and

G   t

have the form shown in (17). Consequently, by applying (21) into (17a), we get

  0  A cos   0  B sin   0  A  0

F

, (22a)

  L  B sin   pL  0

F

. (22b)

We shall reject that B0 in (22b). Thus,

 

L

p n n

pL

pL 0 



 



sin , (23)

for

n  1 , 2 , 3 ,...

. Consequently,

   

x F

 

x

L px n

x

F  _n



 



 





sin

sin , (24)

where we set B1 for simplicity and because it does not vary with

n

. The subscript

n

indicates that the value is associated to

n

. It is important to note here that there are infinite solutions of

F   x

. The result in (23) can be used in (17b) and get

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 





 



 



 



  t

L D cn

L t C cn

t

G_n _n



_n



sin

cos . (25)

Ultimately,

      



 



 



 



 



 



 

 

 



 

 x

L t n

L D cn

L t C cn

t G x F t x

u

_n _n _n _n



_n

 

sin sin

cos

,

. (25)

From linearity theorem, we can conclude that

  



^







 

 



 



 



 



 



 

 

 



 



1 1

sin sin

cos ,

n

n n

n

x

L t n

L D cn

L t C cn

t x u

u   

(26)

To find

C

_n and

D

_n, we shall use the initial conditions in (20). Thus,

       

^x ^f

 

^x

L C n

L x D n

C x

u

n n n

n

n 



 



 



 



 



 

^





1 1

sin sin

0 sin 0

cos 0

,

 

. (27)

We shall see in the future lecture about Fourier series that (27) can be solved and gives

  

^_^ ^_^

 ^L

n x dx

L x n

L f

C 2 0 sin



. (28)

Since



^



 

 



 



 



 



 



 

 

 



 

 



1

sin cos

sin

n

x

L t n

L cn L

D cn L t

cn L

C cn t

u     

, (29)

then,

  

^

   



 

 



 

 

  

 



1

sin 0 cos 0

0 sin ,

n

x

L n L

D cn L

C cn t

x

u   

 

x g L x

n L

D cn

n

n 



 







^ 

1

sin



(30)

Fourier series analysis suggests that

  

^_^ ^_^

 ^L

n x dx

L x n

cn g

D 2 0 sin



^. ⁽³¹⁾

Each value of cn



/L is often called eigenvalue or characteristic value. Its corresponding function

 

x t

u_n , is called eigenfunction or characteristic function.

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Each eigenfunction is essentially a harmonic of the vibration, and the eigenvalue is the frequency (in radian) of the vibration.

The fundamental mode occurs when n1. For n1, the modes are called overtunes.

To understand physical interpretation of the solution, let’s look at (26) again,

  



^







 

 



 



 



 



 



 

 

 



 



1 1

sin sin

cos ,

n

n n

n

x

L t n

L D cn

L t C cn

t x u

u   

(26)

We can expand it as



^







 



 



 



 



 



 

 

 



 



 



 

1

sin sin

sin cos

n

x

L t n

L D cn

L x t n

L C cn

u    

(32)

Now consider the case that the initial velocity or change in

u

is zero, i.e.

g   x  0

. Then, the second term in (32) disappears. Using trigonometric identities, we can rewrite the first term in (32) as

   



^



  

 

 



 



 

 



 



 



1

sin 2 sin

n

x ct

L ct n

L x n

u C  

. (33)

The term

sin  n   x  ct  / L 

is basically a sine wave that shifts its center to the right by

ct

unit. As

t

increases,

ct

increases and the wave shifts even further to the right. In other words, it represents wave travelling to the right.

On the contrary,

sin  n   x  ct  / L 

is a sine wave that shifts its center to the left

ct

unit by. It thus represents wave travelling to the left.

Thus, the resulting wave is a standing wave that does neither move to the left nor right, but only vibrate up and down.

[Fig. 259 and Fig. 263, E. Kreyszig]

One-Dimensional Heat Equation

The one-dimensional heat equation goes as

2 2 2

x c u t u



 



, (34)

where

u  u   x , t

.

Let the boundary conditions be

  0 , t  0

u

and

u   L , t  0

; (35a,b)

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and the initial condition be

  x f   x

u , 0 

. (36)

Notice that we only need one initial condition here because (34) only involves first order time- derivative. For the wave equation shown before, we needed to two initial conditions because it involved second order time-derivative.

To solve (34), let

  x t F     x G t

u , 

. (37)

Then,

   

t t x G t F u



 



and

    G t

x x F x

u

2 2 2 2



 



. (38a,b)

Substituting (38) into (37) yields

      G t x

x c F

t x G

F

₂

2 2



 



. (39)

Rearranging (39) gives

     

2 2 2

1 1

x x F x F t G t G

c 

 



. (40)

Once again, the left and the right hand sides of (40) are exclusively dependent of either

t

or

x

. Thus, we can set (40) equal to a constant, says k. Thus,

   

x k x F x

F 



2

1

2

, (41)

which is the same as (15) in the wave equation.

Since the boundary conditions are also the same, we can conclude right away that the result is the same as (24). That is,

   





 



 

 x

L px n

x

F_n



sin

sin , (42)

for n1,2,3,..., and

L p_ n



. (43)

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Equation for

G   t

is, however, different from the previous case. That is

  c p G   t

t p G t k

G t G c

2 2 2

2

1  



 





 



. (44)

The solution for (44) is

 

n ^^cn ^L^^t t

p c n

n

x A e A e

G 

^ ² ²



^ ^^/ ² . (45)

Thus,

     

n ^^cn ^L^^t

n

x e

L A n

x G x F t x

u , sin 

 ^^/ ²

 

 



 



, (46)

and by linearity

  

^ ^ ^



 



 



 

1

/ ²

sin ,

n

t L cn

n x e

L A n

t x

u



^

. (47)

By applying the initial condition in (36), we get

 

x f

 

x

L A n

x u

n

n 



 







^ 

1

sin 0

,



. (48)

Thus, using Fourier series, we find that

  

^_^ ^_^

 ^L

n x dx

L x n

L f

A 2 0 sin



. (49)

Physically, this is simply a sine function whose amplitude continuously decays as time increases.

[Fig. 267, E. Kreyszig]

Steady-State Two-Dimensional Heat Equation The two-dimensional heat equation is

 

 







 



 



2 2 2 2 2

y u x

c u t

u

, (50)

which involves 3 independent variables. However, when the system reaches steady state, the time- derivative becomes zero, i.e.

 0



 t

u

, (51)

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thus the number of independent variables is reduced to 2.

Equation (50) in steady state becomes

2 2 2

2 2

2 2 2

0

2

y u x

u y

u x

c u

^ss ^ss



 

 

 

 

 







 



 

. (52)

where subscript “ss” indicates that

u

is in steady state, and u_ssu_ss

 

x,y . To solve this equation, we need 4 boundary conditions, 2 for /x and 2 for /y, on the boundary as shown in the figure below.

Let’s call the boundary around the shade area C.

The steady-state two-dimensional heat problem has been categorized and given name based upon the type of its boundary condition:

 Dirichlet problem is the problem that

u

is prescribed on C.

 Neumann problem is the problem that u/n is prescribed on C, where

n

is the normal.

 Mixed problem is the problem that

u

is prescribed on some part of C, and u/n is prescribed on the rest of C.

Let’s start with the Dirichlet problem where

 

x,0 u

 

0,y u

 

a,y 0

u_ss _ss _ss and

u

_ss

  x , b  f   x

. (53a,b)

To solve (52), let

 

x y F

   

xG y

u_ss ,  . (54)

Thus,

y y

x

x y

x

(11)

       

2 2 2

2

y y x G F y x G

x F



 

 

 , (55)

and

   

2 2 2

2

1

y y G y G x

x F x

F 

 

 



(56)

Let (56) be equal to k, then

   

x k x F x

F 



2

1

2

(57)

which is the same as before. With similar boundary condition at x0 and

x  a

, we can readily conclude that

   





 



 

 x

a px n

x

F_n



sin

sin , (58)

and k is negative, i.e.

k   p

², and pn



/a Consequently, the other equation gives

   

2

1

2

p y k

y G y

G   



. (59)

Thus,

 

_n ^py _n ^py _n ^ⁿ ^a^^y _n ^ⁿ ^a^^y

n t Ae Be Ae Be

G   ^  ^^/  ^ ^^/ . (60)

Consequently,

        



 



 

 ^ x

a e n

B e

A y x

u_ss_n _n ⁿ_ ^a ^y _n ⁿ_ ^a ^y



sin

, ^/ ^/

, (61)

and

  

^



^ ^ ^ ^





 



 



 



1

/

/ sin

,

n

y a n n y a n n

ss x

a e n

B e

A y

x

u _ _



. (62)

Since

u

_ss

  x , 0  0

, then

 

^sin ⁰

1





 



 



^

 n

n

n x

a B n

A



. (63)

(12)

The term 



 



 x

a n



sin does not always equal to zero. So for (63) this to be true,

A

_n

 B

_n must be equal to zero. Thus,

n

n A

B  . (64)

Consequently,

  

^



^ ^ ^ ^





 



 



 



1

/

/ sin

,

n

y a n n y a n n

ss x

a e n

A e

A y

x

u _ _



. (65)

But

    



 



 

 ^ y

a e n

eⁿ_ ^a ^y ⁿ_ ^a ^y



sinh

/ 2

/ , (66)

then

  

^





 



 



 



 

1

sin sinh

2 ,

n n

ss x

a y n

a A n

y x

u

 

. (67)

Now, we shall apply the boundary condition in (53b),

  x f   x

a b n

a A n

b x u

n n

ss

 



 



 



 



 

^



1

sin sinh

2

,  

. (68)

Fourier series analysis suggests that

        

 



 





^a

n

x dx

a x n

a f a b

A n

0

sin

sinh 2

2  

, (69)

or

    

^_^ ^_^

 ^a

n x dx

a x n

a f b n A a

0 sin

/ sin

1



^. ⁽⁷⁰⁾

Equations (67) and (70) are the complete solution of the steady-state two-dimensional heat equation.

Two-Dimensional Wave Equation

It is possible to apply method of separation of variables with more than two variables by evaluating two sets of variables at a time.

Consider the two-dimensional wave equation

(13)

 

 







 



 



2 2 2 2 2 2 2

y u x

c u t

u

, (71)

where

u  u  x , y , t 

. To solve this problem, we need 4 boundary conditions, and 2 initial conditions. So let

 0 , y , t    u a , y , t    u x , 0 , t    u x , b , t   0

u

, (72)

 x y    f x y

u , , 0  ,

, (73)

and



^x ^y

  

^g ^x ^y

tu , ,0  ,



 . (74)

Let’s write the solution as

 x y t  F     x y G t

u , ,  ,

. (75)

Then by substituting it in (71), we get

 

 







 



 



 G

y G F x c F t

F G

₂

2 2

2 2 2 2

. (76)

We can arrange (76) as

 

 







 



 



2 2 2

2

1 1

1

y F F x

F c F

t G

G

^. ⁽⁷⁷⁾

Since the left and the right hand sides depend different variables, we can conclude that (77) is equal to a constant. Let

y k F x

F F t

G G

c  



 







 



 



2 2 2 2 2

2 2

1

. (78)

So we have

G t kc

G

2





and

kF

y F x

F 



 



2 2 2 2

(79a,b) Notice that (79b) is now a PDE with 2 independent variables. So let

  x y M     x N y

F , 

. (80)

(14)

Substituting (80) into (79b) gives

y kMN

M N x

N M 



 



2 2 2

2

. (81)

Rearranging gives

2 2 2

2

1

y N k N

x M

M 

 

 



. (82)

The left and the right hand sides now depend on different variables. Thus, we can set (82) to a constant, says l. So, we get

x lM M 



2 2

, (83a)

 k l  N

y

N  



2 2

. (83b)

Since now

u  x , y , t   F     x , y G t  M       x N y G t

and

u  0 , y , t    u a , y , t   0

, then

  0  M   a  0

M

. (84)

It follows that l has to be negative, i.e.

l   p

². Thus,

   





 



 

 x

a px m

x

M_m



sin

sin ; m1,2,3,... (85)

Similarly,

u  x , 0 , t    u x , b , t   0

suggest that

  0  N   b  0

N

, (86)

and kl is negative, i.e. klq². So,

   





 



 

 y

b qy n

y

N_n



sin

sin ; n1,2,3,... (87)

Consequently,

      



 



 



 



 

 y

b x n

a x m

N x M y x

F

_m_n

 

sin sin

,

. (88)

We have seen that

(15)

2

q

p k l

k     

. (89)

Thus,

2

q

p

k   

, (90)

which is negative. Let k r². Then,

2 2

2

p q

r  

. (91)

Consequently,

n

r

m

b n a q m

p

r 

²



²

    

_,

(92) Back to (79a),

G t kc

G

₂

2 2

 



. (79a)

Its solution is then

 

^t ^A



^r ^ct



^B



^r ^ct



G_m_,_n  cos _m_,_n  sin _m_,_n . (93)

Ultimately,

       





 



 



 



 

 y

b x n

a ct m

r B ct r A t y x

u_m_n _m_n _m_n

 

sin sin

sin cos

,

, _, _,

, , (94)

and

     

^

     

















 



 



 



 



1 1

, ,

1 1

, , , cos sin sin sin

, ,

m n

n m n

m

m n

n

m y

b x n

a ct m

r B ct r A t

y x u t

y x

u

 

(95) where

b n a r_m_n m

 



,  ; m1,2,3,... and n1,2,3,... (96)

By applying the initial conditions to (95), we find that

   

^_^ ^_^ ^_^ ^_^

 ^b ^a y dxdy

b x n

a y m

x ab f

A 4 0 0 , sin



sin



, (97)

and

(16)

               



^b ^a

n m

dxdy b y

x n a y m

x abcr g

B

0 0

,

sin sin

4 ,  

. (98)

Two-Dimensional Circular Wave Equation

We have so far considered either the end-point boundary or a square boundary. It is also possible that the boundary has other shapes. One of the most often found shapes is the circular shape. To evaluate PDE with circular boundary, we need to modify our PDE, which is currently limited to Cartesian

coordinate, to work with polar coordinate.

For two-dimensional wave equation, it is

 

 







 



 



 



2 2 2 2

2 2 2

2

1 1

 u r r u r r c u t

u

, (99)

where

u  u  r ,  , t 

.

Let’s assume further that the solution is radially symmetric, so it does not depend on



, i.e.

0





u . (100)

Thus, equation (99) is reduced to

 

 







 



 



r u r r c u t

u 1

2 2 2 2 2

, (101)

where

u  u   r , t

.

Here, we need a boundary condition and 2 initial conditions. Let

  R , t  0

u

, (102)

  r f   r

u , 0 

, (103)

and

 

^g

 

^r

t r

u 



 ,0

. (104)

If we write

  r t F     r G t

u , 

, (105)

then substituting (105) into (101) gives

(17)

 

 







 



 



r F r r G F t c

F G 1

2 2 2 2 2

. (106)

Rearranging gives

r F rF r

F F t

G G

c 

 



 



 1 1

1

2 2 2

2

2 . (107)

Since the two sides depend on different independent variables, we can set them to a constant. Let’s also assume that the constant is negative (as we have seen before). Then,

2 2

1 1

1 k p

r F rF r

F F t

G G

c   



 



 



. (108)

So,

F r p

F r r

F

2

1  



 



, (109a)

G c t p

G

₂ ₂

2 2



 



. (109b)

Notice that we can write (109a) as

2

0

2 2

2

 



 



 r p F

r r F r

r F

. (110)

Substituting r s/ p into (110), we get

2

0

2 2

2

 



 



 s F

s s F s

s F

. (111)

This is essentially the Bessel’s equation with



0. Thus,

 

r AJ

 

s AJ

 

pr

F  ₀  ₀ . (112)

The boundary condition suggests that F

 

R  J₀

 

pR 0. (We let A1 here for simplicity.) This can happen when

,...

6537 . 8 , 5201 . 5 , 4048 .

2

pR (113)

or

(18)

,...

, , 6537,...

. ,8 5201 . ,5 4048 . 2

3 2

1 p p

R p R

p R  . (114)

Consequently,

 

^r ^J

 

^p ^r

F_n  ₀ _n . (115)

Equation (109b) can be easily solved as we have seen before. The solution is

 

t A

 

cp t B

 

cp t

G_n  cos _n  sin _n . (116)

Ultimately, we find that

 

r t



A

 

cp t B

 

cp t



J

 

p r

u_n ,  cos _n  sin _n ₀ _n , (117)

  

^

       







1

sin 0

cos ,

n

n n

nt B cp t J p r

cp A t

r

u , (118)

where

,...

, , 6537,...

. ,8 5201 . ,5 4048 . 2

3 2

1 p p

R p R

p_n  R  . (114)

What is A and B then?