Partial Differential Equations and Boundary Value Problems† National Chiao Tung University Chun-Jen Tsai 12/16/2019 † Chapter 12.1 ~ 12.5 in the textbook.

Partial Differential Equations and Boundary Value Problems

National Chiao Tung University Chun-Jen Tsai 12/16/2019

/49

Partial Differential Equation

 A partial differential equation (PDE) is a differential equation that contains partial derivatives of a

dependent variable that is a function of at least two independent variables.

 Example: one-dimensional heat equation:

 u(x, t) is the temperature function of x (position) and t (time) of a heated rod, k is a constant parameter determined by the material of the rod

2

Linear Partial Differential Equations

 If u is a function of two independent variables x and y, the general form of a linear 2^nd-order PDE is given by:

where A, B, C, D, …, G are functions of x and y.

 Example: one-dimensional heat propagation equation can be described by:

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Solving PDE for Separable Functions

 General solutions for PDE are difficult to find, so in practice, we only look for particular solutions.

 In addition to using initial or boundary conditions to constrain our solutions, we often assume that the solution function is separable, that is:

u(x, y) = X(x)Y(y).

Thus, we have:

4

Example: Solving (1/4)

 Let u(x, y) = X(x)Y(y), we have XY = 4XY, or

where

l

Thus, X and Y must be solutions of X + 4

l

l

These are the eigenvalue problem of ODE’s^†.

Consider the three cases:

l

l

a

l

a

l

 

Y Y X

X

/49

Example: Solving (2/4)

 Case I:

l

The two equations become X = 0 and Y = 0. The general solutions are X(x) = c₁ + c₂ x and Y(y) = c₃, respectively.

Thus, a particular solution of the PDE is u(x, y) = X(x)Y(y) = (c₁ + c₂ x)c₃ = C₁ + C₂ x.

6

Example: Solving (3/4)

 Case II:

l

a

a

The two equations becomes X – 4

a

a

X(x) = c₁ cosh 2

a

a

Thus, a particular solution of the PDE is

u(x, y) = X(x)Y(y) = (c₁ cosh 2

a

a

= C₁ e^a2ycosh 2

a

a

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Example: Solving (4/4)

 Case III:

l

a

a

The two equations becomes X + 4

a

a

a

a

Thus, a particular solution of the PDE is

u(x, y) = X(x)Y(y) = (c₁ cos 2

a

a

= C₁ e^–a2ycos 2

a

a

8

Superposition Principle for PDE

 If u₁, u₂, …, u_k are solutions of a homogeneous linear partial differential equation, then the linear combination

where the c_i, i = 1, 2, …, k, are constants, is also a solution.

The property is true even when k = .



i iu c

1

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Classification of PDE

 The linear 2^nd-order partial differential equation with two independent variables,

where A, B, C, D, …, G are real constants, is said to be:

 Hyperbolic if B² – 4 AC > 0,

 Parabolic if B² – 4 AC = 0,

 Elliptic if B² – 4 AC < 0.

10

2 ,

2 2

2 2

G y Fu

E u x

D u y

C u y

x B u x

A u  



 



 



 





 





Derivation of Classical PDEs

 The derivation of the mathematical model that can be used to explain or predict the behavior of a physical phenomenon is the key to most engineering problems

 Example: the optical flow model.

The motion (dx/dt, dy/dt) of the image pixels E(x, y, t) taken by a camera can be approximated by:

,

 0



 

 

 

 



t E dt

dy y

E dt

dx x

E

/49

Derivation of the Heat Equation (1/3)

 Assume that we have a heated rod:

u(x, t) is the temperature of the rod at x and time t.

 From empirical study of thermodynamics:

 The amount of heat in a element of mass m and temperature u is Q = g mu, g is a constant parameter of the rod.

 The heat flow Q_t = –KAu_x is the flow of heat in the direction of decreasing temperature, K is a constant parameter of the rod.

12

cross section area A

x

heater insulated shell, like a heat conduit

† One calorie is the amount of heat required at a pressure of one atmosphere to raise the temperature of one gram of water by one degree Celsius

x = 0

Derivation of the Heat Equation (2/3)

 The heat content in a segment of the rod is:

Q =

g

g



D

and the heat flow in this segment is

dQ/dt =

g

D

D

 Another way to estimate the heat flow is to compute the difference of amount of heat entering/leaving the segment as

D

Q_t(x+Dx, t) – Q_t(x, t) = KA[u_x(x+

D

x x+Dx

Q_t(x, t) Q_t(x+Dx, t)

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Derivation of the Heat Equation (3/3)

 Eq (1) and (2) should equal each other as

D

D

g

D

D

Therefore

Finally, we obtain the following heat equation:

where k = K/

g

14

2 .

2

t u x

k u



 





t x

x

x u

x

t x u t

x x

u

K 

D

 D

 

 D

)]

, ( )

, (

lim [

g

BVP of the Heat Equation (1/3)

 The solution of a PDE involves arbitrary functions of some dependent variables. For example,

the partial DE

has a general solution u(x, t) = g(x), where g(x) can be any function of x.

Hence, the “initial condition” of a partial DE is a

boundary function. In the case of the heated rod, we may have the boundary function u(x, 0) = f(x), where f(x) is the heat function (of x) at time 0.

) 0 ,

( 



 t

t x u

/49

BVP of the Heat Equation (2/3)

 We may also constrain the temperature function at two ends of the rod and try to solve the PDE. For example,

u(0, t) = u(L, t) = 0, for all t > 0.

A boundary value problem of the heated rod PDE may be as follows:

);

0 ,

0

2 (

2   



 



 x L t

x k u t

u

), 0 (

0 )

, ( )

, 0

( t  u L t  t 

u

).

0 ( )

( )

0 ,

(x f x x L

u   

16

BVP of the Heat Equation (3/3)

 Another possible boundary condition for the heated rod is that no heat will flow through either end (i.e. both

ends are heat-insulated):

u_x(0, t) = u_x(L, t) = 0, for all t.

 Physical intuition tells us that if the initial condition f(x) is a reasonable function, there exists a unique solution u(x, t) for the boundary value problem.

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Derivation of the Wave Equation (1/2)

 A PDE that models the vibrations of a string can be derived with the following assumptions:

 A perfectly flexible uniform string with density  is stretched under a uniform tension force of T between x = 0 and x = L.

 Each point on the string moves only in u direction

 u(x, t) is the shape of the string at time t.

 The slope of the curve is small for all x  sin   tan = u_x(x, t).

18

Derivation of the Wave Equation (2/2)

 Apply Newton’s law to the segment [x, x + Dx], Tsin









= T[u_x(x + Dx, t) – u_x(x, t)]

= (



 So, division by DxT on both side yields

As Dx  0, we have u_xx = (



   

, . ,

tt x

x u

T x

t x u t

x x

u _



D

 D



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BVP of the Wave Equation

 If we set

we have the one-dimensional wave equation that

models the free vibrations of a uniform flexible string:

20

2 ,



a  T

);

0 ,

0

2 (

2 2

2 2   



 



 x L t

t u x

a u

), 0 (

, 0 )

, ( )

, 0

( t  u L t  t  u

), 0

( )

( )

0 ,

(x f x x L

u   

).

0 ( )

( )

0 ,

(x g x x L

u_t   

Laplacian of a 2-D Function u(x, y)

 The Laplacian of the function u(x, y) is defined as

 The Laplace’s equation ²u = 0 is often used to model the steady-state behavior of a 2-D (or higher

dimensional) phenomenon (e.g., temperature of an object).

2 .

2 2

2 2

y u x

u u



 



 



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Modeling of 2-D Heat/Wave Equations

 Given a 2-D thin plate with thermal diffusivity k, its temperature u(x, y, t) at the point (x, y) at time t

satisfies the 2-D heat equation:

 Note that u_t = k²u is the 2-D extension of the 1-D heat equation u_t = ku_xx. Similarly, u_tt = a²²u is the 2-D

extension of the 1-D wave equation u_tt = a²u_xx.

 c k K u

y k u x

k u t

u    



 







 



 



 ₂ ² ,

2 2

2

y

x

R

C

22

Heat/Wave Eqs with Influences

 The 1-D heat/wave equation can be modified to take into account external and internal influences:

and

where G() may be the ambient temperature influences to the heated rod; and F() may represent the external, damping, and restoring forces of the string vibration

,

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Solution to the BVP of Heat Equation

 Note that the BVP of a heated rod is modelled as:

 Note that the heat equation is linear. That is, if u₁ and u₂ satisfy the PDE, w = c₁u₁ + c₂u₂ also satisfies the PDE.

However, a solution of the PDE must also satisfy the boundary conditions.

24

);

0 ,

0

2 (

2   



 



 x L t

x k u t

u

), 0 (

0 )

, ( )

, 0

( t  u L t  t  u

).

0 ( )

( )

0 ,

(x f x x L

u   

Meeting Boundary Conditions (1/2)

 If u₁ and u₂ satisfies the (homogeneous) conditions u(0, t) = u(L, t) = 0, for all t > 0,

w = c₁u₁ + c₂u₂ will also satisfy the condition. However, the general form of w may not satisfy the boundary

condition  only a particular choice of c₁ and c₂ satisfy the non-homogeneous boundary condition:

u(x, 0) = f(x), 0 < x < L.

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Meeting Boundary Conditions (2/2)

 In general, we must find an infinite sequence u₁, u₂, u₃, …, of solutions that satisfies both the PDE and the homogeneous boundary conditions, and assume the general solution form as follows:

Then, determine the coefficients c₁, c₂, c₃, … that satisfy the non-homogeneous boundary condition.

.) , ( )

, (







n

n

nu x t

c t

x u

26

General Solutions of a Linear BVP

 Suppose that each of the functions u₁, u₂, u₃, …,

satisfies both the PDE for 0 < x < L and t > 0 and the homogeneous conditions, and c₁, c₂, c₃, … are chosen to meet the following three criteria:

1. For 0 < x < L and t > 0, the function u(x, t) = c_nu_n(x, t) is

continuous and term-wise differentiable (for /t and ²/x²).

2.

3. The function u(x, t) = c_nu_n(x, t) is continuous within, and at the boundary of the region 0  x  L and t  0.

Then u(x, t) is a solution of the BVP.

. 0

) ( )

0 , (

1

L x

x f x

u c

n

n

n   





for

/49

Separation of Variables (1/4)

 In solving the heated rod problem, Fourier sought for a sequence of solutions u₁, u₂, u₃, …, which are

“separable.” That is for each of u_i, we have u(x, t) = X(x)T(t),

where X(x) and T(t) are functions of x and t,

respectively. Substitution of such u(x, t) into the heat equation u_t = ku_xx yields XT = kX"T, or

where

l

l,



 

 

kT T X

X

28

Separation of Variables (2/4)

 Thus, the solution can be obtained by solving two ODEs for some common value of

l

For X(x), we have u(0, t) = X(0)T(t) = 0, u(L, t) = X(L)T(t) = 0. Thus X(0) = X(L) = 0 if T(t) is nontrivial.

X(x) has a nontrivial solution if and only if

and then

. 0 )

( )

(

, 0 )

( )

(



 



 

t kT t

T

x X x

X

l l

,...

3 , 2 , 1

2 ,

2

2 

 n

L n

n

l 

,...

3 , 2 , 1 ,

sin )

(  n x n 

x

X 

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Separation of Variables (3/4)

 To solve for T(t), substituting the value

l

A nontrivial solution of T_n(t) is

.

2 0

2

2 

  _n

n T

L k T n 





exp )

(t   n^{2 2}kt L² n 

T_n 

30

Separation of Variables (4/4)

 Now, we have sequences of solutions to the PDE u_n(x, t) = X(x)T(t) = exp(–n²





n = 1, 2, 3, …. Each of these functions satisfies the heat equation and the homogeneous conditions.

We want to find c₁, c₂, c₃, … such that c_nu_n(x, t) satisfies

But this is the Fourier series of f(x) on [0, L]. Thus, . 0

), ( sin

) 0 , (

1

L x

x L f

x c n

x u

n

n   









,....

3 , 2 , 1 ,

sin ) 2 (

0 



 ^b _L



c_{n n} ^L 

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Insulated Endpoint Conditions

 When the heated rod is insulated at both ends, the homogeneous boundary condition becomes

u_x(0, t) = u_x(L, t) = 0. We can use the separation of variables approach again to solve this problem.

Solving the ODE of X(x) gives us:

Similarly, solving the ODE of T(t) gives us:

. cos

) (

2 ,

2 2

L x x n

L X n

n n



l   

. exp

)

( ₂

2 2



 



  

L

kt t n

T_n 

32

Heated Rod with Insulated Ends

 For a heated rod with zero endpoint temperatures, the general solution is

where {a_n} are the Fourier cosine coefficients of u(x, 0).





2 exp )

, (

1

2 2

0











n

n L

x L n

kt n

a a t

x

u  

/49

Solution to the BVP of Wave Equation

 The BVP of a vibrating string is modelled as:

Here, we have two non-homogeneous boundary conditions.

34

), 0

( )

( )

0 ,

(x f x x L

u   

).

0 ( )

( )

0 ,

(x g x x L

u_t   

);

0 ,

0

2 (

2 2

2

2   



 



 x L t

t u x

a u

), 0 (

, 0 )

, ( )

, 0

( t  u L t  t  u

Problems with Two Nonzero BCs

 To solve the wave equation, we divide the system into two sub-problems:

 Problem A:

 u_tt = a² u_xx; u(0, t) = u(L, t) = 0, u(x, 0) = f(x), u_t(x, 0) = 0.

 Problem B:

 u_tt = a² y_xx; u(0, t) = u(L, t) = 0, u(x, 0) = 0, u_t(x, 0) = g(x).

The overall solution is the sum of the two sub-problems since

u(x, 0) = u_A(x, 0) + u_B(x, 0) = f(x) + 0 = f(x),

u_t(x, 0) = {u_A}_t(x, 0) + {u_B}_t(x, 0) = 0 + g(x) = g(x).

nonzero initial offset

nonzero initial velocity

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Problem A Solution (1/3)

 By separation of variables, substitution of

u(x, t) = X(x)T(t) in u_tt = a²u_xx yields XT = a²XT for all x and t. Therefore, assume that

 we have a system of ODE:

The first equation is an eigenvalue problem:

.

2  

l

l

 

T a

T X

X

0 . )

0 ( ,

0

0 )

( )

0 ( ,

0

 2



 













T T

a T

L X X

X X

l l

and ...

, 3 , 2 , 1

2 ,

2

2 

 n

L n

n

l 

L x x n

X_n



36

Problem A Solution (2/3)

 Substitute

l

The solution to the IVP is

 Hence,

satisfies all the homogeneous boundary conditions.

 Choose {A_n} to satisfy the non-homogeneous boundary condition

. 0 ) 0 ( ,

2 0

2 2

2   

 _{n n}

n T T

L a T n 

...

, 3 , 2 , 1 ,

cos )

(  n 

L at A n

t

T_{n n} 

L x n L

at A n

t T x X t

x u t

x u

n

n n

n n

n

n



 sin cos

) ( ) ( )

, ( )

, (

1 1

1

 



















. 0

), ( sin

) 0 ,

( n x f x x L

A x

u 



/49

Problem A Solution (3/3)

 If we choose

the condition is simply the Fourier sine series expansion of f(x) on [0, L].

 Example: if

and g(x) = 0, the solution u(x, t) is

. sin

) 2 (



 ^L

n dx

L x x n

L f

A 

. sin

2 cos 1 sin

) 4 , (

1 2

2







 



 

n L

x n L

at n n

n t bL

x

u   



2 , / ),

(

2 / 0

) , ( 









 

L x L

x L b

L x

x bx f

x u

L L/2

u = f(x)

½bL

38

d’Alembert form of Solution (1/2)

 An alternative form of solution of problem A can be obtained by applying trigonometric identity:

If we define we have

u(x, t) = [F(x + at) + F(x – at)]/2.

.) (

2 sin ) 1

( 2 sin

1

sin cos

) , (

1 1

1





























n

n n

n n

n

at L x

A n at

L x A n

L x n L

at A n

t x u









, sin

) (







n

n L

x A n

x

F 

/49

d’Alembert form of Solution (2/2)

 The functions F(x + at) and F(x – at) in d’Alembert form of Solution represents waves moving to the left and right, respectively, along the string with speed a.

x u

/2 

(a) At time t = 0.

x u

/2 

(a) At time t = /8.

x u

/2 

(a) At time t = /4.

x u

/2 

(a) At time t = 3/8.

40

Problem B Solution (1/2)

 Solution for Problem B is similar to that for A, except that

A non-trivial solution is Hence,

. 0 ) 0 ( ,

2 0

2 2 2 2

2 ⁿ  T_n  T_n 

L a n

dt T

d 

...

, 3 , 2 , 1 ,

sin )

(  n 

L at B n

t

T_{n n} 

. sin

sin )

( ) ( )

, (

1

1 L

x n L

at B n

t T x X t

x u

n

n n

n n

















/49

Problem B Solution (2/2)

 Again, the coefficients {B_n} that satisfies the non- homogeneous boundary condition

would be the Fourier sine coefficient b_n of g(x) on [0, L] divided by n



Hence, we choose

. 0

), ( sin

) 0 , (

1

L x

x L g

x n L

a B n

x u

n

n

t 









. sin

) 2 (

0 dx

L x x n

L g L b

a

B_n n^{ } _n ^



. sin

) 2 (

0 dx

L x x n

a g

B_n ^ n_



42

Total Solution to the Wave Equation

 The complete solution is the summation of Problem A and Problem B:

where

, sin

sin cos

) , (

1 L

x n L

at B n

L at A n

t x u

n

n n











 



 



. sin

) 2 (

0 dx

L x x n

a g

B_n ^ n_



, sin

) 2 (



 ^L

n dx

L x x n

L f

A 

/49

Steady-State Temperature

 The steady-state temperature of a plate can be

described by a function u(x, y), i.e., u_t = 0. Thus, we have the 2-D Laplace equation:

 A boundary value problem of the Laplace equation can be formulated as follows (i.e. the Dirichlet problem):

.

2 0

2 2

2 2 



 



 

 y

u x

u u

. ) on is ) , ( if ( )

, ( )

, (

) R within (

2 0

2 2

2 2









 

 



 



C y

x y

x f y

x u

y u x

u u ^y

x

R C

44

Solutions to the Laplace’s Equation

 Suppose we want to find the steady-state temperature u(x, y) in a thin rectangular plate with width a and height b. The problem can be formulated as a BVP problem as follows:

u_xx + u_yy = 0;

u(0, y) = f₁(x), u(a, y) = f₂(x), u(x, b) = f₃(x), u(x, 0) = f₄(x).

This is called the Dirichlet problem.

/49

 Solve the boundary value problem for the rectangle R.

u_xx + u_yy = 0;

u(0, y) = u(a, y) = u(x, b) = 0, u(x, 0) = f(x).

Assume that u(x, y) = X(x)Y(y), we have XY + XY = 0.

Thus,

Example: The Dirichlet Problem (1/4)

0. ) ( )

0 (

0

















 



 

a X X

X X

Y Y X

X l

l

46

y

x

R

(a, b) u = 0

u = 0 u = 0

u = f(x)

Example: The Dirichlet Problem (2/4)

 The eigenvalues and eigenfunctions of X are

As a result,

The general solution of Y_n is

,...

3 , 2 , 1 ,

sin )

(

2 ,

2

2  

 n

a x x n

a X n

n n

 l 

. 0 )

( ,

2 0

2

2  

 Y Y b

a

Y_n n  _{n n}

. sinh

cosh )

( a

y B n

a y A n

y

Y_{n n}  _n 





/49

Example: The Dirichlet Problem (3/4)

 To compute the particular solution, we must solve A_n and B_n using Y_n(b) = 0:



Therefore,

48

).

/ sinh(

/ ) ,

sinh ( c A n b a

a y b

c_n n  _{n  n} 



. 0 sinh

cosh )

(   

a b B n

a b A n

b

Y_{n n}  _n 

. sinh

cosh a

b n a

b A n

B_{n   n}  

a y n a

b n a

b A n

a y A n

y

Y_{n n}  _n   

sinh sinh

cosh cosh

)

( 



 







Example: The Dirichlet Problem (4/4)

 The formal series solution is then

c_n must satisfy the nonhomogeneous condition

Therefore,

). sinh (

sin )

( ) ( )

, (

1

1











 



n

n n

n

n a

y b

n a

x c n

y Y x X y

x

u  

).

( sin

sinh )

0 , (

1

x a f

x n a

b c n

x u

n

n  



 













. sin

) ) (

/ sinh(

2



 ^a

n dx

a x x n

a f b n

c a 



#