25_3_soln_sep_vars.pdf

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Solution usingSeparation of Variables

25.3Introduction

The main topic of this Section is the solution of PDEs using the method of separation of variables.In this method a PDE involving n independent variables is converted into n ordinary differentialequations. (In this introductory account n will always be 2).

You should be aware that other analytical and also numerical methods are available for solv-ing PDEs. However, the separation of variables technique does give some useful solutions toimportant PDEs.

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%Prerequisites

Before starting this Section you should . . .

① some first and second order constantcoefficient ordinary differential equations

Learning Outcomes

After completing this Section you should be

able to . . .

✓ apply the separation of variables methodto obtain solutions of the heat conductionequation, wave equation and 2-D Laplaceequation for specified boundary and/orinitial conditions



1. Solution of Important PDEs

We shall just consider two analytic solution techniques for PDEs:

(a) Direct integration

(b) The method of separation of variables

The latter method is the more important and we will study it in detail shortly.

You should note that many practical problems involving PDEs have to be solved by numerical

methods but that is another story.

The method of direct integration is a straightforward extension of solving very simple ODEs bydirect integration.

Solve the ODEd2y

dx2= x2 + 2

given that y = 1 when x = 0 anddy

dx= 2 when x = 0.

Finddy

dxby integrating once, not forgetting the arbitrary constant of integration

Your solution

d y

d x =

x 3

3 + 2 x + A

Now find y by integrating again, not forgetting to include another arbitrary constant.

Your solution

y = x

4

1 2 + x 2 + A x + B

Now find A and B by inserting the two given initial conditions:

Your solution

y ( 0 ) = 1 g i v e s B = 1 y ( 0 ) = 2 g i v e s A = 2

s o t h e r e q u i r e d s o l u t i o n i s

y = x

4

1 2 + x

2 + 2 x + 1

HELM (VERSION 1: March 18, 2004): Workbook Level 225.3: Solution using Separation of Variables

2



Consider now a similar type of PDE i.e. one that can also be solved by direct integration.Suppose we require the general solution of

∂ 2u

∂x2= 2xet

where u is a function of x and t.Integrating with respect to x gives us

∂u

∂x= x2et + f (t)

where the arbitrary function f (t) replaces the normal “arbitrary constant” of ordinary integra-tion. This function of t only is needed because we are integrating “partially” with respect to x

i.e. we are reversing a partial differentiation with respect to x at constant t.Integrating again with respect to x gives the general solution:

u =x3

3et + x f (t) + g(t)

where g(t) is a second arbitrary function. We have now obtained the general solution of thegiven PDE but to find the arbitrary function we must know two initial conditions.Suppose, for the sake of example, that these conditions are

u(0, t) = t ,∂u

∂x(0, t) = et

Inserting the first of these conditions into the general solution gives

g(t) = t

Inserting the second condition into the general solution gives

f (t) = et

so the final solution is

u =x3

3et + xet + t.

Solve the PDE∂ 2u

∂x∂y= sin x cos y

subject to the conditions

∂u

∂x= 2x at y =

π

2, u = 2 sin y at x = π.

First integrate the PDE with respect to y: (it is equally valid to integrate first

with respect to x). Don’t forget the appropriate arbitrary function.

3 HELM (VERSION 1: March 18, 2004): Workbook Level 225.3: Solution using Separation of Variables



Your solution

R e c a l l t h a t ∂

2 u

∂ x ∂ y =

∂

∂ y

∂ u

∂ x

H e n c e i n t e g r a t i o n w i t h r e s p e c t t o y g i v e s

∂ u

∂ x

= s i n x s i n y + f ( x )

Since one of the given conditions is on∂u

∂xwe impose this condition at this stage to determine

the arbitrary function f (x):

Your solution

A t y = π / 2 t h e c o n d i t i o n g i v e s

s i n x s i n π / 2 + f ( x ) = 2 x i . e . f ( x ) = 2 x − s i n x

S o ∂ u

∂ x = s i n x s i n y + 2 x − s i n x

Now integrate again to determine u

Your solution

I n t e g r a t i n g n o w w i t h r e s p e c t t o x g i v e s

u = − c o s x s i n y + x 2

+ c o s x + g ( y )

Finally, obtain the arbitrary function g(y)

Your solution

T h e c o n d i t i o n u ( π , y ) = 2 s i n y g i v e s

− c o s π s i n y + π 2

+ c o s π + g ( y ) = 2 s i n y

∴ s i n y + π 2 −

1 + g ( y ) = 2 s i n y

∴ g ( y ) = s i n y + 1 − π 2

Now write down the final answer for u(x, y)

Your solution

u = x 2

+ c o s x ( 1 −

s i n y ) + s i n y + 1 −

π 2


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2. Method of Separation of VariablesIn the previous Section we showed that

(a) u(x, y) = sin x cosh y

is a solution of the two-dimensional Laplace Equation

(b) u(x, t) = e−2π2t sin πx

is a solution of the one-dimensional heat conduction equation

(c) u(x, t) = u0 sinπx

cos

πct

is a solution of the one-dimensional wave equation.

All three solutions here have a specific form: in (a) u(x, y) is a product of a function of x alone,viz sin x, and a function of y alone, viz cosh y.

Similarly in both (b) and (c) u(x, t) is a product of a function of x alone and a function of talone.The method of separation of variables involves finding solutions of PDEs which are of thisproduct form. In the method we assume that a solution to a PDE has the form. In a similarmanner to second order differential equations (25.1 p3) if two or more different solutions arepossible, their sum is also a solution.

e.q. if u(x, y) = X 1(x)Y 1(y) and u(x, y) = X 2(x)Y 2(y) are solutions then so is u(x, t) =X 1(x)Y 1(y) + X 2(x)Y 2(y)

u(x, t) = X (x)T (t)

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

where X (x) is a function of x only, T (t) is a function of t only and Y (y) is a function y only.You should note that not all solutions to PDEs are of this type; for example, it is easy to verifythat

u(x, y) = x2− y2

(which is not of the form u(x, y) = X (x)Y (y)) is a solution of the Laplace Equation.However many interesting and useful solutions of PDEs are obtainable which are of the productform.We shall firstly consider the types of solution obtainable for our three basic PDEs using trialsolutions of the product form.

Heat conduction equation

∂ 2u

∂x2=

1

k

∂u

∂tk > 0 . . . (1)

Assuming that

u = X (x)T (t)

= XT for short

. . . (2)




then

∂u

∂x=

dX

dxT = X T for short

∂

2

u∂x2 = d

2

X dx2 T = X T for short

∂u

∂t= X

dT

dt= XT for short

Substituting into the original PDE (1)

X T =1

kXT

which can be re-arranged as

X

X = 1

kT

T . . . (3)

Now the left hand side of (3) involves functions of x only and the right hand side expressioncontain functions of t only.Thus altering the value of t cannot change the left hand side of (3) i.e. it stays constant. Henceso must the r.h.s. be constant. We conclude that T (t) is a function such that

1

k

T

T = K . . . (4)

where K is a constant whose sign is yet to be determined.By a similar argument, altering the value of x cannot change the right hand side of (3) andconsequently the left hand side must be a constant

i.e.X

X = K . . . (5)

We see that the effect of assuming a product trial solution of the form (2) converts the PDE (1)into the two ODEs (4) and (5).Both these ODEs are types whose solution we revised at the beginning of this Unit but we shallnot attempt to solve them yet. In particular the solution of (5) depends on whether the constant

K is positive or negative.

By following a similar procedure to the above, assume a product solution

u(x, t) = X (x)T (t)

for the wave equation∂ 2u

∂x2=

1

c2∂ 2u

∂t2. . . (6)

and find the two ODEs satisfied by X (x) and T (t) respectively.


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First obtain∂ 2u

∂x2and

∂ 2u

∂y2.

Your solution

u = X ( x ) T ( t ) g i v e s ∂

2 u

∂ x 2 = X

T a n d

∂ 2

u

∂ t 2 = X T

Now substitute these results into (6) and transpose so the variables are separated i.e. all func-tions of x are on the left hand side, all funtions of t on the right hand side.

Your solution

W e g e t X T = 1

c 2

X T a n d , t r a n s p o s i n g , X

X = 1

c 2

T

T

Finally, write down the required ODE

Your solution

E q u a t i n g b o t h s i d e s t o t h e s a m e c o n s t a n t K g i v e s

X

X = K o r

d 2

X

d x 2

− K X = 0 . . . ( 7 )

a n d 1

c 2

T

T = K o r

d 2

T

d t 2

− K c 2

T = 0 . . . ( 8 )

Again, the solution of the ODEs (7) and (8) has been revised earlier and in both cases thesolution will depend on the sign of K .

Separating the variables for Laplace’s equation

∂ 2u

∂x2+

∂ 2u

∂y2= 0

follows similar lines. Obtain the ODEs satisfied by X (x) and Y (y).

Your solution




A s s u m i n g u ( x , y ) = X ( x ) Y ( y )

l e a d s t o : ∂

2 u

∂ x 2 = X

Y

∂ 2

u

∂ y 2 = X Y

s o

X

Y + X Y

= 0 o r X

X = −

Y

Y

E q u a t i n g e a c h s i d e t o a c o n s t a n t K

X

X = K o r

d 2

X

d x 2

− K X = 0 . . . ( 9 )

Y

Y = − K o r

d 2

Y

d y 2 + K Y = 0 . . . ( 1 0 )

( N o t e c a r e f u l l y t h e d i ff e r e n t s i g n s i n t h e t w o O D E s . Y e t a g a i n t h e s i g n o f t h e “ s e p a r a t i o n c o n s t a n t ” K w i l l d e t e r m i n e t h e s o l u t i o n s . )

We shall now study some specific problems which can be fully solved by the separation of variables method.

Example Solve the heat conduction equation

∂ 2u

∂x2=

1

2

∂u

∂t

over 0 < x < 3, t > 0 for the boundary conditions

u(0, t) = u(3, t) = 0

and the initial conditionu(x, 0) = 5sin4πx.

Solution

Assuming u(x, t) = X (x)T (t) gives rise to the differential equations (4) and (5) with theparameter k = 2:

dT

dt= 2KT

d2X

dx2= KX

Now the T equation has general solution

T = Ae2Kt

which will increase exponentially with increasing t if K is positive and decrease with t if K isnegative. In any physical problem the latter is the meaningful situation. To emphasise that K


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is being taken as negative we putK = −λ2

soT = Ae−2λ2t.

The X equation then becomesd2X

dx2= −λ2X

which has solutionX (x) = B cos λx + C sin λx.

Hence

u(x, t) = X (x)T (t)

= (D cos λx + E sin λx)e−2λ2t . . . (11)

where D = AB and E = AC .

(You should always try to keep the number of arbitrary constants down to an absolute minimumby multiplying them together in this way.)We now insert the initial and boundary conditions to obtain the constant D and E and also theseparation constant λ.The initial condition u(0, t) = 0 gives

(D cos 0 + E sin0)e−2λ2t = 0 for all t.

Since sin 0 = 0 and cos 0 = 1 this must imply that D = 0.The other initial condition u(3, t) = 0 then gives

E sin(3λ)e−2λ2t

= 0 for all t.We cannot deduce that the constant E has to be zero because then the solution (11) would bethe trivial solution u ≡ 0.The only sensible deduction is that

sin3λ = 0 i.e. 3λ = nπ (where n is some integer).Hence solutions of the form (11) satisfying the 2 boundary conditions have the form

u(x, t) = E n sinnπx

3

e−

2n2π2t

9

where we have written E n for E to allow for the possibility of a different value for the constant

for each different value of n.We obtain the value of n by using the initial condition u(x, 0) = 5sin4πx and forcing thissolution to agree with it.That is,

u(x, 0) = E n sinnπx

3

= 5 sin 4πx

so we must choose n = 12 with E 12 = 5.Hence, finally,

u(x, t) = 5 sin

12πx

3

e−

2

9(12)2π2t

= 5 sin(4πx) e−32π2t.




Solve the 1−dimensional wave equation

∂ 2u

∂x2=

1

16

∂ 2u

∂t2

for 0 < x < 2, t > 0

The boundary conditions are

u(0, t) = u(2, t) = 0

(as in the previous example).

The initial conditions are

(i) u(x, 0) = 6 sin πx − 3sin4πx

(which is similar to, but slightly more complex, than in the previous example.)

(ii) ∂u∂t

(x, 0) = 0

(which has no counterpart in the previous example.)

Firstly using (7) and (8) (or by working from first principles from the productsolution u(x, t) = X (x)T (t)) write down the ODEs satisfied by X (x) and T (t).

Your solution

X

X = K

T

1 6 T = K

Now decide on the appropriate sign for K and then write down the solution to these equations.

Your solution

C h o o s i n g K a s n e g a t i v e ( s a y K = − λ 2 ) w i l l p r o d u c e S i n u s o i d a l s o l u t i o n s f o r X a n d T w h i c h

a r e a p p r o p r i a t e i n t h e c o n t e x t o f t h e w a v e e q u a t i o n w h e r e o s c i l l a t o r y s o l u t i o n s c a n b e e x p e c t e d .

T h e n X

= − λ 2

X g i v e s X = A c o s λ x + B s i n λ x

S i m i l a r l y T

= −

1 6 λ 2

T g i v e s T = C c o s 4 λ t + D s i n 4 λ t

Now obtain the general solution u(x, t) by multiplying X (x) by T (t) and insert the two boundaryconditions to obtain information about two of the constants.

Your solution


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u ( x , t ) = ( A c o s λ x + B s i n λ x ) ( C c o s 4 λ t + D s i n 4 λ t )

u ( 0 , t ) = 0 f o r a l l t g i v e s A ( C c o s 4 λ t + D s i n 4 λ t ) = 0

w h i c h i m p l i e s t h a t A = 0 .

u ( 2 , t ) = 0 f o r a l l t g i v e s B s i n 2 λ ( C c o s 4 λ t + D s i n 4 λ t ) = 0

s o , f o r a n o n - t r i v i a l s o l u t i o n ,

s i n 2 λ = 0 i . e . λ = n π

2 f o r s o m e i n t e g e r n .

A t t h i s s t a g e w e w r i t e t h e s o l u t i o n a s

u ( x , t ) = s i n n π x

2

( E c o s 2 n π t + F s i n 2 n π t )

w h e r e w e h a v e m u l t i p l i e d c o n s t a n t s a n d p u t E = B C a n d F = B D .

Now insert the initial condition

∂u

∂t(x, 0) = 0 for all x 0 < x < 2.

and deduce the value of F .

Your solution

D i ff e r e n t i a t i n g p a r t i a l l y w i t h r e s p e c t t o t

∂ u

∂ t = s i n n π x

2

( − 2 n π E s i n 2 n π t + 2 n π F c o s 2 n π t )

s o a t t = 0 ∂ u

∂ t ( x , 0 ) = s i n n π x

2

2 n π F = 0

f r o m w h i c h w e m u s t h a v e t h a t F = 0 .

Using the other the initial condition

u(x, 0) = 6sin(πx)− 3sin(4πx)

deduce the form of u(x, t).

Your solution




A t t h i s s t a g e t h e s o l u t i o n r e a d s

u ( x , t ) = E s i n n π x

2

c o s ( 2 n π t ) . . . ( 1 2 )

W e n o w h a v e t o i n s e r t t h e fi n a l c o n d i t i o n v i z . t h e i n i t i a l c o n d i t i o n

u ( x , 0 ) = 6 s i n π x −

3 s i n 4 π x . . . ( 1 3 )

T h i s s e e m s s t r a n g e b e c a u s e , p u t t i n g t = 0 i n o u r s o l u t i o n ( 1 2 ) s u g g e s t s

u ( x , 0 ) = E s i n n π x

2

A t t h i s p o i n t w e s e e m t o h a v e i n c o m p a t a b i l i t y b e c a u s e n o s i n g l e v a l u e o f n w i l l e n a b l e u s t o s a t i s f y ( 1 3 ) .

H o w e v e r i n t h e s o l u t i o n ( 1 2 ) , a n y p o s i t i v e i n t e g e r v a l u e o f n i s a c c e p t a b l e a n d w e c a n i n f a c t , s u p e r p o s e s o l u t i o n s o f t h e f o r m ( 1 2 ) a n d s t i l l h a v e a v a l i d s o l u t i o n t o t h e P D E H e n c e w e fi r s t w r i t e , i n s t e a d o f ( 1 2 )

u ( x , t ) =

∞

n = 1

E n s i n n π x

2

c o s ( 2 n π t ) . . . ( 1 4 )

f r o m w h i c h

u ( x , 0 ) =

∞

n = 1

E n s i n n π x

2

. . . ( 1 5 )

( w h i c h l o o k s v e r y m u c h l i k e , a n d i n d e e d i s , a F o u r i e r S e r i e s . ) T o m a k e t h e s o l u t i o n ( 1 5 ) fi t t h e i n i t i a l c o n d i t i o n ( 1 3 ) w e d o n o t r e q u i r e a l l t h e t e r m s i n t h e i n fi n i t e F o u r i e r S e r i e s .

W e n e e d o n l y t h e t e r m s w i t h n = 2 w i t h c o e ffi c i e n t E 2 = 6 a n d t h e t e r m f o r w h i c h n = 8 w i t h E 8 = − 3 . A l l t h e o t h e r c o e ffi c i e n t s E n h a v e t o b e c h o s e n a s z e r o .

U s i n g t h e s e r e s u l t s i n ( 1 4 ) w e o b t a i n t h e s o l u t i o n

u ( x , t ) = 6 s i n π x c o s 4 π t − 3 s i n 4 π x c o s 1 6 π t

The above solution perhaps seems rather involved but there is a definite sequence of logical stepswhich can be readily applied to other similar problems.


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