pdes - solutions (2)

8/3/2019 PDEs - Solutions (2)

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MATH20401(PDEs) Tony Shardlow Answer Sheet Part II

1. (a) X X = 0 with X (0) = X ( ) = 0, = 0 gives X = 0, so X (x) = a + bx for some constants a, b .

BCs give X (0) = a = 0, so a = 0.X ( ) = b = 0 and = 0, so b = 0.Hence X (x) 0. Therefore = 0 is not an eigenvalue as X (x) is zero andtrivial. Eigenfunctions cannot equal zero.

> 0 with = 2 = 0, gives X 2X = 0, so X (x) = aex + be x .BCs give X (0) = a + b = 0 so b = aand X ( ) = ae + be = 0 and hence a(e e ) = 0 so that a = b = 0(since e e = 0).Hence X (x) 0. > 0 are never eigenvalues as they give trivial eigenfunctions.

< 0, with = 2 = 0, gives X + 2X = 0, so X (x) = a cos(x) + bsin(x)for some a, b .

BCs give X (0) = a cos 0 + bsin 0 = a = 0, so a = 0,and X ( ) = bsin( ) = 0,To have b = 0, we assume that sin( ) = 0.That is, if = n for integers n then X (x) = bsin(nx/ ) .

Finally, the eigenvalues n = (n/ )2 for n = 1 , 2, . . . and the correspondingeigenfunctions X n (x) = sin( nx/ ). We drop the constant b, as eigenfunctions areonly dened upto scalar multiplication. Only positive values n = 1 , 2, . . . are usedas n = 0 gives trivial function and n give the same eigenfunction upto scalarmultiplication; i.e., sin( nx/ ) = sin( nx/ )

(b) Y Y = 0 with Y (0) = Y (l) = 0 , taking Y = Y (y): = 0 gives Y = 0, so Y = a + by. BCs give Y (0) = b = 0, so b = 0, and

Y ( ) = b = 0, so b = 0 (again). Thus any value of a = 0 gives a solutionY (y) = a if = 0.

> 0, with = 2 = 0, gives Y 2Y = 0, so Y (y) = aey + be y .

BCs give Y (0) = a b = 0 so b = aand Y ( ) = ae be = 0. Hence and a(e e ) = 0 so that a = b = 0(since e e = 0). Hence Y (y) 0. Not an eigenvalue.

< 0, with = 2 = 0, gives Y + 2Y = 0, so Y (y) = a cos(y) + bsin(y).BCs give Y (0) = a sin 0 + bcos0 = b = 0, so b = 0, andY ( ) = a sin( ) = 0, so it is possible to have a = 0 only if sin( ) = 0.That is, if = n for n = 1 , 2, . . . then Y (y) = a cos(ny/ ) .

Hence we nd the eigenvalues, n = (n/ )2 for n = 0 , 1, 2, . . . and correspondingeigenfunctions, Y n (y) = cos( ny/ ). (Note: this is the constant 1 for n = 0).Eigenfunctions are only dened upto scalar multiplication, so we need only put a = 1.

MATH20401(PDEs): Answer Sheet II: Page 1


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(c) Z Z = 0 with Z (0) = Z (l) = 0 , taking Z = Z (z): = 0 gives Z = 0, so Z (z) = a + bz. BCs give Z (0) = b = 0, so b = 0, and

Z ( ) = a = 0, so a = 0. Hence Z (z) 0. > 0, with = 2 = 0, gives Z 2Z = 0, so Z (z) = aez + be z .

BCs give Z (0) = a b = 0 so b = aand Z ( ) = ae + be = 0. Hence a(e + e ) = 0 so that a = b = 0 (sincee + e = 0). Hence Z 0. Not an eigenvalue.

< 0, with = 2 = 0, gives Z + 2Z = 0, so Z (z) = a cos(z) + bsin(z).BCs give Z (0) = a sin 0 + bcos0 = b = 0, so b = 0,and Z ( ) = a cos( ) = 0, so it is possible to have a = 0 only if cos( ) = 0.

That is, if = ( n +12 ) for integers n , then Z (z) = a cos (n +

12 )z/ .

Hence we nd the eigenvalues, n = (n + 12 )/2 for n = 0 , 1, 2, . . . and the

corresponding eigenfunctions, Z n (z) = cos (n + 12 )z/ . Negative values of n areexcluded as they repeat eigenvalues.

(d) F F = 0 with F (0) = F ( ) = 0 , taking F = F (f ) : = 0 gives F = 0, so F = a + bf . BCs give F (0) = a = 0, so a = 0, and

F ( ) = b = 0, so b = 0. Hence F 0. Not an eigenvalue. > 0, with = 2 = 0, gives F 2F = 0, so F (f ) = aef + be f .

BCs give F (0) = a + b = 0 and F ( ) = ae be = 0, so b = aand a(e + e ) = 0 so that a = b = 0 (since e + e = 0). Hence F 0.Not an eigenvalue.

< 0, with = 2 = 0, gives F + 2F = 0, so F (f ) = a cos(f ) + bsin(f ).BCs give F (0) = a cos 0 + bsin 0 = a = 0, so a = 0, andF ( ) = bcos( ) = 0, so it is possible to have b = 0 only if cos( ) = 0.That is, if = ( n + 12 ) for integers n then F n (f ) = bsin (n +

12 )f/ .

Hence we nd the eigenvalues, = (n + 12 )/2 for n = 1 , 2, . . .

and the corresponding eigenfunctions F n (f ) = sin (n + 12 )f/ . The choice n = 0gives the trivial function and n give the same eigenfunction for the same choice of n .



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2. We have: (1 + t)u t = u xx with the B.C.s: u(t, ) = 0 and u(t, ) = 0 .

Separate Variables Assuming u(x, t ) = T (t)X (x) gives (1 + t)T X = TX or

X (x)X (x) =

1 + t

T (t)T (t) =

with the B.C.s giving T (t)X ( ) = 0 and T (t)X () = 0.

Eigenvalue problem So X X = 0 and (1 + t)T T = 0

with B.C.s: X ( ) = 0 and X ( ) = 0 , since we want to have T (t) 0.

= 0 If = 0 then X (x) = ax + b.BC X ( ) = 0 gives b a = 0,BC X () = 0 gives b + a = 0.Solve simultaneous equations to get a = b = 0.

Not an eigenvalue. > 0 If = 2 > 0 then X (x) = aex + be x .

BC X ( ) = 0 gives ae + be = 0 ,BC X () = 0 gives ae + be = 0Solve simultaneous eqns to get a = be2 and hence be3 + be = 0,leading to a = b = 0 . Not an eigenvalue.

< 0 If = 2 < 0 then X (x) = a cos(x) + bsin(x) and the B.C.s give:

a cos() bsin() = 0a cos() + bsin() = 0

a cos() = 0bsin() = 0

There are two possibilities here: cos( ) = 0, in which case we can nd non-

trivial eigenfunctions with a = 0.Or sin( ) = 0, in which case we can nd non-trivial eigenfunctions with b = 0.Case 1 cos() = 0. Then = ( n + 1 / 2) for integers n . In this case b = 0

and a = 0 so X (x) = a cos(x)Case 2 sin() = 0. Then = n for integers n . In this case a = 0 and

b = 0, so X (x) = bsin(x).Putting everything together by setting m = 2 n + 1 for case 1 and m = 2 n forcase 2, the eigenvalues values are

m = 12

m

2, m = 1 , 2, 3, . . . ,

with corresponding eigenfunctions

X m (x) =sin( 12 mx/ ), m = 2 n evencos( 12 mx/ ), m = 2 n + 1 odd.

Solution for T Correspondingly, we haveT T

=

1 + tso that ln T = ln(1 + t) + const. or T = const. (1 + t) .

Ignoring constant multiples, the solutions are therefore

um (t, x ) = (1 + t) (14 m

2 2 / 2 ) sin( 12 mx/ ) : m evencos( 12 mx/ ) : m odd



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Initial condition With initial condition u(0, x) = A cos(x/ 2), the solution is givenby the case m = 1. The solution

u(t, x ) = A(1 + t)14

2 / 2 cosx2

.

3. (a) The trig identity cos A cos B = (cos( A B ) + cos( A + B ))/ 2 gives

cos(nx/ )cos(mx/ ) =12

cos((n m)x/ ) + cos(( n + m)x/ ) .

It is simple to integrate over [0 , ] and show that

0 cos(nx/ )cos(mx/ ) dx = 0, n = m/2, n = m.The fact that the integral is zero for n = m is known as orthogonality. To determinethe Fourier coefficient an in

x =

n =0an cos

nx

multiply by cos( mx/ ) and integrate over (0 , ):

0 x cos mx dx =

n =0an 0 cos nx cos mx dx.

Notice that when n = m the terms in the sum are zero (because the integral is zerodue to orthogonality). Hence only one term remains from the sum, the case n = mand so

0

x cosmx

dx = am

l

0cos2

mxdx

Using the rst part, for m = 0 ,

0 x dx = 2/ 2 = a0 a0 = /2.and for m > 0

0 x cos(mx ) dx = am 2where

0 x cosmx

dx = x m sinmx

0 0 m sinmx

dx

=m

2cos

mx0

=m

2( 1)m 1 =

2 m2 for m odd

0 for m even.

Writing m = 2 k + 1 for m odd,

2 2

(2k + 1) 22= a2k+1 2



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ora2k+1 =

4(2k + 1) 22

with a2k = 0 for the even case. Finally,

x =2

k=0

4

(2k + 1) 22cos

(2k + 1) x.

(b) The calculation for

0 cos (n + 12 )x cos (m + 12 )x dxis the similar to before, integrating after using the trig identity.Now

=

n =0an cos

(n + 12 )x

Multiply by cos(( m + 1 / 2)x/ ) and integrate over (0 , ) :

0 cos (m+12 )x dx =

n =0an 0 cos (n+

12 )x cos (m+

12 )x dx

= am 0 cos2 (m+ 12 )x dx,where we use orthogonality to eliminate all terms in the sum with n = m . Evaluating:

0 cos (m + 12 )x dx = (m + 12 ) sin (m +12 )x

0

=( 1)m

m + 12.

Hence( 1)n

n + 12= an 2 or an =

2( 1)n

n + 12

and so =

n =0

2( 1)n

n + 12 cos(n + 1

2)x

.

4. Using the trigonometric identities

sin2(A) =12

(1 cos(2A)) , sin(A)sin( B ) =12

(cos(A B ) cos(A + B ))

It is easy to derive the orthogonality relation.Now

f (x) =+1 for 0 x < 1 for x 0

=

n =1bn sin nx,

multiply by sin mx and integrate over [ , ]

f (x)sin( mx ) dx =

n =1an sin nx sin mxdx = am .

Hence, the Fourier coefficients

bn =1 0 sin nxdx + 1 0 sin nxdx

=1

cos nxn

0

+

cos nxn

0

=1

1 ( 1)n

n

( 1)n 1n

=2 [1 ( 1)n ]

n.



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If n be even let n = 2 r then 1 ( 1)2r = 0 , i.e. b2r = 0. If n be odd let n = 2 r +1 then

b2r +1 =4

(2r + 1)

and

f (x) = 4

r =0sin(2r + 1) x

2r + 1

5. Using the trigonometric identities

cos2(A) =12

(1 + cos(2 A)) , cos(A)cos(B ) =12

(cos(A B ) + cos( A + B ))

It is easy to derive the orthogonality relation.

f (x) =x for 0 x < x for x 0

=

n =0an cos nx,

Multiply by cos mx and integrate over [ , ]

f (x)cos(mx ) dx =

n =0an cos nx cos mxdx = a02, m = 0am , m = 1 , 2, . . .

Thus, the Fourier coefficients

a0 =1

2 0 x dx + 12 0 x dx = 12 12x20

+

12

12

x2

0

=1

4 +

1

4 = / 2.

an =1 0 x cos nxdx + 1 0 x cos nxdx

= 1

xsin nx

n

0

0 sin nxn dx

+1

xsin nx

n

0 0 sin nxn dx

=1

n

cos nxn

0

cos nxn

0

= 1n

1 + ( 1)nn

+ ( 1)n 1n

= 2 ( 1)n 1

n2.

If n be even let n = 2 r then 1 ( 1)2r = 0 , i.e. a2r = 0. If n be odd let n = 2 r +1 then

a2r +1 = 4

(2r + 1) 2.

Hence

f (x) =12

4

r =0

cos (2r + 1) x(2r + 1) 2



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6. The relevant orthogonality relations are

20 sin(nx )sin( mx ) dx = 20 sin(nx )cos(mx ) dx = 20 cos(nx )cos(mx ) dx = 0for n = m . Further

20 cos2(nx ) dx = 2, n = 0 ,1, n = 1 , 2, . . . , 20 sin2(nx ) dx = 0, n = 0 ,1, n = 1 , 2, . . .They are veried using trigonometric identities (as in the previous questions).Now, for 0 < x < 2, we have the Fourier series

f (x) = 4 x2 = a0 +

n =1an cos(nx ) + bn sin(nx ).

Multiply by 1 = cos(0) and integrate over [0 , 2] gives

2a0 = 2

04 x2 dx = 4x

13

x32

0=

163

Multiply by cos( nx ) and integrate over [0 , 2] gives

an = 20 4 x2 cos nxdx= 4 x2

sin( nx )n

2

0+ 20 2x sin(nx )n dx

=2

n x

cos(nx )

n

2

0

+

2

0

cos(nx )

ndx

= 4

n22+

2n22

sin(nx )n

2

0=

4n22

,

Multiply by sin( nx ) and integrate over [0 , 2] gives

bn = 20 4 x2 sin nxdx= 4 x2

cos(nx )n

2

0 20 2x cos(nx )n dx

=4

n 2

n xsin( nx )

n

2

0

2

0

sin(nx )n dx

=4

n+

2n22

cos(nx )n

2

0=

4n

.

Hence

4 x2 =83

4

2

n =1

cos(nx )n2

+4

n =1

sin(nx )n

.

7. We have u t = uxx with the B.C.s u(t, 0) = 0 and ux (t, a ) = 0 .



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Separate Variables Assuming u(x, t ) = T (t)X (x) and that T (t) 0 gives T X = T X

orX X

=T T

= with the B.C.s giving T (t)X (0) = 0 and T (t)X (a) = 0.

Eigenvalue problem So T T = 0 and X X = 0 with X (0) = X (a) = 0 .If = 0 then X = ax + b with b = 0 , a = 0 so that a = b = 0. Not an eigenvalue.

If = 2 > 0 then X = aex + be x with a + b = 0 , aea be a = 0.so that a = b = 0. Not an eigenvalue.

If = 2 < 0 then X = a cos(x) + bsin(x) and the B.C.s give:

a = 0 and bcos(a) a sin(a) = 0 so that a = 0 and bcos(a) = 0 , allowingb = 0 only if cos( a) = 0 or a = ( 12 + n) for n = 0 , 1, 2, . . . .

Thus eigenfunctions X n (x) = sin(( 12 + n)x/a ) are found for the eigenvalues n =

(( 12 + n)/a )2, n = 0 , 1, 2, . . . ,Solve for T Also, correspondingly, we have T T = 0 so that T = const. et . Then

un (t, x ) = e ((12 + n )/a )

2 t sin(1/ 2 + n)x

a.

These are all linearly independent so that we have an innite number of suitable solutions.

Superposition An innite series solution is u =

n =0bn e ((

12 + n )/a )

2 t sin(( 12 + n)x/a ),

where bn are to be determined.Boundary condition at t = 0 when u(0, x) = a for 0 < x < a, gives

a =

n =0bn sin(( 12 + n)x/a ).

Sturm-Liouville theory ensures that all of the eigenfunctions X n are orthogonal un-

der the inner product ( p,q) = a0 p(x)q(x) dxso that a0 a sin ( 12 + n)x/a dx = bn a0 sin2 ( 12 + n)x/a dx.Evaluating:

a

0a sin ( 12 + n)x/a dx = a

a

(12 + n)

cos ( 12 + n)x/aa

0

=a2

(12 + n)

and a0 sin2 ( 12 + n)x/a dx = a0 12 1 cos 2( 12 + n)x/a dx= 12 x

a(1 + 2 n)

sin((1 + 2 n)x/a )a

0= 12 a

This gives bn =4a

(1 + 2 n)and nal solutions is

u(t, x ) =

n =0

4a(1 + 2 n)

e ((12 + n )/a )

2t sin ( 12 + n)x/a .



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8.uxx + uyy = 0 with u(0, y) = u(, y) = 0, u(x, 0) = 0 and u(x, 1) = :

Separate Variables Assuming u(x, y) = X (x)Y (y) we can use the PDE with uxx =X (x)Y (y) and uyy = X (x)Y (y) to nd uxx + uyy = X Y + XY = 0.Dividing by XY (assumed not equal to zero) gives X

X= Y

Y = .

Thus we have X X = 0 and Y + Y = 0with constant because XX is independent of y and

Y Y is independent of x .

The homogeneous boundary conditions giveu(0, y) = X (0)Y (y) = 0, u(, y) = X ()Y (y) = 0, u(x, 0) = X (x)Y (0) = 0 .Since we seek XY = 0 this gives X (0) = X () = 0 and Y (0) = 0 .The boundary condition u(x, 1) = X (x)Y (1) = cannot be used being non-homogeneous.

Eigenvalue problem Solving X X = 0 with X (0) = X () = 0: = 0 : we have X = 0 giving X = A + Bx .

X (0) = X () = 0 give A = 0 and B = 0 so that A = B = 0.Only the trivial solution arises for = 0.

= 2 > 0 : we have X 2X = 0 giving X = Aex + Be x . X (0) = X () = 0gives A + B = 0 and Ae + Be = 0,i.e. (substituting) A(e e ) = 0 so that A = B = 0 since e e = 0 .Only trivial solutions arise for > 0.

= 2 < 0 : we have X + 2X = 0 giving X = A cos(x) + B sin(x).X (0) = X () = 0 give A = 0 and B sin() = 0.Thus we can have B = 0 if and only if sin( ) = 0or = n for any n = 1 , 2 , 3 , . . . .

Hence the only eigenvalues are = n = n2 for n Nwith corresponding eigenfunctions X = X n = sin( nx ) .

Solve for Y Solving Y + Y = 0 with Y (0) = 0 :For any = n2 we have Y n2Y = 0. Thus Y = Aeny + Be ny .The condition Y (0) = 0 gives A + B = 0 so, substituting,Y = A(eny e ny ) = 2 A e

ny e ny2

or Y = Y n (y) = sinh( ny), multiplied by any constant.Superposition Since the solutions un (x, y ) = X n (x)Y n (y) all satisfy a homogeneous

PDE with homogeneous boundary conditions, the principle of superposition meansthat any linear combination of such solutions is also a solution. Thus any sum

u =

n =1An X n Y n =

n =1An sinh( ny ) sin(nx )

is also a solution, for constants An .Boundary conditions At y = 1 we have u(x, 1) = so that

n =1An sinh( n)sin( nx ) = .

Since the eigenfunctions X n = sin( nx ) are orthogonal under the inner product(f, g ) = 0 f (x)g(x) dx we have

An sinh( n)

0sin2(nx ) dx =

0sin(nx ) dx.



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Integrating 0 sin2(nx ) dx = 0 12 (1 cos(2nx )) dx = 2 [ 12n sin(2nx )]0 = 2 and 0 sin(nx ) dx = [ 1n cos(nx )]0 = 2n if n is odd or 0 if n is even.Thus An sinh( n) 2 = 2n or 0 so that An = 4n sinh( n ) or 0.Setting n = 2 k + 1, the solution can therefore be written as

u(x, y) =

k=0

42k + 1

sinh (2k + 1) ysinh(2 k + 1) sin (2k + 1) x .

9. These answers are more abbreviated.

(a) uxx + uyy = 0 with ux (0, y) = u(, y) = 0, u(x, 0) = 0 and u(x, 1) = 12 :

Setting u = X (x)Y (y) the PDE becomes X Y + XY = 0 and so, for XY non-zero,XX =

Y Y = with =

XX independent of y and =

Y Y independent of x

so that is constant. Thus X X = 0 and Y + Y = 0. Homogeneous BCsgive ux (0, y) = X (0)Y (y) = 0, u(, y) = X ()Y (y) = 0, u(x, 0) = X (x)Y (0) = 0 sothat, for X (x) and Y (y) non-zero, X (0) = X () = 0 and Y (0) = 0 .

Solving X X = 0 with X (0) = X () = 0 gives eigenvalues = (n + 12)2 for

n = 0, 1, 2, . . . and eigenfunctions X = cos (n + 12 )x .

Solving Y (n + 12 )2Y = 0 with Y (0) = 0 gives Y = Ae(n +

12 )y + Be (n +

12 )y .

BC gives Y (0) = A + B = 0 so that B = A and Y = A(e(n +12 )y e (n +

12 )y) =

2A e( n + 12 ) y e ( n +

12 ) y

2 or Y = sinh (n +12 )y times any constant.

Thus u = sinh (n + 12 )y cos (n +12 )x is a solution for any n

Adding these solutions of the homogeneous problem gives the general solution u =n =0 An sinh (n +

12 )y cos (n +

12 )x .

At y = 1, using u(x, 1) = 12 gives12 =

n =0 An sinh( n +

12 )cos (n +

12 )x so

that, from orthogonality of the eigenfunctions, 12

0 cos (n +12 )x dx = An sinh( n +

12 ) 0 cos2 (n + 12 )x dx . Integrating

0 cos2 (n + 12 )x dx = 2 and 0 cos (n +

12 )x dx =

1n + 12

[sin (n + 12 )x ]0 =

( 1) n

n + 12.

Thus 12 ( 1) n

n + 12= 12 A n sinh( n +

12 ) so that An =

( 1) n

n + 121

sinh( n + 12 )and so

u =

n =0

( 1)n

n + 12

sinh (n + 12 )ysinh( n + 12 )

cos (n + 12 )x .

(b) uxx + uyy = 0 with u(0, y) = ux (, y) = 0, uy(x, 0) = 0 and u(x, 1) = 12 :

Setting u = X (x)Y (y) the PDE becomes X Y + XY = 0 and so, for XY non-zero,X

X= Y

Y = with = X

Xindependent of y and = Y

Y independent of x so

that is constant. Thus X X = 0 and Y + Y = 0. Homogeneous BCs giveu(0, y) = X (0)Y (y) = 0, ux (, y) = X ()Y (y) = 0, uy(x, 0) = X (x)Y (0) = 0 sothat, for X (x) and Y (y) non-zero, X (0) = X () = 0 and Y (0) = 0 .

Solving X X = 0 with X (0) = X () = 0 gives eigenvalues = (n + 12 )2 for

n = 0, 1, 2, . . . and eigenfunctions X = sin (n + 12 )x .

Solving Y (n + 12 )2Y = 0 with Y (0) = 0 gives Y = Ae(n +

12 )y + Be (n +

12 )y . BC gives

Y (0) = ( n + 12 )A (n +12 )B = 0 so that B = A and Y = A(e

(n + 12 )y + e (n +12 )y) =

2A e( n + 12 ) y + e ( n +

12 ) y

2 or Y = cosh (n +12 )y times any constant.

Thus u = cosh (n + 12 )y sin (n +12 )x is a solution for any n



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Adding these solutions of the homogeneous problem gives the general solution u =n =0 An cosh (n +

12 )y sin (n +

12 )x .

At y = 1, using u(x, 1) = 12 gives12 =

n =0 An cosh(n +

12 )sin (n +

12 )x so

that, from orthogonality of the eigenfunctions, 12 0 sin (n + 12 )x dx = An cosh(n +12 )

0 sin

2 (n + 12 )x dx . Integrating

0 sin2 (n + 12 )x dx =

2 and

0 sin (n +12 )x dx = 1n + 12 [cos (n + 12 )x ]0 = 1n + 12 .Thus 12

1n + 12

= 12 A n cosh(n +12 ) so that An =

1/n + 12

1cosh( n + 12 )

and so

u =1

n =0

1n + 12

cosh (n + 12 )ycosh(n + 12 )

sin (n + 12 )x .

10. (a) ( p,q) = p(x) q(x) dx = q(x) p(x) dx = ( q, p).(b) ( p,p) =

( p(x))2 dx which is positive or zero because ( p(x)) 2 is positive or zero

throughout the range of integration.(c) Since p(x) is continuous for all x [, ] it follows that ( p(x))2 is continuous as

well as non-negative. If ( p(a))2 = b > 0 at some point a then, because( p(x)) 2 is continuous, for any > 0 > 0 such that |( p(x)) 2 b| < for allx (a , a+ ) [, ]. Let us choose = b/ 2 then |( p(x))2 b| < b/ 2 for allx (a , a+ ) [, ], an interval of length at least , over which ( p(x)) 2 > b/ 2.This would contribute at least b/2 > 0 to ( p,p), making it non-zero. Hence if ( p,p) = 0 then ( p(x)) 2 , and therefore p(x), must be zero at all points x [, ].

(d) ( c1 p1 + c2 p2, q) = c1 p1(x) q(x) dx + c2 p2(x) q(x) dx = c1( p1, q) + c2( p2, q).To show linear independence we assume that c1 p1 + c2 p2 = 0 for nonzero c1 and c2 andtry to show that c1 = c2 = 0, so that we have a contradiction. Taking the inner productwith p1 gives

0 = ( c1 p1 + c2 p2, p1) = c1( p1, p1) + c2( p2, p1) = c1( p1, p1)

since ( p2, p1) = 0. Also ( p1, p1) > 0 from (b) and (c) and hence c1 = 0. Taking the innerproduct with p2 and applying the same argument shows that c2 = 0 .

11. By assumption, u and v satisfy ( ) :

( pu ) + qu = 0 , x (0, 1); u (0) = 0 , u(1) = 0 ,

( pv ) + qv = 0 , x (0, 1); v (0) = 0 , v(1) = 0 .(a) Integrating by parts gives

(Lu,v ) = 10 ( pu ) v dx + 10 quv dx= 10 ( pu )v dx pu v 10 + 10 quv dx= 10 pu v dx + 10 quv dx,



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where the boundary term [ pu v]10 disappears because v(1) = 0 and u (0) = 0. Simi-larly

(u,Lv ) = 10 ( pv ) u dx + 10 quv dx=

1

0( pv )u dx pv u 10 +

1

0quv dx

= 10 pu v dx + 10 quv dx = ( Lu,v ). (b) From above,

(Lu,u ) = 10 p(u )2 dx 0

+ 10 qu2 dx 10 qu2 dx.now since u = 0 we have u2 > 0, so that qu2 > 0 in (0,1). This means that

1

0 qu2 dx > 0 and thus ( Lu,u ) > 0. Moreover since u satises ( ) , (Lu,u ) =

(wu, u ) = 1

0wu2 dx

> 0> 0. Thus > 0.

(c) We know that Ln = n wn and Lm = m wm . Thus, using Lagranges identity

(n , wm ) =1

m(n , Lm ) =

1m

(Ln , m ) =nm

(wn , m ) =nm

(n , wm ),

So that1

nm

(n , wm ) = 0 ,

since n m = 1 we must have that ( n , wm ) = 0 . (d) Given that f (x) = n =1 cn n (x). Multiplying by the weight function w and taking

the inner product with k gives, using orthogonality,

(f, w k ) = (

n =1cn n , wk ) =

n =1cn (n , wk) = ck(k , wk),

thusck =

(f, w k )(k , w k )

.

12. We are looking for solutions to the following problem:

u tt urr 2rur = 0 with lim

r 0u(r, t ) < and u(1, t ) = 0, for all t > 0.

Assuming u = R(r )T (t) we see that u tt urr 2r ur = RT R T 2r R T = 0 .

Dividing by RT (assumed not equal to zero) gives T T =RR +

2r

RR = .

Thus we have R + 2r R = R and T = T . To solve R + 2r R = R with lim r 0 R(r ) < and R(1) = 0, we multiply by r

to give rR + 2 R = rR and make the change of variable X = rR . This givesX = R + rR and X = 2 R + rR , so that X = X .Only trivial solutions arise for 0, so we set = 2 < 0. Thus we haveX = 2X giving X (r ) = A cos(r ) + B sin(r ) or R(r ) = Ar cos(r ) +

Br sin(r ).



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The condition lim r 0 R(r ) < implies that A = 0, thus R(r ) = Br sin(r ). Thecondition R(1) = 0 then gives B sin() = 0. Thus we can have B = 0 if and only if sin() = 0or = n for any n = 1 , 2 , 3 , . . . .Hence the only eigenvalues are = n = (n )2 for n N

with corresponding eigenfunctions R = Rn =sin( nr )

r . Solving T = n T :

For any n = n22 we have T n = n22T n . Thus T n (t) = An sin(nt ) +Bn cos(nt ) .

Since the solutions u = Rn (r )T n (t) all satisfy a homogeneous PDE with homogeneousboundary conditions, the principle of superposition means that any linear combina-tion of such solutions is also a solution. Thus a convergent sum

u(r, t ) =

n =1(an sin(nt ) + bn cos(nt ))

sin(nr )r

is the general solution, with the constants an and bn uniquely determined by theinitial conditions.

pdes - solutions (2)

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