PDEs - Solutions (2)
-
Upload
scribd6289 -
Category
Documents
-
view
214 -
download
0
Transcript of PDEs - Solutions (2)
-
8/3/2019 PDEs - Solutions (2)
1/13
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
-
8/3/2019 PDEs - Solutions (2)
2/13
(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 .
MATH20401(PDEs): Answer Sheet II: Page 2
-
8/3/2019 PDEs - Solutions (2)
3/13
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
MATH20401(PDEs): Answer Sheet II: Page 3
-
8/3/2019 PDEs - Solutions (2)
4/13
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
MATH20401(PDEs): Answer Sheet II: Page 4
-
8/3/2019 PDEs - Solutions (2)
5/13
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.
MATH20401(PDEs): Answer Sheet II: Page 5
-
8/3/2019 PDEs - Solutions (2)
6/13
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
MATH20401(PDEs): Answer Sheet II: Page 6
-
8/3/2019 PDEs - Solutions (2)
7/13
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 .
MATH20401(PDEs): Answer Sheet II: Page 7
-
8/3/2019 PDEs - Solutions (2)
8/13
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 .
MATH20401(PDEs): Answer Sheet II: Page 8
-
8/3/2019 PDEs - Solutions (2)
9/13
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.
MATH20401(PDEs): Answer Sheet II: Page 9
-
8/3/2019 PDEs - Solutions (2)
10/13
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
MATH20401(PDEs): Answer Sheet II: Page 10
-
8/3/2019 PDEs - Solutions (2)
11/13
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,
MATH20401(PDEs): Answer Sheet II: Page 11
-
8/3/2019 PDEs - Solutions (2)
12/13
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 ).
MATH20401(PDEs): Answer Sheet II: Page 12
-
8/3/2019 PDEs - Solutions (2)
13/13
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.