HW2, Math 322, Fall 2016
Nasser M. Abbasi
December 30, 2019
Contents
0.1 Summary table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.2 section 2.3.1 (problem 1) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
0.2.1 part (a) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40.2.2 Part (b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50.2.3 Part (d) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
0.3 section 2.3.2 (problem 2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70.3.1 Part (d) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70.3.2 Part (f) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90.3.3 Part (g) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
0.4 section 2.3.3 (problem 3) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130.4.1 Part (b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130.4.2 Part (d) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
0.5 section 2.3.4 (problem 4) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180.5.1 Part (a) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180.5.2 Part (b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190.5.3 Part (c) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
0.6 section 2.3.5 (problem 5) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200.7 section 2.3.7 (problem 6) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
0.7.1 part (a) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220.7.2 Part (b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220.7.3 Part c . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250.7.4 Part d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 260.7.5 Part (e) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
0.8 section 2.3.8 (problem 7) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270.8.1 part (a) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270.8.2 Part (b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
0.9 section 2.3.10 (problem 8) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 300.9.1 Part b . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 310.9.2 Part c . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
0.10 section 2.4.1 (problem 9) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
1
2
0.10.1 Part (b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 320.10.2 Part (c) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
0.11 section 2.4.2 (problem 10) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 340.12 section 2.4.3 (problem 11) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 370.13 section 2.4.6 (problem 12) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
0.13.1 Part (a) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 420.13.2 Part (b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
3
0.1 Summary table
For 1π· bar
Left Right π = 0 π > 0 π’ (π₯, π‘)
π’ (0) = 0 π’ (πΏ) = 0 Noππ = οΏ½
πππΏοΏ½2, π = 1, 2, 3,β―
ππ = π΅π sin οΏ½βπππ₯οΏ½ββπ=1 π΅π sin οΏ½βπππ₯οΏ½ πβππππ‘
π’ (0) = 0 ππ’(πΏ)ππ₯ = 0 No
ππ = οΏ½ππ2πΏοΏ½2, π = 1, 3, 5,β―
ππ = π΅π sin οΏ½βπππ₯οΏ½ββπ=1,3,5,β― π΅π sin οΏ½βπππ₯οΏ½ πβππππ‘
ππ’(0)ππ₯ = 0 π’ (πΏ) = 0 No
ππ = οΏ½ππ2πΏοΏ½2, π = 1, 3, 5,β―
ππ = π΄π cos οΏ½βπππ₯οΏ½ββπ=1,3,5β―π΄π cos οΏ½βπππ₯οΏ½ πβππππ‘
π’ (0) = 0 π’ (πΏ) + ππ’(πΏ)ππ₯ = 0
π0 = 0π0 = π΄0
tan οΏ½βπππΏοΏ½ = βππππ = π΅π sin οΏ½βπππ₯οΏ½
π΄0 +ββπ=1 π΅π sin οΏ½βπππ₯οΏ½ πβππππ‘
ππ’(0)ππ₯ = 0 ππ’(πΏ)
ππ₯ = 0π0 = 0π0 = π΄0
ππ = οΏ½πππΏοΏ½2, π = 1, 2, 3,β―
ππ = π΄π cos οΏ½βπππ₯οΏ½π΄0 +β
βπ=1π΄π cos οΏ½βπππ₯οΏ½ πβππππ‘
For periodic conditions π’ (βπΏ) = π’ (πΏ) and ππ’(βπΏ)ππ₯ = ππ’(πΏ)
ππ₯
ππ = οΏ½πππΏοΏ½2, π = 1, 2, 3,β―
π’ (π₯, π‘) =π=0βπ0 +
π>0
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½βοΏ½π=1
π΄π cos οΏ½οΏ½πππ₯οΏ½ πβππππ‘ +βοΏ½π=1
π΅π sin οΏ½οΏ½πππ₯οΏ½ πβππππ‘
Note on notation When using separation of variables π (π‘) is used for the time function andπ (π₯) , π (π) , Ξ (π) etc. for the spatial functions. This notation is more common in other booksand easier to work with as the dependent variable π,π,β― and the independent variableπ‘, π₯,β― are easier to match (one is upper case and is one lower case) and this produces lesssymbols to remember and less chance of mixing wrong letters.
4
0.2 section 2.3.1 (problem 1)2.3. Heat Equation With Zero Temperature Ends
EXERCISES 2.3
55
2.3.1. For the following partial differential equations, what ordinary differentialequations are implied by the method of separation of variables?
(a) au ka (r2u)
* at r ar &
a2u a2u*
(C) 09x2 + ft2 = o
&U 04U
*(e) at = ka 44
a2u 2 02U
ate ax
2.3.2. Consider the differential equation
z2+A0=0.
Determine the eigenvalues \ (and corresponding eigenfunctions) if 0 satisfiesthe following boundary conditions. Analyze three cases (.\ > 0, A = 0, A <0). You may assume that the eigenvalues are real.
(a) 0(0) = 0 and 0(-,r) = 0*(b) 0(0) = 0 and 5(1) = 0
(c) !LO (0) = 0 and LO (L) = 0 (If necessary, see Sec. 2.4.1.)
*(d) 0(0) = 0 and O (L) = 0
(e) LO (0) = 0 and O(L) = 0
*(f) O(a) = 0 and O(b) = 0 (You may assume that A > 0.)
(g) Β’(0) = 0 and LO(L)
+ cb(L) = 0 (If necessary, see Sec. 5.8.)
2.3.3. Consider the heat equation
OU 82U
at - kax2subject to the boundary conditions
u(0,t) = 0 and u(L,t) = 0.
Solve the initial value problem if the temperature is initially
(a) u(x, 0) = 6 sin s (b) u(x, 0) = 3 sin i - sin i
(b) -` = k09x22
- v0 ax
(d)
* (f)=c
* (c) u(x, 0) = 2 cos lmE (d) u(x, 0)1 0 < x < L/22 L/2<x<L
0.2.1 part (a)
1πππ’ππ‘
=1ππππ οΏ½
πππ’ππ οΏ½
(1)
Let
π’ (π‘, π) = π (π‘) π (π)
Thenππ’ππ‘
= πβ² (π‘) π (π)
Andπππ οΏ½
πππ’ππ οΏ½
=ππ’ππ
+ ππ2π’ππ2
= ππ β² (π) + πππ β²β² (π)
Hence (1) becomes1ππβ² (π‘) π (π) =
1π(ππ β² (π) + πππ β²β² (π))
Note From now on πβ² (π‘) is written as just πβ² and similarly for π β² (π) = π β² and π β²β² (π) = π β²β² tosimplify notations and make it easier and more clear to read. The above is reduced to
1ππβ²π =
1πππ β² + ππ β²β²
Dividing throughout 1 by π (π‘) π (π) gives1ππβ²
π=1ππ β²
π +π β²β²
π Since each side in the above depends on a diοΏ½erent independent variable and both are equal
1π (π‘) π (π) can not be zero, as this would imply that either π (π‘) = 0 or π (π) = 0 or both are zero, in whichcase there is only the trivial solution.
5
to each others, then each side is equal to the same constant, say βπ. Therefore1ππβ²
π=1ππ β²
π +π β²β²
π = βπ
The following diοΏ½erential equations are obtained
πβ² + πππ = 0ππ β²β² + π β² + πππ = 0
In expanded form, the above isππππ‘+ πππ (π‘) = 0
ππ2π ππ2
+ππ ππ‘+ πππ (π) = 0
0.2.2 Part (b)
1πππ’ππ‘
=π2π’ππ₯2
βπ£0πππ’ππ₯
(1)
Let
π’ (π₯, π‘) = ππ
Thenππ’ππ‘
= πβ²π
Andππ’ππ₯
= πβ²π
π2π’ππ₯2
= πβ²β²π
Substituting these in (1) gives1ππβ²π = πβ²β²π β
π£0ππβ²π
Dividing throughout by ππ β 0 gives1ππβ²
π=πβ²β²
πβπ£0ππβ²
πSince each side in the above depends on a diοΏ½erent independent variable and both are equalto each others, then each side is equal to the same constant, say βπ. Therefore
1ππβ²
π=πβ²β²
πβπ£0ππβ²
π= βπ
The following diοΏ½erential equations are obtained
πβ² + πππ = 0
πβ²β² βπ£0ππβ² + ππ = 0
6
The above in expanded form isππππ‘+ πππ (π‘) = 0
π2πππ₯2
βπ£0πππππ₯
+ ππ (π₯) = 0
0.2.3 Part (d)
1πππ’ππ‘
=1π2πππ οΏ½
π2ππ’ππ οΏ½
(1)
Let
π’ (π‘, π) β‘ ππ
Thenππ’ππ‘
= πβ²π
Andπππ οΏ½
π2ππ’ππ οΏ½
= 2πππ’ππ
+ π2π2π’ππ2
= 2πππ β² + π2ππ β²β²
Substituting these in (1) gives1ππβ²π =
1π2οΏ½2πππ β² + π2ππ β²β²οΏ½
=2πππ β² + ππ β²β²
Dividing throughout by ππ β 0 gives1ππβ²
π=2ππ β²
π +π β²β²
π Since each side in the above depends on a diοΏ½erent independent variable and both are equalto each others, then each side is equal to the same constant, say βπ. Therefore
1ππβ²
π=2ππ β²
π +π β²β²
π = βπ
The following diοΏ½erential equations are obtained
πβ² + πππ = 0ππ β²β² + 2π β² + πππ = 0
The above in expanded form isππππ‘+ πππ (π‘) = 0
ππ2π ππ2
+ 2ππ ππ
+ πππ (π) = 0
7
0.3 section 2.3.2 (problem 2)
2.3. Heat Equation With Zero Temperature Ends
EXERCISES 2.3
55
2.3.1. For the following partial differential equations, what ordinary differentialequations are implied by the method of separation of variables?
(a) au ka (r2u)
* at r ar &
a2u a2u*
(C) 09x2 + ft2 = o
&U 04U
*(e) at = ka 44
a2u 2 02U
ate ax
2.3.2. Consider the differential equation
z2+A0=0.
Determine the eigenvalues \ (and corresponding eigenfunctions) if 0 satisfiesthe following boundary conditions. Analyze three cases (.\ > 0, A = 0, A <0). You may assume that the eigenvalues are real.
(a) 0(0) = 0 and 0(-,r) = 0*(b) 0(0) = 0 and 5(1) = 0
(c) !LO (0) = 0 and LO (L) = 0 (If necessary, see Sec. 2.4.1.)
*(d) 0(0) = 0 and O (L) = 0
(e) LO (0) = 0 and O(L) = 0
*(f) O(a) = 0 and O(b) = 0 (You may assume that A > 0.)
(g) Β’(0) = 0 and LO(L)
+ cb(L) = 0 (If necessary, see Sec. 5.8.)
2.3.3. Consider the heat equation
OU 82U
at - kax2subject to the boundary conditions
u(0,t) = 0 and u(L,t) = 0.
Solve the initial value problem if the temperature is initially
(a) u(x, 0) = 6 sin s (b) u(x, 0) = 3 sin i - sin i
(b) -` = k09x22
- v0 ax
(d)
* (f)=c
* (c) u(x, 0) = 2 cos lmE (d) u(x, 0)1 0 < x < L/22 L/2<x<L
0.3.1 Part (d)
π2πππ₯2
+ ππ = 0
π (0) = 0ππππ₯
(πΏ) = 0
Substituting an assumed solution of the form π = π΄πππ₯ in the above ODE and simplifyinggives the characteristic equation
π2 + π = 0π2 = βπ
π = Β±ββπ
Assuming π is real. The following cases are considered.
case π < 0 In this case, βπ and also ββπ, are positive. Hence both roots Β±ββπ are real andpositive. Let
ββπ = π
8
Where π > 0. Therefore the solution is
π (π₯) = π΄ππ π₯ + π΅πβπ π₯
ππππ₯
= π΄π ππ π₯ β π΅π πβπ π₯
Applying the first boundary conditions (B.C.) gives
0 = π (0)= π΄ + π΅
Applying the second B.C. gives
0 =ππππ₯
(πΏ)
= π΄π β π΅π = π (π΄ β π΅)= π΄ β π΅
The last step above was after dividing by π since π β 0. Therefore, the following two equationsare solved for π΄,π΅
0 = π΄ + π΅0 = π΄ β π΅
The second equation implies π΄ = π΅ and the first gives 2π΄ = 0 or π΄ = 0. Hence π΅ = 0.Therefore the only solution is the trivial solution π (π₯) = 0. π < 0 is not an eigenvalue.
case π = 0 In this case the ODE becomes
π2πππ₯2
= 0
The solution is
π (π₯) = π΄π₯ + π΅ππππ₯
= π΄
Applying the first B.C. gives
0 = π (0)= π΅
Applying the second B.C. gives
0 =ππππ₯
(πΏ)
= π΄
Hence π΄,π΅ are both zero in this case as well and the only solution is the trivial one π (π₯) = 0.π = 0 is not an eigenvalue.
case π > 0 In this case, βπ is negative, therefore the roots are both complex.
π = Β±πβπ
9
Hence the solution is
π (π₯) = π΄ππβππ₯ + π΅πβπβππ₯
Which can be writing in terms of cos, sin using Euler identity as
π (π₯) = π΄ cos οΏ½βππ₯οΏ½ + π΅ sin οΏ½βππ₯οΏ½
Applying first B.C. gives
0 = π (0)= π΄ cos (0) + π΅ sin (0)
0 = π΄
The solution now is π (π₯) = π΅ sin οΏ½βππ₯οΏ½ . Hence
ππππ₯
= βππ΅ cos οΏ½βππ₯οΏ½
Applying the second B.C. gives
0 =ππππ₯
(πΏ)
= βππ΅ cos οΏ½βππΏοΏ½
= βππ΅ cos οΏ½βππΏοΏ½
Since π β 0 then either π΅ = 0 or cos οΏ½βππΏοΏ½ = 0. But π΅ = 0 gives trivial solution, therefore
cos οΏ½βππΏοΏ½ = 0
This implies
βππΏ =ππ2
π = 1, 3, 5,β―
In other words, for all positive odd integers. π < 0 can not be used since π is assumedpositive.
π = οΏ½ππ2πΏ οΏ½2
π = 1, 3, 5,β―
The eigenfunctions associated with these eigenvalues are
ππ (π₯) = π΅π sin οΏ½ππ2πΏπ₯οΏ½ π = 1, 3, 5,β―
0.3.2 Part (f)
π2πππ₯2
+ ππ = 0
π (π) = 0π (π) = 0
It is easier to solve this if one boundary condition was at π₯ = 0. (So that one constant drops
10
out). Let π = π₯ β π and the ODE becomes (where now the independent variable is π)π2π (π)ππ2
+ ππ (π) = 0 (1)
With the new boundary conditions π (0) = 0 and π (π β π) = 0. Assuming the solution isπ = π΄πππ, the characteristic equation is
π2 + π = 0π2 = βπ
π = Β±ββπ
Assuming π is real and also assuming π > 0 (per the problem statement) then βπ is negative,and both roots are complex.
π = Β±πβπ
This gives the solution
π (π) = π΄ cos οΏ½βπποΏ½ + π΅ sin οΏ½βπποΏ½
Applying first B.C.
0 = π (0)= π΄ cos 0 + π΅ sin 0= π΄
Therefore the solution is π (π) = π΅ sin οΏ½βπποΏ½. Applying the second B.C.
0 = π (π β π)
= π΅ sin οΏ½βπ (π β π)οΏ½
π΅ = 0 leads to trivial solution. Choosing sin οΏ½βπ (π β π)οΏ½ = 0 gives
οΏ½ππ (π β π) = ππ
οΏ½ππ =ππ
(π β π)π = 1, 2, 3β―
Or
ππ = οΏ½πππβποΏ½2
π = 1, 2, 3,β―
The eigenfunctions associated with these eigenvalue are
ππ (π) = π΅π sin οΏ½οΏ½ππποΏ½
= π΅π sin οΏ½ππ
(π β π)ποΏ½
Transforming back to π₯
ππ (π₯) = π΅π sin οΏ½ππ
(π β π)(π₯ β π)οΏ½
11
0.3.3 Part (g)
π2πππ₯2
+ ππ = 0
π (0) = 0ππππ₯
(πΏ) + π (πΏ) = 0
Assuming solution is π = π΄πππ₯, the characteristic equation is
π2 + π = 0π2 = βπ
π = Β±ββπ
The following cases are considered.
case π < 0 In this case βπ and also ββπ are positive. Hence the roots Β±ββπ are both real.Let
ββπ = π
Where π > 0. This gives the solution
π (π₯) = π΄0ππ π₯ + π΅0πβπ π₯
Which can be manipulated using sinh (π π₯) = ππ π₯βπβπ π₯
2 , cosh (π π₯) = ππ π₯+πβπ π₯
2 to the following
π (π₯) = π΄ cosh (π π₯) + π΅ sinh (π π₯)Where π΄,π΅ above are new constants. Applying the left boundary condition gives
0 = π (0)= π΄
The solution becomes π (π₯) = π΅ sinh (π π₯) and henceππππ₯ = π cosh (π π₯) . Applying the right
boundary conditions gives
0 = π (πΏ) +ππππ₯
(πΏ)
= π΅ sinh (π πΏ) + π΅π cosh (π πΏ)= π΅ (sinh (π πΏ) + π cosh (π πΏ))
But π΅ = 0 leads to trivial solution, therefore the other option is that
sinh (π πΏ) + π cosh (π πΏ) = 0But the above is
tanh (π πΏ) = βπ Since it was assumed that π > 0 then the RHS in the above is a negative quantity. Howeverthe tanh function is positive for positive argument and negative for negative argument.The above implies then that π πΏ < 0. Which is invalid since it was assumed π > 0 and πΏis the length of the bar. Hence π΅ = 0 is the only choice, and this leads to trivial solution.π < 0 is not an eigenvalue.
12
case π = 0
In this case, the ODE becomes
π2πππ₯2
= 0
The solution is
π (π₯) = π1π₯ + π2Applying left B.C. gives
0 = π (0)= π2
The solution becomes π (π₯) = π1π₯. Applying the right B.C. gives
0 = π (πΏ) +ππππ₯
(πΏ)
= π1πΏ + π1= π1 (1 + πΏ)
Since π1 = 0 leads to trivial solution, then 1 + πΏ = 0 is the only other choice. But thisinvalid since πΏ > 0 (length of the bar). Hence π1 = 0 and this leads to trivial solution.π = 0 is not an eigenvalue.
case π > 0
This implies that βπ is negative, and therefore the roots are both complex.
π = Β±πβπ
This gives the solution
π (π₯) = π΄ππβππ₯ + π΅πβπβππ₯
= π΄ cos οΏ½βππ₯οΏ½ + π΅ sin οΏ½βππ₯οΏ½
Applying first B.C. gives
π (0) = 0 = π΄ cos (0) + π΅ sin (0)0 = π΄
The solution becomes π (π₯) = π΅ sin οΏ½βππ₯οΏ½ andππππ₯
= βππ΅ cos οΏ½βππ₯οΏ½
Applying the second B.C.
0 =ππππ₯
(πΏ) + π (πΏ)
= βππ΅ cos οΏ½βππΏοΏ½ + π΅ sin οΏ½βππΏοΏ½ (1)
Dividing (1) by cos οΏ½βππΏοΏ½ , which can not be zero, because if cos οΏ½βππΏοΏ½ = 0, then π΅ sin οΏ½βππΏοΏ½ =
13
0 from above, and this means the trivial solution, results in
π΅ οΏ½βπ + tan οΏ½βππΏοΏ½οΏ½ = 0
But π΅ β 0, else the solution is trivial. Therefore
tan οΏ½βππΏοΏ½ = ββπ
The eigenvalue π is given by the solution to the above nonlinear equation. The text book, insection 5.4, page 196 gives the following approximate (asymptotic) solution which becomesaccurate only for large π and not used here
οΏ½ππ βΌππΏ οΏ½π β
12οΏ½
Therefore the eigenfunction is
ππ (π₯) = π΅ sin οΏ½βππ₯οΏ½
Where π is the solution to tan οΏ½βππΏοΏ½ = ββπ.
0.4 section 2.3.3 (problem 3)
2.3. Heat Equation With Zero Temperature Ends
EXERCISES 2.3
55
2.3.1. For the following partial differential equations, what ordinary differentialequations are implied by the method of separation of variables?
(a) au ka (r2u)
* at r ar &
a2u a2u*
(C) 09x2 + ft2 = o
&U 04U
*(e) at = ka 44
a2u 2 02U
ate ax
2.3.2. Consider the differential equation
z2+A0=0.
Determine the eigenvalues \ (and corresponding eigenfunctions) if 0 satisfiesthe following boundary conditions. Analyze three cases (.\ > 0, A = 0, A <0). You may assume that the eigenvalues are real.
(a) 0(0) = 0 and 0(-,r) = 0*(b) 0(0) = 0 and 5(1) = 0
(c) !LO (0) = 0 and LO (L) = 0 (If necessary, see Sec. 2.4.1.)
*(d) 0(0) = 0 and O (L) = 0
(e) LO (0) = 0 and O(L) = 0
*(f) O(a) = 0 and O(b) = 0 (You may assume that A > 0.)
(g) Β’(0) = 0 and LO(L)
+ cb(L) = 0 (If necessary, see Sec. 5.8.)
2.3.3. Consider the heat equation
OU 82U
at - kax2subject to the boundary conditions
u(0,t) = 0 and u(L,t) = 0.
Solve the initial value problem if the temperature is initially
(a) u(x, 0) = 6 sin s (b) u(x, 0) = 3 sin i - sin i
(b) -` = k09x22
- v0 ax
(d)
* (f)=c
* (c) u(x, 0) = 2 cos lmE (d) u(x, 0)1 0 < x < L/22 L/2<x<L
0.4.1 Part (b)
ππ’ππ‘
= ππ2π’ππ₯2
Let π’ (π₯, π‘) = π (π‘) π (π₯), and the PDE becomes1ππβ²π = πβ²β²π
14
Dividing by ππ β 01ππβ²
π=πβ²β²
πSince each side depends on diοΏ½erent independent variable and both are equal, they mustbe both equal to same constant, say βπ where π is assumed to be real.
1ππβ²
π=πβ²β²
π= βπ
The two ODEβs are
πβ² + πππ = 0 (1)
πβ²β² + ππ = 0 (2)
Starting with the space ODE equation (2), with corresponding boundary conditions π (0) =0, π (πΏ) = 0. Assuming the solution is π (π₯) = πππ₯, Then the characteristic equation is
π2 + π = 0π2 = βπ
π = Β±ββπ
The following cases are considered.
case π < 0 In this case, βπ and also ββπ are positive. Hence the roots Β±ββπ are both real.Let
ββπ = π
Where π > 0. This gives the solution
π (π₯) = π΄ cosh (π π₯) + π΅ sinh (π π₯)Applying the left B.C. π (0) = 0 gives
0 = π΄ cosh (0) + π΅ sinh (0)= π΄
The solution becomes π (π₯) = π΅ sinh (π π₯). Applying the right B.C. π’ (πΏ, π‘) = 0 gives
0 = π΅ sinh (π πΏ)We want π΅ β 0 (else trivial solution). This means sinh (π πΏ) must be zero. But sinh (π πΏ) is zeroonly when its argument is zero. This means either πΏ = 0 which is not possible or π = 0, butwe assumed π β 0 in this case, therefore we run out of options to satisfy this case. Henceπ < 0 is not an eigenvalue.
case π = 0
The ODE becomesπ2πππ₯2
= 0
The solution is
π (π₯) = π1π₯ + π2
15
Applying left boundary conditions π (0) = 0 gives
0 = π (0)= π2
Hence the solution becomes π (π₯) = π1π₯. Applying the right B.C. gives
0 = π (πΏ)= π1πΏ
Hence π1 = 0. Hence trivial solution. π = 0 is not an eigenvalue.
case π > 0
Hence βπ is negative, and the roots are both complex.
π = Β±πβπ
The solution is
π (π₯) = π΄ cos οΏ½βππ₯οΏ½ + π΅ sin οΏ½βππ₯οΏ½
The boundary conditions are now applied. The first B.C. π (0) = 0 gives
0 = π΄ cos (0) + π΅ sin (0)= π΄
The ODE becomes π (π₯) = π΅ sin οΏ½βππ₯οΏ½. Applying the second B.C. gives
0 = π΅ sin οΏ½βππΏοΏ½
π΅ β 0 else the solution is trivial. Therefore taking
sin οΏ½βππΏοΏ½ = 0
οΏ½πππΏ = ππ π = 1, 2, 3,β―
Hence eigenvalues are
ππ =π2π2
πΏ2π = 1, 2, 3,β―
The eigenfunctions associated with these eigenvalues are
ππ (π₯) = π΅π sin οΏ½πππΏπ₯οΏ½
The time domain ODE is now solved. πβ² + ππππ = 0 has the solution
ππ (π‘) = πβππππ‘
For the same set of eigenvalues. Notice that there is no need to add a new constant in theabove as it will be absorbed in the π΅π when combined in the following step below. Thesolution to the PDE becomes
π’π (π₯, π‘) = ππ (π‘) ππ (π₯)
16
But for linear system the sum of eigenfunctions is also a solution, therefore
π’ (π₯, π‘) =βοΏ½π=1
π’π (π₯, π‘)
=βοΏ½π=1
π΅π sin οΏ½πππΏπ₯οΏ½ πβποΏ½
πππΏ οΏ½
2π‘
Initial conditions are now applied. Setting π‘ = 0, the above becomes
π’ (π₯, 0) = 3 sin ππ₯πΏβ sin 3ππ₯
πΏ=
βοΏ½π=1
π΅π sin οΏ½πππΏπ₯οΏ½
As the series is unique, the terms coeοΏ½cients must match for those shown only, and all otherπ΅π terms vanish. This means that by comparing terms
3 sin οΏ½ππ₯πΏοΏ½ β sin οΏ½
3ππ₯πΏ οΏ½ = π΅1 sin οΏ½ππ₯
πΏοΏ½ + π΅3 sin οΏ½
3ππΏπ₯οΏ½
Therefore
π΅1 = 3π΅3 = β1
And all other π΅π = 0. The solution is
π’ (π₯, π‘) = 3 sin οΏ½ππΏπ₯οΏ½ πβποΏ½
ππΏ οΏ½
2π‘ β sin οΏ½
3ππΏπ₯οΏ½ π
βποΏ½ 3ππΏ οΏ½2π‘
0.4.2 Part (d)
Part (b) found the solution to be
π’ (π₯, π‘) =βοΏ½π=1
π΅π sin οΏ½πππΏπ₯οΏ½ πβποΏ½
πππΏ οΏ½
2π‘
The new initial conditions are now applied.
π (π₯) =βοΏ½π=1
π΅π sin οΏ½πππΏπ₯οΏ½ (1)
Where
π (π₯) =
β§βͺβͺβ¨βͺβͺβ©1 0 < π₯ β€ πΏ/22 πΏ/2 < π₯ < πΏ
Multiplying both sides of (1) by sin οΏ½πππΏ π₯οΏ½ and integrating over the domain gives
οΏ½πΏ
0sin οΏ½ππ
πΏπ₯οΏ½ π (π₯) ππ₯ = οΏ½
πΏ
0οΏ½βοΏ½π=1
π΅π sin οΏ½πππΏπ₯οΏ½ sin οΏ½ππ
πΏπ₯οΏ½οΏ½ ππ₯
Interchanging the order of integration and summation
οΏ½πΏ
0sin οΏ½ππ
πΏπ₯οΏ½ π (π₯) ππ₯ =
βοΏ½π=1
οΏ½π΅π οΏ½οΏ½πΏ
0sin οΏ½ππ
πΏπ₯οΏ½ sin οΏ½ππ
πΏπ₯οΏ½ ππ₯οΏ½οΏ½
17
But β«πΏ
0sin οΏ½πππΏ π₯οΏ½ sin οΏ½πππΏ π₯οΏ½ ππ₯ = 0 for π β π, hence only one term survives
οΏ½πΏ
0sin οΏ½ππ
πΏπ₯οΏ½ π (π₯) ππ₯ = π΅ποΏ½
πΏ
0sin2 οΏ½ππ
πΏπ₯οΏ½ ππ₯
Renaming π back to π and since β«πΏ
0sin2 οΏ½πππΏ π₯οΏ½ ππ₯ =
πΏ2 the above becomes
οΏ½πΏ
0sin οΏ½ππ
πΏπ₯οΏ½ π (π₯) ππ₯ =
πΏ2π΅π
π΅π =2πΏ οΏ½
πΏ
0sin οΏ½ππ
πΏπ₯οΏ½ π (π₯) ππ₯
=2πΏ
ββββββοΏ½
πΏ2
0sin οΏ½ππ
πΏπ₯οΏ½ π (π₯) ππ₯ +οΏ½
πΏ
πΏ2
sin οΏ½πππΏπ₯οΏ½ π (π₯) ππ₯
ββββββ
=2πΏ
ββββββοΏ½
πΏ2
0sin οΏ½ππ
πΏπ₯οΏ½ ππ₯ + 2οΏ½
πΏ
πΏ2
sin οΏ½πππΏπ₯οΏ½ ππ₯
ββββββ
=2πΏ
ββββββββββ
β cos οΏ½πππΏ π₯οΏ½πππΏ
οΏ½
πΏ2
0
+ 2β cos οΏ½πππΏ π₯οΏ½
πππΏ
οΏ½
πΏ
πΏ2
ββββββββββ
=2ππ
ββββββοΏ½β cos οΏ½ππ
πΏπ₯οΏ½οΏ½
πΏ2
0+ 2 οΏ½β cos οΏ½ππ
πΏπ₯οΏ½οΏ½
πΏ
πΏ2
ββββββ
=2ππ οΏ½οΏ½
β cos οΏ½πππΏπΏ2οΏ½+ cos (0)οΏ½ + 2 οΏ½β cos (ππ) + cos οΏ½ππ
2οΏ½οΏ½οΏ½
=2ππ
οΏ½β cos οΏ½ππ2οΏ½ + 1 β 2 cos (ππ) + 2 cos οΏ½ππ
2οΏ½οΏ½
=2ππ
οΏ½cos οΏ½ππ2οΏ½ + 1 β 2 cos (ππ)οΏ½
Hence the solution is
π’ (π₯, π‘) =βοΏ½π=1
π΅π sin οΏ½πππΏπ₯οΏ½ πβποΏ½
πππΏ οΏ½
2π‘
With
π΅π =2ππ
οΏ½cos οΏ½ππ2οΏ½ β 2 cos (ππ) + 1οΏ½
=2ππ
οΏ½1 β 2 (β1)π + cos οΏ½ππ2οΏ½οΏ½
18
0.5 section 2.3.4 (problem 4)56 Chapter 2. Method of Separation of Variables
[Your answer in part (c) may involve certain integrals that do not need tobe evaluated.]
2.3.4. Consider
k02,
subject to u(0, t) = 0, u(L, t) = 0, and u(x, 0) = f (x).
*(a) What is the total heat energy in the rod as a function of time?
(b) What is the flow of heat energy out of the rod at x = 0? at x = L?
*(c) What relationship should exist between parts (a) and (b)?
2.3.5. Evaluate (be careful if n = m)
L nzrx m7rxsin L sin L dx forn>0,m>0.
Use the trigonometric identity
*2.3.6. Evaluate
sin asin b = 2 [cos(a - b) - cos(a + b)] .
L n7rx m7rxcog L cc
Ldx for n > O, m > 0.
Use the trigonometric identity
cos a cos b = 2 [cos(a + b) + cos(a - b)] .
(Be careful if a - b = 0 or a + b = 0.)
2.3.7. Consider the following boundary value problem (if necessary, see Sec. 2.4.1):
= k82U
with au (0, t)=O, au (L, t) = 0, and u(x, 0) = f (x).at ax2 ax ax
(a) Give a one-sentence physical interpretation of this problem.
(b) Solve by the method of separation of variables. First show that thereare no separated solutions which exponentially grow in time. [Hint:The answer is
u(x, t) = Ao + > cos nix .
n=1
What is An?
0.5.1 Part (a)
By definition the total heat energy is
πΈ = οΏ½ππππ’ (π₯, π‘) ππ£
Assuming constant cross section area π΄, the above becomes (assuming all thermal propertiesare constant)
πΈ = οΏ½πΏ
0πππ’ (π₯, π‘) π΄ππ₯
But π’ (π₯, π‘) was found to be
π’ (π₯, π‘) =βοΏ½π=1
π΅π sin οΏ½πππΏπ₯οΏ½ πβποΏ½
πππΏ οΏ½
2π‘
For these boundary conditions from problem 2.3.3. Where π΅π was found from initial condi-tions. Substituting the solution found into the energy equation gives
πΈ = πππ΄οΏ½πΏ
0οΏ½βοΏ½π=1
π΅π sin οΏ½πππΏπ₯οΏ½ πβποΏ½
πππΏ οΏ½
2π‘οΏ½ ππ₯
= πππ΄βοΏ½π=1
οΏ½π΅ππβποΏ½ πππΏ οΏ½
2π‘οΏ½
πΏ
0sin οΏ½ππ
πΏπ₯οΏ½ ππ₯οΏ½
= πππ΄βοΏ½π=1
π΅ππβποΏ½ πππΏ οΏ½
2π‘
ββββββββ cos οΏ½πππΏ π₯οΏ½
πππΏ
βββββββ
πΏ
0
= πππ΄βοΏ½π=1
π΅ππβποΏ½ πππΏ οΏ½
2π‘ πΏππ
οΏ½β cos οΏ½πππΏπΏοΏ½ + cos (0)οΏ½
= πππ΄βοΏ½π=1
π΅ππβποΏ½ πππΏ οΏ½
2π‘ πΏππ
(1 β cos (ππ))
=πΏπππ΄π
βοΏ½π=1
οΏ½π΅ππ(1 β cos (ππ)) πβποΏ½
πππΏ οΏ½
2π‘οΏ½
19
0.5.2 Part (b)
By definition, the flux is the amount of heat flow per unit time per unit area. Assuming thearea is π΄, then heat flow at π₯ = 0 into the rod per unit time (call it π» (π₯)) is
π»|π₯=0 = π΄ ποΏ½π₯=0
= βπ΄πππ’ππ₯οΏ½π₯=0
Similarly, heat flow at π₯ = πΏ out of the rod per unit time is
π»|π₯=πΏ = π΄ ποΏ½π₯=πΏ
= βπ΄πππ’ππ₯οΏ½π₯=πΏ
To obtain heat flow at π₯ = 0 leaving the rod, the sign is changed and it becomes π΄π ππ’ππ₯ οΏ½π₯=0
.
Since π’ (π₯, π‘) = ββπ=1 π΅π sin οΏ½πππΏ π₯οΏ½ π
βποΏ½ πππΏ οΏ½2π‘ then
ππ’ππ₯
=βοΏ½π=1
π΅ππππΏ
cos οΏ½πππΏπ₯οΏ½ πβποΏ½
πππΏ οΏ½
2π‘
Then at π₯ = 0 then heat flow leaving of the rod becomes
π΄πππ’ππ₯οΏ½π₯=0
= π΄πβοΏ½π=1
πππΏπ΅ππ
βποΏ½ πππΏ οΏ½2π‘
And at π₯ = πΏ, the heat flow out of the bar
βπ΄πππ’ππ₯οΏ½π₯=πΏ
= βπ΄πβοΏ½π=1
π΅ππππΏ
cos οΏ½πππΏπΏοΏ½ πβπ οΏ½
πππΏ οΏ½
2π‘
= βπ΄πβοΏ½π=1
π΅ππππΏ
cos (ππ) πβπ οΏ½πππΏ οΏ½
2π‘
= βπ΄πβοΏ½π=1
(β1)π π΅ππππΏπβπ οΏ½
πππΏ οΏ½
2π‘
0.5.3 Part (c)
Total πΈ inside the bar at time π‘ is given by initial energy πΈπ‘=0 and time integral of flow ofheat energy into the bar. Since from part (a)
πΈ = πΏπππ΄π
βοΏ½π=1
π΅πππβποΏ½
πππΏ οΏ½
2π‘ (1 β cos (ππ))
Then initial energy is
πΈπ‘=0 = πΏπππ΄π
βοΏ½π=1
π΅ππ(1 β cos (ππ))
20
And total heat flow into the rod (per unit time) is οΏ½βπ΄πππ’ππ₯ οΏ½π₯=0
+ π΄π ππ’ππ₯ οΏ½π₯=πΏ
οΏ½, therefore
πΏπππ΄π
βοΏ½π=1
π΅πππβποΏ½
πππΏ οΏ½
2π‘ (1 β cos (ππ)) = οΏ½
π‘
0οΏ½βπ΄π
ππ’ππ₯οΏ½π₯=0
+ π΄πππ’ππ₯οΏ½π₯=πΏοΏ½ ππ₯
= π΄ποΏ½π‘
0οΏ½ππ’ (πΏ)ππ₯
βππ’ (0)ππ₯ οΏ½ ππ₯
Butππ’ (πΏ)ππ₯
βππ’ (0)ππ₯
=ππΏ
βοΏ½π=1
ππ΅π (β1)π πβποΏ½
πππΏ οΏ½
2π‘ β
ππΏ
βοΏ½π=1
ππ΅ππβποΏ½ πππΏ οΏ½
2π‘
=ππΏ οΏ½
βοΏ½π=1
ππ΅π (β1)π πβποΏ½
πππΏ οΏ½
2π‘ β
βοΏ½π=1
ππ΅ππβποΏ½ πππΏ οΏ½
2π‘οΏ½
HenceπΏπππ΄π
βοΏ½π=1
π΅ππ
expβποΏ½πππΏ οΏ½
2π‘ (1 β cos (ππ)) = π΄ππ
πΏ οΏ½π‘
0οΏ½βοΏ½π=1
ππ΅π (β1)π πβποΏ½
πππΏ οΏ½
2π‘ β
βοΏ½π=1
ππ΅ππβποΏ½ πππΏ οΏ½
2π‘οΏ½ ππ₯
0.6 section 2.3.5 (problem 5)
56 Chapter 2. Method of Separation of Variables
[Your answer in part (c) may involve certain integrals that do not need tobe evaluated.]
2.3.4. Consider
k02,
subject to u(0, t) = 0, u(L, t) = 0, and u(x, 0) = f (x).
*(a) What is the total heat energy in the rod as a function of time?
(b) What is the flow of heat energy out of the rod at x = 0? at x = L?
*(c) What relationship should exist between parts (a) and (b)?
2.3.5. Evaluate (be careful if n = m)
L nzrx m7rxsin L sin L dx forn>0,m>0.
Use the trigonometric identity
*2.3.6. Evaluate
sin asin b = 2 [cos(a - b) - cos(a + b)] .
L n7rx m7rxcog L cc
Ldx for n > O, m > 0.
Use the trigonometric identity
cos a cos b = 2 [cos(a + b) + cos(a - b)] .
(Be careful if a - b = 0 or a + b = 0.)
2.3.7. Consider the following boundary value problem (if necessary, see Sec. 2.4.1):
= k82U
with au (0, t)=O, au (L, t) = 0, and u(x, 0) = f (x).at ax2 ax ax
(a) Give a one-sentence physical interpretation of this problem.
(b) Solve by the method of separation of variables. First show that thereare no separated solutions which exponentially grow in time. [Hint:The answer is
u(x, t) = Ao + > cos nix .
n=1
What is An?
πΌ = οΏ½πΏ
0sin οΏ½πππ₯
πΏοΏ½ sin οΏ½πππ₯
πΏοΏ½ ππ₯
Considering first the case π = π. The integral becomes
πΌ = οΏ½πΏ
0sin2 οΏ½πππ₯
πΏοΏ½ ππ₯ =
πΏ2
For the case where π β π, using
sin π sin π = 12(cos (π β π) β cos (π + π))
21
The integral πΌ becomes 2
πΌ =12 οΏ½
πΏ
0cos οΏ½πππ₯
πΏβπππ₯πΏ
οΏ½ β cos οΏ½πππ₯πΏ
+πππ₯πΏ
οΏ½ ππ₯
=12 οΏ½
πΏ
0cos οΏ½
ππ₯ (π β π)πΏ οΏ½ β cos οΏ½
ππ₯ (π + π)πΏ οΏ½ ππ₯
=12
ββββββββββ
sin οΏ½ππ₯(πβπ)πΏοΏ½
π(πβπ)πΏ
ββββββββββ
πΏ
0
β12
ββββββββββ
sin οΏ½ππ₯(π+π)πΏοΏ½
π(π+π)πΏ
ββββββββββ
πΏ
0
=πΏ
2π (π β π) οΏ½sin οΏ½
ππ₯ (π β π)πΏ οΏ½οΏ½
πΏ
0β
πΏ2π (π + π) οΏ½
sin οΏ½ππ₯ (π + π)
πΏ οΏ½οΏ½πΏ
0(1)
But
οΏ½sin οΏ½ππ₯ (π β π)
πΏ οΏ½οΏ½πΏ
0= sin (π (π β π)) β sin (0)
And since π β π is integer, then sin (π (π β π)) = 0, therefore οΏ½sin οΏ½ππ₯(πβπ)πΏοΏ½οΏ½πΏ
0= 0. Similarly
οΏ½sin οΏ½ππ₯ (π + π)
πΏ οΏ½οΏ½πΏ
0= sin (π (π + π)) β sin (0)
Since π + π is integer then sin (π (π + π)) = 0 and οΏ½sin οΏ½ππ₯(π+π)πΏοΏ½οΏ½πΏ
0= 0. Therefore
οΏ½πΏ
0sin οΏ½πππ₯
πΏοΏ½ sin οΏ½πππ₯
πΏοΏ½ ππ₯ =
β§βͺβͺβ¨βͺβͺβ©
πΏ2 π = π0 otherwise
2Note that the term (π β π) showing in the denominator is not a problem now, since this is the case whereπ β π.
22
0.7 section 2.3.7 (problem 6)
56 Chapter 2. Method of Separation of Variables
[Your answer in part (c) may involve certain integrals that do not need tobe evaluated.]
2.3.4. Consider
k02,
subject to u(0, t) = 0, u(L, t) = 0, and u(x, 0) = f (x).
*(a) What is the total heat energy in the rod as a function of time?
(b) What is the flow of heat energy out of the rod at x = 0? at x = L?
*(c) What relationship should exist between parts (a) and (b)?
2.3.5. Evaluate (be careful if n = m)
L nzrx m7rxsin L sin L dx forn>0,m>0.
Use the trigonometric identity
*2.3.6. Evaluate
sin asin b = 2 [cos(a - b) - cos(a + b)] .
L n7rx m7rxcog L cc
Ldx for n > O, m > 0.
Use the trigonometric identity
cos a cos b = 2 [cos(a + b) + cos(a - b)] .
(Be careful if a - b = 0 or a + b = 0.)
2.3.7. Consider the following boundary value problem (if necessary, see Sec. 2.4.1):
= k82U
with au (0, t)=O, au (L, t) = 0, and u(x, 0) = f (x).at ax2 ax ax
(a) Give a one-sentence physical interpretation of this problem.
(b) Solve by the method of separation of variables. First show that thereare no separated solutions which exponentially grow in time. [Hint:The answer is
u(x, t) = Ao + > cos nix .
n=1
What is An?
0.7.1 part (a)
This PDE describes how temperature π’ changes in a rod of length πΏ as a function of time π‘and location π₯. The left and right end are insulated, so no heat escapes from these boundaries.Initially at π‘ = 0, the temperature distribution in the rod is described by the function π (π₯).
0.7.2 Part (b)
ππ’ππ‘
= ππ2π’ππ₯2
Let π’ (π₯, π‘) = π (π‘) π (π₯), then the PDE becomes1ππβ²π = πβ²β²π
Dividing by ππ β 01ππβ²
π=πβ²β²
πSince each side depends on diοΏ½erent independent variable and both are equal, they mustbe both equal to same constant, say βπ. Where π is assumed real.
1ππβ²
π=πβ²β²
π= βπ
The two ODEβs generated are
πβ² + πππ = 0 (1)
πβ²β² + ππ = 0 (2)
23
Starting with the space ODE equation (2), with corresponding boundary conditions ππππ₯(0) =
0, ππππ₯ (πΏ) = 0. Assuming the solution is π (π₯) = πππ₯, Then the characteristic equation is
π2 + π = 0π2 = βπ
π = Β±ββπ
The following cases are considered.
case π < 0 In this case, βπ and also ββπ are positive. Hence the roots Β±ββπ are both real.Let
ββπ = π
Where π > 0. This gives the solution
π (π₯) = π΄ cosh (π π₯) + π΅ sinh (π π₯)ππππ₯
= π΄ sinh (π π₯) + π΅ cosh (π π₯)
Applying the left B.C. gives
0 =ππππ₯
(0)
= π΅ cosh (0)= π΅
The solution becomes π (π₯) = π΄ cosh (π π₯) and henceππππ₯ = π΄ sinh (π π₯). Applying the right B.C.gives
0 =ππππ₯
(πΏ)
= π΄ sinh (π πΏ)π΄ = 0 result in trivial solution. Therefore assuming sinh (π πΏ) = 0 implies π πΏ = 0 whichis not valid since π > 0 and πΏ β 0. Hence only trivial solution results from this case.π < 0 is not an eigenvalue.
case π = 0
The ODE becomesπ2πππ₯2
= 0
The solution is
π (π₯) = π1π₯ + π2ππππ₯
= π1
24
Applying left boundary conditions gives
0 =ππππ₯
(0)
= π1
Hence the solution becomes π (π₯) = π2. Thereforeππππ₯ = 0. Applying the right B.C. provides
no information.
Therefore this case leads to the solution π (π₯) = π2. Associated with this one eigenvalue,the time equation becomes ππ0
ππ‘ = 0 hence π0 is constant, say πΌ. Hence the solution π’0 (π₯, π‘)associated with this π = 0 is
π’0 (π₯, π‘) = π0π0= π2πΌ= π΄0
where constant π2πΌwas renamed toπ΄0 to indicate it is associated with π = 0. π = 0 is an eigenvalue.
case π > 0
Hence βπ is negative, and the roots are both complex.
π = Β±πβπ
The solution is
π (π₯) = π΄ cos οΏ½βππ₯οΏ½ + π΅ sin οΏ½βππ₯οΏ½
ππππ₯
= βπ΄βπ sin οΏ½βππ₯οΏ½ + π΅βπ cos οΏ½βππ₯οΏ½
Applying the left B.C. gives
0 =ππππ₯
(0)
= π΅βπ cos (0)
= π΅βπ
Therefore π΅ = 0 as π > 0. The solution becomes π (π₯) = π΄ cos οΏ½βππ₯οΏ½ and ππππ₯ = βπ΄βπ sin οΏ½βππ₯οΏ½.
Applying the right B.C. gives
0 =ππππ₯
(πΏ)
= βπ΄βπ sin οΏ½βππΏοΏ½
π΄ = 0 gives a trivial solution. Selecting sin οΏ½βππΏοΏ½ = 0 gives
βππΏ = ππ π = 1, 2, 3,β―
Or
ππ = οΏ½πππΏοΏ½2
π = 1, 2, 3,β―
25
Therefore the space solution is
ππ (π₯) = π΄π cos οΏ½πππΏπ₯οΏ½ π = 1, 2, 3,β―
The time solution is found by solvingπππππ‘
+ πππππ = 0
This has the solution
ππ (π‘) = πβππππ‘
= πβποΏ½πππΏ οΏ½
2π‘ π = 1, 2, 3,β―
For the same set of eigenvalues. Notice that no need to add a constant here, since it willbe absorbed in the π΄π when combined in the following step below. Since for π = 0 the
time solution was found to be constant, and for π > 0 the time solution is πβποΏ½πππΏ οΏ½
2π‘, then
no time solution will grow with time. Time solutions always decay with time as the exponent
βπ οΏ½πππΏ οΏ½2π‘ is negative quantity. The solution to the PDE for π > 0 is
π’π (π₯, π‘) = ππ (π‘) ππ (π₯) π = 0, 1, 2, 3,β―
But for linear system sum of eigenfunctions is also a solution. Hence
π’ (π₯, π‘) = π’π=0 (π₯, π‘) +βοΏ½π=1
π’π (π₯, π‘)
= π΄0 +βοΏ½π=1
π΄π cos οΏ½πππΏπ₯οΏ½ πβποΏ½
πππΏ οΏ½
2π‘
0.7.3 Part c
From the solution found above, setting π‘ = 0 gives
π’ (π₯, 0) = π΄0 +βοΏ½π=1
π΄π cos οΏ½πππΏπ₯οΏ½
Therefore, π (π₯) must satisfy the above
π (π₯) = π΄0 +βοΏ½π=1
π΄π cos οΏ½πππΏπ₯οΏ½
26
0.7.4 Part d
Multiplying both sides with cos οΏ½πππΏ π₯οΏ½ where in this problem π = 0, 1, 2,β― (since there wasan eigenvalue associated with π = 0), and integrating over the domain gives
οΏ½πΏ
0π (π₯) cos οΏ½ππ
πΏπ₯οΏ½ ππ₯ = οΏ½
πΏ
0cos οΏ½ππ
πΏπ₯οΏ½ οΏ½π΄0 +
βοΏ½π=1
π΄π cos οΏ½πππΏπ₯οΏ½οΏ½ ππ₯
= οΏ½πΏ
0π΄0 cos οΏ½ππ
πΏπ₯οΏ½ ππ₯ +οΏ½ cos οΏ½ππ
πΏπ₯οΏ½
βοΏ½π=1
π΄π cos οΏ½πππΏπ₯οΏ½ ππ₯
= οΏ½πΏ
0π΄0 cos οΏ½ππ
πΏπ₯οΏ½ ππ₯ +οΏ½
πΏ
0
βοΏ½π=1
π΄π cos οΏ½πππΏπ₯οΏ½ cos οΏ½ππ
πΏπ₯οΏ½ ππ₯
Interchanging the order of summation and integration
οΏ½πΏ
0π (π₯) cos οΏ½ππ
πΏπ₯οΏ½ ππ₯ = οΏ½
πΏ
0π΄0 cos οΏ½ππ
πΏπ₯οΏ½ ππ₯ +
βοΏ½π=1
π΄ποΏ½πΏ
0cos οΏ½ππ
πΏπ₯οΏ½ cos οΏ½ππ
πΏπ₯οΏ½ ππ₯ (1)
case π = 0
When π = 0 then cos οΏ½πππΏ π₯οΏ½ = 1 and the above simplifies to
οΏ½πΏ
0π (π₯) ππ₯ = οΏ½
πΏ
0π΄0ππ₯ +
βοΏ½π=1
π΄ποΏ½πΏ
0cos οΏ½ππ
πΏπ₯οΏ½ ππ₯
But β«πΏ
0cos οΏ½πππΏ π₯οΏ½ ππ₯ = 0 and the above becomes
οΏ½πΏ
0π (π₯) ππ₯ = οΏ½
πΏ
0π΄0ππ₯
= π΄0πΏ
Therefore
π΄0 =1πΏβ«πΏ0π (π₯) ππ₯
case π > 0
From (1), one term survives in the integration when only π = π, hence
οΏ½πΏ
0π (π₯) cos οΏ½ππ
πΏπ₯οΏ½ ππ₯ = π΄0οΏ½
πΏ
0cos οΏ½ππ
πΏπ₯οΏ½ ππ₯ + π΄ποΏ½
πΏ
0cos2 οΏ½ππ
πΏπ₯οΏ½ ππ₯
But β«πΏ
0cos οΏ½πππΏ π₯οΏ½ ππ₯ = 0 and the above becomes
οΏ½πΏ
0π (π₯) cos οΏ½ππ
πΏπ₯οΏ½ ππ₯ = π΄π
πΏ2
Therefore
π΄π =2πΏβ«πΏ0π (π₯) cos οΏ½πππΏ π₯οΏ½ ππ₯
For π = 1, 2, 3,β―
27
0.7.5 Part (e)
The solution was found to be
π’ (π₯, π‘) = π΄0 +βοΏ½π=1
π΄π cos οΏ½πππΏπ₯οΏ½ πβποΏ½
πππΏ οΏ½
2π‘
In the limit as π‘ β β the term πβποΏ½πππΏ οΏ½
2π‘ β 0. What is left is π΄0. But π΄0 =
1πΏβ«πΏ0π (π₯) ππ₯ from
above. This quantity is the average of the initial temperature.
0.8 section 2.3.8 (problem 7)
2.3. Heat Equation With Zero Temperature Ends 57
(c) Show that the initial condition, u(x, 0) = f (x), is satisfied if
f (x) = Ao + E A. cos00
n=1
(d) Using Exercise 2.3.6, solve for AO and An(n > 1).(e) What happens to the temperature distribution as t -+ oo? Show that
it approaches the steady-state temperature distribution (see Sec. 1.4).
*2.3.8. Consider8u 02u& = kax2 - au.
This corresponds to a one-dimensional rod either with heat loss through thelateral sides with outside temperature 0Β° (a > 0, see Exercise 1.2.4) or withinsulated lateral sides with a heat sink proportional to the temperature.Suppose that the boundary conditions are
u(0,t) = 0 and u(L,t) = 0.
(a) What are the possible equilibrium temperature distributions if a > 0?(b) Solve the time-dependent problem [u(x, 0) = f (x)] if a > 0. Analyze
the temperature for large time (t --+ oo) and compare to part (a).
*2.3.9. Redo Exercise 2.3.8 if a < 0. [Be especially careful if -a/k = (n7r/L)2.]
2.3.10. For two- and three-dimensional vectors, the fundamental property of dotproducts, A B = IAI[BI cos9, implies that
IA - BI < IAIIBI. (2.3.44)
In this exercise we generalize this to n-dimensional vectors and functions,in which case (2.3.44) is known as Schwarz's inequality. [The names ofCauchy and Buniakovsky are also associated with (2.3.44).]
(a) Show that IA - -yBi2 > 0 implies (2.3.44), where ry = A B/B B.(b) Express the inequality using both
00 00 b
n.n=1 n=1 Cn
*(c) Generalize (2.3.44) to functions. [Hint: Let A A. B mean the integralJ L A(x)B(x) dx.]
2.3.11. Solve Laplace's equation inside a rectangle:
2=
02u 02uV U
axe+ 8y2 = 0
subject to the boundary conditions
u(0,y) = g(y) u(x,0) = 0u(L, y) = 0 u(x, H) = 0.
(Hint: If necessary, see Sec. 2.5.1.)
0.8.1 part (a)
Equilibrium is at steady state, which implies ππ’ππ‘ = 0 and the PDE becomes an ODE, since
π’ β‘ π’ (π₯) at steady state. Hence
π2π’ππ₯2
βπΌππ’ = 0
The characteristic equation is π2 = πΌπ or π = Β±οΏ½
πΌπ . Since πΌ > 0 and π > 0 then the roots are
real, and the solution is
π’ = π΄0ποΏ½πΌπ π₯ + π΅0π
βοΏ½πΌπ π₯
This can be rewritten as
π’ (π₯) = π΄ cosh οΏ½οΏ½πΌππ₯οΏ½ + π΅ sinh οΏ½
οΏ½πΌππ₯οΏ½
Applying left B.C. gives
0 = π’ (0)= π΄ cosh (0)= π΄
28
The solution becomes π’ (π₯) = π΅ sinh οΏ½οΏ½πΌπ π₯οΏ½. Applying the right boundary condition gives
0 = π’ (πΏ)
= π΅ sinh οΏ½οΏ½πΌππΏοΏ½
π΅ = 0 leads to trivial solution. Setting sinh οΏ½οΏ½πΌππΏοΏ½ = 0 implies οΏ½
πΌππΏ = 0. But this is not
possible since πΏ β 0. Hence the only solution possible is
π’ (π₯) = 0
0.8.2 Part (b)
ππ’ππ‘
= ππ2π’ππ₯2
β πΌπ’
ππ’ππ‘
+ πΌπ’ = ππ2π’ππ₯2
Assuming π’ (π₯, π‘) = π (π₯) π (π‘) and substituting in the above gives
ππβ² + πΌππ = πππβ²β²
Dividing by πππ β 0πβ²
ππ+πΌπ=πβ²β²
πSince each side depends on diοΏ½erent independent variable and both are equal, they mustbe both equal to same constant, say βπ. Where π is assumed real.
1ππβ²
π+πΌπ=πβ²β²
π= βπ
The two ODEβs are1ππβ²
π+πΌπ= βπ
πβ²β²
π= βπ
Or
πβ² + (πΌ + ππ) π = 0πβ²β² + ππ = 0
The solution to the space ODE is the familiar (where π > 0 is only possible case, As foundin problem 2.3.3, part d. Since it has the same B.C.)
ππ = π΅π sin οΏ½πππΏπ₯οΏ½ π = 1, 2, 3,β―
Where ππ = οΏ½πππΏοΏ½2. The time ODE is now solved.
πππππ‘
+ (πΌ + πππ) ππ = 0
29
This has the solution
ππ (π‘) = πβ(πΌ+πππ)π‘
= πβπΌπ‘πβοΏ½πππΏ οΏ½
2ππ‘
For the same eigenvalues. Notice that no need to add a constant here, since it will beabsorbed in the π΅π when combined in the following step below. Therefore the solution tothe PDE is
π’π (π₯, π‘) = ππ (π‘) ππ (π₯)
But for linear system sum of eigenfunctions is also a solution. Hence
π’ (π₯, π‘) =βοΏ½π=1
π’π (π₯, π‘)
=βοΏ½π=1
π΅π sin οΏ½πππΏπ₯οΏ½ πβπΌπ‘πβοΏ½
πππΏ οΏ½
2ππ‘
= πβπΌπ‘βοΏ½π=1
π΅π sin οΏ½πππΏπ₯οΏ½ πβοΏ½
πππΏ οΏ½
2ππ‘
Where πβπΌπ‘ was moved outside since it does not depend on π. From initial condition
π’ (π₯, 0) = π (π₯) =βοΏ½π=1
π΅π sin οΏ½πππΏπ₯οΏ½
Applying orthogonality of sin as before to find π΅π results in
π΅π =2πΏ οΏ½
πΏ
0sin οΏ½ππ
πΏπ₯οΏ½ π (π₯) ππ₯
Hence the solution becomes
π’ (π₯, π‘) =2πΏπβπΌπ‘ οΏ½
βοΏ½π=1
οΏ½οΏ½πΏ
0sin οΏ½ππ
πΏπ₯οΏ½ π (π₯) ππ₯οΏ½ sin οΏ½ππ
πΏπ₯οΏ½ πβοΏ½
πππΏ οΏ½
2ππ‘οΏ½
Hence it is clear that in the limit as π‘ becomes large π’ (π₯, π‘) β 0 since the sum is multipliedby πβπΌπ‘ and πΌ > 0
limπ‘ββ
π’ (π₯, π‘) = 0
This agrees with part (a)
30
0.9 section 2.3.10 (problem 8)
2.3. Heat Equation With Zero Temperature Ends 57
(c) Show that the initial condition, u(x, 0) = f (x), is satisfied if
f (x) = Ao + E A. cos00
n=1
(d) Using Exercise 2.3.6, solve for AO and An(n > 1).(e) What happens to the temperature distribution as t -+ oo? Show that
it approaches the steady-state temperature distribution (see Sec. 1.4).
*2.3.8. Consider8u 02u& = kax2 - au.
This corresponds to a one-dimensional rod either with heat loss through thelateral sides with outside temperature 0Β° (a > 0, see Exercise 1.2.4) or withinsulated lateral sides with a heat sink proportional to the temperature.Suppose that the boundary conditions are
u(0,t) = 0 and u(L,t) = 0.
(a) What are the possible equilibrium temperature distributions if a > 0?(b) Solve the time-dependent problem [u(x, 0) = f (x)] if a > 0. Analyze
the temperature for large time (t --+ oo) and compare to part (a).
*2.3.9. Redo Exercise 2.3.8 if a < 0. [Be especially careful if -a/k = (n7r/L)2.]
2.3.10. For two- and three-dimensional vectors, the fundamental property of dotproducts, A B = IAI[BI cos9, implies that
IA - BI < IAIIBI. (2.3.44)
In this exercise we generalize this to n-dimensional vectors and functions,in which case (2.3.44) is known as Schwarz's inequality. [The names ofCauchy and Buniakovsky are also associated with (2.3.44).]
(a) Show that IA - -yBi2 > 0 implies (2.3.44), where ry = A B/B B.(b) Express the inequality using both
00 00 b
n.n=1 n=1 Cn
*(c) Generalize (2.3.44) to functions. [Hint: Let A A. B mean the integralJ L A(x)B(x) dx.]
2.3.11. Solve Laplace's equation inside a rectangle:
2=
02u 02uV U
axe+ 8y2 = 0
subject to the boundary conditions
u(0,y) = g(y) u(x,0) = 0u(L, y) = 0 u(x, H) = 0.
(Hint: If necessary, see Sec. 2.5.1.)
οΏ½οΏ½ΜοΏ½ β πΎοΏ½ΜοΏ½οΏ½2 = οΏ½οΏ½ΜοΏ½ β πΎοΏ½ΜοΏ½οΏ½ β οΏ½οΏ½ΜοΏ½ β πΎοΏ½ΜοΏ½οΏ½
Since οΏ½οΏ½ΜοΏ½ β πΎοΏ½ΜοΏ½οΏ½2 β₯ 0 then
οΏ½οΏ½ΜοΏ½ β πΎοΏ½ΜοΏ½οΏ½ β οΏ½οΏ½ΜοΏ½ β πΎοΏ½ΜοΏ½οΏ½ β₯ 0
Expanding
οΏ½οΏ½ΜοΏ½ β οΏ½ΜοΏ½οΏ½ β πΎ οΏ½οΏ½ΜοΏ½ β οΏ½ΜοΏ½οΏ½ β πΎ οΏ½οΏ½ΜοΏ½ β οΏ½ΜοΏ½οΏ½ + πΎ2 οΏ½οΏ½ΜοΏ½ β οΏ½ΜοΏ½οΏ½ β₯ 0
But οΏ½ΜοΏ½ β οΏ½ΜοΏ½ = οΏ½ΜοΏ½ β οΏ½ΜοΏ½, henceοΏ½οΏ½ΜοΏ½ β οΏ½ΜοΏ½οΏ½ β 2πΎ οΏ½οΏ½ΜοΏ½ β οΏ½ΜοΏ½οΏ½ + πΎ2 οΏ½οΏ½ΜοΏ½ β οΏ½ΜοΏ½οΏ½ β₯ 0
Using the definition of πΎ = οΏ½ΜοΏ½β οΏ½ΜοΏ½οΏ½ΜοΏ½β οΏ½ΜοΏ½ into the above gives
οΏ½οΏ½ΜοΏ½ β οΏ½ΜοΏ½οΏ½ β 2οΏ½ΜοΏ½ β οΏ½ΜοΏ½οΏ½ΜοΏ½ β οΏ½ΜοΏ½
οΏ½οΏ½ΜοΏ½ β οΏ½ΜοΏ½οΏ½ +οΏ½οΏ½ΜοΏ½ β οΏ½ΜοΏ½οΏ½
2
οΏ½οΏ½ΜοΏ½ β οΏ½ΜοΏ½οΏ½2 οΏ½οΏ½ΜοΏ½ β οΏ½ΜοΏ½οΏ½ β₯ 0
οΏ½οΏ½ΜοΏ½ β οΏ½ΜοΏ½οΏ½ β 2οΏ½οΏ½ΜοΏ½ β οΏ½ΜοΏ½οΏ½
2
οΏ½ΜοΏ½ β οΏ½ΜοΏ½+οΏ½οΏ½ΜοΏ½ β οΏ½ΜοΏ½οΏ½
2
οΏ½ΜοΏ½ β οΏ½ΜοΏ½β₯ 0
οΏ½οΏ½ΜοΏ½ β οΏ½ΜοΏ½οΏ½ βοΏ½οΏ½ΜοΏ½ β οΏ½ΜοΏ½οΏ½
2
οΏ½ΜοΏ½ β οΏ½ΜοΏ½β₯ 0
οΏ½οΏ½ΜοΏ½ β οΏ½ΜοΏ½οΏ½ οΏ½οΏ½ΜοΏ½ β οΏ½ΜοΏ½οΏ½ β οΏ½οΏ½ΜοΏ½ β οΏ½ΜοΏ½οΏ½2β₯ 0
οΏ½οΏ½ΜοΏ½ β οΏ½ΜοΏ½οΏ½ οΏ½οΏ½ΜοΏ½ β οΏ½ΜοΏ½οΏ½ β₯ οΏ½οΏ½ΜοΏ½ β οΏ½ΜοΏ½οΏ½2
But οΏ½οΏ½ΜοΏ½ β οΏ½ΜοΏ½οΏ½2= οΏ½οΏ½ΜοΏ½ β π΅οΏ½2 since οΏ½ΜοΏ½ β οΏ½ΜοΏ½ is just a number. The above becomes
οΏ½οΏ½ΜοΏ½ β οΏ½ΜοΏ½οΏ½ οΏ½οΏ½ΜοΏ½ β οΏ½ΜοΏ½οΏ½ β₯ οΏ½οΏ½ΜοΏ½ β π΅οΏ½2
31
And οΏ½ΜοΏ½ β οΏ½ΜοΏ½ = οΏ½οΏ½ΜοΏ½οΏ½2 and οΏ½οΏ½ΜοΏ½ β οΏ½ΜοΏ½οΏ½ = οΏ½οΏ½ΜοΏ½οΏ½2 by definition as well. Therefore the above becomes
οΏ½οΏ½ΜοΏ½ β π΅οΏ½2 β€ οΏ½οΏ½ΜοΏ½οΏ½2 οΏ½οΏ½ΜοΏ½οΏ½2
Taking square root gives
οΏ½οΏ½ΜοΏ½ β π΅οΏ½ β€ οΏ½οΏ½ΜοΏ½οΏ½ οΏ½οΏ½ΜοΏ½οΏ½
Which is Schwarzβs inequality.
0.9.1 Part b
From the norm definition
οΏ½οΏ½ΜοΏ½οΏ½ = οΏ½οΏ½π₯2 + π¦2 + π§2
Then
οΏ½οΏ½ΜοΏ½ β οΏ½ΜοΏ½οΏ½ = οΏ½οΏ½ΜοΏ½οΏ½2 =οΏ½π₯2 + π¦2 + π§2
Hence
οΏ½οΏ½ΜοΏ½οΏ½2 =βοΏ½π=1
π2π
οΏ½οΏ½ΜοΏ½οΏ½2 =βοΏ½π=1
π2π
And
οΏ½ΜοΏ½ β οΏ½ΜοΏ½ =βοΏ½π=1
ππππ
Therefore the inequality can be written as
οΏ½οΏ½ΜοΏ½ β οΏ½ΜοΏ½οΏ½2β€ οΏ½οΏ½ΜοΏ½οΏ½2 οΏ½οΏ½ΜοΏ½οΏ½2
οΏ½βοΏ½π=1
πππποΏ½2
β€ οΏ½βοΏ½π=1
π2ποΏ½ οΏ½βοΏ½π=1
π2ποΏ½
0.9.2 Part c
Using οΏ½ΜοΏ½ β οΏ½ΜοΏ½ for functions to mean β«πΏ
0π΄ (π₯) π΅ (π₯) ππ₯ then inequality for functions becomes
οΏ½οΏ½πΏ
0π΄ (π₯) π΅ (π₯) ππ₯οΏ½
2
β€ οΏ½οΏ½πΏ
0π΄2 (π₯) ππ₯οΏ½ οΏ½οΏ½
πΏ
0π΅2 (π₯) ππ₯οΏ½
32
0.10 section 2.4.1 (problem 9)
2.4. Worked Examples with the Heat Equation
Table 2.4.1: Boundary Value Problemsford2o
e= -a0
69
(0) = 0 m(-L) _ (L)Boundary 46(0) = 0conditions 46(L) = 0
(L) = 0 dx (-L) = dx (L)
Eigenvaluesnn) a )( (nvr) 2
An( L
n = 1, 2, 3,...
TL
n = 0, 1, 2, 3,...L
n = 0, 1, 2, 3,...
Eigenfunctionsn7rx
sin L naxcos L
naxsin
Land cos L
00x1(x) EΒ°n cos n
Series00
f ( x ) _ E Bn sinnrx f(x) A. cos
Ln=0
n=1 L n=0 L 00 n x6 in s n+E
n=1
1 La0 = 2L /-L'(.) ds
1 L !(:) dxA0 = L ICoefficients
/r`L
Bn = f(z).in n-dx
2
/O
1an - IL f(.)- nrsdzL 0 L
Z nxsLAn J /(z) toy dz
L L I.
L O LEn - 1 /L !(s)sfn nva ds
1L L L
EXERCISES 2.4
*2.4.1. Solve the heat equation 8u/8t = k82u/8x2, 0 < x < L, t > 0, subject to
8x(O,t)0 t>0
(L, t)0 t>0.
(a) u(x,0) =0 x < L/21 x>L/2
(c) u(x, 0) = -2 sin L
(b)u(x,0)=6+4cos31rx
(d) u(x, 0) = -3 cos jLx
The same boundary conditions was encountered in problem 2.3.7, therefore the solutionused here starts from the same general solution already found, which is
π0 = 0
ππ = οΏ½πππΏοΏ½2
π = 1, 2, 3,β―
π’ (π₯, π‘) = π΄0 +βοΏ½π=1
π΄π cos οΏ½πππΏπ₯οΏ½ πβποΏ½
πππΏ οΏ½
2π‘
0.10.1 Part (b)
π’ (π₯, 0) = 6 + 4 cos 3ππ₯πΏ
Comparing terms with the general solution at π‘ = 0 which is
π’ (π₯, 0) = π΄0 +βοΏ½π=1
π΄π cos οΏ½πππΏπ₯οΏ½
results in
π΄0 = 6π΄3 = 4
And all other π΄π = 0. Hence the solution is
π’ (π₯, π‘) = 6 + 4 cos οΏ½3ππΏπ₯οΏ½ π
βποΏ½ 3ππΏ οΏ½2π‘
33
0.10.2 Part (c)
π’ (π₯, 0) = β2 sin ππ₯πΏ
Hence
β2 sin ππ₯πΏ= π΄0 +
βοΏ½π=1
π΄π cos οΏ½πππΏπ₯οΏ½ (1)
Multiplying both sides of (1) by cos οΏ½πππΏ π₯οΏ½ and integrating gives
οΏ½πΏ
0β2 sin οΏ½ππ₯
πΏοΏ½ cos οΏ½ππ
πΏπ₯οΏ½ ππ₯ = οΏ½
πΏ
0οΏ½π΄0 cos οΏ½ππ
πΏπ₯οΏ½ + cos οΏ½ππ
πΏπ₯οΏ½
βοΏ½π=1
π΄π cos οΏ½πππΏπ₯οΏ½οΏ½ ππ₯
= οΏ½πΏ
0π΄0 cos οΏ½ππ
πΏπ₯οΏ½ ππ₯ +οΏ½
πΏ
0
βοΏ½π=1
π΄π cos οΏ½πππΏπ₯οΏ½ cos οΏ½ππ
πΏπ₯οΏ½ ππ₯
Interchanging the order of integration and summation
οΏ½πΏ
0β2 sin οΏ½ππ₯
πΏοΏ½ cos οΏ½ππ
πΏπ₯οΏ½ ππ₯ = οΏ½
πΏ
0π΄0 cos οΏ½ππ
πΏπ₯οΏ½ ππ₯ +
βοΏ½π=1
π΄ποΏ½πΏ
0cos οΏ½ππ
πΏπ₯οΏ½ cos οΏ½ππ
πΏπ₯οΏ½ ππ₯
Case π = 0
The above becomes
οΏ½πΏ
0β2 sin οΏ½ππ₯
πΏοΏ½ ππ₯ = οΏ½
πΏ
0π΄0ππ₯ +
βοΏ½π=1
π΄ποΏ½πΏ
0cos οΏ½ππ
πΏπ₯οΏ½ ππ₯
But β«πΏ
0cos οΏ½πππΏ π₯οΏ½ ππ₯ = 0 hence
οΏ½πΏ
0β2 sin οΏ½ππ₯
πΏοΏ½ ππ₯ = οΏ½
πΏ
0π΄0ππ₯
π΄0πΏ = β2οΏ½πΏ
0sin οΏ½ππ₯
πΏοΏ½ ππ₯
π΄0πΏ = β2
ββββββββ
cos οΏ½ππ₯πΏ οΏ½ππΏ
βββββββ
πΏ
0
= β2πΏπ οΏ½β cos οΏ½
ππΏπΏ οΏ½
+ cos οΏ½π0πΏ οΏ½οΏ½
= β2πΏπ(β (β1) + 1)
= β4πΏπ
Hence
π΄0 =β4π
Case π > 0
34
οΏ½πΏ
0β2 sin οΏ½ππ₯
πΏοΏ½ cos οΏ½ππ
πΏπ₯οΏ½ ππ₯ = οΏ½
πΏ
0π΄0 cos οΏ½ππ
πΏπ₯οΏ½ ππ₯ +
βοΏ½π=1
π΄ποΏ½πΏ
0cos οΏ½ππ
πΏπ₯οΏ½ cos οΏ½ππ
πΏπ₯οΏ½ ππ₯
One term survives the summation resulting in
οΏ½πΏ
0β2 sin οΏ½ππ₯
πΏοΏ½ cos οΏ½ππ
πΏπ₯οΏ½ ππ₯ =
β4π οΏ½
πΏ
0cos οΏ½ππ
πΏπ₯οΏ½ ππ₯ + π΄ποΏ½
πΏ
0cos2 οΏ½ππ
πΏπ₯οΏ½ ππ₯
But β«πΏ
0cos οΏ½πππΏ π₯οΏ½ ππ₯ = 0 and β«
πΏ
0cos2 οΏ½πππΏ π₯οΏ½ ππ₯ =
πΏ2 , therefore
οΏ½πΏ
0β2 sin οΏ½ππ₯
πΏοΏ½ cos οΏ½ππ
πΏπ₯οΏ½ ππ₯ = π΄π
πΏ2
π΄π =β4πΏ οΏ½
πΏ
0sin οΏ½ππ₯
πΏοΏ½ cos οΏ½ππ
πΏπ₯οΏ½ ππ₯
But
οΏ½πΏ
0sin οΏ½ππ₯
πΏοΏ½ cos οΏ½ππ
πΏπ₯οΏ½ ππ₯ =
βπΏ (1 + cos (ππ))π οΏ½π2 β 1οΏ½
Therefore
π΄π = 4(1 + cos (ππ))π οΏ½π2 β 1οΏ½
= 4(β1)π + 1π οΏ½π2 β 1οΏ½
π = 1, 2, 3,β―
Hence the solution becomes
π’ (π₯, π‘) =β4π+4π
βοΏ½π=1
(β1)π + 1οΏ½π2 β 1οΏ½
cos οΏ½πππΏπ₯οΏ½ πβποΏ½
πππΏ οΏ½
2π‘
0.11 section 2.4.2 (problem 10)
70 Chapter 2. Method of Separation of Variables
*2.4.2. Solvez
= k8-z with 8 (0, t) = 0
u(L, t) = 0
u(x,0) = f(x)
For this problem you may assume that no solutions of the heat equationexponentially grow in time. You may also guess appropriate orthogonalityconditions for the eigenfunctions.
*2.4.3. Solve the eigenvalue problem
d2,0
dx2- _AO
subject to
0(0) = 0(27r) and ;jj(O) =
dx
(21r).
2.4.4. Explicitly show that there are no negative eigenvalues for
d2O
x_ -A subject to dz (0) = 0 and (L) = 0.
2.4.5. This problem presents an alternative derivation of the heat equation for athin wire. The equation for a circular wire of finite thickness is the two-dimensional heat equation (in polar coordinates). Show that this reducesto (2.4.25) if the temperature does not depend on r and if the wire is verythin.
2.4.6. Determine the equilibrium temperature distribution for the thin circularring of Section 2.4.2:
(a) Directly from the equilibrium problem (see Sec. 1.4)
(b) By computing the limit as t - oo of the time-dependent problem
2.4.7. Solve Laplace's equation inside a circle of radius a,
I .92UV 2U
r Or (r 8r) + rz 902 = 0,
subject to the boundary condition
u(a,9) = f(9).
(Hint: If necessary, see Sec. 2.5.2.)
ππ’ππ‘
= π π2π’ππ₯2
35
Let π’ (π₯, π‘) = π (π‘) π (π₯), then the PDE becomes1π πβ²π = πβ²β²π
Dividing by ππ1π πβ²
π=πβ²β²
πSince each side depends on diοΏ½erent independent variable and both are equal, they mustbe both equal to same constant, say βπ. Where π is real.
1π πβ²
π=πβ²β²
π= βπ
The two ODEβs are
πβ² + πππ = 0 (1)
πβ²β² + ππ = 0 (2)
Per problem statement, π β₯ 0, so only two cases needs to be examined.
Case π = 0
The space equation becomes πβ²β² = 0 with the solution
π = π΄π₯ + π
Hence left B.C. implies πβ² (0) = 0 or π΄ = 0. Therefore the solution becomes π = π. Theright B.C. implies π (πΏ) = 0 or π = 0. Therefore this leads to π = 0 as the only solution. Thisresults in trivial solution. Therefore π = 0 is not an eigenvalue.
Case π > 0
Starting with the space ODE, the solution is
π (π₯) = π΄ cos οΏ½βππ₯οΏ½ + π΅ sin οΏ½βππ₯οΏ½
ππππ₯
= βπ΄βπ sin οΏ½βππ₯οΏ½ + π΅βπ cos οΏ½βππ₯οΏ½
Left B.C. gives
0 =ππππ₯
(0)
= π΅βπ
Hence π΅ = 0 since it is assumed π β 0 and π > 0. Solution becomes
π (π₯) = π΄ cos οΏ½βππ₯οΏ½
Applying right B.C. gives
0 = π (πΏ)
= π΄ cos οΏ½βππΏοΏ½
36
π΄ = 0 leads to trivial solution. Therefore cos οΏ½βππΏοΏ½ = 0 or
βπ =ππ2πΏ
π = 1, 3, 5,β―
=(2π β 1) π
2πΏπ = 1, 2, 3β―
Hence
ππ = οΏ½ππ2πΏοΏ½2
π = 1, 3, 5,β―
=(2π β 1)2 π2
4πΏ2π = 1, 2, 3β―
Therefore
ππ (π₯) = π΄π cos οΏ½ππ2πΏπ₯οΏ½ π = 1, 3, 5,β―
And the corresponding time solution
ππ = πβποΏ½ ππ2πΏ οΏ½
2π‘ π = 1, 3, 5,β―
Hence
π’π (π₯, π‘) = ππππ
π’ (π₯, π‘) =βοΏ½
π=1,3,5,β―π΄π cos οΏ½ππ
2πΏπ₯οΏ½ πβποΏ½
ππ2πΏ οΏ½
2π‘
=βοΏ½π=1
π΄π cos οΏ½(2π β 1) π
2πΏπ₯οΏ½ π
βποΏ½ (2πβ1)π2πΏ οΏ½2π‘
From initial conditions
π (π₯) =βοΏ½
π=1,3,5,β―π΄π cos οΏ½ππ
2πΏπ₯οΏ½
Multiplying both sides by cos οΏ½ππ2πΏ π₯οΏ½ and integrating
οΏ½πΏ
0π (π₯) cos οΏ½ππ
2πΏπ₯οΏ½ ππ₯ = οΏ½
βββββ
βοΏ½
π=1,3,5,β―π΄π cos οΏ½ππ
2πΏπ₯οΏ½ cos οΏ½ππ
2πΏπ₯οΏ½βββββ ππ₯
Interchanging order of summation and integration and applying orthogonality results in
οΏ½πΏ
0π (π₯) cos οΏ½ππ
2πΏπ₯οΏ½ ππ₯ = π΄π
πΏ2
π΄π =2πΏ οΏ½
πΏ
0π (π₯) cos οΏ½ππ
2πΏπ₯οΏ½ ππ₯
Therefore the solution is
π’ (π₯, π‘) =2πΏ
βοΏ½
π=1,3,5,β―οΏ½οΏ½
πΏ
0π (π₯) cos οΏ½ππ
2πΏπ₯οΏ½ ππ₯οΏ½ cos οΏ½ππ
2πΏπ₯οΏ½ πβποΏ½
ππ2πΏ οΏ½
2π‘
or
37
π’ (π₯, π‘) =2πΏ
βοΏ½π=1
οΏ½οΏ½πΏ
0π (π₯) cos οΏ½
(2π β 1) π2πΏ
π₯οΏ½ ππ₯οΏ½ cos οΏ½(2π β 1) π
2πΏπ₯οΏ½ π
βποΏ½ (2πβ1)π2πΏ οΏ½2π‘
0.12 section 2.4.3 (problem 11)
70 Chapter 2. Method of Separation of Variables
*2.4.2. Solvez
= k8-z with 8 (0, t) = 0
u(L, t) = 0
u(x,0) = f(x)
For this problem you may assume that no solutions of the heat equationexponentially grow in time. You may also guess appropriate orthogonalityconditions for the eigenfunctions.
*2.4.3. Solve the eigenvalue problem
d2,0
dx2- _AO
subject to
0(0) = 0(27r) and ;jj(O) =
dx
(21r).
2.4.4. Explicitly show that there are no negative eigenvalues for
d2O
x_ -A subject to dz (0) = 0 and (L) = 0.
2.4.5. This problem presents an alternative derivation of the heat equation for athin wire. The equation for a circular wire of finite thickness is the two-dimensional heat equation (in polar coordinates). Show that this reducesto (2.4.25) if the temperature does not depend on r and if the wire is verythin.
2.4.6. Determine the equilibrium temperature distribution for the thin circularring of Section 2.4.2:
(a) Directly from the equilibrium problem (see Sec. 1.4)
(b) By computing the limit as t - oo of the time-dependent problem
2.4.7. Solve Laplace's equation inside a circle of radius a,
I .92UV 2U
r Or (r 8r) + rz 902 = 0,
subject to the boundary condition
u(a,9) = f(9).
(Hint: If necessary, see Sec. 2.5.2.)
ππ2
ππ₯2+ ππ = 0
π (0) = π (2π)ππππ₯
(0) =ππππ₯
(2π)
First solution using transformation
Let π = π₯ β π, hence the above system becomes
ππ2
ππ2+ ππ = 0
π (βπ) = π (π)ππππ
(βπ) =ππππ
(π)
The characteristic equation is π2 + π = 0 or π = Β±ββπ. Assuming π is real. There are threecases to consider.
Case π < 0
Let π = ββπ > 0
π (π) = π1 cosh (π π) + π2 sinh (π π)πβ² (π) = π π1 sinh (π π) + π π2 cosh (π π)
38
Applying first B.C. gives
π (βπ) = π (π)π1 cosh (π π) β π2 sinh (π π) = π1 cosh (π π) + π2 sinh (π π)
2π2 sinh (π π) = 0π2 sinh (π π) = 0 (1)
Applying second B.C. gives
πβ² (βπ) = πβ² (π)βπ π1 sinh (π π) + π π2 cosh (π π) = π π1 sinh (π π) + π π2 cosh (π π)
2π1 sinh (π π) = 0π1 sinh (π π) = 0 (2)
Since sinh (π π) is zero only for π π = 0 and π π is not zero because π > 0. Then the only otheroption is that both π1 = 0 and π2 = 0 in order to satisfy equations (1)(2). Hence trivialsolution. Hence π < 0 is not an eigenvalue.
Case π = 0
The space equation becomesππ2
ππ2 = 0 with the solution π (π) = π΄π+π΅. Applying the first B.C.gives
π (βπ) = π (π)βπ΄π + π΅ = π΄π + π΅
0 = 2π΄π
Hence π΄ = 0. The solution becomes π (π) = π΅. And πβ² (π) = 0. The second B.C. just gives0 = 0. Therefore the solution is
π (π) = πΆ
Where πΆ is any constant. Hence π = 0 is an eigenvalue.
Case π > 0
π (π) = π1 cos οΏ½βπποΏ½ + π2 sin οΏ½βπποΏ½
πβ² (π) = βπ1βπ sin οΏ½βπποΏ½ + π2βπ cos οΏ½βπποΏ½
Applying first B.C. gives
π (βπ) = π (π)
π1 cos οΏ½βπποΏ½ β π2 sin οΏ½βπποΏ½ = π1 cos οΏ½βπποΏ½ + π2 sin οΏ½βπποΏ½
2π2 sin οΏ½βπποΏ½ = 0
π2 sin οΏ½βπποΏ½ = 0 (3)
39
Applying second B.C. gives
πβ² (βπ) = πβ² (π)
π1βπ sin οΏ½βπποΏ½ + π2βπ cos οΏ½βπποΏ½ = βπ1βπ sin οΏ½βπποΏ½ + π2βπ cos οΏ½βπποΏ½
2π1βπ sin οΏ½βπποΏ½ = 0
π1 sin οΏ½βπποΏ½ = 0 (2)
Both (3) and (2) can be satisfied for non-zero βππ. The trivial solution is avoided. Thereforethe eigenvalues are
sin οΏ½βπποΏ½ = 0
οΏ½πππ = ππ π = 1, 2, 3,β―ππ = π2 π = 1, 2, 3,β―
Hence the corresponding eigenfunctions are
οΏ½cos οΏ½οΏ½ππποΏ½ , sin οΏ½οΏ½ππποΏ½οΏ½ = {cos (ππ) , sin (ππ)}
Transforming back to π₯ using π = π₯ β π
{cos (π (π₯ β π)) , sin (π (π₯ β π))} = {cos (ππ₯ β ππ) , sin (ππ₯ β ππ)}
But cos (π₯ β π) = β cos π₯ and sin (π₯ β π) = β sin π₯, hence the eigenfunctions are
{β cos (ππ₯) , β sin (ππ₯)}
The signs of negative on an eigenfunction (or eigenvector) do not aοΏ½ect it being such as thisis just a multiplication by β1. Hence the above is the same as saying the eigenfunctions are
{cos (ππ₯) , sin (ππ₯)}
Summary
eigenfunctions
π = 0 arbitrary constant
π > 0 {cos (ππ₯) , sin (ππ₯)} for π = 1, 2, 3β―
Second solution without transformation
(note: Using transformation as shown above seems to be easier method than this below).
40
The characteristic equation is π2 + π = 0 or π = Β±ββπ. Assuming π is real. There are threecases to consider.
Case π < 0
In this case βπ is positive and the roots are both real. Assuming ββπ = π where π > 0, thenthe solution is
π (π₯) = π΄ππ π₯ + π΅πβπ π₯
πβ² (π₯) = π΄π ππ π₯ β π΅π πβπ π₯
First B.C. gives
π (0) = π (2π)π΄ + π΅ = π΄π2π π + π΅πβ2π π
π΄οΏ½1 β π2π ποΏ½ + π΅ οΏ½1 β πβ2π ποΏ½ = 0 (1)
The second B.C. gives
πβ² (0) = πβ² (2π)π΄π β π΅π = π΄π π2π π β π΅π πβ2π π
π΄οΏ½1 β π2π ποΏ½ + π΅ οΏ½β1 + πβ2π ποΏ½ = 0 (2)
After dividing by π since π β 0. Now a 2 by 2 system is setup from (1),(2)ββββββοΏ½1 β π2π ποΏ½ οΏ½1 β πβ2π ποΏ½οΏ½1 β π2π ποΏ½ οΏ½β1 + πβ2π ποΏ½
ββββββ
ββββββπ΄π΅
ββββββ =
ββββββ00
ββββββ
Since this is ππ₯ = π with π = 0 then for non-trivial solution |π| must be zero. Checking thedeterminant to see if it is zero or not:
οΏ½οΏ½1 β π2π ποΏ½ οΏ½1 β πβ2π ποΏ½οΏ½1 β π π2π ποΏ½ οΏ½β1 + π πβ2π ποΏ½
οΏ½ = οΏ½1 β π2π ποΏ½ οΏ½β1 + πβ2π ποΏ½ β οΏ½1 β πβ2π ποΏ½ οΏ½1 β π2π ποΏ½
= οΏ½β1 + πβ2π π + π2π π β 1οΏ½ β οΏ½1 β π2π π β πβ2π π + 1οΏ½
= β1 + πβ2π π + π2π π β 1 β 1 + π2π π + πβ2π π β 1= β4 + 2π2π π + 2πβ2π π
= β4 + 2 οΏ½π2π π + πβ2π ποΏ½
= β4 + 4 cosh (2π π)Hence for the determinant to be zero (so that non-trivial solution exist) then β4+4 cosh (2π π) =0 or cosh (2π π) = 1 which has the solution 2π π = 0. Which means π = 0. But the assumptionwas that π > 0. This implies only a trivial solution exist and π < 0 is not an eigenvalue.
case π = 0
The space equation becomesππ2
ππ₯2 = 0 with the solution π (π₯) = π΄π₯ + π΅. Applying the first B.C.
41
gives
π΅ = 2π΄π + π΅0 = 2π΄π
Hence π΄ = 0. The solution becomes π (π₯) = π΅. And πβ² (π₯) = 0. The second B.C. just gives0 = 0. Therefore the solution is
π (π₯) = πΆ
Where πΆ is any constant. Hence π = 0 is an eigenvalue.
Case π > 0
In this case the solution is
π (π₯) = π΄ cos οΏ½βππ₯οΏ½ + π΅ sin οΏ½βππ₯οΏ½
πβ² (π₯) = βπ΄βπ sin οΏ½βππ₯οΏ½ + π΅βπ cos οΏ½βππ₯οΏ½
Applying first B.C. gives
π (0) = π (2π)
π΄ = π΄ cos οΏ½2πβποΏ½ + π΅ sin οΏ½2πβποΏ½
π΄ οΏ½1 β cos οΏ½2πβποΏ½οΏ½ β π΅ sin οΏ½2πβποΏ½ = 0
Applying second B.C. gives
πβ² (0) = πβ² (2π)
π΅βπ = βπ΄βπ sin οΏ½2πβποΏ½ + π΅βπ cos οΏ½2πβποΏ½
π΄βπ sin οΏ½2πβποΏ½ + π΅ οΏ½βπ β βπ cos οΏ½2πβποΏ½οΏ½ = 0
π΄ sin οΏ½2πβποΏ½ + π΅ οΏ½1 β cos οΏ½2πβποΏ½οΏ½ = 0
Therefore ββββββ1 β cos οΏ½2πβποΏ½ β sin οΏ½2πβποΏ½
sin οΏ½2πβποΏ½ 1 β cos οΏ½2πβποΏ½
ββββββ
ββββββπ΄π΅
ββββββ =
ββββββ00
ββββββ (3)
Setting |π| = 0 to obtain the eigenvalues gives
οΏ½1 β cos οΏ½2πβποΏ½οΏ½ οΏ½1 β cos οΏ½2πβποΏ½οΏ½ + sin οΏ½2πβποΏ½ sin οΏ½2πβποΏ½ = 0
1 β cos οΏ½2πβποΏ½ = 0
42
Hence
cos οΏ½2πβποΏ½ = 1
2ποΏ½ππ = ππ π = 2, 4,β―
οΏ½ππ =π2
π = 2, 4,β―
Or
οΏ½ππ = π π = 1, 2, 3,β―ππ = π2 π = 1, 2, 3,β―
Therefore the eigenfunctions are
ππ (π₯) = {cos (ππ₯) , sin (ππ₯)}Summary
eigenfunctions
π = 0 arbitrary constant
π > 0 {cos (ππ₯) , sin (ππ₯)} for π = 1, 2, 3β―
0.13 section 2.4.6 (problem 12)
70 Chapter 2. Method of Separation of Variables
*2.4.2. Solvez
= k8-z with 8 (0, t) = 0
u(L, t) = 0
u(x,0) = f(x)
For this problem you may assume that no solutions of the heat equationexponentially grow in time. You may also guess appropriate orthogonalityconditions for the eigenfunctions.
*2.4.3. Solve the eigenvalue problem
d2,0
dx2- _AO
subject to
0(0) = 0(27r) and ;jj(O) =
dx
(21r).
2.4.4. Explicitly show that there are no negative eigenvalues for
d2O
x_ -A subject to dz (0) = 0 and (L) = 0.
2.4.5. This problem presents an alternative derivation of the heat equation for athin wire. The equation for a circular wire of finite thickness is the two-dimensional heat equation (in polar coordinates). Show that this reducesto (2.4.25) if the temperature does not depend on r and if the wire is verythin.
2.4.6. Determine the equilibrium temperature distribution for the thin circularring of Section 2.4.2:
(a) Directly from the equilibrium problem (see Sec. 1.4)
(b) By computing the limit as t - oo of the time-dependent problem
2.4.7. Solve Laplace's equation inside a circle of radius a,
I .92UV 2U
r Or (r 8r) + rz 902 = 0,
subject to the boundary condition
u(a,9) = f(9).
(Hint: If necessary, see Sec. 2.5.2.)
The PDE for the thin circular ring is
ππ’ππ‘
= ππ2π’ππ₯2
π’ (βπΏ, π‘) = π’ (πΏ, π‘)ππ’ (βπΏ, π‘)
ππ‘=ππ’ (πΏ, π‘)ππ‘
π’ (π₯, 0) = π (π₯)
0.13.1 Part (a)
At equilibrium ππ’ππ‘ = 0 and the PDE becomes
0 =π2π’ππ₯2
43
As it now has one independent variable, it becomes the following ODE to solve
π2π’ (π₯)ππ₯2
= 0
π’ (βπΏ) = π’ (πΏ)ππ’ππ₯(βπΏ) =
ππ’ππ₯(πΏ)
Solution to π2π’ππ₯2 = 0 is
π’ (π₯) = π1π₯ + π2Where π1, π2 are arbitrary constants. From the first B.C.
π’ (βπΏ) = π’ (πΏ)βπ1πΏ + π2 = π1πΏ + π2
2π1πΏ = 0π1 = 0
Hence the solution becomes
π’ (π₯) = π2The second B.C. adds nothing as it results in 0 = 0. Hence the solution at equilibrium is
π’ (π₯) = π2
This means at equilibrium the temperature in the ring reaches a constant value.
0.13.2 Part (b)
The time dependent solution was derived in problem 2.4.3 and also in section 2.4, page 62in the book, given by
π’ (π₯, π‘) = π0 +βοΏ½π=1
ππ cos οΏ½πππ₯πΏοΏ½ πβποΏ½
πππ₯πΏ οΏ½
2π‘ +
βοΏ½π=1
ππ sin οΏ½πππ₯πΏοΏ½ πβποΏ½
πππ₯πΏ οΏ½
2π‘
As π‘ β β the terms πβποΏ½πππ₯πΏ οΏ½
2π‘ β 0 and the above reduces to
π’ (π₯,β) = π0Since π0 is constant, this is the same result found in part (a).