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Unformatted text preview: Section 10.5, Problem 2 Problem. Determine whether separation of variables can be used to replace
the partial differential equation tum + amt = 0 by a pair of ordinary differential
equations; if so, ﬁnd the equations. Solution. To use the method of separation of variables, one begins with the
assumption that u(x,t) can be written as the product of a function of x and a
function of t, that is, that u(:z:,t) = X Then um = X’ (w)T(t) and um =
X”(m)T(t), and at = X(a:)T’(t). Substituting this into the differential equation, tX”(x)T(t) + xX (x)T’ (t) = 0.
Then one may divide by X (m)T (t) (in separation of variables problems one always,
eventually, divides by the original function, in this case X This gives one
75%? + 1101111410?)2 = 0. Bringing one of the two expressions to the other side, the equation can be written t§—I§I(f)l = ~xTTlég, which is nearly in the most useful form. One would like to have only functions of x on one side of the equation, and only
functions of t on the other; this can be accomplished by dividing both sides of the
equation by x and t. Then one has 2%? = 5%, a fully separated equation. As one has a function
purely of x on the left, and a function purely of t on the right, the only way this
equation can hold true is if both left and right sides are equal to some constant.
Call it A. Then inf—£3 = A, and —%I(% = A, which is a pair of differential equations. One
would most likely write the pair as X” — MEX = 0 and T’ (t) + AtT(t) = 0,
however, as those forms look most like ordinary differential equations that one knows how to solve. Section 10.5, Problem 5 Problem. Determine whether separation of variables can be used to replace
the partial differential equation um + (a:+y)uyy = 0 by a pair of ordinary differential
equations; if so, ﬁnd the equations. Solution. To use the method of separation of variables, one begins with
the assumption that u(:c, y) can be written as the product of a function of a: and.
a function of 3/, that is, u(x, y) = X Then um = X’ and um =
X”($)Y(y), and uy = X(x)Y’(y) and um, = X(x)Y”(y). ' Substituting this into the differential equation, X ” (x)Y(y) + (w+y) X (x)Y” (y)
= 0, or X” + xX (m)Y” (y) + yX(x)Y” = 0. One then divides the entire
expression by X (in separation of variables problems one always, eventually,
divides by the original function, in this case X This gives one égf)! +
xY—HQ—Jl + 111%? = 0. Then one tries to move all the expressions containing a: to one Y(y)
side of the equals sign, and all those containing y to the other side — and a problem develops.
One can write that $23)) = —x%%2 — 31%, but one cannot get rid of the a: on the right hand side of the equation — at best it can be moved to the denominator
of the yyvuéi—Sl term, which is no better than having it where it already is. The conclusion is that the separation of variables method cannot be used for this problem. Section 10.5, Problem 7 Problem. Find the solution of the heat conduction problem 100%,, = ut, on
O < :c < 1, for t > 0; with the boundary conditions u(0, t) = 0, u(1,t) = 0 for t > 0;
and initial condition u(m, 0) = sin(27r:1:) —— sin(57ra:). Solution. Suppose the problem can be solved by the method of separation of
variables. Then suppose u(a:, t) is the product of a function just of a: and a function
just of t, that is, u(:c,t) = X(m)T(t). Then u,m = X”(:c)T(t), and ut = X(x)T’ The differential equation can
then be written as 100X”(x)T(t) = X(cr)T’ One can then divide the entire
equation by the original function u(:c,t) = X (33)T(t) (as one always must do in a
separation of variables problem), turning the equation into 32%,? = ﬁgé). (For
heat conduction problems, it turns out to be very convenient if a constant — such as
the 100 in this problem — is kept with the time, rather than the space, coordinate.) As the left side of the equation is a function exclusively of ac, and the right
side exclusively of t, the only way this equation can hold true is if both sides are
equal to some constant. Call this constant —A (the negative sign proves to make
the expressions below easier to work with). Then the left side, the X equation, is X” + AX = 0. The right side,
the T(t) equation, is T’ (t) + 100AT(t) = 0. I The general solution to the spatial equation, the X equation, is X =
(Smash/Xx) + Cgsin(\/X:I:). To satisfy the left boundary condition, one ﬁnds that
X (0) = Clcos(0) +Cgsin(0) = 01 = 0. To satisfy the right boundary condition, one
ﬁnds that X(1) = Czsin(\/Xl) = Czsin(\/X) = 0. This can be trueonly if 02 = 0
e making the solution the boring case of the trivial solution, for which X = 0
for all m — or if = 0. For to equal zero, then x/X = km", for some
integer k. So the eigenvalues are A], = (19702, with the corresponding eigenfunctions
uk 2 sin(k7mc). The general solutions for the time equation, T' (t) + 100AT(t) = 0, have the
form T(t) = 6—100”. The values of A must be the eigenvalues from the spatial
equation, so that the eigenfunctions for Tk(t) = e‘100k27'2t. The fundamental solutions of the heat conduction equation in this problem CONTINUED then are uk(a:, t) = e‘100k2"2‘sin(k7rx). Therefore the general solution to the entire differential equation has the form: x
u(z,t) = Z Cke‘100k2W2tsin(k7rx)
lc=1 . The coefficients Ck are be chosen to satisfy the initial condition, the require
ment that u(a:, O) = 2211 Cksin(k7rx) = sin(27rm) —— sin(57r$) The only way the quantity on the left can equal the quantity on the right, for
all values of as, is for the coefﬁcient in front of the sin(27rx) term on the left to equal
the coefﬁcient in front of the sin(27rw) term on the right — so that 02 = 1 — and for
the sin(57mc) coefﬁcient on the left to equal the sin(57rx) coefﬁcient on the right ~
so that C5 = —1 — and for all the other coefficients Ck to equal zero, as there are no.
sin(k7m:) terms on the right for any values of k other than k = 2 and k = 5. This
is from the orthogonality of the sine functions. Therefore the solution to the given heat conduction problem is —4007r2t 45007th u(a:, t) = e sin(27rw) — e sin(57rac) Section 10.5, Problem 8 Problem. Find the solution of the heat conduction problem umc = 41», for
0 < a: < 2 and t > 0; with the boundary conditions u(0, t) = 0 and u(2,t) = 0
for t > 0; and the initial condition u(a:,0) = 25in(§m) — stn(7ra:) + 4stn(27rx) on
03x32 Solution. Suppose the problem can be solved by the method of separation of
variables. Then suppose u(w, t) 'is the product of a function just of x and a function
just of t, that is, u(x,t) = X(x)T(t). Then um = X”(m)T(t), and u, = X(m)T’ The differential equation can
then be written as X ” (x)T (t) = 4X(m)T' One can then divide the entire equation
by the original function u(m, t) = X (x)T(t) (as one always must do in a separation
of variables problem), turning the equation into 5%? = 4%. (For heat conduction
problems, it turns out to be very convenient if a constant — such as the 4 in this
problem — is kept with the time, rather than the space, coordinate.) As the left side of the equation is a function exclusively of 2:, and the right
side exclusively of t, the only way this equation can hold true is if both sides are
equal to some constant. Call this constant —A (the negative sign proves to make
the expressions below easier to work with). Then the left side, the X equation, is X” + /\X = 0. The right side,
the T(t) equation, is 4T’ (t) + AT (t) = 0. The general solution to the spatial equation, the X equation, is X (ac) =
Clcos(\/X$) + Cgsin(\/Xx). To satisfy the left boundary condition, one ﬁnds that
X (0) = Clcos(0) +Czsz'n(0) = Cl = 0. To satisfy the right boundary condition, one
ﬁnds that X (2) = Cgsin(2\/X) = This can be true only if 02 = 0 — making the
solution the boring case of the trivial solution, for which X = 0 for all a: — or if
sin(2\/X) = 0. For sin(2\/X) to equal zero, then 2x5 2 [mt for some integer k. Then the eigenvalues A = 52442 for the integers k, and the corresponding eigenfunctions are Xk(x) = The general solutions for the time equation, 4T’ (t) + AT(t) = 0, have the form T(t) = e‘it. The values of A must be the eigenvalues from the spatial equation, so 2W2
that the eigenfunctions for Tk(t) = e‘LIF‘t. CONTINUED The fundamental solutions of the heat conduction equation in this problem 2W2 .  .
Lle‘tsmwfx). Therefore the general solution to the entire differential equation has the form: then are uk(x,t) = 6‘ 0° 2 2
u(a:,t) = Z Cke_k_1g'tsin(%7:m)
16:1 The coefﬁcients Ck must be chosen to satisfy the initial conditions, that
u(z; 0) ‘= 2311 Cksin(%’‘x) = 23in(%x) — sin(7ra:) + 4sin(27r$). The only way the quantity on the left can equal the quantity on the right, for
all values of :13, is for the coefficient in front of the term on the left to equal
the coefﬁcient in front of the term on the right —— so that 0'1 = 1 —— and for
the coefﬁcient in front of the sin(7r$) term on the left to equal the coefﬁcient in front
of the sin(7rac) term on the right — so that 02 = —1 — and for the coefficient in front
of the sin(27m:) term on the left to equal the coefﬁcient in front of the sin(27rx) term
on the right — so that C4 = 4 — and for all the other coefﬁcients Ck to equal zero, as
there are no terms on the right for any other values of k. This is from the orthogonality of the sine functions. Therefore the solution to the given heat conduction problem is 2 ",2 _7T2 t u(x, t) = 26—116—t3’in(22:$) — e‘TtsinUrac) + 46 sin(27m:) Section 10.5, Problem 9
Problem. Consider the conduction of heat in a rod 40cm in length Whose ends are maintained at 0 degrees Celsius for all t > 0. Suppose that a2 = 1 and ﬁnd an expression for the temperature u(m, t) if the initial temperature distribution
in the rod is the function u(ac, 0) = 50 for 0 < a: < 40. Solution. The heat conduction problem given, translated into symbols, is
the differential equation um = at for O < :10 < 40 and t > 0 (the heat conduction
problem), with the boundary conditions u(0, t) = 0 and u(40, t) = O for t > 0 (the
ends of the bar are kept at 0 degrees), and the initial condition u(a:, 0) = 50 for
0 < a: < 40 (the bar is initially at a temperature of 50 degrees). Suppose the solutionis of the form u(w, t) = X Then the differential
equation can be written as X”(a:)T(t) = X (m)T ’ (t), and can then be separated into
%ﬂ(3 = For this equation to be true, both sides must equal a constant —A. Following the routine as seen in Problems 7 and 8 in this section, the eigen—
values Ak equal f—Z—gand the associated eigenvalues for X are = sin(%x).
The corresponding solutions for T (t) are Tk(t) = e‘%%t. So the general solution of
the problem is ' 00 2,2
u(m, t) ‘= Z Olga—ism
k=1 To satisfy the initial condition, the coefficients Ck must satisfy:
°° _, k7r
u(9:, 0) = Z Ckszn(Zb—m) 2 50,0 < a: < 40
k=1 Notice that this means the coefficients Ck are the Fourier sine series coeﬂicients
for u(x,0) = 50. So the coefﬁcients Ck = 2% 510 503in(ﬁ—gx)dm = g 610 sing—gm) = 100%(1 — cos(k7r)). So the solution of the given heat conduction problem is [W 100 °° 1— cos [M 1313
us, 0 = 7,— 2: —# 15w)
k=1 t .
e 1600 k ( Section 10.5, Problem 13
Problem. Consider again the rod in Problem 9. For t = 5 and t = 20 determine how many terms are needed to ﬁnd the solution correct to three decimal places. A reasonable way to do this is to ﬁnd it so that including one more term
does not change the ﬁrst three decimal places of u(20, 5). Repeat for t = 20 and
t = 80. Form a conclusion about the speed of convergence of the series for u(.7;, t). Solution. Using the solution from Problem 9 and evaluating it at the point
2 2
:r = 20, one ﬁnds u(20,t) = ¥ 23:1 :clzime‘ilfs‘aftsimif). This can be simpliﬁed to u(x, t) = ¥ 233:1 %¥ie‘gﬂfeﬁgﬁt So u(20, 5) = 2% °° girlie—W. This is the sum of a series whose
sign alternates, so — as one may check by referring back to the sequences and series
chapter in a freshman Calculus book — one knows that the difference between the
sum of the inﬁnite series and the sum of the ﬁrst n terms cannot differ by more than
the absolute value of the (71+ 1)st term in the series. So to ﬁnd the number of terms
necessary to get the ﬁrst three digits of this summation correct, one need only ﬁnd
the ﬁrst term in the series whose absolute value is smaller than 0.0005. This ﬁrst
occurs when m = 17, that is, when one adds the ﬁrst 9 terms together. . For t = 20, following the same algebra, one has the equation u(20, 20) =
399 Z” we‘W—gtllﬂ. The ﬁrst term in this series whose absolute value is
smaller than 0.0005 is the term when m = 9, that is, when one adds the ﬁrst 4
terms together. For t = 80, following the same algebra, one has the equation u(20, 80) =
272—0 2%,.1 gg—TlnTLfe‘W. The ﬁrst term in this series whose absolute value is
smaller than 0.0005 is the therm when m = 5, that is, when one adds the ﬁrst 3
terms together. As t increases, the series converges more rapidly; it requires fewer terms to get to a given accuracy. Section 10.5, Problem 14 Problem. For the rod in Problem 9: a. Plot u versus a: for t = 5, 10, 20, 40, 100, and 200. Put all the graphs on the
same set of axes and thereby obtain a picture of the way in which the temperature
distribution changes with time. ‘ b. Plot u versus t for x = 5,10, 15, and 20. 0. Draw a threedimensional plot of u versus m and t. d. How long does it take for the entire rod to cool off to a. temperature of no more than 1 degree Celsius? 0 Solution. The solution of the given heat conduction problem, as seen above, is
_ 100 °° 1—cos(k7r) _k2_7r_2_t , kw
u(x,t)~ 7T :4; k e 1600 szn(40m) a. Plotting u versus a: for t = 5, 10, 20, 40, 100, 200: b. Plotting u versus t for :1: = 5, 10, 15, 20: 52
50
48
46
‘44
42
40 as
as
34
32
so
28
26
24
22
20 c. Plotting u versus a: and t: CONTINUED d. Plotting a time when u($, t) < 1 for all as, for example, at t = 675 seconds: 1 0.8 0.6 0.4 0.2 Section 10.5, Problem 18
Problem. Let a metallic rod 20 cm long be heated to a uniform temperature of 100 degrees Celsius. Suppose that at t = 0 the ends of the bar are plunged into an ice bath at 0 degrees Celsius, and thereafter maintained at this temperature, but that
‘ no heat is allowed to escape through the lateral surface. Find an expression for the
temperature at any point in the bar at any later time. Determine the temperature at the center of the bar at time t = 30 seconds if the bar is made of (a) silver, (b) aluminum, or (0) cast iron. Solution. This problem represents the heat conduction problem a2 for 0 < :8 < 20 as t > 0, with boundary conditions u(0, t) = 0 and u(20, t) = 0 for “9:2 = “t t > 0, and initial conditions u($, 0) = 100 for 0 < a: < 20. Using the separation of
variables technique, assuming u(x, t) = X (x)T(t), and the same procedure used in Problems 7, 8, and 9, one ﬁnds the solution to this heat conduction problem is: D 2 °° 1 — k 21:2 k
u(w,t) = 729— ; ——————c:s( 7r) e'%a2tsin(§%x) The temperature at the center of the bar, at time t = 30, is 200‘ '°° 1 — 008(k7r) e42 k21r2 a'z' k7r u(10, 30) = 7 kgl —l;"—’_ 40 3271(7) Table 10.5.1 gives the values for a2 for the metals used in the problem —
for silver, (12 = 1.71 and u(10, 30) is approximately equal to 35.9 degrees. For
aluminum, (12 = 0.86 and u(10, 30) is approximately equal to 67.2 degrees. For cast
iron, a2 = 0.12 and u(10, 30) is approximately equal to 100 degrees still. E 105191;? Section 10.5, Problem 20
Problem. In solving differential equations the computations can almost al—
ways be simpliﬁed by the use of dimensionless variables. Show that if the dimen sionless variable g = f— is introduced, the heat conduction equation becomes 8% L261;
6—62—EE,0<§<1,t>0 Since 2—: has the units of time, it is convenient to use this quantity to deﬁne a
dimensionless time variable 7' = Then show that the heat conduction equation
reduces to (9211 Bu ¥=E,O<€<1,T>0 Solution. Using the chain rule of differentiation, if E = % then 32% = 35%.
2 2
As w = 5L, then % = L. So 3—; = Lg—g, and 3—5 = nggg.
So the heat conduction equation is at = 05211“ = a2u§§%, so that ugg = (5)21”.
The original limits on a: were 0 < :1: < L; the corresponding limits on E are 0 < é < 1.
To change the time coordinate, one sets the new variable 7' 2 (%)2t. By the may. chain rule of dlfferentlatlon, at — 8T dt, and so 8T — ( a at '
'61:. SO the Conduction equation is uég = ($21“ = E, or ugg : 2—2. The old limit
was t > 0; this converts just to ’7' > 0.
The dimensionless heat conduction equation is Ugg = %, for 0 < E < 1 and 7'>0. I 105/7. 13 Section 10.5, Problem 22 Problem. The heat conduction equation in two space dimensions is 0:2 (um +
uyy) = ut. Assuming that u(x, y, t) = X (m)Y(y)T(t), ﬁnd ordinary differential equa—
tions satisﬁed by X (as), Y(y), and T(t). 1 Solution. From the assumption that u(x,y,t) = X (x)Y(y)T(t), one takes
the partial derivatives and can rewrite the differential equation as X ” (x)Y(y)T(t) +
X (x)Y” (y)T(t) = SEX (x)Y(y)T’ (t). (The coefﬁcient a2 is put over to the time
coordinate’s side of the equation, as this proved convenient for functions in one
space dimension.) ‘ Dividing the entire equation by u(x, y, t) = X (x)Y(y)T(t), one can rewrite the
system as 3%; + w = In order for this equation to hold true, both sides
— the left, a function of a: and y, and the right, a function just of t — must equal the same constant; call it a2. XII ) Y” __ 2 1 TI 1; 2 . .
Then 71%— + 76;} — A , and —5 Ta) — —A . nght away one can write the differential equation for T(t): T' (t) = —a2)\2T(t). The other side of the equation, %ﬂ;‘—)l + YVHéE)1 = —)\2, does not appear at ﬁrst
to lend itself to the same sort of solution — one'is required to look further. One
, would like the, equation to be written as a function exclusively of x on oneside of
the equation, and one exclusively of y on the other. This, it turns out, is possible. Bring the m to the right side of the equation, and one has the equation I/ II %f)2 = —1;7é%2 —— A2. Now one has an expression just of cc on the left and another
expression just of g on the right. For this equation to hold true, both expressions must be equal to some constant; call it —u2. Then %:—)l = —,u2 and so X”(ac) + ,uZX(:r) = 0. This is the second of the
differential equations. The remaining expression is —Xi%f—’)l—A2 = —/r2, which can be rewritten $551 =
(/\2 — M2) 30 Y”(y) + (/\2 — M2)Y(y) = 0. The differential equations for the heat conduction equation in two variables are: X”(x) + u2X(:r) = 0, Y”(y) + (A2 — p2)Y(y) = 0, and T’(t) + a2/\2T(t) = 0. {105/5191 ...
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 Spring '04
 Yoon
 Differential Equations, Equations, Partial differential equation

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