A rod of 30cm long has its end a and b are kept at 20 and 80 respectively until steady state condition prevail. The temperature at each end is suddenly reduced to 0 and kept so. Find the resulting temperature u(x,t) from the end a.
Answers
The ends A and B of a rod 20cm long have the temperature at 30 degree centigrade and 80 degree
centigrade respectively until steady state prevails. The temperatures of the ends are changed to 40 degree
centigrade to 60 degree centigrade respectively. Find the temperature distribution in the rod at time t.
Solution
At time = 0 steady state prevails:
2(,0)
2 = 0, for 0 ≤ ≤ 20,
(0,0) = 30 and (20,0) = 80.
The solution to this boundary value problem is easily found, since the general solution of the differential
equation is () = + ; where A and B are arbitrary constants. Then the boundary conditions reduce to
(0) = = 30 and (20) = ∙ 20 + = 80.
We conclude that = 30 and =
80−30
20
=
5
2
.
So the steady-state initial temperature is
(, 0) =
5
2
+ 30.
We will start this section by solving the initial/boundary value problem
(,) = (,) for > 0 0 < < 20,
(0,) = 40 (20,) = 60, for > 0,
(, 0) =
5
2
+ 30, for 0 ≤ ≤ 20.
It is useful for both mathematical and physical purposes to split the problem into two parts. We first find
the steady-state temperature that satisfies the boundary conditions. A steady-state temperature is one
that does not depend on time. Then = 0, so the heat equation simplifies to = 0. : Hence we are
looking for a function
() defined for 0 ≤ ≤ 20 such that
2
()
2 = 0, for 0 < < 20,
(0,) = 40 and
(20,) = 60 for > 0.
The solution to this boundary value problem is easily found, since the general solution of the differential
equation is
() = + ; where A and B are arbitrary constants. Then the boundary conditions reduce
to
(0) = = 40 and
(20) = ∙ 20 + = 60.
We conclude that = 40 and =
60−40
20
= 1.
So the steady-state temperature is
(, 0) = + 40.
It remains to find = −
. It will be a solution to the heat equation, since both and are, and the
heat equation is linear. The boundary and initial conditions that satisfies can be calculated from those for
and
. Thus, = −
, must satisfy
hope it helps..
The heat equation is
Let u = X(x) . T(t) be the solution of (1), where „X‟ is a function of „x‟ alone and „T‟ is a function of „t‟ alone.
Substituting these in (1), we get
Now the left side of (2) is a function of „x‟ alone and the right side is a function of „t‟ alone. Since „x‟ and „t‟ are independent variables, (2) can be true only if each side is equal to a constant.
Hence, we get X′′ - kX = 0 and T′ -a2kT=0.-------------- (3).
Solving equations (3), we get
(i) when „k‟, is say positive and k = l2
X = c1 elx + c2 e - lx
(iii) when „k‟ is zero.
X = c7 x + c8
T = c9
Thus the various possible solutions of the heat equation (1) are
Of these three solutions, we have to choose that solution which suits the physical nature of the problem and the given boundary conditions. As we are dealing with problems on heat flow, u(x,t) must be a transient solution such that „u‟ is to decrease with the increase of time „t‟.
Therefore, the solution given by (5),
is the only suitable solution of the heat equation.