Let u = X(x) . T(t) be the solution of (1), where â€žXâ€Ÿ is a function of â€žxâ€Ÿ alone and â€žTâ€Ÿ is a function of â€žtâ€Ÿ alone.

**Solution of the heat equation**

The heat equation is

Let u = X(x) . T(t) be the solution of (1), where â€žXâ€Ÿ is a function of â€žxâ€Ÿ alone and â€žTâ€Ÿ is a function of â€žtâ€Ÿ alone.

Substituting these in (1), we get

Now the left side of (2) is a function of â€žxâ€Ÿ alone and the right side is a function of â€žtâ€Ÿ alone. Since â€žxâ€Ÿ and â€žtâ€Ÿ are independent variables, (2) can be true only if each side is equal to a constant.

Hence, we get Xâ€²â€² - kX = 0 and Tâ€² -a2kT=0.-------------- (3).

Solving equations (3), we get

(i) when â€žkâ€Ÿ, is say positive and k =

X = c1 elx + c2 e - lx

(iii) when â€žkâ€Ÿ 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 â€žuâ€Ÿ is to decrease with the increase of time â€žtâ€Ÿ.

Therefore, the solution given by (5),

is the only suitable solution of the heat equation.

**Illustrative Examples**

**Example 7**

A rod â€žâ„“â€Ÿ cm with insulated lateral surface is initially at temperature f(x) at an inner point of distance x cm from one end. If both the ends are kept at zero temperature, find the temperature at any point of the rod at any subsequent time.

The boundary conditions are

(i)u (0,t) = 0, " t â‰¥ 0

(ii) u (â„“,t) "t > 0 = 0,

(iii) u (x,0) = f (x), 0 < x < â„“

The solution of equation (1) is given by

neglecting radiation. Find an expression for u, if the ends of the bar are maintained at zero temperature and if, initially, the temperature is T at the centre of the bar and falls uniformly to zero at its ends.

Let u be the temperature at P, at a distance x from the end A at time t.

The temperature function u (x,t) is given by the equation

The boundary conditions are

(i) u (0,t) = 0, "t __>__ 0.

(ii) u (â„“=0,,t)"__>__ 0.

The solution of (1) is of the form

Applying conditions (i) and (ii) in (2), we get

**Steady - state conditions and zero boundary conditions Example 9**

A rod of length â€žâ„“â€Ÿ has its ends A and B kept at 0Â°C and 100Â°C until steady state conditions prevails. If the temperature at B is reduced suddenly to 0Â°C and kept so while that of A is maintained, find the temperature u(x,t) at a distance x from A and at time â€žtâ€Ÿ.

The heat-equation is given by

Prior to the temperature change at the end B, when t = 0, the heat flow was independent of time (steady state condition).

When the temperature u depends only on x, equation(1) reduces to

**Example 10**

A rod, 30 c.m long, has its ends A and B kept at 20Â°C and 80Â°C respectively, until steady state conditions prevail. The temperature at each end is then suddenly reduced to 0Â°C and kept so. Find the resulting temperature function u (x,t) taking x = 0 at A.

The one dimensional heat flow equation is given by

The initial conditions, in steady â€“state, are

u = 20, when x = 0

u = 80, when x = 30

Therefore, (3) gives b = 20, a = 2.

u (x) = 2x + 20 ------------- (4)

Hence the boundary conditions are

(i) u (0,t) = 0, " t > 0

(ii) u (30,t) = 0, " t > 0

(iii) u (x,0) = 2x + 20, for 0 < x < 30

The solution of equation (1) is given by

Applying conditions (i) and (ii), we get

**Steadyâ€“state conditions and nonâ€“zero boundary conditions**

**Example 11**

The ends A and B of a rod 30cm. long have their temperatures kept at 20Â°C and 80Â°C, until steadyâ€“state conditions prevail. The temperature of the end B is suddenly reduced to 60Â°C and kept so while the end A is raised to 40Â°C. Find the temperature distribution in the rod after time t.

Let the equation for the heat- flow be

The initial conditions, in steadyâ€“state, are

u = 20, when x = 0

u = 80, when x = 30

From (2), b = 20 & a = 2.

Thus the temperature function in steadyâ€“state is

u (x) = 2x + 20-------------- (3)

Hence the boundary conditions in the transientâ€“state are

(i) u (0,t) = 40, " t > 0

(ii) u (30,t) = 60, " t > 0

(iii) u (x,0) = 2x + 20, for 0 < x < 30

we break up the required funciton u (x,t) into two parts and write

u (x,t) = us (x) + ut (x,t)--------------- (4)

where us (x) is a solution of (1), involving x only and satisfying the boundary condition (i) and (ii). ut (x,t) is then a function defined by (4) satisfying (1).

Thus us(x) is a steady state solution of (1) and ut(x,t) may therefore be regarded as a transient solution which decreases with increase of t.

To find us(x)

we have to solve the equation

Solving, we get us(x) = ax + b ------------- (5)

Here us(0) = 40, us(30) = 60.

Using the above conditions, we get b = 40, a = 2/3.

To find ut(x,t)

ut ( x,t) = u (x,t) â€“us (x)

Now putting x = 0 and x = 30 in (4), we have

ut (0,t) = u (0,t) â€“us (0) = 40â€“40 = 0

and ut (30,t) = u (30,t) â€“us (30) = 60â€“60 = 0

Also ut (x,0) = u (x,0) â€“us (x)

Hence the boundary conditions relative to the transient solution ut (x,t) are

ut (0,t) = 0

ut (30,t) = 0 --------------(v)

and ut (x,0) = (4/3) x â€“20 ------------- (vi)

**Exercises**

(1) Solve Â¶u*/* Â¶t = a2 (Â¶2u */* Â¶x2) subject to the boundary conditions u(0,t) = 0, u(l,t) = 0, u(x,0) = x, 0<x<l.

(2) Find the solution to the equation Â¶u*/* Â¶t = a2 (Â¶2u */* Â¶x2) that satisfies the conditions

i. u(0,t) = 0,

ii. u(l,t) = 0, "t >0,

iii. u(x,0) = x for 0<x<l*/* 2.

= l â€“x for l*/* 2<x<l.

(3) Solve the equation Â¶u*/* Â¶t = a2 (Â¶2u */* Â¶x2) subject to the boundary conditions

i. u(0,t) = 0,

ii. u(l,t) = 0, "t >0,

iii. u(x,0) = kx(l â€“x), k >0, 0 Â£x Â£l.

(4) A rod of length â€žlâ€Ÿ has its ends A and B kept at 0 o C and 120 o C respectively until steady state conditions prevail. If the temperature at Bis reduced to 0 o C and kept so while that of A is maintained, find the temperature distribution in the rod.

(5) A rod of length â€žlâ€Ÿ has its ends A and B kept at 0o C and 120 o C respectively until steady state conditions prevail. If the temperature at Bis reduced to 0 o C and kept so while 10 o C and at the same instant that at A is suddenly raised to 50 o C. Find the temperature distribution in the rod after time â€žtâ€Ÿ.

(6) A rod of length â€žlâ€Ÿ has its ends A and B kept at 0 o C and 100 o C respectively until steady state conditions prevail. If the temperature of A is suddenly raised to 50 o C and that of B to 150 o C, find the temperature distribution at the point of the rod and at any time.

(7) A rod of length 10 cm. has the ends A and B kept at temperatures 30o C and 100o C, respectively until the steady state conditions prevail. After some time, the temperature at A is lowered to 20o C and that of B to 40o C, and then these temperatures are maintained. Find the subsequent temperature distribution.

(8) The two ends A and B of a rod of length 20 cm. have the temperature at 30o C and 80o C respectively until th steady state conditions prevail. Then the temperatures at the ends A and B are changed to 40o C and 60o C respectively. Find u(x,t).

(9) A bar 100 cm. long, with insulated sides has its ends kept at 0o C and 100o C until steady state condition prevail. The two ends are then suddenly insulated and kept so. Find the temperature distribution

(10) Solve the equation Â¶u*/* Â¶t = a2 (Â¶2u */* Â¶x2) ) subject to the conditions (i) â€žuâ€Ÿ is not infinite

as t Â®Â¥ (ii) u = 0 for x = 0 and x = p, "t (iii) u = px -x2 for t = 0 in (0, p).

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Mathematics (maths) : Applications of Partial Differential Equations : Solution of the heat equation |

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