Partial Differential Equation - Heat Equation Question

Question

Let T(x,t) denote temperature distribution in a slab 0<x<1, initially at temperature f(x), when the surface at x=0 in insulated and surface heat transfer take places at surface x=1 into a medium at temperature zero. According to Newton's law of cooling, the condition on the surface x=1 is T_x(1,t) = -hT(1,t), where h is positive constant. The boundary value problem to be solved is, then,

 

T_t(x,t) = kT_xx(x,t)    ,            0<x<1,         t>0

 

with,   T_x(0,t) = 0            and            T_x(1,t) = -hT(1,t)

 

and, 

         T(x,0) = f(x)

 

Obtain the temperature distribution is the slab 0<x<1.

 

 

Answer 1

Start by supposing a solution of separable variables exists:

Instead of using T, let's use u, so, let u = TX, where T is the time dependent function and X is the space dependent function.

We then get u_t(x,t) = ku_xx(x,t) is given by:

XT' = kX''T

If we pull everything to its own side we get:

kX''/X = T'/T, but since the LHS only depends on X, and the RHS only depends on T, then they must be equal to some constant, call it L

So we solve the ODEs:

Case 1: L = 0:

kX'' = 0

X' = C_1

X = C_1x + C_2

X'(0) = 0 and X'(1) = -hX(1)

so C_1 = 0

and C_1 = -h(C_1 + C_2) = 0 = -hC_2 = C_2

So C_1 = C_2 = 0, so X(x) = 0, so u(x,t) = X(x)T(t) = 0

Case 2: k = 0:

If k = 0:

X = 0

so u(x,t) = 0

Case 3: L/k = m^2 > 0

X'' - m^2X = 0

X'' - m^2X = 0

X = C_1e^(mx) + C_2e^(-mx)

X'(0) = mC_1 - mC_2 = 0

so C_1 = C_2

and X'(1) = -hX(1)

so C_1me^(mx) - C_2me^(-mx) = -hC_1e^(mx) - hC_2e^(mx)

so mC_1 = -hC_1 and -mC_2 = -mC_1 = -hC_1 = -hC_2, which is only true for C_1 = 0

so C_2 = 0 as well

so X(x) = 0, and finally u(x,t) = 0

Case 4: L/k = -m^2 < 0

X = C_1sin(mx) + C_2cos(mx)

X'(0) = 0

so mC_1 = 0 = C_1

X'(1) = -hX(1)

so -mC_2sin(m) = -hC_2cos(m) so C_2(msin(m) - hcos(m)) = 0

If C_2 = 0

the solution is X(x) = 0

so u(x,t) = 0

or mtan(m) = h, but mtan(m) has no closed form solution, so we have to check L = -km^2 for every m which satisfies mtan(m) = h.

Then we have X(x) = C_2cos(mx)

Then T'(t) + km^2T = 0

so T(t) = C_1 e^(-km^2t)

and T(0) = f(0)

so C_1 = f(0).

Therefore T(t) = f(0) e^(Lt) for some constant L = -km^2 for all m which satisfy mtan(m) = h

So we get:

u(x,t) = X(x)T(t) = C_2cos(mx)f(0)e^(-km^2t) for all m which satisfy mtan(m) = h

So overall using the linear combination we get:

u(x,t) = c_1*cos(m_1x)*e^(-k((m_1)^2)t) + c_2*cos(m_2x)*e^(-k((m_2)^2)t) + ...

For constants c_n and m_n satisfies m_n * tan(m_n) = h

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