Question

A rod of 30cm length has its ends P and Q kept 20°C and 80°C respectively until steady state condition prevail. The temperature at each point end is suddenly reduced to 0°C and kept so. Find the conditions for solving the equation.

(a) u(0,t) = 0 = u(30,t) and u(x,0) = 20 + 60/10 x

(b) ux(0,t) = 0 = ux(30,t) and u(x,0) = 20 + 60/30 x

(c) ut(0,t) = 0 = ut(30,t) and u(x,0) = 20 + 60/10 x

(d) u(0,t) = 0 = u(30,t) and u(x,0) = 20 + 60/30 x

The question was posed to me in examination.

Enquiry is from Solution of 1D Heat Equation topic in chapter Applications of Partial Differential Equations of Engineering Mathematics

Answer 1

Right choice is (d) u(0,t) = 0 = u(30,t) and u(x,0) = 20 + 60/30 x

The explanation: 0 and 30 are the end points. So, at these points the function is zero. Hence u(0,t) = 0 = u(30,t). Next due to steady state conditions, at the beginning that is initial conditions, we have u(x,0) = 20 + 60/30 x.