Answer to Question #172771 in Differential Equations for Shahil Singh

Question #172771

 A bar, 10 cm long with insulated sides, has its ends A and B kept at 20°C and 40°C respectively until steady state conditions prevail. The temperature at A is then suddenly raised to 50°C and at the same instant that at B is lowered to 10°C. Find the subsequent temperature at any point of the bar at any time. (16)

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1
Expert's answer
2021-03-22T06:56:00-0400

The heat equation:

ut=α22ux2\frac{\partial u}{\partial t}=\alpha^2\frac{\partial^2u}{\partial x^2}

In steady state:

2ux2=0\frac{\partial^2u}{\partial x^2}=0

Solving, we get

u=ax+bu=ax+b

The initial conditions, in steady–state, are 

u=20,x=0u=20, x=0

u=40,x=Lu=40, x=L

Then:

b=20,a=20/Lb=20, a=20/L

The temperature function in steady–state is

u(x)=20x/L+20u(x)=20x/L+20


The boundary conditions in the transient state are 

u(0,t)=50,u(L,t)=10,u(x,0)=20x/L+20u(0,t)=50, u(L,t)=10, u(x,0)=20x/L+20


We break up the required funciton u (x,t) into two parts:

u(x,t)=us(x)+ut(x,t)u(x,t)=u_s(x)+u_t(x,t)

us(x)u_s(x) is a steady state solution, ut(x,t)u_t(x,t) is a transient solution


us(x)=ax+bu_s(x)=ax+b

b=50,a=40/Lb=50, a=-40/L

us(x)=40x/L+50u_s(x)=-40x/L+50


ut(x,t)=u(x,t)us(x)u_t(x,t)=u(x,t)-u_s(x)

Putting boundary conditions, we have the boundary conditions relative to the transient solution :

ut(0,t)=u(0,t)us(0)=5050=0u_t(0,t)=u(0,t)-u_s(0)=50-50=0

ut(L,t)=u(L,t)us(L)=1010=0u_t(L,t)=u(L,t)-u_s(L)=10-10=0

ut(x,0)=u(x,0)us(x)=20x/L+20+40x/L50=60x/L30u_t(x,0)=u(x,0)-u_s(x)=20x/L+20+40x/L-50=60x/L-30


We have:

ut(x,t)=(Acosλx+Bsinλx)eα2λ2tu_t(x,t)=(Acos\lambda x+Bsin\lambda x)e^{-\alpha^2\lambda^2t}

Using boundary conditions, we get:

A=0,λ=πn/LA=0, \lambda=\pi n/L

Then:

ut(x,t)=BsinπnxLeα2π2n2L2tu_t(x,t)=Bsin\frac{\pi nx}{L}e^{-\frac{\alpha^2\pi^2 n^2}{L^2}t}

The most general solution:

ut(x,t)=n=1BnsinπnxLeα2π2n2L2tu_t(x,t)=\displaystyle\sum_{n=1}^{\infin}B_nsin\frac{\pi nx}{L}e^{-\frac{\alpha^2\pi^2 n^2}{L^2}t}

Using boundary conditions:

ut(x,0)=n=1BnsinπnxL=60x/L30,0<x<Lu_t(x,0)=\displaystyle\sum_{n=1}^{\infin}B_nsin\frac{\pi nx}{L}=60x/L-30, 0<x<L

Bn=2L0L(60x/L30)sinπnxLdx=B_n=\frac{2}{L}\displaystyle\intop_0^L(60x/L-30)sin\frac{\pi nx}{L}dx=

=2L((60x/L30)cos(πnx/L)πn/L60Lsin(πnx/L)π2n2/L2)0L==\frac{2}{L}((60x/L-30)\frac{-cos(\pi nx/L)}{\pi n/L}-\frac{60}{L}\cdot\frac{-sin(\pi nx/L)}{\pi^2n^2/L^2})|^L_0=

=2L(30Lcos(πn)πn30Lπn)=60πn(cos(πn)+1)=\frac{2}{L}(-\frac{30Lcos(\pi n)}{\pi n}-\frac{30L}{\pi n})=-\frac{60}{\pi n}(cos(\pi n)+1)


Bn=60(1+(1)n)πnB_n=-\frac{60(1+(-1)^n)}{\pi n}


Bn=0B_n=0 when nn is odd

Bn=120πnB_n=-\frac{120}{\pi n} when nn is even


ut(x,t)=n=2,4,6,..(120πnsinπnxL)eα2π2n2L2tu_t(x,t)=\displaystyle\sum_{n=2,4,6,..}^{\infin}(-\frac{120}{\pi n}sin\frac{\pi nx}{L})e^{-\frac{\alpha^2\pi^2n^2}{L^2}t}


u(x,t)=40x/L+50+n=2,4,6,..(120πnsinπnxL)eα2π2n2L2tu(x,t)=-40x/L+50+\displaystyle\sum_{n=2,4,6,..}^{\infin}(-\frac{120}{\pi n}sin\frac{\pi nx}{L})e^{-\frac{\alpha^2\pi^2n^2}{L^2}t}



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