Question #91732
Solve the heat conduction equation:
for the following boundary and initial conditions:
8(d^2u/dt^2)=(du/dt), for 0<X< 5 and t> 0,
u(0,t) = u(5,t) = 0,
u(x,0) =2sin(πx)− 4sin(2πx)
1
Expert's answer
2019-07-24T10:30:23-0400
uT=82ux2\frac{\partial u}{\partial T} = 8 \frac{\partial^2 u}{\partial x^2}u(0,t)=u(5,t)=0u(x,0)=2sinπx4sin2πxu(0, t) = u(5, t) = 0 \qquad u(x,0) = 2 \sin \pi x - 4 \sin 2\pi x

Let's find the solution in the form

u(x,t)=X(x)T(t)u(x,t) = X(x) T(t)XT=8XTX T' = 8 X'' TTT(t)=8XX(x)=8λ=const\frac{T'}{T}(t) = 8\frac{X''}{X}(x) = -8\lambda = \mathrm{const}

where lambda is an unknown constant.

From the boundary conditions

X+λX=0X(0)=X(5)=0X'' + \lambda X = 0 \qquad X(0) = X(5) = 0X(x)=Csinnπx5λ=(πn5)2X(x) = C \sin \frac{n \pi x}{5} \qquad \lambda = \bigg(\frac{\pi n}{5}\bigg)^2

where n is integer.

Hence

T=8λTT(t)=Cexp(8λt)T' = -8\lambda T \rightarrow T(t) = C \exp(-8\lambda t)

The heat equation is linear, so the solution can be written in the form

u(x,t)=n=1CnXn(x)Tn(t)u(x,t) = \sum_{n=1}^{\infty} C_n X_n (x) T_n(t)

where Cn are constants determined from the initial conditions and Xn, Tn are the functions found above with the integer parameter n (for n = 0 X0(x)= 0).

From the initial conditions

u(x,t)=2exp(8π25252t)sinπx4(8π210252t)sin2πxu(x,t) = 2 \exp \bigg(-8\frac{\pi^2 5^2}{5^2}t\bigg)\sin \pi x - 4 \bigg(-8\frac{\pi^2 10^2}{5^2}t\bigg) \sin 2\pi xu(x,t)=2e8π2tsinπx4e32π2tsin2πxu(x,t) = 2 e^{-8\pi^2 t}\sin \pi x - 4 e^{-32\pi^2 t} \sin 2\pi x


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