Answer to Question #86618 in Differential Equations for Evol

Question #86618
solve the heat conduction equation: 8*∂^2u(x,t) /∂x^2 = ∂u(x,t)/∂t; for0<x<5 and t>0 the following boundary and initial conditions: u(0,t) = u(5,t) = 0, u(x,0) = 2sin(πx) − 4sin(2πx)
1
Expert's answer
2019-04-09T09:58:41-0400
"\\partial u\/\\partial t=8\\ (\\partial^2\\ u)\/(\\partial x^2\\ )\\ \\ \\ \\ \\ \\ (0<x<5,\\ t>0)"

"u(0,t)=u(5,t)=0\\ \\ \\ \\ (\\ t\\geq0),\\ u(x,0)=2\\sin \\pi x-4\\sin 2\\pi x\\ \\ \\ \\ \\ \\ (0<x<5)."

Using separation of variables, with "u(x,t)=X(x)T(t)" :

"X(x)\\ T\\dot (t)=kT(t) X^{\\prime\\prime} (x)\\Rightarrow(T \\dot(t))\/kT(t)\\ =(X^{\\prime\\prime} (x))\/X(x)\\ =-\\lambda"

"u(0,t)=u(5,t)=0\\ \\ \\ \\ (\\ t\\geq0)\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ X(0)=X(5)=0"

We got an eigenvalue problem for X (x):

"X^{\\prime\\prime} (x)+\\lambda X(x)=0, X(0)=X(5)=0 \\Rightarrow X_n (x)=\\sin{\\pi nx\/5} ,\\ \\ \\lambda_n=(n^2 \\pi^2)\/25,\\ n=1,2,\\dots"

On substituting eigenvalue to differential equation, we obtain


"(\\dot T_n(t))\/(8T_n(t))=-\\lambda_n=\\frac{n^2\\pi^2}{25} \\Rightarrow \\dot T(t)+8 \\frac{n^2\\pi^2}{25} T_n (t)=0""T_n(t) = a_n \\exp{\\left(-8 \\frac{n^2\\pi^2}{25}t\\right)}"

General Solution:


"u\\left(x,t\\right)=\\sum_{n=0}^{\\infty\\ \\ }{T_n\\left(t\\right)X_n\\left(x\\right)}=\\sum_{n=0}^{\\infty\\ \\ }{a_n\\exp{\\left(-8 \\frac{n^2\\pi^2}{25}t\\right)}\\sin{\\left(\\pi nx\/5\\right)}}"

Applying initial condition:


"u\\left(x,0\\right)=\\sum_{n=0}^{\\infty\\ \\ }{a_n\\sin{\\left(\\pi nx\/5\\right)}} = 2\\sin{(\\pi x)} - 4 \\sin{(2\\pi x)}"

As X_N(x) is orthonormal basis, "a_5 = 2, a_{10} = -4,\\, and \\, a_n=0" all other possible values of n.

Finally:


"u\\left(x,t\\right)=2\\exp{(-8\\pi^2 t)} \\sin{\\left(\\pi x\\right)} - 4\\exp{(-32\\pi^2 t)} \\sin{\\left(2 \\pi x\\right)}"


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