Answer to Question #152924 in Differential Equations for kashif ali

Question

solve the following bondry value problems by the method of separation of variables

U_t = U_xx 0<x<1 , t>0

U(0,t)= 0 ,U(pi ,t)= 0

U(x , 0) = sinx + 2sin3x

Answer 1

$\displaystyle \frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}\\ \textsf{Let}\,\, u(x, t) = X(x)Y(t)\\ X''Y = XY' = 0,\,\, X''Y= XY'\\ X''Y= XY' = K\\ \frac{X''}{X} = \frac{Y'}{Y} = K\\ \frac{X''}{X} = K,\,\, \frac{Y'}{Y} = K\\ \textsf{If}\,\, K\,\, \textsf{is chosen to be negative}\\ \textsf{say}\,\, K = -\lambda^2,\,\,\textsf{then}\,\,\\ X'' = -\lambda^2 X\\ X = A\cos(\lambda x) + B\sin(\lambda x)\\ Y' = -\lambda^2 Y\\ Y = Ce^{-\lambda^2 t}\\ u(x,t) = \left( A\cos(\lambda x) + B\sin(\lambda x)\right)\left(Ce^{-\lambda^2 t}\right)\\ u(x,0) = C\left( A\cos(\lambda x) + B\sin(\lambda x)\right) = \sin(x) + 2\sin(3x)\\ u(0,0) = AC = 0\\ u(0,t) = ACe^{-\lambda^2 t} = 0\\ u(0,0) = AC = 0\\ \textsf{Avoiding trivial zeros, we select}\,\, A = 0 \\ u(\pi,t) = \left( A\cos(\pi\lambda) + B\sin(\pi\lambda)\right)\left(Ce^{-\lambda^2 t}\right) = 0\\ u(\pi,t) = \left(B\sin(\pi\lambda)\right)\left(Ce^{-\lambda^2 t}\right) = 0\\ \implies B\sin(\pi\lambda) = 0,\,\, \sin(\pi\lambda) = \sin(n \pi) \\ \therefore \lambda = n \\ u(x,t) = \left(B\sin(nx)\right)\left(Ce^{-n^2 t}\right) = D\sin(nx)e^{-n^2 t}\\ u(x,t)\,\,\textsf{can be superposed as;}\\ \begin{aligned} u(x,t) &=\sum_{n = 1}^{\infty} D_n\sin(nx)e^{-n^2 t} \end{aligned}\\ u(x,0) = \sin(x) + 2\sin(3x)\\ \begin{aligned} u(x,0) = \sin(x) + 2\sin(3x) &=\sum_{n = 1}^{\infty} D_n\sin(nx) \end{aligned}\\ \textsf{We have here a half-range Fourier sine}\\ \textsf{series representation of a function}\,\, u(x,0)\\ \textsf{defined over}\,\ 0< x < 1, \,\,\textsf{Extending}\,\, u(x,0)\,\, \textsf{as an}\\ \textsf{even periodic function with period}\,\, 2\,\, \textsf{and}\\ \textsf{using standard Fourier series theory gives}\\ \begin{aligned} D_n &= 2\int_0^1 \left(\sin(x) + 2\sin(3x)\right) \sin(n x)\, \mathrm{d}x \\&= \int_0^1 \left(\cos(x - nx) - \cos(x + nx) + 2\cos(3x - nx) - 2\cos(3x + nx)\right) \\&= \left(\frac{\sin(x - nx)}{1 - n} - \frac{\sin(x + nx)}{1 + n} + \frac{2\sin(3x - nx)}{3 - n} - \frac{2\sin(3 + nx)}{3 + n}\right)\biggr\vert_0^1 \\&= \frac{\sin(1 - n)}{1 - n} - \frac{\sin(1 + n)}{1 + n} + \frac{2\sin(3 - n)}{3 - n} - \frac{2\sin(3 + n)}{3 + n} \end{aligned}\\ \begin{aligned} \therefore u(x,t) =\sum_{n = 1}^{\infty}\left(\frac{\sin(1 - n)}{1 - n} - \frac{\sin(1 + n)}{1 + n} + \frac{2\sin(3 - n)}{3 - n} - \frac{2\sin(3 + n)}{3 + n}\right)\sin(nx)e^{-n^2 t}\\ \end{aligned}\\$