Question

find the solution and ususal about the solution uxx=1/100ut , u(0,t)=Ae-dt u(x,0)=0 and ux approaches to 0 as x approahes to zero

Answer 1

$\displaystyle u(x, t) = X(x)T(t)\\ u_{xx}(x, t) = X''(x)T(t)\\ u_{t} = X(x)T'(t)\\ X''(x)T(t) = \frac{1}{100}X(x)T'(t)\\ \frac{100X''(x)}{X(x)} = \frac{T'(t)}{T(t)} = K\\ \int \frac{T'(t)}{T(t)}\,\mathrm{d}t = \int K \,\mathrm{d}t\\ \ln\left(T(t)\right) = Kt + C\\ T(t) = e^{Kt + C} = Ae^{Kt}\\ \frac{100X''(x)}{X(x)} = K\\ \frac{X''(x)}{X(x)} = \frac{K}{100}\\ X''(x) - \frac{KX(x)}{100} = 0\\ \textsf{If}\,\, K\,\, \textsf{is chosen to be negative}\\ \textsf{say}\,\, K = -\lambda^2,\,\,\textsf{then}\,\,\\ X'' = -\frac{\lambda^2X(x)}{100}\\ \therefore X = B\cos\left(\frac{x}{10}\right) + C\sin\left(\frac{x}{10}\right)\\ \begin{aligned} u(x, t) &= Ae^{-\lambda^2t}\left(B\cos\left(\frac{x}{10}\right) + C\sin\left(\frac{x}{10}\right)\right)\\ &= e^{-\lambda^2t}\left(D\cos\left(\frac{x}{10}\right) + E\sin\left(\frac{x}{10}\right)\right) \end{aligned} \\ u(0, t) = De^{-\lambda^2t}= Ae^{-dt} \\ \therefore A = D,\,\, \textsf{and}\,\, d = \lambda^2\\ u(x, 0) = D\cos\left(\frac{x}{10}\right) + E\sin\left(\frac{x}{10}\right) = 0 \\ \lim_{x \to 0} u(x, 0) = \lim_{x \to 0} \left(D\cos\left(\frac{x}{10}\right) + E\sin\left(\frac{x}{10}\right)\right) = 0 \\ \implies D = 0.\\ \begin{aligned} \therefore u(x, t) &= Ee^{-dt}\sin\left(\frac{x}{10}\right) \end{aligned}$