Question

By the method of separation of variables solve the boundary value problem

∂u/∂x-5(∂u/∂t)=u given u(x,0)=e^-2x

Answer 1

$\displaystyle \frac{\partial u}{\partial x} - 5\frac{\partial u}{\partial t} = u, u(x,0)= e^{-2x}\\ \text{Let $u(x,t) = P(x)G(t)$}\\ \frac{dP(x)G(t)}{dx} - 5\frac{d P(x)G(t)}{d t} = P(x)G(t)-(1)\\ \text{Divide 1 by P(x)G(t), we get}\\ \frac{d P(x)}{d x}\cdot \frac{1}{P} - 5\frac{dG(t)}{d t}\frac{1}{G} = 1\\ = \frac{d P(x)}{d x}\cdot \frac{1}{P} = 1+ 5\frac{dG(t)}{d t}\frac{1}{G}=\lambda\\ \text{where $\lambda$ is the separation constant}\\ \text{Next, we consider 3 cases, $\lambda = 0, \lambda>0, \lambda < 0$}\\ \text{When $\lambda = 0$, we have that}\\ p'(x)= 0 \implies p(x) = c_1, \text{ where $c_1$ is constant. Also }\\ \frac{G'(t)}{G(t)}=-\frac{1}{5} , \text{ integrating we have that}\\ G(t) = e^{-\frac{1}{5}t}\\ u(x,t) = c_1e^{-\frac{1}{5}t}\\ \text{ Applying initial conditions, we have that}\\ u(x,0) = c_1 = e^{-2x}\\ \therefore u(x,t) = e^{-2x}e^{-\frac{1}{5}t}\\ \text{when $\lambda < 0$, let $\lambda = -k^2$}\\ \implies \frac{P'(x)}{P(x)}=-k^2, \text{ integrating we have}\\ \ln P(x) = -k^2x \implies P(x) = e^{-k^2x}\\ \text{Also, } 5\frac{G'(t)}{G(t)} = -k^2-1 \implies \frac{G'(t)}{G(t)} = \frac{-k^2-1}{5}\\ \ln G(t)=\frac{-k^2-1}{5}t\\ \implies G(t)=e^{\frac{-k^2-1}{5}t}\\ \therefore u(x,t) = e^{-k^2x}e^{\frac{-k^2-1}{5}t}\\ \text{Applying initial conditions, we have }\\ u(x,0)= e^{-k^2x}= e^{-2x}\\ \implies k = \sqrt{2}\\ \implies u(x,t)=e^{-2x}e^{-\frac{3}{5}t}\\ \text{when $\lambda > 0$, let $\lambda = k^2$}\\ \implies \frac{P'(x)}{P(x)}=k^2, \text{ integrating we have}\\ \ln P(x) = k^2x \implies P(x) = e^{k^2x}\\ \text{Also, } 5\frac{G'(t)}{G(t)} = k^2-1 \implies \frac{G'(t)}{G(t)} = \frac{k^2-1}{5}\\ \ln G(t)=\frac{k^2-1}{5}t\\ \implies G(t)=e^{\frac{k^2-1}{5}t}\\ \therefore u(x,t) = e^{k^2x}e^{\frac{k^2-1}{5}t}\\ \text{Applying initial conditions, we have }\\ u(x,0)= e^{k^2x}= e^{-2x}\\ \implies k = \sqrt{2}i\\ \implies u(x,t)=e^{-2x}e^{-\frac{3}{5}t}$