Answer to Question #252332 in Differential Equations for pde

Question

Using the Laplace transform method, solve the partial differential equation

∂u/∂t=u-∂u/∂x. Subject to the initial condition u(x,0)=e^-7x, x>0,t>0

Answer 1

Answer:-

Laplace transform:

$L(u(x,t))=\int^{\infin}_0 e^{-st}u(x,t)dt=U(x,s)$

$L(u_t(x,t))=\int^{\infin}_0 e^{-st}u_t(x,t)dt=e^{-st}u(x,t)|^{\infin}_0+s\int^{\infin}_0 e^{-st}u(x,t)dt=$

$=sU(x,s)-U(x,0)$

$L(u_x(x,t))=\int^{\infin}_0 e^{-st}u_x(x,t)dt=U_x(x,s)$

Then:

$sU(x,s)-U(x,0)=U(x,s)-U_x(x,s)$

$\frac{dU}{dx}(x,s)+(s-1)U(x,s)=e^{-7x}$

integrating factor:

$\mu=e^{\int(s-1)dx}=e^{(s-1)x}$

Thus we have:

$\frac{d}{dx}[e^{(s-1)x}U(x,s)]=e^{(s-1)x}e^{-7x}$

$U(x,s)=e^{-(s-1)x}\int^x_0 e^{(s-1)r}e^{-7r}dr+ce^{-(s-1)x}=$

$=e^{-(s-1)x}\cdot\frac{e^{(s-8)r}}{s-8}|^x_0+ce^{-(s-1)x}=\frac{e^{-7x}-e^{-(s-1)x}}{s-8}+ce^{-(s-1)x}$

Taking the inverse Laplace transform we have:

$u(x,t)=e^{-7x}e^{8t}-e^{x+8(t-x)}H(t-x)+ce^t\delta(t-x)$

where H is Heaviside function:

$H(x)=\begin{cases} 0 &x<0 \\ 1 &x\ge0 \end{cases}$