Answer to Question #252332 in Differential Equations for pde

Question #252332

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


1
Expert's answer
2021-11-03T11:57:40-0400

Answer:-

Laplace transform:

L(u(x,t))=0estu(x,t)dt=U(x,s)L(u(x,t))=\int^{\infin}_0 e^{-st}u(x,t)dt=U(x,s)


L(ut(x,t))=0estut(x,t)dt=estu(x,t)0+s0estu(x,t)dt=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)=sU(x,s)-U(x,0)


L(ux(x,t))=0estux(x,t)dt=Ux(x,s)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)Ux(x,s)sU(x,s)-U(x,0)=U(x,s)-U_x(x,s)


dUdx(x,s)+(s1)U(x,s)=e7x\frac{dU}{dx}(x,s)+(s-1)U(x,s)=e^{-7x}


integrating factor:

μ=e(s1)dx=e(s1)x\mu=e^{\int(s-1)dx}=e^{(s-1)x}


Thus we have:

ddx[e(s1)xU(x,s)]=e(s1)xe7x\frac{d}{dx}[e^{(s-1)x}U(x,s)]=e^{(s-1)x}e^{-7x}


U(x,s)=e(s1)x0xe(s1)re7rdr+ce(s1)x=U(x,s)=e^{-(s-1)x}\int^x_0 e^{(s-1)r}e^{-7r}dr+ce^{-(s-1)x}=


=e(s1)xe(s8)rs80x+ce(s1)x=e7xe(s1)xs8+ce(s1)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)=e7xe8tex+8(tx)H(tx)+cetδ(tx)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)={0x<01x0H(x)=\begin{cases} 0 &x<0 \\ 1 &x\ge0 \end{cases}

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