Using the Laplace transform, solve the initial boundary value problem $\frac{\partial u}{\partial t} + x\frac{\partial u}{\partial x} = x$

Question

Consider a domain

$$D = \{(x,t)|x \gt 0, t \gt 0\}$$

Using the Laplace transform, solve the initial boundary value problem

$$\frac{\partial u}{\partial t} + x\frac{\partial u}{\partial x} = x$$

subject to

$$u(x,0) = 0 \space \space\text{for}\space x \gt 0$$

and

$$u(0,t) = 0\space \space\text{for}\space t\gt0$$

Answer 1

We begin with the PDE

$$u_t(x,t)+xu_x(x,t)=x \tag1$$

for $x>0$ and $t>0$, subject to the initial and boundary conditions $u(x,0)=0$ and $u(0,t)=0$.

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Taking the Laplace Transform of $(1)$ with respect to $t$, and subject to the initial condition, yields the ODE

$$xU_x(x,s)+sU(x,s)=\frac xs\tag 2$$

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We can solve the ODE in $(2)$, subject to the initial condition $U(0,s)=0$, by using the integrating factor $x^s$. Proceeding, we find that

$$U(x,s)=\frac{x}{s(s+1)}\tag 3$$

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Finally, taking the inverse Laplace Transform of $(3)$, through use of partial fraction expansion, reveals

$$\bbox[5px,border:2px solid #C0A000]{u(x,t)=x(1-e^{-t})}$$

And we are done!