# Solving a Linear Laplace Transformed 1st Order PDE

I'm trying to solve this PDE using the Laplace Transform: $$x\frac{\partial u}{\partial t} + \frac{\partial u}{\partial x}=x$$ with the following conditions : $$u(x,0)=0 , x > 0$$ $$u(0,t)=0 , t > 0$$

After performing my Laplace Transform of the PDE then I end with this : $$\frac{\partial U(x,s)}{\partial x} + xs U(x,s) = \frac{x}{s}$$

This is where I am stuck, how can I deal this the PDE (or is it ODE?), I don't know if I can use an integrating factor here or maybe the method of variation of parameters ?

Thank you !

• From the last equation it is clear that $U(x,s)$ is not a function of variable $t$ hence the derivative $\frac{\partial{U(x,s)}}{\partial{x}} = \frac{dU(x,s)}{dx}$. By doing this the equation turns into linear equation of form$\frac{dy}{dx}+P(x)y= Q(x)$. Which can be solved easily, then invert the $U(x,s)$ to get $u(x,t)$ Dec 4, 2022 at 14:52
• I see so the fact that there is a dependence on "s" does not impact our ability to use the integrating factor ? Dec 4, 2022 at 14:59
• Since the linear equation so formed does not depends on $s$ it is treated as constant in calculating the 'integrating factor'. Dec 4, 2022 at 15:03
• Ok I got it, thanks ! Dec 4, 2022 at 15:12
• See my answer below Dec 4, 2022 at 16:08

The equation formed by operating the Laplace transform on the given PDE is $$\frac{\partial U(x,s)}{\partial x}+ xsU(x,s) = \frac{x}{s}$$ Since the $$U(x,s)$$ no longer is a function of variable $$t$$ is can be written as $$\frac{\partial U(x,s)}{\partial x}= \frac{dU(x,s)}{dx}$$
The equation thus formed is a linear equation of the form $$\frac{dy}{dx}+P(x)y=Q(x)$$ The integrating factor is $$I.F = e^{\int{xs}dx} = e^{\frac{sx^2}{2}}$$ And the solution $$U(x,s)e^{\frac{sx^2}{2}} = \int \left(\frac{x}{s}\right) e^{\frac{sx^2}{2}} dx +C$$ On solving $$U(x,s) = \frac{1}{s^2}+ Ce^{\frac{-sx^2}{2}}$$ inverting this $$\mathcal{L^{-1}}(U(x,s)) = \mathcal{L^{-1}}(\frac{1}{s^2}) + \mathcal{L^{-1}}(Ce^{\frac{-sx^2}{2}})$$ $$u(x,t) = t + CH\left(t-\frac{x^2}{2}\right)$$ Where $$H(t-a)$$ is Heaviside unit step function defined as $$H(t-a) = \begin{cases} 0, & \text{t a} \end{cases}$$ From the boundary condition $$u(0,t)=0, t>0$$ we can write, $$u(0,t)=0= t+CH(t-0)$$ since
from the defination $$H(t) =1$$ so the value of constant $$C = -t$$ The solution for the PDE is written as $$u(x,t) = t\left[1-H \left(t-\frac{x^2}{2}\right)\right]$$ Where $$H \left(t-\frac{-x^2}{2} \right) = \begin{cases} 0, & \text{t<\frac{-x^2}{2}} \\ 1, & \text{t \geq \frac{-x^2}{2}} \end{cases}$$