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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 !

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  • $\begingroup$ 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 $u(x,t)$ $\endgroup$ Dec 4, 2022 at 14:52
  • $\begingroup$ I see so the fact that there is a dependence on "s" does not impact our ability to use the integrating factor ? $\endgroup$
    – bsaoptima
    Dec 4, 2022 at 14:59
  • $\begingroup$ Since the linear equation so formed does not depends on $s$ it is treated as constant in calculating the 'integrating factor'. $\endgroup$ Dec 4, 2022 at 15:03
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    $\begingroup$ Ok I got it, thanks ! $\endgroup$
    – bsaoptima
    Dec 4, 2022 at 15:12
  • $\begingroup$ See my answer below $\endgroup$ Dec 4, 2022 at 16:08

1 Answer 1

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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 $$ H(t-a) = \begin{cases} 0, & \text{t<a} \\ 1, & \text{t> a} \end{cases}$$ 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 $$H \left(t-\frac{-x^2}{2} \right) = \begin{cases} 0, & \text{t<$\frac{-x^2}{2}$} \\ 1, & \text{t $\geq \frac{-x^2}{2}$} \end{cases}$$

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  • $\begingroup$ Thank you so much ! I actually management to get to the Heaviside Unit Step Function part but I just added an additional constant of integration without realising so I also had a term which Inverse Laplace Transform is the Dirac Delta Function $\endgroup$
    – bsaoptima
    Dec 4, 2022 at 16:39
  • $\begingroup$ Your welcome :) $\endgroup$ Dec 4, 2022 at 16:42

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