Laplace of the PDE $ x frac{partial(w)}{partial(x)} + frac{partial(w)}{partial(t)}=xt$.












1












$begingroup$


Question is find the Laplace transform of this equation:
$$ x frac{partial(w)}{partial(x)} + frac{partial(w)}{partial(t)}=xt$$



Boundary conditions : $$ w(x,0) = 0 qquad x geq0 $$
$$w(0,t)=0 qquad t geq 0$$



I've got as far as
$$ frac{d(W)}{d(x)}+frac{s}{x}W=frac{1}{s^2}$$



I'm not even sure this is correct but even if it is, I dont know how to integrate this to get $W(x,s)$






pde laplace-transform







share|cite|improve this question











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  • $begingroup$
    Try with $W = a x + b x^{-s}$ for some $a$ and $b$ which depend on $s$
    $endgroup$
    – Federico
    Dec 13 '18 at 14:57










  • $begingroup$
    The idea is that after you multiply by $x^s$, you get $x^sfrac{dW}{dx}+sx^{s-1}W$, which is $frac{d(x^sW)}{dx}$...
    $endgroup$
    – Federico
    Dec 13 '18 at 14:59


















1












$begingroup$


Question is find the Laplace transform of this equation:
$$ x frac{partial(w)}{partial(x)} + frac{partial(w)}{partial(t)}=xt$$



Boundary conditions : $$ w(x,0) = 0 qquad x geq0 $$
$$w(0,t)=0 qquad t geq 0$$



I've got as far as
$$ frac{d(W)}{d(x)}+frac{s}{x}W=frac{1}{s^2}$$



I'm not even sure this is correct but even if it is, I dont know how to integrate this to get $W(x,s)$






pde laplace-transform







share|cite|improve this question











$endgroup$












  • $begingroup$
    Try with $W = a x + b x^{-s}$ for some $a$ and $b$ which depend on $s$
    $endgroup$
    – Federico
    Dec 13 '18 at 14:57










  • $begingroup$
    The idea is that after you multiply by $x^s$, you get $x^sfrac{dW}{dx}+sx^{s-1}W$, which is $frac{d(x^sW)}{dx}$...
    $endgroup$
    – Federico
    Dec 13 '18 at 14:59
















1












1








1





$begingroup$


Question is find the Laplace transform of this equation:
$$ x frac{partial(w)}{partial(x)} + frac{partial(w)}{partial(t)}=xt$$



Boundary conditions : $$ w(x,0) = 0 qquad x geq0 $$
$$w(0,t)=0 qquad t geq 0$$



I've got as far as
$$ frac{d(W)}{d(x)}+frac{s}{x}W=frac{1}{s^2}$$



I'm not even sure this is correct but even if it is, I dont know how to integrate this to get $W(x,s)$






pde laplace-transform







share|cite|improve this question











$endgroup$




Question is find the Laplace transform of this equation:
$$ x frac{partial(w)}{partial(x)} + frac{partial(w)}{partial(t)}=xt$$



Boundary conditions : $$ w(x,0) = 0 qquad x geq0 $$
$$w(0,t)=0 qquad t geq 0$$



I've got as far as
$$ frac{d(W)}{d(x)}+frac{s}{x}W=frac{1}{s^2}$$



I'm not even sure this is correct but even if it is, I dont know how to integrate this to get $W(x,s)$






pde laplace-transform




pde laplace-transform






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 19:55









Nosrati

26.5k62354




26.5k62354










asked Dec 13 '18 at 14:51









Elliot SilverElliot Silver

244




244












  • $begingroup$
    Try with $W = a x + b x^{-s}$ for some $a$ and $b$ which depend on $s$
    $endgroup$
    – Federico
    Dec 13 '18 at 14:57










  • $begingroup$
    The idea is that after you multiply by $x^s$, you get $x^sfrac{dW}{dx}+sx^{s-1}W$, which is $frac{d(x^sW)}{dx}$...
    $endgroup$
    – Federico
    Dec 13 '18 at 14:59




















  • $begingroup$
    Try with $W = a x + b x^{-s}$ for some $a$ and $b$ which depend on $s$
    $endgroup$
    – Federico
    Dec 13 '18 at 14:57










  • $begingroup$
    The idea is that after you multiply by $x^s$, you get $x^sfrac{dW}{dx}+sx^{s-1}W$, which is $frac{d(x^sW)}{dx}$...
    $endgroup$
    – Federico
    Dec 13 '18 at 14:59


















$begingroup$
Try with $W = a x + b x^{-s}$ for some $a$ and $b$ which depend on $s$
$endgroup$
– Federico
Dec 13 '18 at 14:57




$begingroup$
Try with $W = a x + b x^{-s}$ for some $a$ and $b$ which depend on $s$
$endgroup$
– Federico
Dec 13 '18 at 14:57












$begingroup$
The idea is that after you multiply by $x^s$, you get $x^sfrac{dW}{dx}+sx^{s-1}W$, which is $frac{d(x^sW)}{dx}$...
$endgroup$
– Federico
Dec 13 '18 at 14:59






$begingroup$
The idea is that after you multiply by $x^s$, you get $x^sfrac{dW}{dx}+sx^{s-1}W$, which is $frac{d(x^sW)}{dx}$...
$endgroup$
– Federico
Dec 13 '18 at 14:59












1 Answer
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$begingroup$

$$frac{dW}{dx}+frac{s}{x}W=frac{1}{s^2}$$



Following the hints in the comments, the solution of the equation is



$$W(x,s)=dfrac{x}{s^2(s+1)}+c_1x^{-s}$$



But $W(0,s)=0$ and because it holds for any $s$



$$0=dfrac{c_1}{x^s}implies c_1=0$$



$$W(x,s)=dfrac{x}{s^2(s+1)}$$



and



$$w(x,t)=x(e^{-t}+t-1)$$






share|cite|improve this answer











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    3












    $begingroup$

    $$frac{dW}{dx}+frac{s}{x}W=frac{1}{s^2}$$



    Following the hints in the comments, the solution of the equation is



    $$W(x,s)=dfrac{x}{s^2(s+1)}+c_1x^{-s}$$



    But $W(0,s)=0$ and because it holds for any $s$



    $$0=dfrac{c_1}{x^s}implies c_1=0$$



    $$W(x,s)=dfrac{x}{s^2(s+1)}$$



    and



    $$w(x,t)=x(e^{-t}+t-1)$$






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      $$frac{dW}{dx}+frac{s}{x}W=frac{1}{s^2}$$



      Following the hints in the comments, the solution of the equation is



      $$W(x,s)=dfrac{x}{s^2(s+1)}+c_1x^{-s}$$



      But $W(0,s)=0$ and because it holds for any $s$



      $$0=dfrac{c_1}{x^s}implies c_1=0$$



      $$W(x,s)=dfrac{x}{s^2(s+1)}$$



      and



      $$w(x,t)=x(e^{-t}+t-1)$$






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        $$frac{dW}{dx}+frac{s}{x}W=frac{1}{s^2}$$



        Following the hints in the comments, the solution of the equation is



        $$W(x,s)=dfrac{x}{s^2(s+1)}+c_1x^{-s}$$



        But $W(0,s)=0$ and because it holds for any $s$



        $$0=dfrac{c_1}{x^s}implies c_1=0$$



        $$W(x,s)=dfrac{x}{s^2(s+1)}$$



        and



        $$w(x,t)=x(e^{-t}+t-1)$$






        share|cite|improve this answer











        $endgroup$



        $$frac{dW}{dx}+frac{s}{x}W=frac{1}{s^2}$$



        Following the hints in the comments, the solution of the equation is



        $$W(x,s)=dfrac{x}{s^2(s+1)}+c_1x^{-s}$$



        But $W(0,s)=0$ and because it holds for any $s$



        $$0=dfrac{c_1}{x^s}implies c_1=0$$



        $$W(x,s)=dfrac{x}{s^2(s+1)}$$



        and



        $$w(x,t)=x(e^{-t}+t-1)$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 14 '18 at 10:08

























        answered Dec 13 '18 at 19:45









        Rafa BudríaRafa Budría

        5,7601825




        5,7601825






























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