Laplace of the PDE $ x frac{partial(w)}{partial(x)} + frac{partial(w)}{partial(t)}=xt$.
$begingroup$
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
edited Dec 13 '18 at 19:55
Nosrati
26.6k62354
asked Dec 13 '18 at 14:51
Elliot SilverElliot Silver
244
$endgroup$
add a comment |
$begingroup$
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
edited Dec 13 '18 at 19:55
Nosrati
26.6k62354
asked Dec 13 '18 at 14:51
Elliot SilverElliot Silver
244
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Dec 13 '18 at 19:55
Nosrati
26.6k62354
asked Dec 13 '18 at 14:51
Elliot SilverElliot Silver
244
$endgroup$
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
edited Dec 13 '18 at 19:55
Nosrati
26.6k62354
asked Dec 13 '18 at 14:51
Elliot SilverElliot Silver
244
edited Dec 13 '18 at 19:55
Nosrati
26.6k62354
asked Dec 13 '18 at 14:51
Elliot SilverElliot Silver
244
edited Dec 13 '18 at 19:55
Nosrati
26.6k62354
edited Dec 13 '18 at 19:55
Nosrati
26.6k62354
edited Dec 13 '18 at 19:55
Nosrati
26.6k62354
26.6k62354
asked Dec 13 '18 at 14:51
Elliot SilverElliot Silver
244
asked Dec 13 '18 at 14:51
Elliot SilverElliot Silver
244
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$frac{dW}{dx}+frac{s}{x}W=frac{1}{s^2}$$
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
$$w(x,t)=x(e^{-t}+t-1)$$
answered Dec 13 '18 at 19:45
Rafa BudríaRafa Budría
5,8151825
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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oldest
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votes
$begingroup$
$$frac{dW}{dx}+frac{s}{x}W=frac{1}{s^2}$$
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
$$w(x,t)=x(e^{-t}+t-1)$$
answered Dec 13 '18 at 19:45
Rafa BudríaRafa Budría
5,8151825
$endgroup$
add a comment |
$begingroup$
$$frac{dW}{dx}+frac{s}{x}W=frac{1}{s^2}$$
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
$$w(x,t)=x(e^{-t}+t-1)$$
answered Dec 13 '18 at 19:45
Rafa BudríaRafa Budría
5,8151825
$endgroup$
add a comment |
$begingroup$
$$frac{dW}{dx}+frac{s}{x}W=frac{1}{s^2}$$
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
$$w(x,t)=x(e^{-t}+t-1)$$
answered Dec 13 '18 at 19:45
Rafa BudríaRafa Budría
5,8151825
$endgroup$
$$frac{dW}{dx}+frac{s}{x}W=frac{1}{s^2}$$
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
$$w(x,t)=x(e^{-t}+t-1)$$
answered Dec 13 '18 at 19:45
Rafa BudríaRafa Budría
5,8151825
edited Dec 14 '18 at 10:08
answered Dec 13 '18 at 19:45
Rafa BudríaRafa Budría
5,8151825
answered Dec 13 '18 at 19:45
Rafa BudríaRafa Budría
5,8151825
answered Dec 13 '18 at 19:45
Rafa BudríaRafa Budría
5,8151825
5,8151825
add a comment |
add a comment |
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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