Forced wave equation using Green's functions and Laplace transforms
Solve $$boxed{frac{partial^2 u}{partial t^2}-c^2frac{partial^2u}{partial x^2}=F(x,t),0<x<infty} $$ with the boundary conditions $frac{partial u}{partial x}=0$ at $x=0$ and $uto 0$ as $x to infty$ and initial conditions $u(x,0)=0,frac{partial u(x,0)}{partial t}=0$.
My attempt:
Applying a Green's function gives
$$frac{partial^2 G}{partial t^2}-c^2frac{partial^2 G}{partial x^2}=delta(x-xi)(t-tau)$$
and setting $x-xi:=x, t-tau:=t$ and taking Laplace transforms give $$s^2overline{G}(s)-sG(0)-G'(0)-c^2frac{partial^2 overline{G}}{partial x^2}=boxed{s^2overline{G}(s)-c^2frac{partial^2 overline{G}}{partial x^2}=delta(x)}$$ using the initial conditions.
For $x neq 0 $ we have a standard ODE and I get $overline{G}=Ae^{s/cx}+Be^{-s/cx} $. Since $u to 0$, $A=0$ and $boxed{overline{G}=Be^{-s/cx}}$.
My problem is how do I find the constant $B$? For the 3D case, I have been shown to use the Gauss theorem to integrate a volume around $0$. I know the integral of the right hand side is 1 and I tried to integrate both sides to give $[s^2overline{G}x]_0^infty-c^2[frac{partial overline{G}}{partial x}]_0^infty=1$ which substituting $Be^{-s/cx}$ into gave me $-s^2+s/cB=0$ but then $B$ depends on s.
pde laplace-transform greens-function
asked Nov 27 '18 at 15:52
user30523user30523
52959
add a comment |
Solve $$boxed{frac{partial^2 u}{partial t^2}-c^2frac{partial^2u}{partial x^2}=F(x,t),0<x<infty} $$ with the boundary conditions $frac{partial u}{partial x}=0$ at $x=0$ and $uto 0$ as $x to infty$ and initial conditions $u(x,0)=0,frac{partial u(x,0)}{partial t}=0$.
My attempt:
Applying a Green's function gives
$$frac{partial^2 G}{partial t^2}-c^2frac{partial^2 G}{partial x^2}=delta(x-xi)(t-tau)$$
and setting $x-xi:=x, t-tau:=t$ and taking Laplace transforms give $$s^2overline{G}(s)-sG(0)-G'(0)-c^2frac{partial^2 overline{G}}{partial x^2}=boxed{s^2overline{G}(s)-c^2frac{partial^2 overline{G}}{partial x^2}=delta(x)}$$ using the initial conditions.
For $x neq 0 $ we have a standard ODE and I get $overline{G}=Ae^{s/cx}+Be^{-s/cx} $. Since $u to 0$, $A=0$ and $boxed{overline{G}=Be^{-s/cx}}$.
My problem is how do I find the constant $B$? For the 3D case, I have been shown to use the Gauss theorem to integrate a volume around $0$. I know the integral of the right hand side is 1 and I tried to integrate both sides to give $[s^2overline{G}x]_0^infty-c^2[frac{partial overline{G}}{partial x}]_0^infty=1$ which substituting $Be^{-s/cx}$ into gave me $-s^2+s/cB=0$ but then $B$ depends on s.
pde laplace-transform greens-function
asked Nov 27 '18 at 15:52
user30523user30523
52959
initial conditions?
– phdmba7of12
Nov 27 '18 at 15:54
1
@phdmba7of12, $u(0)=0$ and $u_t(0)=0$
– user30523
Nov 27 '18 at 16:20
$B$ is allowed to depend on $s$. It's only constant w.r.t $x$
– Dylan
Nov 28 '18 at 11:08
add a comment |
Solve $$boxed{frac{partial^2 u}{partial t^2}-c^2frac{partial^2u}{partial x^2}=F(x,t),0<x<infty} $$ with the boundary conditions $frac{partial u}{partial x}=0$ at $x=0$ and $uto 0$ as $x to infty$ and initial conditions $u(x,0)=0,frac{partial u(x,0)}{partial t}=0$.
My attempt:
Applying a Green's function gives
$$frac{partial^2 G}{partial t^2}-c^2frac{partial^2 G}{partial x^2}=delta(x-xi)(t-tau)$$
and setting $x-xi:=x, t-tau:=t$ and taking Laplace transforms give $$s^2overline{G}(s)-sG(0)-G'(0)-c^2frac{partial^2 overline{G}}{partial x^2}=boxed{s^2overline{G}(s)-c^2frac{partial^2 overline{G}}{partial x^2}=delta(x)}$$ using the initial conditions.
For $x neq 0 $ we have a standard ODE and I get $overline{G}=Ae^{s/cx}+Be^{-s/cx} $. Since $u to 0$, $A=0$ and $boxed{overline{G}=Be^{-s/cx}}$.
My problem is how do I find the constant $B$? For the 3D case, I have been shown to use the Gauss theorem to integrate a volume around $0$. I know the integral of the right hand side is 1 and I tried to integrate both sides to give $[s^2overline{G}x]_0^infty-c^2[frac{partial overline{G}}{partial x}]_0^infty=1$ which substituting $Be^{-s/cx}$ into gave me $-s^2+s/cB=0$ but then $B$ depends on s.
pde laplace-transform greens-function
asked Nov 27 '18 at 15:52
user30523user30523
52959
Solve $$boxed{frac{partial^2 u}{partial t^2}-c^2frac{partial^2u}{partial x^2}=F(x,t),0<x<infty} $$ with the boundary conditions $frac{partial u}{partial x}=0$ at $x=0$ and $uto 0$ as $x to infty$ and initial conditions $u(x,0)=0,frac{partial u(x,0)}{partial t}=0$.
My attempt:
Applying a Green's function gives
$$frac{partial^2 G}{partial t^2}-c^2frac{partial^2 G}{partial x^2}=delta(x-xi)(t-tau)$$
and setting $x-xi:=x, t-tau:=t$ and taking Laplace transforms give $$s^2overline{G}(s)-sG(0)-G'(0)-c^2frac{partial^2 overline{G}}{partial x^2}=boxed{s^2overline{G}(s)-c^2frac{partial^2 overline{G}}{partial x^2}=delta(x)}$$ using the initial conditions.
For $x neq 0 $ we have a standard ODE and I get $overline{G}=Ae^{s/cx}+Be^{-s/cx} $. Since $u to 0$, $A=0$ and $boxed{overline{G}=Be^{-s/cx}}$.
My problem is how do I find the constant $B$? For the 3D case, I have been shown to use the Gauss theorem to integrate a volume around $0$. I know the integral of the right hand side is 1 and I tried to integrate both sides to give $[s^2overline{G}x]_0^infty-c^2[frac{partial overline{G}}{partial x}]_0^infty=1$ which substituting $Be^{-s/cx}$ into gave me $-s^2+s/cB=0$ but then $B$ depends on s.
pde laplace-transform greens-function
pde laplace-transform greens-function
asked Nov 27 '18 at 15:52
user30523user30523
52959
asked Nov 27 '18 at 15:52
user30523user30523
52959
asked Nov 27 '18 at 15:52
user30523user30523
52959
asked Nov 27 '18 at 15:52
user30523user30523
52959
asked Nov 27 '18 at 15:52
user30523user30523
52959
52959
initial conditions?
– phdmba7of12
Nov 27 '18 at 15:54
1
@phdmba7of12, $u(0)=0$ and $u_t(0)=0$
– user30523
Nov 27 '18 at 16:20
$B$ is allowed to depend on $s$. It's only constant w.r.t $x$
– Dylan
Nov 28 '18 at 11:08
add a comment |
initial conditions?
– phdmba7of12
Nov 27 '18 at 15:54
1
@phdmba7of12, $u(0)=0$ and $u_t(0)=0$
– user30523
Nov 27 '18 at 16:20
$B$ is allowed to depend on $s$. It's only constant w.r.t $x$
– Dylan
Nov 28 '18 at 11:08
initial conditions?
– phdmba7of12
Nov 27 '18 at 15:54
initial conditions?
– phdmba7of12
Nov 27 '18 at 15:54
1
1
@phdmba7of12, $u(0)=0$ and $u_t(0)=0$
– user30523
Nov 27 '18 at 16:20
@phdmba7of12, $u(0)=0$ and $u_t(0)=0$
– user30523
Nov 27 '18 at 16:20
$B$ is allowed to depend on $s$. It's only constant w.r.t $x$
– Dylan
Nov 28 '18 at 11:08
$B$ is allowed to depend on $s$. It's only constant w.r.t $x$
– Dylan
Nov 28 '18 at 11:08
add a comment |
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initial conditions?
– phdmba7of12
Nov 27 '18 at 15:54
1
@phdmba7of12, $u(0)=0$ and $u_t(0)=0$
– user30523
Nov 27 '18 at 16:20
$B$ is allowed to depend on $s$. It's only constant w.r.t $x$
– Dylan
Nov 28 '18 at 11:08