Solve using Laplace transforms The Next CEO of Stack OverflowInverse Laplace Transform of $bar p_D = fracK_0(sqrt[]s r_D)sK_0(sqrt[]s)$Solving Black scholes PDE using Laplace transformSolve pde using laplace?Solve the PDE $fracpartial upartial t+fracpartial upartial x=x$ using Laplace transform in $t$Using Laplace Transforms to solve a PDEIs my approach to solve this PDE using Fourier transform right?How to do laplace transform on time dependent domain?Laplace transform of heat equation in spherical coordinatesHeat equation using the Laplace TransformSolving Heat Transfer PDE using Laplace Transform
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Solve using Laplace transforms
The Next CEO of Stack OverflowInverse Laplace Transform of $bar p_D = fracK_0(sqrt[]s r_D)sK_0(sqrt[]s)$Solving Black scholes PDE using Laplace transformSolve pde using laplace?Solve the PDE $fracpartial upartial t+fracpartial upartial x=x$ using Laplace transform in $t$Using Laplace Transforms to solve a PDEIs my approach to solve this PDE using Fourier transform right?How to do laplace transform on time dependent domain?Laplace transform of heat equation in spherical coordinatesHeat equation using the Laplace TransformSolving Heat Transfer PDE using Laplace Transform
$begingroup$
I want to solve the PDE
$$xfracpartial upartial t +fracpartial upartial x=x$$
$$u(x,0)=0, x>0$$
$$u(0,t)=0, t>0$$
for $x>0,t>0$ using a Laplace transform in t.
I Laplace transform the equation and get
$$xshatu(x,s)+fracpartialpartial xhatu(x,s)=fracxs$$
where $hatu(x,s)=L(u(x,t))=int_t=0^infty e^-stu(x,t)dt$.
Solving this ODE, I obtain
$$hatu(x,s)=frac1s^2 + A(s)e^frac-sx^22.$$
I then apply the second boundary condition to get
$$hatu(x,s)=frac1s^2-frac1s^2e^frac-sx^22.$$
Now I need to inverse transform to get $u(x,t)$. I know that
$$L^-1(frac1s^2)=t$$
but I don't know how to find
$$L^-1(frac1s^2e^frac-sx^22).$$
pde laplace-transform
asked Mar 19 at 21:29
vladr10vladr10
423
$endgroup$
add a comment |
$begingroup$
I want to solve the PDE
$$xfracpartial upartial t +fracpartial upartial x=x$$
$$u(x,0)=0, x>0$$
$$u(0,t)=0, t>0$$
for $x>0,t>0$ using a Laplace transform in t.
I Laplace transform the equation and get
$$xshatu(x,s)+fracpartialpartial xhatu(x,s)=fracxs$$
where $hatu(x,s)=L(u(x,t))=int_t=0^infty e^-stu(x,t)dt$.
Solving this ODE, I obtain
$$hatu(x,s)=frac1s^2 + A(s)e^frac-sx^22.$$
I then apply the second boundary condition to get
$$hatu(x,s)=frac1s^2-frac1s^2e^frac-sx^22.$$
Now I need to inverse transform to get $u(x,t)$. I know that
$$L^-1(frac1s^2)=t$$
but I don't know how to find
$$L^-1(frac1s^2e^frac-sx^22).$$
pde laplace-transform
asked Mar 19 at 21:29
vladr10vladr10
423
$endgroup$
3
$begingroup$
I get $(t-x^2/2),u(t-x^2/2),$ where $u$ is the Heaviside step function.
$endgroup$
– Adrian Keister
Mar 19 at 21:37
1
$begingroup$
Use the shift theorem
$endgroup$
– Dylan
Mar 21 at 3:28
add a comment |
$begingroup$
I want to solve the PDE
$$xfracpartial upartial t +fracpartial upartial x=x$$
$$u(x,0)=0, x>0$$
$$u(0,t)=0, t>0$$
for $x>0,t>0$ using a Laplace transform in t.
I Laplace transform the equation and get
$$xshatu(x,s)+fracpartialpartial xhatu(x,s)=fracxs$$
where $hatu(x,s)=L(u(x,t))=int_t=0^infty e^-stu(x,t)dt$.
Solving this ODE, I obtain
$$hatu(x,s)=frac1s^2 + A(s)e^frac-sx^22.$$
I then apply the second boundary condition to get
$$hatu(x,s)=frac1s^2-frac1s^2e^frac-sx^22.$$
Now I need to inverse transform to get $u(x,t)$. I know that
$$L^-1(frac1s^2)=t$$
but I don't know how to find
$$L^-1(frac1s^2e^frac-sx^22).$$
pde laplace-transform
asked Mar 19 at 21:29
vladr10vladr10
423
$endgroup$
I want to solve the PDE
$$xfracpartial upartial t +fracpartial upartial x=x$$
$$u(x,0)=0, x>0$$
$$u(0,t)=0, t>0$$
for $x>0,t>0$ using a Laplace transform in t.
I Laplace transform the equation and get
$$xshatu(x,s)+fracpartialpartial xhatu(x,s)=fracxs$$
where $hatu(x,s)=L(u(x,t))=int_t=0^infty e^-stu(x,t)dt$.
Solving this ODE, I obtain
$$hatu(x,s)=frac1s^2 + A(s)e^frac-sx^22.$$
I then apply the second boundary condition to get
$$hatu(x,s)=frac1s^2-frac1s^2e^frac-sx^22.$$
Now I need to inverse transform to get $u(x,t)$. I know that
$$L^-1(frac1s^2)=t$$
but I don't know how to find
$$L^-1(frac1s^2e^frac-sx^22).$$
pde laplace-transform
pde laplace-transform
asked Mar 19 at 21:29
vladr10vladr10
423
asked Mar 19 at 21:29
vladr10vladr10
423
asked Mar 19 at 21:29
vladr10vladr10
423
asked Mar 19 at 21:29
vladr10vladr10
423
asked Mar 19 at 21:29
vladr10vladr10
423
423
3
$begingroup$
I get $(t-x^2/2),u(t-x^2/2),$ where $u$ is the Heaviside step function.
$endgroup$
– Adrian Keister
Mar 19 at 21:37
1
$begingroup$
Use the shift theorem
$endgroup$
– Dylan
Mar 21 at 3:28
add a comment |
3
$begingroup$
I get $(t-x^2/2),u(t-x^2/2),$ where $u$ is the Heaviside step function.
$endgroup$
– Adrian Keister
Mar 19 at 21:37
1
$begingroup$
Use the shift theorem
$endgroup$
– Dylan
Mar 21 at 3:28
3
3
$begingroup$
I get $(t-x^2/2),u(t-x^2/2),$ where $u$ is the Heaviside step function.
$endgroup$
– Adrian Keister
Mar 19 at 21:37
$begingroup$
I get $(t-x^2/2),u(t-x^2/2),$ where $u$ is the Heaviside step function.
$endgroup$
– Adrian Keister
Mar 19 at 21:37
1
1
$begingroup$
Use the shift theorem
$endgroup$
– Dylan
Mar 21 at 3:28
$begingroup$
Use the shift theorem
$endgroup$
– Dylan
Mar 21 at 3:28
add a comment |
0
active
oldest
votes
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3
$begingroup$
I get $(t-x^2/2),u(t-x^2/2),$ where $u$ is the Heaviside step function.
$endgroup$
– Adrian Keister
Mar 19 at 21:37
1
$begingroup$
Use the shift theorem
$endgroup$
– Dylan
Mar 21 at 3:28