heat equation with exponentially time dependent boundary condition

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Consider the one dimensional heat equation on a semi-infinite bar such that
$u_t=u_{xx}$
with initial condition of
$u(x,0)=0$.
What is the solution for all x and t if the boundaries are
$u(0,t)=e^{-t}$
and
$u(x,t)=0$ as x approaches infinity?
heat-equation
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up vote
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down vote
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Consider the one dimensional heat equation on a semi-infinite bar such that
$u_t=u_{xx}$
with initial condition of
$u(x,0)=0$.
What is the solution for all x and t if the boundaries are
$u(0,t)=e^{-t}$
and
$u(x,t)=0$ as x approaches infinity?
heat-equation
You must take the 'Laplace Transform of your equation' but, in that way, it's not clear hot to handle "$displaystylemathrm{u}left(x,tright) = 0$ as x approaches infinity".
– Felix Marin
Nov 14 at 16:58
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider the one dimensional heat equation on a semi-infinite bar such that
$u_t=u_{xx}$
with initial condition of
$u(x,0)=0$.
What is the solution for all x and t if the boundaries are
$u(0,t)=e^{-t}$
and
$u(x,t)=0$ as x approaches infinity?
heat-equation
Consider the one dimensional heat equation on a semi-infinite bar such that
$u_t=u_{xx}$
with initial condition of
$u(x,0)=0$.
What is the solution for all x and t if the boundaries are
$u(0,t)=e^{-t}$
and
$u(x,t)=0$ as x approaches infinity?
heat-equation
heat-equation
asked Nov 13 at 20:23
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You must take the 'Laplace Transform of your equation' but, in that way, it's not clear hot to handle "$displaystylemathrm{u}left(x,tright) = 0$ as x approaches infinity".
– Felix Marin
Nov 14 at 16:58
add a comment |
You must take the 'Laplace Transform of your equation' but, in that way, it's not clear hot to handle "$displaystylemathrm{u}left(x,tright) = 0$ as x approaches infinity".
– Felix Marin
Nov 14 at 16:58
You must take the 'Laplace Transform of your equation' but, in that way, it's not clear hot to handle "$displaystylemathrm{u}left(x,tright) = 0$ as x approaches infinity".
– Felix Marin
Nov 14 at 16:58
You must take the 'Laplace Transform of your equation' but, in that way, it's not clear hot to handle "$displaystylemathrm{u}left(x,tright) = 0$ as x approaches infinity".
– Felix Marin
Nov 14 at 16:58
add a comment |
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You must take the 'Laplace Transform of your equation' but, in that way, it's not clear hot to handle "$displaystylemathrm{u}left(x,tright) = 0$ as x approaches infinity".
– Felix Marin
Nov 14 at 16:58