heat equation with exponentially time dependent boundary condition
$begingroup$
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
asked Nov 13 '18 at 20:23
catcat
63
$endgroup$
add a comment |
$begingroup$
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
asked Nov 13 '18 at 20:23
catcat
63
$endgroup$
$begingroup$
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".
$endgroup$
– Felix Marin
Nov 14 '18 at 16:58
add a comment |
$begingroup$
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
asked Nov 13 '18 at 20:23
catcat
63
$endgroup$
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 '18 at 20:23
catcat
63
asked Nov 13 '18 at 20:23
catcat
63
asked Nov 13 '18 at 20:23
catcat
63
asked Nov 13 '18 at 20:23
catcat
63
asked Nov 13 '18 at 20:23
catcat
63
63
$begingroup$
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".
$endgroup$
– Felix Marin
Nov 14 '18 at 16:58
add a comment |
$begingroup$
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".
$endgroup$
– Felix Marin
Nov 14 '18 at 16:58
$begingroup$
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".
$endgroup$
– Felix Marin
Nov 14 '18 at 16:58
$begingroup$
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".
$endgroup$
– Felix Marin
Nov 14 '18 at 16:58
add a comment |
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$begingroup$
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".
$endgroup$
– Felix Marin
Nov 14 '18 at 16:58