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?










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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

















up vote
0
down vote

favorite
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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?










share|cite|improve this question






















  • 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















up vote
0
down vote

favorite
2









up vote
0
down vote

favorite
2






2





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?










share|cite|improve this question













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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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




















  • 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

















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