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Heat equation with non zero BC

1

1

$begingroup$

Assume I have a heat equation on $[0,pi]$ with 0 value on the boundaries and say 1 initial value, constant. I can see that I can write the solution as a series. Now, I want to change the boundary condition value to 2 and 4 on the left and on the right. How would I approach solving this problem?

pde numerical-methods

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asked Aug 27 '16 at 17:15

MedanMedan

227619

$endgroup$

add a comment | 

1

1

$begingroup$

Assume I have a heat equation on $[0,pi]$ with 0 value on the boundaries and say 1 initial value, constant. I can see that I can write the solution as a series. Now, I want to change the boundary condition value to 2 and 4 on the left and on the right. How would I approach solving this problem?

pde numerical-methods

share|cite|improve this question

asked Aug 27 '16 at 17:15

MedanMedan

227619

$endgroup$

add a comment | 

$begingroup$

Assume I have a heat equation on $[0,pi]$ with 0 value on the boundaries and say 1 initial value, constant. I can see that I can write the solution as a series. Now, I want to change the boundary condition value to 2 and 4 on the left and on the right. How would I approach solving this problem?

pde numerical-methods

share|cite|improve this question

asked Aug 27 '16 at 17:15

MedanMedan

227619

$endgroup$

share|cite|improve this question

asked Aug 27 '16 at 17:15

MedanMedan

227619

share|cite|improve this question

asked Aug 27 '16 at 17:15

MedanMedan

227619

asked Aug 27 '16 at 17:15

MedanMedan

227619

asked Aug 27 '16 at 17:15

MedanMedan

227619

1 Answer
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oldest

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0

$begingroup$

The system that you want to solve is
beginsplit
partial_tu &= partial_xxu, xin[0,pi], t>0\
u (0,t) &= 2, \
u (pi,t) &= 4 \
u(x,0) &= u_0
endsplit
where you have probably found that using the ansatz $u(x,t) = X(x)T(t)$ and proceeding by separation of variables doesn't work.

The strategy is to rewrite the solution $u(x,t)$ in terms of a new variable $v(x,t)$ such that the new problem has homogeneous boundary conditions.
We start by defining $v$
beginalign*
v(x,t) = u(x,t) - u_E(x,t)
endalign*
where $u_E(x,t)$ is the solution at equilibrium.
Hence one would firstly solve
beginalign
beginsplit
partial_xxu_E &= 0, xin[0,pi]\
u_E (0) &= 2 \
u_E (pi) &= 4
endsplit
endalign
for which the solution is
beginalign
u_E(x) = frac2pix + 2.
endalign
Then we can write the new system for $v(x,t) = u(x,t) - frac2pix - 2$ as the following
beginalign
beginsplit
partial_tv &= partial_xxv, xin[0,pi]\
v (0,t) &= 0 \
v (pi,t) &= 0 \
v(x,0) &= u_0 - frac2pix - 2, xin[0,pi].
endsplit
endalign
and solve using separation of variables using the ansatz $v(x,t) = X(x)T(t)$ in the usual way.

Then one can finally write the solution as $u(x,t) = v(x,t) + u_E(x)$

share|cite|improve this answer

answered Apr 2 at 13:40

D.ShanksD.Shanks

12

$endgroup$

add a comment | 

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0

$begingroup$

The system that you want to solve is
beginsplit
partial_tu &= partial_xxu, xin[0,pi], t>0\
u (0,t) &= 2, \
u (pi,t) &= 4 \
u(x,0) &= u_0
endsplit
where you have probably found that using the ansatz $u(x,t) = X(x)T(t)$ and proceeding by separation of variables doesn't work.

The strategy is to rewrite the solution $u(x,t)$ in terms of a new variable $v(x,t)$ such that the new problem has homogeneous boundary conditions.
We start by defining $v$
beginalign*
v(x,t) = u(x,t) - u_E(x,t)
endalign*
where $u_E(x,t)$ is the solution at equilibrium.
Hence one would firstly solve
beginalign
beginsplit
partial_xxu_E &= 0, xin[0,pi]\
u_E (0) &= 2 \
u_E (pi) &= 4
endsplit
endalign
for which the solution is
beginalign
u_E(x) = frac2pix + 2.
endalign
Then we can write the new system for $v(x,t) = u(x,t) - frac2pix - 2$ as the following
beginalign
beginsplit
partial_tv &= partial_xxv, xin[0,pi]\
v (0,t) &= 0 \
v (pi,t) &= 0 \
v(x,0) &= u_0 - frac2pix - 2, xin[0,pi].
endsplit
endalign
and solve using separation of variables using the ansatz $v(x,t) = X(x)T(t)$ in the usual way.

Then one can finally write the solution as $u(x,t) = v(x,t) + u_E(x)$

share|cite|improve this answer

answered Apr 2 at 13:40

D.ShanksD.Shanks

12

$endgroup$

add a comment | 

0

$begingroup$

The system that you want to solve is
beginsplit
partial_tu &= partial_xxu, xin[0,pi], t>0\
u (0,t) &= 2, \
u (pi,t) &= 4 \
u(x,0) &= u_0
endsplit
where you have probably found that using the ansatz $u(x,t) = X(x)T(t)$ and proceeding by separation of variables doesn't work.

The strategy is to rewrite the solution $u(x,t)$ in terms of a new variable $v(x,t)$ such that the new problem has homogeneous boundary conditions.
We start by defining $v$
beginalign*
v(x,t) = u(x,t) - u_E(x,t)
endalign*
where $u_E(x,t)$ is the solution at equilibrium.
Hence one would firstly solve
beginalign
beginsplit
partial_xxu_E &= 0, xin[0,pi]\
u_E (0) &= 2 \
u_E (pi) &= 4
endsplit
endalign
for which the solution is
beginalign
u_E(x) = frac2pix + 2.
endalign
Then we can write the new system for $v(x,t) = u(x,t) - frac2pix - 2$ as the following
beginalign
beginsplit
partial_tv &= partial_xxv, xin[0,pi]\
v (0,t) &= 0 \
v (pi,t) &= 0 \
v(x,0) &= u_0 - frac2pix - 2, xin[0,pi].
endsplit
endalign
and solve using separation of variables using the ansatz $v(x,t) = X(x)T(t)$ in the usual way.

Then one can finally write the solution as $u(x,t) = v(x,t) + u_E(x)$

share|cite|improve this answer

answered Apr 2 at 13:40

D.ShanksD.Shanks

12

$endgroup$

add a comment | 

$begingroup$

The system that you want to solve is
beginsplit
partial_tu &= partial_xxu, xin[0,pi], t>0\
u (0,t) &= 2, \
u (pi,t) &= 4 \
u(x,0) &= u_0
endsplit
where you have probably found that using the ansatz $u(x,t) = X(x)T(t)$ and proceeding by separation of variables doesn't work.

The strategy is to rewrite the solution $u(x,t)$ in terms of a new variable $v(x,t)$ such that the new problem has homogeneous boundary conditions.
We start by defining $v$
beginalign*
v(x,t) = u(x,t) - u_E(x,t)
endalign*
where $u_E(x,t)$ is the solution at equilibrium.
Hence one would firstly solve
beginalign
beginsplit
partial_xxu_E &= 0, xin[0,pi]\
u_E (0) &= 2 \
u_E (pi) &= 4
endsplit
endalign
for which the solution is
beginalign
u_E(x) = frac2pix + 2.
endalign
Then we can write the new system for $v(x,t) = u(x,t) - frac2pix - 2$ as the following
beginalign
beginsplit
partial_tv &= partial_xxv, xin[0,pi]\
v (0,t) &= 0 \
v (pi,t) &= 0 \
v(x,0) &= u_0 - frac2pix - 2, xin[0,pi].
endsplit
endalign
and solve using separation of variables using the ansatz $v(x,t) = X(x)T(t)$ in the usual way.

Then one can finally write the solution as $u(x,t) = v(x,t) + u_E(x)$

share|cite|improve this answer

answered Apr 2 at 13:40

D.ShanksD.Shanks

12

$endgroup$

share|cite|improve this answer

answered Apr 2 at 13:40

D.ShanksD.Shanks

12

answered Apr 2 at 13:40

D.ShanksD.Shanks

12

answered Apr 2 at 13:40

D.ShanksD.Shanks

12