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How to solve these kind nonhomogeneous heat equation [on hold]
Heat-Equation with different initial valuesNonhomogeneous heat equationHow to solve this nonhomogeneous heat equationHow to solve this instance of the heat equation?Solve the initial value problem, heat equation. What method to use.Fundamental solution for 1D nonhomogeneous wave equationHow to solve heat equationHow solve next heat equation?Solving the Heat Equation using the Fourier TransformSolving heat equation variant
$begingroup$
Solve
$$begincases
u_t-u_xx=frac12xt &textfor 0lt xlt pi, tgt0\
u(0,t)=u(pi,t)=0 \
u(x,0) = sin x \
endcases$$
How to transform it to the normal case?
pde
asked Mar 11 at 9:20
Jaqen ChouJaqen Chou
460110
$endgroup$
put on hold as off-topic by Eevee Trainer, Cesareo, Gibbs, José Carlos Santos, Parcly Taxel Mar 11 at 14:01
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Cesareo, Gibbs, José Carlos Santos, Parcly Taxel
add a comment |
$begingroup$
Solve
$$begincases
u_t-u_xx=frac12xt &textfor 0lt xlt pi, tgt0\
u(0,t)=u(pi,t)=0 \
u(x,0) = sin x \
endcases$$
How to transform it to the normal case?
pde
asked Mar 11 at 9:20
Jaqen ChouJaqen Chou
460110
$endgroup$
put on hold as off-topic by Eevee Trainer, Cesareo, Gibbs, José Carlos Santos, Parcly Taxel Mar 11 at 14:01
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Cesareo, Gibbs, José Carlos Santos, Parcly Taxel
add a comment |
$begingroup$
Solve
$$begincases
u_t-u_xx=frac12xt &textfor 0lt xlt pi, tgt0\
u(0,t)=u(pi,t)=0 \
u(x,0) = sin x \
endcases$$
How to transform it to the normal case?
pde
asked Mar 11 at 9:20
Jaqen ChouJaqen Chou
460110
$endgroup$
Solve
$$begincases
u_t-u_xx=frac12xt &textfor 0lt xlt pi, tgt0\
u(0,t)=u(pi,t)=0 \
u(x,0) = sin x \
endcases$$
How to transform it to the normal case?
pde
pde
asked Mar 11 at 9:20
Jaqen ChouJaqen Chou
460110
asked Mar 11 at 9:20
Jaqen ChouJaqen Chou
460110
asked Mar 11 at 9:20
Jaqen ChouJaqen Chou
460110
asked Mar 11 at 9:20
Jaqen ChouJaqen Chou
460110
asked Mar 11 at 9:20
Jaqen ChouJaqen Chou
460110
460110
put on hold as off-topic by Eevee Trainer, Cesareo, Gibbs, José Carlos Santos, Parcly Taxel Mar 11 at 14:01
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Cesareo, Gibbs, José Carlos Santos, Parcly Taxel
put on hold as off-topic by Eevee Trainer, Cesareo, Gibbs, José Carlos Santos, Parcly Taxel Mar 11 at 14:01
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Cesareo, Gibbs, José Carlos Santos, Parcly Taxel
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Try to find a solution of the form
$$ u(x,t) = sum_n T_n(t)X_n(x) $$
where
begincases
X'' = -lambda^2 X \
X(0) = X(pi) = 0
endcases
You'll get
$$ u(x,t) = sum_n=1^infty T_n(t)sin(nx) $$
and the PDE becomes
$$ sum_n=1^inftybig[T_n'(t) + n^2T_n(t)big]sin(nx) = frac12 tx $$
Decompose the RHS into the corresponding Fourier series
$$ frac12 tx = sum_n=1^infty a_n(t) sin(nx) $$
You'll get a family of IVPs
begincases
T_n''(t) + n^2 T_n(t) = a_n(t) \
T_1(0) = 1 \
T_n>1(0) = 0
endcases
answered Mar 11 at 10:08
DylanDylan
13.9k31027
$endgroup$
$begingroup$
There is one thing I'm confused, in fact we can get $X''=-lambda^2 X$ in homogeneous case because $fracX''X=fracT'T$ has to be a constant. But in this nonhomogeneous case, I can only get $T'X=X''T+frac12xt$.
$endgroup$
– Jaqen Chou
Mar 11 at 10:45
$begingroup$
The full explanation is a bit complicated. This method isn't separation of variables, just related to it. It just so happens that the eigenfunctions in $x$ span the full function space, so we use it as a basis of our solution. Think of it like an eigenvalue-eigenvector decomposition.
$endgroup$
– Dylan
Mar 11 at 10:51
$begingroup$
Okay, think I'll find it in my following study. Thanks a lot
$endgroup$
– Jaqen Chou
Mar 11 at 10:55
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Try to find a solution of the form
$$ u(x,t) = sum_n T_n(t)X_n(x) $$
where
begincases
X'' = -lambda^2 X \
X(0) = X(pi) = 0
endcases
You'll get
$$ u(x,t) = sum_n=1^infty T_n(t)sin(nx) $$
and the PDE becomes
$$ sum_n=1^inftybig[T_n'(t) + n^2T_n(t)big]sin(nx) = frac12 tx $$
Decompose the RHS into the corresponding Fourier series
$$ frac12 tx = sum_n=1^infty a_n(t) sin(nx) $$
You'll get a family of IVPs
begincases
T_n''(t) + n^2 T_n(t) = a_n(t) \
T_1(0) = 1 \
T_n>1(0) = 0
endcases
answered Mar 11 at 10:08
DylanDylan
13.9k31027
$endgroup$
$begingroup$
There is one thing I'm confused, in fact we can get $X''=-lambda^2 X$ in homogeneous case because $fracX''X=fracT'T$ has to be a constant. But in this nonhomogeneous case, I can only get $T'X=X''T+frac12xt$.
$endgroup$
– Jaqen Chou
Mar 11 at 10:45
$begingroup$
The full explanation is a bit complicated. This method isn't separation of variables, just related to it. It just so happens that the eigenfunctions in $x$ span the full function space, so we use it as a basis of our solution. Think of it like an eigenvalue-eigenvector decomposition.
$endgroup$
– Dylan
Mar 11 at 10:51
$begingroup$
Okay, think I'll find it in my following study. Thanks a lot
$endgroup$
– Jaqen Chou
Mar 11 at 10:55
add a comment |
$begingroup$
Try to find a solution of the form
$$ u(x,t) = sum_n T_n(t)X_n(x) $$
where
begincases
X'' = -lambda^2 X \
X(0) = X(pi) = 0
endcases
You'll get
$$ u(x,t) = sum_n=1^infty T_n(t)sin(nx) $$
and the PDE becomes
$$ sum_n=1^inftybig[T_n'(t) + n^2T_n(t)big]sin(nx) = frac12 tx $$
Decompose the RHS into the corresponding Fourier series
$$ frac12 tx = sum_n=1^infty a_n(t) sin(nx) $$
You'll get a family of IVPs
begincases
T_n''(t) + n^2 T_n(t) = a_n(t) \
T_1(0) = 1 \
T_n>1(0) = 0
endcases
answered Mar 11 at 10:08
DylanDylan
13.9k31027
$endgroup$
$begingroup$
There is one thing I'm confused, in fact we can get $X''=-lambda^2 X$ in homogeneous case because $fracX''X=fracT'T$ has to be a constant. But in this nonhomogeneous case, I can only get $T'X=X''T+frac12xt$.
$endgroup$
– Jaqen Chou
Mar 11 at 10:45
$begingroup$
The full explanation is a bit complicated. This method isn't separation of variables, just related to it. It just so happens that the eigenfunctions in $x$ span the full function space, so we use it as a basis of our solution. Think of it like an eigenvalue-eigenvector decomposition.
$endgroup$
– Dylan
Mar 11 at 10:51
$begingroup$
Okay, think I'll find it in my following study. Thanks a lot
$endgroup$
– Jaqen Chou
Mar 11 at 10:55
add a comment |
$begingroup$
Try to find a solution of the form
$$ u(x,t) = sum_n T_n(t)X_n(x) $$
where
begincases
X'' = -lambda^2 X \
X(0) = X(pi) = 0
endcases
You'll get
$$ u(x,t) = sum_n=1^infty T_n(t)sin(nx) $$
and the PDE becomes
$$ sum_n=1^inftybig[T_n'(t) + n^2T_n(t)big]sin(nx) = frac12 tx $$
Decompose the RHS into the corresponding Fourier series
$$ frac12 tx = sum_n=1^infty a_n(t) sin(nx) $$
You'll get a family of IVPs
begincases
T_n''(t) + n^2 T_n(t) = a_n(t) \
T_1(0) = 1 \
T_n>1(0) = 0
endcases
answered Mar 11 at 10:08
DylanDylan
13.9k31027
$endgroup$
Try to find a solution of the form
$$ u(x,t) = sum_n T_n(t)X_n(x) $$
where
begincases
X'' = -lambda^2 X \
X(0) = X(pi) = 0
endcases
You'll get
$$ u(x,t) = sum_n=1^infty T_n(t)sin(nx) $$
and the PDE becomes
$$ sum_n=1^inftybig[T_n'(t) + n^2T_n(t)big]sin(nx) = frac12 tx $$
Decompose the RHS into the corresponding Fourier series
$$ frac12 tx = sum_n=1^infty a_n(t) sin(nx) $$
You'll get a family of IVPs
begincases
T_n''(t) + n^2 T_n(t) = a_n(t) \
T_1(0) = 1 \
T_n>1(0) = 0
endcases
answered Mar 11 at 10:08
DylanDylan
13.9k31027
answered Mar 11 at 10:08
DylanDylan
13.9k31027
answered Mar 11 at 10:08
DylanDylan
13.9k31027
answered Mar 11 at 10:08
DylanDylan
13.9k31027
13.9k31027
$begingroup$
There is one thing I'm confused, in fact we can get $X''=-lambda^2 X$ in homogeneous case because $fracX''X=fracT'T$ has to be a constant. But in this nonhomogeneous case, I can only get $T'X=X''T+frac12xt$.
$endgroup$
– Jaqen Chou
Mar 11 at 10:45
$begingroup$
The full explanation is a bit complicated. This method isn't separation of variables, just related to it. It just so happens that the eigenfunctions in $x$ span the full function space, so we use it as a basis of our solution. Think of it like an eigenvalue-eigenvector decomposition.
$endgroup$
– Dylan
Mar 11 at 10:51
$begingroup$
Okay, think I'll find it in my following study. Thanks a lot
$endgroup$
– Jaqen Chou
Mar 11 at 10:55
add a comment |
$begingroup$
There is one thing I'm confused, in fact we can get $X''=-lambda^2 X$ in homogeneous case because $fracX''X=fracT'T$ has to be a constant. But in this nonhomogeneous case, I can only get $T'X=X''T+frac12xt$.
$endgroup$
– Jaqen Chou
Mar 11 at 10:45
$begingroup$
The full explanation is a bit complicated. This method isn't separation of variables, just related to it. It just so happens that the eigenfunctions in $x$ span the full function space, so we use it as a basis of our solution. Think of it like an eigenvalue-eigenvector decomposition.
$endgroup$
– Dylan
Mar 11 at 10:51
$begingroup$
Okay, think I'll find it in my following study. Thanks a lot
$endgroup$
– Jaqen Chou
Mar 11 at 10:55
$begingroup$
There is one thing I'm confused, in fact we can get $X''=-lambda^2 X$ in homogeneous case because $fracX''X=fracT'T$ has to be a constant. But in this nonhomogeneous case, I can only get $T'X=X''T+frac12xt$.
$endgroup$
– Jaqen Chou
Mar 11 at 10:45
$begingroup$
There is one thing I'm confused, in fact we can get $X''=-lambda^2 X$ in homogeneous case because $fracX''X=fracT'T$ has to be a constant. But in this nonhomogeneous case, I can only get $T'X=X''T+frac12xt$.
$endgroup$
– Jaqen Chou
Mar 11 at 10:45
$begingroup$
The full explanation is a bit complicated. This method isn't separation of variables, just related to it. It just so happens that the eigenfunctions in $x$ span the full function space, so we use it as a basis of our solution. Think of it like an eigenvalue-eigenvector decomposition.
$endgroup$
– Dylan
Mar 11 at 10:51
$begingroup$
The full explanation is a bit complicated. This method isn't separation of variables, just related to it. It just so happens that the eigenfunctions in $x$ span the full function space, so we use it as a basis of our solution. Think of it like an eigenvalue-eigenvector decomposition.
$endgroup$
– Dylan
Mar 11 at 10:51
$begingroup$
Okay, think I'll find it in my following study. Thanks a lot
$endgroup$
– Jaqen Chou
Mar 11 at 10:55
$begingroup$
Okay, think I'll find it in my following study. Thanks a lot
$endgroup$
– Jaqen Chou
Mar 11 at 10:55
add a comment |
