method of separation of variables for heat transfer with mixed conditions
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$u_t=c^2 u_xx, quad forall quad 0<x<L quad t>0$
$u_x(L,t)=0, quad u(0,t)=0$
$u(x,0)=f(x)$
Can anyone please tell me how to apply the boundary conditions to this problem and arrive at a general solution.
Any help is much appreciated.
differential-equations pde heat-equation
edited Aug 15 at 13:23
Perelman Jr
1198
asked Aug 15 at 13:03
Fahad Pervaiz
61
add a comment |Â
up vote
-1
down vote
favorite
$u_t=c^2 u_xx, quad forall quad 0<x<L quad t>0$
$u_x(L,t)=0, quad u(0,t)=0$
$u(x,0)=f(x)$
Can anyone please tell me how to apply the boundary conditions to this problem and arrive at a general solution.
Any help is much appreciated.
differential-equations pde heat-equation
edited Aug 15 at 13:23
Perelman Jr
1198
asked Aug 15 at 13:03
Fahad Pervaiz
61
what is the progress?
– Perelman Jr
Aug 15 at 13:09
T(x,t)=∑_(n=0)^∞▒〖Bn.cos[(n+1/2)À/L x)〗]〖e^-〗^(((n+1/2)^2 tÀ^2 )/L^2 )
– Fahad Pervaiz
Aug 15 at 14:33
I have find out this final general form for T(x,y) but dont know its correct or not
– Fahad Pervaiz
Aug 15 at 14:34
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
$u_t=c^2 u_xx, quad forall quad 0<x<L quad t>0$
$u_x(L,t)=0, quad u(0,t)=0$
$u(x,0)=f(x)$
Can anyone please tell me how to apply the boundary conditions to this problem and arrive at a general solution.
Any help is much appreciated.
differential-equations pde heat-equation
edited Aug 15 at 13:23
Perelman Jr
1198
asked Aug 15 at 13:03
Fahad Pervaiz
61
$u_t=c^2 u_xx, quad forall quad 0<x<L quad t>0$
$u_x(L,t)=0, quad u(0,t)=0$
$u(x,0)=f(x)$
Can anyone please tell me how to apply the boundary conditions to this problem and arrive at a general solution.
Any help is much appreciated.
differential-equations pde heat-equation
edited Aug 15 at 13:23
Perelman Jr
1198
asked Aug 15 at 13:03
Fahad Pervaiz
61
edited Aug 15 at 13:23
Perelman Jr
1198
edited Aug 15 at 13:23
Perelman Jr
1198
edited Aug 15 at 13:23
Perelman Jr
1198
1198
asked Aug 15 at 13:03
Fahad Pervaiz
61
asked Aug 15 at 13:03
Fahad Pervaiz
61
asked Aug 15 at 13:03
Fahad Pervaiz
61
61
what is the progress?
– Perelman Jr
Aug 15 at 13:09
T(x,t)=∑_(n=0)^∞▒〖Bn.cos[(n+1/2)À/L x)〗]〖e^-〗^(((n+1/2)^2 tÀ^2 )/L^2 )
– Fahad Pervaiz
Aug 15 at 14:33
I have find out this final general form for T(x,y) but dont know its correct or not
– Fahad Pervaiz
Aug 15 at 14:34
add a comment |Â
what is the progress?
– Perelman Jr
Aug 15 at 13:09
T(x,t)=∑_(n=0)^∞▒〖Bn.cos[(n+1/2)À/L x)〗]〖e^-〗^(((n+1/2)^2 tÀ^2 )/L^2 )
– Fahad Pervaiz
Aug 15 at 14:33
I have find out this final general form for T(x,y) but dont know its correct or not
– Fahad Pervaiz
Aug 15 at 14:34
what is the progress?
– Perelman Jr
Aug 15 at 13:09
what is the progress?
– Perelman Jr
Aug 15 at 13:09
T(x,t)=∑_(n=0)^∞▒〖Bn.cos[(n+1/2)À/L x)〗]〖e^-〗^(((n+1/2)^2 tÀ^2 )/L^2 )
– Fahad Pervaiz
Aug 15 at 14:33
T(x,t)=∑_(n=0)^∞▒〖Bn.cos[(n+1/2)À/L x)〗]〖e^-〗^(((n+1/2)^2 tÀ^2 )/L^2 )
– Fahad Pervaiz
Aug 15 at 14:33
I have find out this final general form for T(x,y) but dont know its correct or not
– Fahad Pervaiz
Aug 15 at 14:34
I have find out this final general form for T(x,y) but dont know its correct or not
– Fahad Pervaiz
Aug 15 at 14:34
add a comment |Â
1 Answer
1
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Since you specify "separation of variables", start by looking for solutions of the form u(x,t)= X(x)T(t). The differential equation becomes $XT'= c^2TX''$ which we can write as $fracT'4T= c^2fracX''X$. If we were to vary t while holding x constant, the right side would not change so the left side must be constant for varying t. Similarly, if we were to vary x while holding t constant, the left side would not change so the right side must be constant for varying x. The two sides must be equal to the same constant: we must have $fracT'T= lambda$ and $c^2fracX''X= lambda$ for some constant $lambda$. The boundary conditions are $u(0, t)= X(0)T(t)= 0$ for all t so we must have X(0)= 0, and $u_x(L, t)= X'(L)T(t)= 0$ for all t so we must have X'(L)= 0.
That is, you want to solve $c^2X''= lambda X$ with boundary conditions X(0)= 0, X'(L)= 0.
What kinds of solutions you get depends upon what $lambda$ is. You should be able to recognize that, in order to get a non-trivial solution for X, $lambda$ must be negative. Write $lambda= -omega^2$. The differential equation becomes $c^2X''= -omega^2 X$. That has general solution $X(x)= A cosleft(fracomegacxright)+ Bsinleft(fracomegacxright)$. The condition that $X(0)= 0$ give $A= 0$ so we must have $X(x)= Bsinleft(fracomegacxright)$. Then $X'(x)= Bfracomegaccosleft(fracomegacxright)$ so that $X'(L)= Bfracomegaccosleft(fracomegacright)= 0$.
We cannot have B= 0 also- that would give the "trivial" solution for X so that u could not satisfy the initial condition. Instead we must have $cosleft(fracomegacLright)= 0$. Cosine is 0 only for odd multiples of $fracpi2$ so we must have $fracomegacL= (2n+1)fracpi2$. We must have $omega= (2n+1)fraccpi2L$ so $lambda= -(2n+1)^2fracc^2pi^24L^2$.
Now put that $lambda$ into $T'= 4lambda T$ and solve for T(t). The solution to the problem is the sum of all such X(x)T(t) summed over n.
answered Aug 15 at 14:40
user247327
9,7781515
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Since you specify "separation of variables", start by looking for solutions of the form u(x,t)= X(x)T(t). The differential equation becomes $XT'= c^2TX''$ which we can write as $fracT'4T= c^2fracX''X$. If we were to vary t while holding x constant, the right side would not change so the left side must be constant for varying t. Similarly, if we were to vary x while holding t constant, the left side would not change so the right side must be constant for varying x. The two sides must be equal to the same constant: we must have $fracT'T= lambda$ and $c^2fracX''X= lambda$ for some constant $lambda$. The boundary conditions are $u(0, t)= X(0)T(t)= 0$ for all t so we must have X(0)= 0, and $u_x(L, t)= X'(L)T(t)= 0$ for all t so we must have X'(L)= 0.
That is, you want to solve $c^2X''= lambda X$ with boundary conditions X(0)= 0, X'(L)= 0.
What kinds of solutions you get depends upon what $lambda$ is. You should be able to recognize that, in order to get a non-trivial solution for X, $lambda$ must be negative. Write $lambda= -omega^2$. The differential equation becomes $c^2X''= -omega^2 X$. That has general solution $X(x)= A cosleft(fracomegacxright)+ Bsinleft(fracomegacxright)$. The condition that $X(0)= 0$ give $A= 0$ so we must have $X(x)= Bsinleft(fracomegacxright)$. Then $X'(x)= Bfracomegaccosleft(fracomegacxright)$ so that $X'(L)= Bfracomegaccosleft(fracomegacright)= 0$.
We cannot have B= 0 also- that would give the "trivial" solution for X so that u could not satisfy the initial condition. Instead we must have $cosleft(fracomegacLright)= 0$. Cosine is 0 only for odd multiples of $fracpi2$ so we must have $fracomegacL= (2n+1)fracpi2$. We must have $omega= (2n+1)fraccpi2L$ so $lambda= -(2n+1)^2fracc^2pi^24L^2$.
Now put that $lambda$ into $T'= 4lambda T$ and solve for T(t). The solution to the problem is the sum of all such X(x)T(t) summed over n.
answered Aug 15 at 14:40
user247327
9,7781515
add a comment |Â
up vote
0
down vote
Since you specify "separation of variables", start by looking for solutions of the form u(x,t)= X(x)T(t). The differential equation becomes $XT'= c^2TX''$ which we can write as $fracT'4T= c^2fracX''X$. If we were to vary t while holding x constant, the right side would not change so the left side must be constant for varying t. Similarly, if we were to vary x while holding t constant, the left side would not change so the right side must be constant for varying x. The two sides must be equal to the same constant: we must have $fracT'T= lambda$ and $c^2fracX''X= lambda$ for some constant $lambda$. The boundary conditions are $u(0, t)= X(0)T(t)= 0$ for all t so we must have X(0)= 0, and $u_x(L, t)= X'(L)T(t)= 0$ for all t so we must have X'(L)= 0.
That is, you want to solve $c^2X''= lambda X$ with boundary conditions X(0)= 0, X'(L)= 0.
What kinds of solutions you get depends upon what $lambda$ is. You should be able to recognize that, in order to get a non-trivial solution for X, $lambda$ must be negative. Write $lambda= -omega^2$. The differential equation becomes $c^2X''= -omega^2 X$. That has general solution $X(x)= A cosleft(fracomegacxright)+ Bsinleft(fracomegacxright)$. The condition that $X(0)= 0$ give $A= 0$ so we must have $X(x)= Bsinleft(fracomegacxright)$. Then $X'(x)= Bfracomegaccosleft(fracomegacxright)$ so that $X'(L)= Bfracomegaccosleft(fracomegacright)= 0$.
We cannot have B= 0 also- that would give the "trivial" solution for X so that u could not satisfy the initial condition. Instead we must have $cosleft(fracomegacLright)= 0$. Cosine is 0 only for odd multiples of $fracpi2$ so we must have $fracomegacL= (2n+1)fracpi2$. We must have $omega= (2n+1)fraccpi2L$ so $lambda= -(2n+1)^2fracc^2pi^24L^2$.
Now put that $lambda$ into $T'= 4lambda T$ and solve for T(t). The solution to the problem is the sum of all such X(x)T(t) summed over n.
answered Aug 15 at 14:40
user247327
9,7781515
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Since you specify "separation of variables", start by looking for solutions of the form u(x,t)= X(x)T(t). The differential equation becomes $XT'= c^2TX''$ which we can write as $fracT'4T= c^2fracX''X$. If we were to vary t while holding x constant, the right side would not change so the left side must be constant for varying t. Similarly, if we were to vary x while holding t constant, the left side would not change so the right side must be constant for varying x. The two sides must be equal to the same constant: we must have $fracT'T= lambda$ and $c^2fracX''X= lambda$ for some constant $lambda$. The boundary conditions are $u(0, t)= X(0)T(t)= 0$ for all t so we must have X(0)= 0, and $u_x(L, t)= X'(L)T(t)= 0$ for all t so we must have X'(L)= 0.
That is, you want to solve $c^2X''= lambda X$ with boundary conditions X(0)= 0, X'(L)= 0.
What kinds of solutions you get depends upon what $lambda$ is. You should be able to recognize that, in order to get a non-trivial solution for X, $lambda$ must be negative. Write $lambda= -omega^2$. The differential equation becomes $c^2X''= -omega^2 X$. That has general solution $X(x)= A cosleft(fracomegacxright)+ Bsinleft(fracomegacxright)$. The condition that $X(0)= 0$ give $A= 0$ so we must have $X(x)= Bsinleft(fracomegacxright)$. Then $X'(x)= Bfracomegaccosleft(fracomegacxright)$ so that $X'(L)= Bfracomegaccosleft(fracomegacright)= 0$.
We cannot have B= 0 also- that would give the "trivial" solution for X so that u could not satisfy the initial condition. Instead we must have $cosleft(fracomegacLright)= 0$. Cosine is 0 only for odd multiples of $fracpi2$ so we must have $fracomegacL= (2n+1)fracpi2$. We must have $omega= (2n+1)fraccpi2L$ so $lambda= -(2n+1)^2fracc^2pi^24L^2$.
Now put that $lambda$ into $T'= 4lambda T$ and solve for T(t). The solution to the problem is the sum of all such X(x)T(t) summed over n.
answered Aug 15 at 14:40
user247327
9,7781515
Since you specify "separation of variables", start by looking for solutions of the form u(x,t)= X(x)T(t). The differential equation becomes $XT'= c^2TX''$ which we can write as $fracT'4T= c^2fracX''X$. If we were to vary t while holding x constant, the right side would not change so the left side must be constant for varying t. Similarly, if we were to vary x while holding t constant, the left side would not change so the right side must be constant for varying x. The two sides must be equal to the same constant: we must have $fracT'T= lambda$ and $c^2fracX''X= lambda$ for some constant $lambda$. The boundary conditions are $u(0, t)= X(0)T(t)= 0$ for all t so we must have X(0)= 0, and $u_x(L, t)= X'(L)T(t)= 0$ for all t so we must have X'(L)= 0.
That is, you want to solve $c^2X''= lambda X$ with boundary conditions X(0)= 0, X'(L)= 0.
What kinds of solutions you get depends upon what $lambda$ is. You should be able to recognize that, in order to get a non-trivial solution for X, $lambda$ must be negative. Write $lambda= -omega^2$. The differential equation becomes $c^2X''= -omega^2 X$. That has general solution $X(x)= A cosleft(fracomegacxright)+ Bsinleft(fracomegacxright)$. The condition that $X(0)= 0$ give $A= 0$ so we must have $X(x)= Bsinleft(fracomegacxright)$. Then $X'(x)= Bfracomegaccosleft(fracomegacxright)$ so that $X'(L)= Bfracomegaccosleft(fracomegacright)= 0$.
We cannot have B= 0 also- that would give the "trivial" solution for X so that u could not satisfy the initial condition. Instead we must have $cosleft(fracomegacLright)= 0$. Cosine is 0 only for odd multiples of $fracpi2$ so we must have $fracomegacL= (2n+1)fracpi2$. We must have $omega= (2n+1)fraccpi2L$ so $lambda= -(2n+1)^2fracc^2pi^24L^2$.
Now put that $lambda$ into $T'= 4lambda T$ and solve for T(t). The solution to the problem is the sum of all such X(x)T(t) summed over n.
answered Aug 15 at 14:40
user247327
9,7781515
answered Aug 15 at 14:40
user247327
9,7781515
answered Aug 15 at 14:40
user247327
9,7781515
answered Aug 15 at 14:40
user247327
9,7781515
9,7781515
add a comment |Â
add a comment |Â
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what is the progress?
– Perelman Jr
Aug 15 at 13:09
T(x,t)=∑_(n=0)^∞▒〖Bn.cos[(n+1/2)À/L x)〗]〖e^-〗^(((n+1/2)^2 tÀ^2 )/L^2 )
– Fahad Pervaiz
Aug 15 at 14:33
I have find out this final general form for T(x,y) but dont know its correct or not
– Fahad Pervaiz
Aug 15 at 14:34