Well-posedness for Heat Equation with Robin Boundary Condition
$begingroup$
Can anyone help me prove the well-posedness of the following heat equation with Robin boundary condition?
$u_t(x,t)=u_{xx}(x,t)$
$u(0,t)=0$
$u_x(1,t)=-au(1,t)$
where $a>0$.
The existence of the solution may be simply obtained by separation of variable. Are there any good references on this problem?
pde heat-equation
asked Dec 8 '18 at 1:08
D. ZhD. Zh
61
$endgroup$
add a comment |
$begingroup$
Can anyone help me prove the well-posedness of the following heat equation with Robin boundary condition?
$u_t(x,t)=u_{xx}(x,t)$
$u(0,t)=0$
$u_x(1,t)=-au(1,t)$
where $a>0$.
The existence of the solution may be simply obtained by separation of variable. Are there any good references on this problem?
pde heat-equation
asked Dec 8 '18 at 1:08
D. ZhD. Zh
61
$endgroup$
add a comment |
$begingroup$
Can anyone help me prove the well-posedness of the following heat equation with Robin boundary condition?
$u_t(x,t)=u_{xx}(x,t)$
$u(0,t)=0$
$u_x(1,t)=-au(1,t)$
where $a>0$.
The existence of the solution may be simply obtained by separation of variable. Are there any good references on this problem?
pde heat-equation
asked Dec 8 '18 at 1:08
D. ZhD. Zh
61
$endgroup$
Can anyone help me prove the well-posedness of the following heat equation with Robin boundary condition?
$u_t(x,t)=u_{xx}(x,t)$
$u(0,t)=0$
$u_x(1,t)=-au(1,t)$
where $a>0$.
The existence of the solution may be simply obtained by separation of variable. Are there any good references on this problem?
pde heat-equation
pde heat-equation
asked Dec 8 '18 at 1:08
D. ZhD. Zh
61
asked Dec 8 '18 at 1:08
D. ZhD. Zh
61
asked Dec 8 '18 at 1:08
D. ZhD. Zh
61
asked Dec 8 '18 at 1:08
D. ZhD. Zh
61
asked Dec 8 '18 at 1:08
D. ZhD. Zh
61
61
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your problem is missing an initial condition such as $u(x,0)=u_0(x)$, which is necessary. Assuming $u_1,u_2$ are two such solutions satisfying the same initial condition and conditions you have specified, then $v=u_1-u_2$ satisfies
$$
v_t = v_{xx} \
v(x,0)=0 \
v(0,t)=0,;;v_x(1,t)=-av(1,t).
$$
Then,
begin{align}
&frac{d}{dt}int_{0}^{1}v(x,t)^2dx \
&=2int_0^1v(x,t)v_t(x,t)dx \
&= 2int_0^1v(x,t)v_{xx}(x,t)dx \
&= 2v(x,t)v_{x}(x,t)|_{x=0}^{1}-2int_0^1 v_x(x,t)^2dx \
&= 2v(1,t)v_x(1,t)-2v(0,t)v_x(0,t)-2int_0^1 v_x^2x \
&= 2v(1,t)v_x(1,t)-2int_0^1 v_x^2dx \
&= -2av_x(1,t)^2-2int_0^1 v_x^2 dx le 0.
end{align}
Because $int_0^1v(x,t)^2dx = 0$ for $t=0$, the above forces
$$int_0^1v(x,t)^2dx=0,;;; t ge 0.$$
This is enough to give $v(x,t)=0$ for all $tge 0$. So $u_1=u_2$.
answered Dec 8 '18 at 4:12
DisintegratingByPartsDisintegratingByParts
59.2k42580
$endgroup$
$begingroup$
One might say that the initial condition is just arbitrary if none is chosen..I.E $u(x,0)= g(x)$. Even if it isn’t stated, it is somewhat implied, no?
$endgroup$
– DaveNine
Dec 8 '18 at 20:26
$begingroup$
@DaveNine : This question is all about having a well-posed problem. So I don't think skipping a discussion of conditions that lead to "well-posedness" is called for in this case.
$endgroup$
– DisintegratingByParts
Dec 8 '18 at 21:04
$begingroup$
@DisintegratingByParts : Thank you. Does the uniqueness of the solution imply well-posedness of the problem?
$endgroup$
– D. Zh
Dec 9 '18 at 18:46
add a comment |
$begingroup$
your problem can be written under the form $$u'(t)=Au\u(0)=u_0$$
where $$A:D(A) subset L^2(0,1) to L^2(0,1)$$
$$D(A)=leftlbrace vin H^1(0,1), v(0)=0, v_x(1)+av(1)=0rightrbrace $$
It is straitforward to see that $(Au,u)_{L^2(0,1)} leqslant 0$, and $I-A$ is maximal, ($A$ is the second derevative), therefore, by semigroup theory, there exists one solution in $C(0,T;L^2(0,1)$ (if your initial state is $L^2(0,1)$)
answered Dec 8 '18 at 9:54
GustaveGustave
729211
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
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active
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$begingroup$
Your problem is missing an initial condition such as $u(x,0)=u_0(x)$, which is necessary. Assuming $u_1,u_2$ are two such solutions satisfying the same initial condition and conditions you have specified, then $v=u_1-u_2$ satisfies
$$
v_t = v_{xx} \
v(x,0)=0 \
v(0,t)=0,;;v_x(1,t)=-av(1,t).
$$
Then,
begin{align}
&frac{d}{dt}int_{0}^{1}v(x,t)^2dx \
&=2int_0^1v(x,t)v_t(x,t)dx \
&= 2int_0^1v(x,t)v_{xx}(x,t)dx \
&= 2v(x,t)v_{x}(x,t)|_{x=0}^{1}-2int_0^1 v_x(x,t)^2dx \
&= 2v(1,t)v_x(1,t)-2v(0,t)v_x(0,t)-2int_0^1 v_x^2x \
&= 2v(1,t)v_x(1,t)-2int_0^1 v_x^2dx \
&= -2av_x(1,t)^2-2int_0^1 v_x^2 dx le 0.
end{align}
Because $int_0^1v(x,t)^2dx = 0$ for $t=0$, the above forces
$$int_0^1v(x,t)^2dx=0,;;; t ge 0.$$
This is enough to give $v(x,t)=0$ for all $tge 0$. So $u_1=u_2$.
answered Dec 8 '18 at 4:12
DisintegratingByPartsDisintegratingByParts
59.2k42580
$endgroup$
$begingroup$
One might say that the initial condition is just arbitrary if none is chosen..I.E $u(x,0)= g(x)$. Even if it isn’t stated, it is somewhat implied, no?
$endgroup$
– DaveNine
Dec 8 '18 at 20:26
$begingroup$
@DaveNine : This question is all about having a well-posed problem. So I don't think skipping a discussion of conditions that lead to "well-posedness" is called for in this case.
$endgroup$
– DisintegratingByParts
Dec 8 '18 at 21:04
$begingroup$
@DisintegratingByParts : Thank you. Does the uniqueness of the solution imply well-posedness of the problem?
$endgroup$
– D. Zh
Dec 9 '18 at 18:46
add a comment |
$begingroup$
Your problem is missing an initial condition such as $u(x,0)=u_0(x)$, which is necessary. Assuming $u_1,u_2$ are two such solutions satisfying the same initial condition and conditions you have specified, then $v=u_1-u_2$ satisfies
$$
v_t = v_{xx} \
v(x,0)=0 \
v(0,t)=0,;;v_x(1,t)=-av(1,t).
$$
Then,
begin{align}
&frac{d}{dt}int_{0}^{1}v(x,t)^2dx \
&=2int_0^1v(x,t)v_t(x,t)dx \
&= 2int_0^1v(x,t)v_{xx}(x,t)dx \
&= 2v(x,t)v_{x}(x,t)|_{x=0}^{1}-2int_0^1 v_x(x,t)^2dx \
&= 2v(1,t)v_x(1,t)-2v(0,t)v_x(0,t)-2int_0^1 v_x^2x \
&= 2v(1,t)v_x(1,t)-2int_0^1 v_x^2dx \
&= -2av_x(1,t)^2-2int_0^1 v_x^2 dx le 0.
end{align}
Because $int_0^1v(x,t)^2dx = 0$ for $t=0$, the above forces
$$int_0^1v(x,t)^2dx=0,;;; t ge 0.$$
This is enough to give $v(x,t)=0$ for all $tge 0$. So $u_1=u_2$.
answered Dec 8 '18 at 4:12
DisintegratingByPartsDisintegratingByParts
59.2k42580
$endgroup$
$begingroup$
One might say that the initial condition is just arbitrary if none is chosen..I.E $u(x,0)= g(x)$. Even if it isn’t stated, it is somewhat implied, no?
$endgroup$
– DaveNine
Dec 8 '18 at 20:26
$begingroup$
@DaveNine : This question is all about having a well-posed problem. So I don't think skipping a discussion of conditions that lead to "well-posedness" is called for in this case.
$endgroup$
– DisintegratingByParts
Dec 8 '18 at 21:04
$begingroup$
@DisintegratingByParts : Thank you. Does the uniqueness of the solution imply well-posedness of the problem?
$endgroup$
– D. Zh
Dec 9 '18 at 18:46
add a comment |
$begingroup$
Your problem is missing an initial condition such as $u(x,0)=u_0(x)$, which is necessary. Assuming $u_1,u_2$ are two such solutions satisfying the same initial condition and conditions you have specified, then $v=u_1-u_2$ satisfies
$$
v_t = v_{xx} \
v(x,0)=0 \
v(0,t)=0,;;v_x(1,t)=-av(1,t).
$$
Then,
begin{align}
&frac{d}{dt}int_{0}^{1}v(x,t)^2dx \
&=2int_0^1v(x,t)v_t(x,t)dx \
&= 2int_0^1v(x,t)v_{xx}(x,t)dx \
&= 2v(x,t)v_{x}(x,t)|_{x=0}^{1}-2int_0^1 v_x(x,t)^2dx \
&= 2v(1,t)v_x(1,t)-2v(0,t)v_x(0,t)-2int_0^1 v_x^2x \
&= 2v(1,t)v_x(1,t)-2int_0^1 v_x^2dx \
&= -2av_x(1,t)^2-2int_0^1 v_x^2 dx le 0.
end{align}
Because $int_0^1v(x,t)^2dx = 0$ for $t=0$, the above forces
$$int_0^1v(x,t)^2dx=0,;;; t ge 0.$$
This is enough to give $v(x,t)=0$ for all $tge 0$. So $u_1=u_2$.
answered Dec 8 '18 at 4:12
DisintegratingByPartsDisintegratingByParts
59.2k42580
$endgroup$
Your problem is missing an initial condition such as $u(x,0)=u_0(x)$, which is necessary. Assuming $u_1,u_2$ are two such solutions satisfying the same initial condition and conditions you have specified, then $v=u_1-u_2$ satisfies
$$
v_t = v_{xx} \
v(x,0)=0 \
v(0,t)=0,;;v_x(1,t)=-av(1,t).
$$
Then,
begin{align}
&frac{d}{dt}int_{0}^{1}v(x,t)^2dx \
&=2int_0^1v(x,t)v_t(x,t)dx \
&= 2int_0^1v(x,t)v_{xx}(x,t)dx \
&= 2v(x,t)v_{x}(x,t)|_{x=0}^{1}-2int_0^1 v_x(x,t)^2dx \
&= 2v(1,t)v_x(1,t)-2v(0,t)v_x(0,t)-2int_0^1 v_x^2x \
&= 2v(1,t)v_x(1,t)-2int_0^1 v_x^2dx \
&= -2av_x(1,t)^2-2int_0^1 v_x^2 dx le 0.
end{align}
Because $int_0^1v(x,t)^2dx = 0$ for $t=0$, the above forces
$$int_0^1v(x,t)^2dx=0,;;; t ge 0.$$
This is enough to give $v(x,t)=0$ for all $tge 0$. So $u_1=u_2$.
answered Dec 8 '18 at 4:12
DisintegratingByPartsDisintegratingByParts
59.2k42580
answered Dec 8 '18 at 4:12
DisintegratingByPartsDisintegratingByParts
59.2k42580
answered Dec 8 '18 at 4:12
DisintegratingByPartsDisintegratingByParts
59.2k42580
answered Dec 8 '18 at 4:12
DisintegratingByPartsDisintegratingByParts
59.2k42580
59.2k42580
$begingroup$
One might say that the initial condition is just arbitrary if none is chosen..I.E $u(x,0)= g(x)$. Even if it isn’t stated, it is somewhat implied, no?
$endgroup$
– DaveNine
Dec 8 '18 at 20:26
$begingroup$
@DaveNine : This question is all about having a well-posed problem. So I don't think skipping a discussion of conditions that lead to "well-posedness" is called for in this case.
$endgroup$
– DisintegratingByParts
Dec 8 '18 at 21:04
$begingroup$
@DisintegratingByParts : Thank you. Does the uniqueness of the solution imply well-posedness of the problem?
$endgroup$
– D. Zh
Dec 9 '18 at 18:46
add a comment |
$begingroup$
One might say that the initial condition is just arbitrary if none is chosen..I.E $u(x,0)= g(x)$. Even if it isn’t stated, it is somewhat implied, no?
$endgroup$
– DaveNine
Dec 8 '18 at 20:26
$begingroup$
@DaveNine : This question is all about having a well-posed problem. So I don't think skipping a discussion of conditions that lead to "well-posedness" is called for in this case.
$endgroup$
– DisintegratingByParts
Dec 8 '18 at 21:04
$begingroup$
@DisintegratingByParts : Thank you. Does the uniqueness of the solution imply well-posedness of the problem?
$endgroup$
– D. Zh
Dec 9 '18 at 18:46
$begingroup$
One might say that the initial condition is just arbitrary if none is chosen..I.E $u(x,0)= g(x)$. Even if it isn’t stated, it is somewhat implied, no?
$endgroup$
– DaveNine
Dec 8 '18 at 20:26
$begingroup$
One might say that the initial condition is just arbitrary if none is chosen..I.E $u(x,0)= g(x)$. Even if it isn’t stated, it is somewhat implied, no?
$endgroup$
– DaveNine
Dec 8 '18 at 20:26
$begingroup$
@DaveNine : This question is all about having a well-posed problem. So I don't think skipping a discussion of conditions that lead to "well-posedness" is called for in this case.
$endgroup$
– DisintegratingByParts
Dec 8 '18 at 21:04
$begingroup$
@DaveNine : This question is all about having a well-posed problem. So I don't think skipping a discussion of conditions that lead to "well-posedness" is called for in this case.
$endgroup$
– DisintegratingByParts
Dec 8 '18 at 21:04
$begingroup$
@DisintegratingByParts : Thank you. Does the uniqueness of the solution imply well-posedness of the problem?
$endgroup$
– D. Zh
Dec 9 '18 at 18:46
$begingroup$
@DisintegratingByParts : Thank you. Does the uniqueness of the solution imply well-posedness of the problem?
$endgroup$
– D. Zh
Dec 9 '18 at 18:46
add a comment |
$begingroup$
your problem can be written under the form $$u'(t)=Au\u(0)=u_0$$
where $$A:D(A) subset L^2(0,1) to L^2(0,1)$$
$$D(A)=leftlbrace vin H^1(0,1), v(0)=0, v_x(1)+av(1)=0rightrbrace $$
It is straitforward to see that $(Au,u)_{L^2(0,1)} leqslant 0$, and $I-A$ is maximal, ($A$ is the second derevative), therefore, by semigroup theory, there exists one solution in $C(0,T;L^2(0,1)$ (if your initial state is $L^2(0,1)$)
answered Dec 8 '18 at 9:54
GustaveGustave
729211
$endgroup$
add a comment |
$begingroup$
your problem can be written under the form $$u'(t)=Au\u(0)=u_0$$
where $$A:D(A) subset L^2(0,1) to L^2(0,1)$$
$$D(A)=leftlbrace vin H^1(0,1), v(0)=0, v_x(1)+av(1)=0rightrbrace $$
It is straitforward to see that $(Au,u)_{L^2(0,1)} leqslant 0$, and $I-A$ is maximal, ($A$ is the second derevative), therefore, by semigroup theory, there exists one solution in $C(0,T;L^2(0,1)$ (if your initial state is $L^2(0,1)$)
answered Dec 8 '18 at 9:54
GustaveGustave
729211
$endgroup$
add a comment |
$begingroup$
your problem can be written under the form $$u'(t)=Au\u(0)=u_0$$
where $$A:D(A) subset L^2(0,1) to L^2(0,1)$$
$$D(A)=leftlbrace vin H^1(0,1), v(0)=0, v_x(1)+av(1)=0rightrbrace $$
It is straitforward to see that $(Au,u)_{L^2(0,1)} leqslant 0$, and $I-A$ is maximal, ($A$ is the second derevative), therefore, by semigroup theory, there exists one solution in $C(0,T;L^2(0,1)$ (if your initial state is $L^2(0,1)$)
answered Dec 8 '18 at 9:54
GustaveGustave
729211
$endgroup$
your problem can be written under the form $$u'(t)=Au\u(0)=u_0$$
where $$A:D(A) subset L^2(0,1) to L^2(0,1)$$
$$D(A)=leftlbrace vin H^1(0,1), v(0)=0, v_x(1)+av(1)=0rightrbrace $$
It is straitforward to see that $(Au,u)_{L^2(0,1)} leqslant 0$, and $I-A$ is maximal, ($A$ is the second derevative), therefore, by semigroup theory, there exists one solution in $C(0,T;L^2(0,1)$ (if your initial state is $L^2(0,1)$)
answered Dec 8 '18 at 9:54
GustaveGustave
729211
answered Dec 8 '18 at 9:54
GustaveGustave
729211
answered Dec 8 '18 at 9:54
GustaveGustave
729211
answered Dec 8 '18 at 9:54
GustaveGustave
729211
729211
add a comment |
add a comment |
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