关于热传导方程
这是二阶线性偏微分方程。我知道它的解是不可逆的，跟热力学第二定律相关。原方程若作替换t->-t，则方程改变了。
我的问题是：如何直观理解这种不可逆性？任意给定一个态，其后的演化总是确定的。这一系列的态构成一个链。链上前面的态构成后面的态的原因，后面的态是前面的态的结果。为什么只能由原因找到结果，而不能从结果找到原因呢？
或者一个结果可能有多种原因？那这就意味着有可能多种初始条件演化后得到同一个态（中间态，非平衡态）。事实是否如此呢？
| 论坛嘉宾: blackhole zzzwp917 |
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blackhole  发表文章数: 196
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 中国是一个从上往下煽耳光,从下往上磕头的社会。
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卢昌海  发表文章数: 768
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热传导方程的解不可逆，其含义并非是说不能在数学上通过态的逆向演化来执果求因，而是说态的逆向演化不满足原热传导方程，因而不是宏观热力学所允许的自发物理过程。
宠辱不惊，看庭前花开花落
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星空浩淼 发表文章数: 799
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热传导方程的解不可逆，其含义并非是说不能在数学上通过态的逆向演化来执果求因，而是说态的逆向演化不满足原热传导方程，因而不是宏观热力学所允许的自发物理过程。
－－－－－－－－－－－－－－－－－－－－－－－－－－－ 帮昌海兄更加通俗化一下：比如热只能自发地从高温物体传递到低温物体，而不能反过来自发地从低温物体传递到高温物体，但是传递到了低温物体的热量，可以逆向地探知它的来源是来自高温物体。不可逆的其他例子：杯子可以掉在地上摔成碎片，碎片不能自发由地上飞上桌子变成完整的杯子，但是我们可以由地上的碎片逆向追索它的来源：是由杯子摔碎而来的。 有些解，比如随时间呈指数衰减的解是允许的，而随时间呈指数增加的解虽然也满足原方程，却不符合物理实际，是不允许的 One may view the world with the p-eye and one may view it with the q-eye but if one opens both eyes simultaneously then one gets crazy
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卢昌海 发表文章数: 768
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谢谢星空兄的通俗说明。
:: 随时间呈指数增加的解虽然也满足原方程，却不符合物理实际 热传导方程本身已带有时间箭头，将热传导方程中随时间衰减的解逆转（即t->-t）而得到的随时间增加的解不再满足原热传导方程。 宠辱不惊，看庭前花开花落
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sage 发表文章数: 359
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In another word, if you are thinking about a microscopic (or pure) state, not its statistical approximation, you should not use the heat equation since it is only a statistical statement.
So far, there is no such a thing as arrow of time in fundamental laws of physics. There is only the statistical statement that entropy tends to grow.
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星空浩淼 发表文章数: 799
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So far, there is no such a thing as arrow of time in fundamental laws of physics. There is only the statistical statement that entropy tends to grow.
－－－－－－－－－－－－－－ 的确，一些宏观统计力学中的不可逆，本质上属于概率论的规律在起作用，比如如果系统处于状态A的概率远远小于系统处于状态B的概率，则系统由A向B的转化概率远远大于由B向A的转化概率，甚至宏观上体现为系统可以有A自发变成B，却不能由B转化为A——然而在微观上，从物理规律本身而言，系统或许原则上总是存在非零的概率由B转化为A。例如，假定状态A和B所包含的每个基本状态是等几率的，只是B所包含的基本状态远远多于A所包含的基本状态，使得系统处于状态B的概率远远大于系统处于状态A的概率，那么尽管由B转化为A的概率虽然远远小于A转化成B的概率，原则上总是可以转化的。 因此，宏观经典力学中的不可逆，并不一定意味着微观层次上或者量子力学的水平上的不可逆。有人甚至试图把CP/T对称破缺用下一个结构层次的物理规律来解释，即认为物理在最本质的层面上，仍然是CP/T对称的。 One may view the world with the p-eye and one may view it with the q-eye but if one opens both eyes simultaneously then one gets crazy
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blackhole 发表文章数: 196
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谢谢各位大牛的回复。
然而我的问题是：给出一个初态，可以由方程获得一小段时间后的状态。但若仅仅知道后一状态，为什么不能由方程得到一小段时间之前的状态？从前一观点来看，“一小段时间之前的状态”是确定的啊。为什么根据方程本身不能得到呢？ 中国是一个从上往下煽耳光,从下往上磕头的社会。
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卢昌海 发表文章数: 768
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:: 但若仅仅知道后一状态，为什么不能由方程得到一小段时间之前的状态？
我第一个回帖回答的就是这个问题啊。热传导方程的解是可以在数学上逆推的（因此由方程可以得到一小段时间之前的状态）。所谓“不可逆”只是说逆向过程不是自发的物理过程。 宠辱不惊，看庭前花开花落
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blackhole 发表文章数: 196
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我第一个回帖回答的就是这个问题啊。热传导方程的解是可以在数学上逆推的（因此由方程可以得到一小段时间之前的状态）。
～～～～～～～～～～～～～～～～～～～～～～～· sorry，没仔细看。 但问题就在于：热传导方程的解在数学上似乎是不能逆推的。我曾就此问题查过许多资料，没看到逆推的情况，都要求t>0。不管是从其级数解还是其他形式的解看，t都不能小于0。 中国是一个从上往下煽耳光,从下往上磕头的社会。
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卢昌海 发表文章数: 768
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从t=2逆推t=1也是逆推。一个微分方程的解在满足存在及唯一性定理的区间中都是可以逆推的。
宠辱不惊，看庭前花开花落
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blackhole 发表文章数: 196
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从t=2逆推t=1也是逆推。
~~~~~~~~~~~~~~~~~~~~~~~` 但从t=0到t=-1也是逆推啊。两者有何不同？而且，实际上，说t=2时的状态，其实已经暗含了已知t=0时的状态。若真的只是知道t=2时的状态，那么从t=2逆推到t=1仍是做不到的。 我想，也许有以下几种可能： 1、初态是可以任意给定的，但中间态不是。一个态要想成为中间态也许要满足一定的条件。这样，当任意给定一中间态时，它不一定有原因。 2、一个中间态可能对应几个初态，即有几个原因。这样也能导致不能逆推。 但从数学上想了一下，好像这两条都不对。 中国是一个从上往下煽耳光,从下往上磕头的社会。
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卢昌海 发表文章数: 768
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What kind of initial condition are you using at t=0? What is the solution given by the method you mentioned?
宠辱不惊，看庭前花开花落
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Night  发表文章数: 15
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quote "为什么只能由原因找到结果，而不能从结果找到原因呢"
热传导方程的具体形式记得不清了 但是应该可以从———拿你举的情况来说吧——— t=2的时刻推出t=0的具体分布函数吧 。 按照数学物理方程的常用解法， 给定t=2的两个初始条件 能够唯一的得到场关于时间 位置的分布函数 然后 只要这个函数取t=0就能唯一的得到t=0时刻分布函数 也就是在这个例子中 我觉得能从结果中找到原因 决定论和时间反演是两个不同的概念吧 决定论的不一定是时间反演的 时间反演的也不一定是经典意义上的决定论的 。
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星空浩淼 发表文章数: 799
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谢谢各位大牛的回复。
－－－－－－－－－－－－－ 郑重声明一下：昌海兄和sage兄是大牛，本人只是一只小绵羊，与他们二位不可同日而语:-) One may view the world with the p-eye and one may view it with the q-eye but if one opens both eyes simultaneously then one gets crazy
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blackhole 发表文章数: 196
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What kind of initial condition are you using at t=0?
~~~~~~~~~~~~~~~~~~~~~~~~~~~ arbitrary, of course. Except for some abnormal case. What is the solution given by the method you mentioned? ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ There are two kinds of solutions. One is the series solution, in which each term contains something like Exp[-k t] (k=0,1,2,3...). Obviusly, if t<0, then Exp[-k t] will grow unlimitedly as k increases. The other one uses the fact that the solution can be put into a function of x/Sqrt[t] only and has the form of an integration with upper bound is somthing like x/Sqrt[t]. This means this solution doesn't apply for the case of t<0. And I tried the latter case, but could not find a similar solution. 但是应该可以从———拿你举的情况来说吧——— t=2的时刻推出t=0的具体分布函数吧 。 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 前面说了，能够这样的前提其实是先知道了t=0时状态。如果真的只知道t=2时的状态，无法由方程得到t=0时的状态。不信可以试试。这正是我感到非常困惑的原因。 中国是一个从上往下煽耳光,从下往上磕头的社会。
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卢昌海 发表文章数: 768
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The integration-type of solution - I may remember this wrong - might come from the use of a Green function. That can't be extended to t<0 because delta source has been used in the derivation.
I suggest you find a special case when you can have a closed form solution as oppose to a series or integration. Then check whether it can be extended to t<0 (in case it can't, you might be able to see the reason clearly because you have an explicit solution). A series or integration may fail in a region where actual solution still exists. 宠辱不惊，看庭前花开花落
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blackhole 发表文章数: 196
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The integration-type of solution might come from the use of a Green function.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Yes, it is connected with Green function. But it can be obtained from the view as I stated. It is something like general solution. As we know, the general solution for wave equation is f(x+at)+g(x-at). And later I know, for Laplace equation (1-d), there is also a general solution, which is f(x+iat)+g(x-iat). I suggest you find a special case when you can have a closed form solution as oppose to a series or integration. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ The reason why I find the solution of the second type is just that I want to find a closed form solution. And I didn't find other kind of solution. In fact, there does be some closed form solutions for some special cases, which are just the series solution with finite terms. This demand that the initial distribution must have the Fourier expansion with finite terms. It seems that only in these special cases can we let t<0. 中国是一个从上往下煽耳光,从下往上磕头的社会。
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卢昌海 发表文章数: 768
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There is nothing intrinsically special for t=0, the limitation of the solution you see in the textbooks is artificial (unless one introduces unrealistic initial condition at t=0).
For instance, you can use the time translation invariance of the heat conduction equation (homogeneous one) to find solution for t<0: Suppose T(t, x) is the general solution you got from those textbooks, instead of let T(0, x) matches the initial condition (suppose it's a realistic condition, not a singular one such as delta function), let T(a, x) matches it. Then the solution can be used to calculate for interval (-a, 0), as long as physical conditions permit. 宠辱不惊，看庭前花开花落
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blackhole 发表文章数: 196
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There is nothing intrinsically special for t=0,
～～～～～～～～～～～～～～～～～ Of course. But it does be special for "initial" when heat conduction equation is concerned, although it is not special for the other two kinds of equations. This is the fact, and I don't know why. Then the solution can be used to calculate for interval (-a, 0) ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ If so, you still can't know the situation when t<-a. The problem is not solved but just shifted. 中国是一个从上往下煽耳光,从下往上磕头的社会。
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bamboo 发表文章数: 9
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正时间方向的热传导方程的解若往以前外推，则在一般的情况下，方程的解会在负的有限时间内出现奇点。例如给定时间t=0时的某一初始非均匀温度分布，那么热传导方程的解f(t,x)在t=-T时的某些点x=a会出现|f(-T,a)|=∞。
用常微分方程作比较会更容易明白。方程y’+y^2=0从表面看不出有奇点，加上初始条件y(0)=1/b>0后，解是y(t)=1/(t+b)，其定义域向后可以到t=+∞，但向前只能到t>-b。 换言之，用热传导方程「倒果求因」会得到一个包含奇点的「原因」。从物理的观点看，奇点的出现意味这不可能是物理状况的完全描述。例如说，若解的奇点在t=-1hr出现，那么我们根本不能用热传导方程逆推t=-2hr的情况。
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卢昌海 发表文章数: 768
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:: 正时间方向的热传导方程的解若往以前外推，则在一般的情况下，
:: 方程的解会在负的有限时间内出现奇点 This is a very good point (although I no longer remember whether the statement itself is correct or not - haven't touched this equation for more than 10 years :-). If temperature goes to infinite at a certain time, that's the case when physical conditions no longer permit backward extension (of course, heat conduction equation itself also no longer provide a good description of the phenomena when one moves close to that point). 宠辱不惊，看庭前花开花落
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blackhole 发表文章数: 196
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bamboo的说法有新意，但对于热传导方程，有以下问题：
1、不能逆推的点不是由方程本身决定的某种奇点，而是人为给定的初始点。用你第一段的例子来说，并非需要到了-T才不能逆推，而是在t=0时就已经不能逆推了。 2、热传导方程是线性方程，不会有奇点或类似于奇点的东西存在。 中国是一个从上往下煽耳光,从下往上磕头的社会。
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萍踪浪迹  发表文章数: 1051
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本人只是一只小绵羊，与他们二位不可同日而语:-)
============================================ 嘿嘿，有装嫩嫌疑了 漫漫长夜不知晓 日落云寒苦终宵
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卢昌海 发表文章数: 768
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:: the series solution, in which each term contains something
:: like Exp[-k t] (k=0,1,2,3...). Obviusly, if t<0, then Exp[-k t] :: will grow unlimitedly as k increases. Why would that be a problem? That is nothing but a x^k type of series with x=e^(-t)>1 for t<0. Whether such a power series divergent or not depends on coefficients. Can you show me why you think it must divergent? 宠辱不惊，看庭前花开花落
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blackhole 发表文章数: 196
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That is nothing but a x^k type of series with x=e^(-t)>1 for t<0. Whether such a power series divergent or not depends on coefficients.
～～～～～～～～～～～～～～～～～～～～～～～～～～～～～～ Yes, you are right. I never realize this before. Now I can say that the initial condition determines the coefficients, and the coefficients determine the covergence range of x=e^(-t)>1, which defines the reasonable range of t. This means that time can go back to some extends in most cases. That's good and resolves my problem partially. So there does exist some singularity for heat conduction equation which depends on the initial condition. 中国是一个从上往下煽耳光,从下往上磕头的社会。
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blackhole 发表文章数: 196
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Sorry, I made a mistake. Each term in the series solution contains something like Exp[-k^2 t], not Exp[-k t]. This changes something but not basically.
Another problem arises. If the coefficients can give the sigularity at t=-b<0 (the range is -b<t<infinity) in some cases, they can also give sigularity at t=d>0 (d<t<infinity) in some other cases. This means this situation could happen: given the initial condition at t=0, the time between 0 and d is not physical; only later time is reasonable. But this CAN NOT happen! 中国是一个从上往下煽耳光,从下往上磕头的社会。
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卢昌海 发表文章数: 768
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:: If the coefficients can give the sigularity at t=-b<0 (the
:: range is -b<t<infinity) in some cases, they can also give :: sigularity at t=d>0 (d Why? 宠辱不惊，看庭前花开花落
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blackhole 发表文章数: 196
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Because the set of coefficients dtermined by initial condition is arbitrary in some sense. (In what sense I don't figure out, and I'll talk aout it later.) So the location of sigularity, if really exists, should also be arbitrary.
中国是一个从上往下煽耳光,从下往上磕头的社会。
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blackhole 发表文章数: 196
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In other words, even if x=e^(-t)<1, there also is possibility for the series to be divergent.
Now another issue. Though the initial distribution is arbitrary, its Fourier coefficients are not. Because an arbitrary triangular series need not to be a Fourier series of some function, just like an arbitrary power series need not to be a Taylor series of some function. This may confine the coefficients. I guess that this confinement may result in that the series solution must be convergent for any x=e^(-t)<1 or t>0. As for the case of x>1 (t<0), I don't know how to deal, because the convergence of triangular series is far more complicated than that of power series and is far less talked about. 中国是一个从上往下煽耳光,从下往上磕头的社会。
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卢昌海 发表文章数: 768
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:: Because the set of coefficients dtermined by initial condition is
:: arbitrary in some sense. No, they are not - at least not in the sense that they can lead to arbitrary location of singularity. I will leave it for you to figure out why yourself. 宠辱不惊，看庭前花开花落
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Re: 关于热传导方程