中国学人，到了研究生以上，很多都要看英文原著，究竟能看懂多少，存疑，我认为，除非英语特别的N次方牛，否则，不一句一句翻，很难真懂，就算今天懂了，明天基本也忘了。缺20年的基础。英语+数学物理，叠加共振，会让国人感觉很痛苦。因为，中国人是在滚铅球，老外是在滚雪球。我有个帖子，分析过这个事情：
把脉双旗镇：论中国发展的最大瓶颈，恐为英语https://mp.weixin.qq.com/s/dhQ4oZXvrWWvQufVGW9NvA

Our first statement was that
我们的第一个声明就是：
L(x+y)= L(x)+ L (y), (25.3)
which of course follows from the fact that a(x+y)=ax+ay, d(x+y)/dt=dx/dt+dy/dt, etc.
它当然遵循着以下事实：a(x+y)=ax+ay , d(x+y)/dt=dx/dt+dy/dt,等等。
Our second statement was, for constant a,
我们的第二个声明就是，对于常数a：
L (ax)=a L (x). (25.4)
[Actually, (25.3)and (25.4)are very closely related, because if we put x+x into (25.3),this is the same as setting a=2 in (25.4),and so on.]
实际上，（25.3）与（25.4），关系密切，因为，如果我们把x+x带入（25.3），则与在（25.4）中，设a=2，是一样的。

In more complicated problems, there may bemore derivatives, and more terms in L ; the question of interest is whether the two equations (25.3)and (25.4)are maintained or not. If they are, we call such a problem a linearproblem. In this chapter we shall discuss some of the properties that existbecause the system is linear, to appreciate the generality of some of theresults that we have obtained in our special analysis of a special equation.
在更复杂的问题中，可能会有更多的导数，L中的项，也可能更多；我们感兴趣的问题就是，两个方程（25.3）和（25.4）是否能维持？如果它们能维持，我们就称这种问题，为线性问题。有些属性，只存在于线性系统中，这一章，我们将讨论它们，目的是，把我们在对特殊方程的特殊分析中，所得到的一些结果，普遍化。

Now let us study some of the properties oflinear differential equations, having illustrated them already with thespecific equation (25.1)that we have studied so closely. The first property of interest is this: supposethat we have to solve the differential equation for a transient, the freeoscillation with no driving force. That is, we want to solve
现在，让我们研究线性方程的一些属性，我们已经非常仔细地研究过具体方程（25.1），在其中，对这些属性，我们已经说明过了。第一个感兴趣的属性就是这个：假设我们为了一个瞬态，一个没有任何外力的自由震荡，必须解微分方程。也就是说，我们想解方程：
L (x)=0. (25.5)
Suppose that, by some hook or crook, we have found a particularsolution, which we shall call x1 . That is, we have an x1 for which L (x1)=0 . Now we notice that ax1 is also a solution to the same equation; we can multiply this specialsolution by any constant whatever, and get a new solution. In other words, ifwe had a motion of a certain “size,” then a motion twice as “big” is again asolution. Proof: L (ax1)= a L (x1)= a⋅0=0 .
假设通过某些钩子或骗子{？}，我们已经找到了具体的解，我们称其为x1。也就是说，我们有x1，满足L (x1)=0。现在，我们注意到，ax1也是同一方程的解；我们可以用任何常数，乘以这个特殊的解，就可得到一个新的解。换句话说，如果我们有一个确定“大小”的运动，那么，一个大小是其一倍的运动，就又是一个解。证明: L (a x1)=a L (x1)= a⋅0=0 .

Next, suppose that, by hook or by crook, wehave not only found one solution x1 , but also another solution, x2 . (Remember that when we substituted x=eiαt for finding the transients, we found two values for α , that is, two solutions, x1 and x2 .) Now let us show that the combination (x1+x2)is also a solution. In other words, if we put x=x1+x2, x is again a solution of the equation. Why? Because, if L (x1)=0and L (x2)=0 , then L (x1+x2)=L (x1)+ L (x2)=0+0=0 . So if we have found a number of solutions for the motion of a linearsystem we can add them together.
接下来，假设通过钩子或骗子{？}，不仅找到了一个解 x1，还找到了另一个解x2。(还记得吧，当我们为了找到瞬态，代换x=eiαt时，我们为α，找到了两个值，也就是说，两个解，x1和 x2 。)现在，让我们指出，组合(x1+x2)，也是一个解。换句话说，如果我们设x=x1+x2，那么，x就又是方程的一个解。为什么？因为，如果L (x1)=0 及L (x2)=0 , 那么L (x1+x2)= L (x1)+L (x2)=0+0=0 。所以，如果对于一个线性系统的运动，我们找到了若干解，那么，我们可以把它们加起来。

Combining these two ideas, we see, ofcourse, that we can also add six of one and two of the other: if x1is a solution, so is αx1 . Therefore any sum of these two solutions, such as (αx1+βx2), is also a solution. If we happen to be able to find three solutions,then we find that any combination of the three solutions is again a solution,and so on. It turns out that the number of what we call independentsolutions1that we have obtained for our oscillator problem is only two. The numberof independent solutions that one finds in the general case depends upon whatis called the number of degrees of freedom. We shall not discuss this indetail now, but if we have a second-order differential equation, there are onlytwo independent solutions, and we have found both of them; so we have the mostgeneral solution.
把这两个想法，组合在一起，当然我们就可以看到，我们可以把一个或两个加起来{？}：如果x1是一个解，那么，αx1也是。因此，任何这两个解的和，例如 (αx1+βx2)，也是一个解。如果碰巧，我们找到了三个解，那么，我们发现，这三个解的任意组合，又是一个解，如此等等。结果就是，我们为震荡问题找到的、被称为独立解（脚注1）的个数，只有两个。在普遍情况中，人们所能发现的独立解的个数，依赖于所谓的自由度的数目。现在，我们将不仔细地讨论它，但是，如果我们有一个二阶微分方程，那么，将只有两个独立的解，我们已经发现了它们；于是，我们就有了最普遍的解{？}。

Now let us go on to another proposition,which applies to the situation in which the system is subjected to an outsideforce. Suppose the equation is
现在，让我们看另外一个命题，它适用的情况为：系统从属于一个外力。假设方程为：
L (x)=F(t), (25.6)
and suppose that we have found a special solution of it. Let us say thatJoe’s solution is xJ , and that L (xJ)=F(t) . Suppose we want to find yet another solution; suppose we add toJoe’s solution one of those that was a solution of the free equation (25.5),say x1 . Then we see by (25.3)that
并且假设，我们为它找到了一个特殊的解。让我们说，Joe的解是xJ , 且 L (xJ)=F(t)。假设我们想找到另一个解；假设我们给Joe的解，加上自由方程（25.5）的一个解，比如说x1。那么，通过(25.3)，我们看到：
L (xJ+x1)=L (xJ)+ L (x1)=F(t)+0=F(t). (25.7)
Therefore, to the “forced” solution we can add any “free” solution,and we still have a solution. The free solution is called a transientsolution.
因此，我们可以给“强迫的”解。加上一个“自由的”解，然后，我们仍有一个解。这个自由的解，就被称为瞬态解。

When we have no force acting, and suddenlyturn one on, we do not immediately get the steady solution that we solved forwith the sine wave solution, but for a while there is a transient which sooneror later dies out, if we wait long enough. The “forced” solution does not dieout, since it keeps on being driven by the force. Ultimately, for long periodsof time, the solution is unique, but initially the motions are different fordifferent circumstances, depending on how the system was started.
当没有力作用时，然后突然的加一个力，我们不会立即得到稳定解，就像我们为正弦波所做的那样，但是，过一会儿，就会有一个瞬态，如果我们等的时间足够长的话，那么，它或迟或早会消失。“强迫的”解，并不会衰减，由于它一直被力所驱动。最终，对长时间而言，解就是唯一的，但是，对于不同的情形，最初的运动，是不一样的，这依赖于系统是如何启动的。

25–2Superposition of solutions 25-2 解的叠加
Now we come to another interestingproposition. Suppose that we have a certain particular driving force Fa(let us say an oscillatory one with a certain ω=ωa, but our conclusions will be true for any functional form of Fa) and we have solved for the forced motion (with or without thetransients; it makes no difference). Now suppose some other force is acting,let us say Fb , and we solve the same problem, but for this different force. Thensuppose someone comes along and says, “I have a new problem for you to solve; Ihave the force Fa+Fb
.” Can we do it? Of course we can do it, because the solution is thesum of the two solutions xa and xb for the forces taken separately—a most remarkable circumstance indeed.If we use (25.3),we see that
现在，我们来到另外一个有趣的命题。假设我们有某个具体的驱动力 Fa，（让我们说，一个震荡，有确定的ω=ωa，但是，我们的结论，对于Fa的任何函数形式，都为真。）且我们为强制的运动（有或没有瞬态，没有区别），解了方程。现在，假设有另外的一个力,比如说叫Fb，在起作用，且我们为这个不同的力，解了同样的问题。然后，假设有人过来说：“我有一个新的问题，让你们解，我有力 Fa+Fb”。这个我们能做吗？当然，我们能做，因为，解就是两个解xa和 xb的和，它们分别是当力单独起作用时的解--这确实是一个最值得说明的情形。如果我们利用（25.3），我们就看到:
L (xa+xb)= L (xa)+L (xb)=Fa (t)+Fb(t). (25.8)

This is an example of what is called the principle of superpositionfor linear systems, and it is very important. It means the following: if wehave a complicated force which can be broken up in any convenient manner into asum of separate pieces, each of which is in some way simple, in the sense thatfor each special piece into which we have divided the force we can solve theequation, then the answer is available for the whole force, because wemay simply add the pieces of the solution back together, in the samemanner as the total force is compounded out of pieces (Fig. 25–1).
对于线性系统，有所谓的叠加原理，这就是该原理的一个例子，它很重要。其意思为：如果我们有一个复杂的力，它可以用任何方便的方式，被分解成几个分力之和，每个分力，从某种方式看，都是简单的，意义为，对我们所分成的每一个特殊的分力，我们都可以用它来解此方程，因此，对于整个力的答案，就可以得到了，因为，我们只需简单地把各个分力的解，加起来，就像总力，是由各个分力，组合而成一样（图25-1）。

Fig. 25–1.An example of the principle ofsuperposition for linear systems. 图25-1 一个线性系统的叠加原理的例子。

Let us give another example of the principleof superposition. In Chapter 12we said that it was one of the great facts of the laws of electricity that ifwe have a certain distribution of charges qa and calculate the electric field Eaarising from these charges at a certain place P , and if, on the other hand, we have another set of charges qband we calculate the field Eb due to these at the corresponding place, then if both chargedistributions are present at the same time, the field E at P is the sum of Ea due to one set plus Eb due to the other. In other words, if we know the field due to acertain charge, then the field due to many charges is merely the vector sum ofthe fields of these charges taken individually. This is exactly analogous tothe above proposition that if we know the result of two given forces taken atone time, then if the force is considered as a sum of them, the response is asum of the corresponding individual responses.
关于叠加原理，我们再给一个例子。在第12章，我们说过，电的规律的伟大的事实之一，就是如果我们有一个确定的电荷qa的分布，并计算出了，它在某个位置P所产生的电场Ea，另一方面，如果我们有另外一组电荷qb，根据它，我们可以计算出，在相应位置所产生的电场Eb，因此，如果两个电荷的分布，在同一时间在场，那么，P处的电场E，就是Ea与Eb之和。换句话说，如果我们知道，由不同的电荷所产生的场，那么，由很多电荷所产生的场，就是这些电荷分别所产生的场的矢量和。这可以与上面的命题，准确类比，即如果我们知道了，在同一时间被给予的两个力的结果，那么，如果总力，被考虑为是两个力的和，那么，总的反应，就是相应的分别反应之和。

Fig. 25–2.The principle of superposition inelectrostatics. 图25-2 静电学中的叠加原理。

The reason why this is true in electricityis that the great laws of electricity, Maxwell’s equations, which determine theelectric field, turn out to be differential equations which are linear,i.e., which have the property (25.3).What corresponds to the force is the charge generating the electricfield, and the equation which determines the electric field in terms of thecharge is linear.
在电学中，为什么这一点为真，原因就是，电学的伟大规律，麦克斯尔方程，决定了电场，此规律，结果也是线性的微分方程，亦即，它们具有（25.3）的属性。与力相应的，是电荷，电荷生了电场，规定电场的方程，是用电荷来描述的，这些方程，是线性的。

As another interesting example of thisproposition, let us ask how it is possible to “tune in” to a particular radiostation at the same time as all the radio stations are broadcasting. The radiostation transmits, fundamentally, an oscillating electric field of very highfrequency which acts on our radio antenna. 作为这个命题的另外一个有趣的例子，让我们追问，当所有广播电台都在广播时，如何可以“调谐”到某一个电台呢？从根本上说，电台在发射一个非常高频的振荡电场，它会作用于我们收音机的天线。It is true that the amplitude of the oscillation of the field ischanged, modulated, to carry the signal of the voice, but that is very slow,and we are not going to worry about it. When one hears “This station isbroadcasting at a frequency of 780 kilocycles,” this indicates that 780,000 oscillations per second is the frequency of the electric fieldof the station antenna, and this drives the electrons up and down at thatfrequency in our antenna. 确实，此电场的震荡的波幅，被改变了，调制了，以承载声音的信号，但是，这是很慢的，我们无需担忧此事。当一个人听到：“这个电台，正在广播，频率为780千赫”，这就表示，此电台天线的电场的频率，是每秒震荡780，000次，它驱动着电子，以这个频率，在我们的天线中，上下震荡。Now at the same time we may have another radio station in the sametown radiating at a different frequency, say 550 kilocycles per second; then the electrons in our antenna arealso being driven by that frequency. Now the question is, how is it that we canseparate the signals coming into the one radio at 780 kilocycles from those coming in at 550 kilocycles? We certainly do not hear both stations at the sametime. 现在，在同一时间，在同一城市，我们可以有另外一家电台，发射着不同的频率，比如说550千赫每秒；那么，我们天线中的电子，也被这个频率驱动着。现在的问题就是，来到一个收音机的信号，既有780千赫，又有550千赫，我们如何区分它们呢？我们当然并没有同时听到两个电台。

Now what about the tuning? How do we tuneit? We change ω0 by changing the L or the C of the circuit, because the frequency of the circuit has to do with thecombination of L and C . In particular, most radios are built so that one can change thecapacitance. When we retune the radio, we can make a new setting of the dial,so that the natural frequency of the circuit is shifted, say, to ωc. In those circumstances we hear neither one station nor the other; weget silence, provided there is no other station at frequency ωc. If we keep on changing the capacitance until the resonance curve isat ωb , then of course we hear the other station. That is how radio tuningworks; it is again the principle of superposition, combined with a resonantresponse.2
现在，调台又是怎么回事呢？我们如何调台的呢？我们通过改变电路的L或C，来改变ω0，因为，电路的频率，与L和C的组合有关。尤其是，大部分收音机的电容可变。当我们重新调收音机时，我们可以造成新的拨号设置，这样电路的自然频率，就被漂移到了ωc，比如说。在这些情形下，假设在频率ωc，没有其他电台，我们一个台也收不到，我们的得到的，就是静音。如果我们继续改变电容，直到共振曲线在ωb，这时，当然，我们就能听其他台了。这就是收音机调台如何工作的；它又是叠加原理，与共振反应的组合（脚注2）。