To Mary Somerville   [8 March 1831]

Slough March 8 1831

Dear Madam,

If you have any doubt about Pontecoulant’s method of indeterminate coefficients, (I have not his book), your better way will be to abandon it altogether and throw yourself on the broad principles of Laplace as laid down in N^o43 of the 2^nd book of the Mec. Cel. , or as he has explained it rather more clearly I think, than in the condensed and general form in which it is stated in the Mec. Cel. in the Mem Acad. Sci. 1772 P373 in a very beautiful memoir, which I would recommend you to look at.

The outline of the principle is this. — Direct approximation leads to an integral consisting of terms of 2 kinds, periodical and algebraic — the former of the form $sin$ or $cos(A+Bt)$ The latter of $t,t^2$ &c But never any term of the form $sin$, $cos(A+Bt+C{t}^2)$ (If such a term as the latter could occur, the method would be inapplicable, but by tracing the approximation step by step we shall easily see that no such term can arise. This ought to be very plainly stated in illustrating the principles to elementary readers).

By thus having one or two steps of approx[imatio]n we discover the General form of the integral of $\frac{d^{i}y}{d{t}^i}+P+\alpha Q=0$ to be¹ $$y=\&c+(A+Bt+C{t}^2+\&c)\cdot sin[cos](m+NT)+\&c$$

Now therefore let such a general series be substituted for $y$ and Since $t$ only enters in the periodical terms of the 1^st degree, no power of $t$ can be produced by differentiation but what arose from parts without the $sin$ & $cos$ nor can any of the terms out of the [illegible] $sin$, $cos$ be ever introduced into them by such an operation. Consequently, in this substitution we may (for a moment) regard the $t$ out of and the $t$ in the sines & cosines as independent and denote them by different letters such as $\nu$ and $t$ and then we have $$y=\sum (A+B\nu +C{\nu }^2+\&c)\times sin[cos](m+nt)$$ [Where $\frac{dt}{dt}=\frac{d\nu }{dt}$ ].

2) Now this substitution being made as above in $\frac{{\partial }^{i}y}{\partial t^i}+P+\alpha Q=0$ an equation of the form $0=k+k\mathrm{\&\#x2032;}\nu +k\mathrm{\&\#x2033;}{\nu }^2+\&c$ must evidently arise, where $k,k\mathrm{\&\#x2032;},\&c$ are terms composed of $sin[cos](m+NT)\&c$ and it is evident that if we can determine the assumed series so as to make $k=0,k\mathrm{\&\#x2032;}=0,k\mathrm{\&\#x2033;}=0$ &c all will be right and $\nu$ will go out.

Instead of putting $y=\sum (A+Bt+C{t}^2...)\cdot sin[cos](m+Nt)$ we may write it as Laplace does $y=X+tY+t^{2}Z+\&c$ where $X,Y,Z$ are periodic terms of the form $sin$ or $cos(m+NT)$ or series of such terms, and the same reasoning still subsists — when substituted the result will still be an equation such as $k+k\mathrm{\&\#x2032;}\nu +k\mathrm{\&\#x2033;}{\nu }^2+\&c=0$ and we may then suppose $y=X+\nu Y+{\nu }^{2}Z+\&c$ provided after the differentiations we make $\partial \nu =\partial t$. Now this comes to the same as supposing $\nu =t+const$ or $(t-\theta )$ and thus if $y=X+tY+t^{2}Z+\&c$ satisfy the eqn then $y=X+(t-\theta )Y+{(t-\theta )}^{2}Z+\&c$ will satisfy it also

The next step is this. $\theta$ is an arbitrary constant. But by supposition $y=X+tY+t^{2}Z+\&c$ is the complete integral — therefore the zero arbitrary constant $\theta$ is only apparently so — therefore it must be a function of the other arbitraries $c,c\mathrm{\&\#x2032;},c\mathrm{\&\#x2033;}$ &c contained in $y$ — Therefore, reciprocally, if $X+(t-\theta )Y+{(t-\theta )}^{2}Z+\&c$ be capable of being regarded as a transformation of $X+tY+t\mathrm{\&\#x2032;}Z+\&c$, then $c,c\mathrm{\&\#x2032;},\&c$ must be certain functions of $\theta$, and $X,Y,Z$ being functions of $c,c\mathrm{\&\#x2032;},\&c$ must also be functions of $\theta$ — such functions may always found, or supposed.

It is therefore [illegible] to assume for $c,c\mathrm{\&\#x2032;},\&c$ such functions of $\theta$ as shall make $X+Y(t-\theta )+Z{(t-\theta )}^2+\&c$ a legitimate transformation of $X+Yt+Z{t}^2+\&c$.

3rd step Now if this transformation be generally possible as an algebraic truth it will be still true (being independent of $\theta$’s particular value) when $\theta =t$ & in that case we have $y=X$; observing that in this expression $X$ which is a function of $c,c\mathrm{\&\#x2032;},\&c$ and thereby a function of $\theta$ becomes now a function of $t$ of a quite different nature from what it was before by writing $t$ for $\theta$.

If then the nature of $X$ regarded as a function of $\theta$, as well as of periodical functions of $t$ can be found — such as $X=\varphi (t,\theta )$ then will the [illegible] value of $y$ be $y=\varphi (t,t)$.

The problem is reduced to this

$XYZ\&c$ & are functions of $cc\mathrm{\&\#x2032;}c\mathrm{\&\#x2033;}\&c$ of a given form $cc\mathrm{\&\#x2032;}c\mathrm{\&\#x2033;}\&$ are functions of $\theta$ of an unknown form (to be found by the condition that $$X+tY+t^{2}Z\&c=X+(t-\theta )Y+{(t-\theta )}^{2}Z+\&c$$ is an equation to be verified independent of particular values of $t$ & $\theta$. Req[d] the forms of $X,c,c\mathrm{\&\#x2032;},c\mathrm{\&\#x2033;}\&c$ in $\theta ,t$

now it will be observed that this equation being to be made identical in $\theta$, the 1^st member does not explicitly contain it, and therefore the 2^d when developed in $\theta$ may have all the differential coefficients $=0$ to that we see developing

$$\begin{array}{c}\hfill X+Yt+Z{t}^2+\&c=X+Yt+Z{t}^2+\&c+\hfill \\ \hfill \theta \{\frac{dX}{d\theta }-Y\}+{\theta }^2\{\frac{d^{2}X}{1\cdot 2\cdot d{\theta }^2}-\frac{dY}{1\cdot dt}+Z\}+\&c\hfill \end{array}$$

Comparing Like terms and destroying the powers of $\theta$ we see $Y=\frac{dX}{d\theta }$ ; whence $\frac{\partial Y}{\partial t}=\frac{d^{2}X}{d{\theta }^2}$ so that $2Z=\frac{dY}{d\theta }$ , &c which are the same equations Laplace gets by a somewhat different consideration.

Step 5. Thus $Y,Z,\&c$ are given in functions of $\theta$ when $X$ is once so expressed, and its differential coefficients. These diff[erentia]l coefficients however it will be observed will involve terms arising from $c,c\mathrm{\&\#x2032;},c\mathrm{\&\#x2033;},\&c$ as follows $$\frac{dX}{d\theta }=\frac{dX}{dc}\cdot \frac{dc}{d\theta }+\frac{dX}{dc\mathrm{\&\#x2032;}}\cdot \frac{dc\mathrm{\&\#x2032;}}{d\theta }+\&c$$

$$\frac{dY}{d\theta }=\frac{dY}{dc}\cdot \frac{dc}{d\theta }+\frac{dY}{dc\mathrm{\&\#x2032;}}\cdot \frac{dc\mathrm{\&\#x2032;}}{d\theta }+\&c$$

Now if any of the of the $c$ enter into a periodic term as a constant as for instance in the term $cos$ or $sin(ct+const)$ and if $X,Y,\&c$ either of them contain such a term, there will arise a term in $\frac{dX}{d\theta }$ of the form $t\times sin\mathrm{\text{or}}cos(ct+const)\times \frac{dc}{d\theta }$ containing the arc $t$ and therefore the general form of $\frac{dX}{d\theta }$ will be $$\frac{dX}{d\theta }=X\mathrm{\&\#x2032;}+T\cdot X\mathrm{\&\#x2033;}$$ and so of the rest.

So the equations determining the identity in question take the form

$$Y=X\mathrm{\&\#x2032;}+\theta \cdot X\mathrm{\&\#x2033;}$$ $$2Z=Y\mathrm{\&\#x2032;}+\theta \cdot Y\mathrm{\&\#x2033;}+X\mathrm{\&\#x2033;}$$

(for the way in which $\theta$ gets in here instead of $t$ see Laplace. There is no difficulty about it).

Step 6. In these $X\mathrm{\&\#x2032;},X\mathrm{\&\#x2033;},Y\mathrm{\&\#x2032;},Y\mathrm{\&\#x2033;},\&c$ are composed entirely of periodic terms in $t$ and linear terms of the form $\frac{dc}{d\theta },\frac{dc\mathrm{\&\#x2032;}}{d\theta \mathrm{\&\#x2032;}},\&c$

Now the object is 1^st to get the values of these & thence by substitution in $X,Y,\&c$ those of the latter in terms of $\theta$. To this end, we observe, 1^st that $t$ and $\theta$ are independent 2^nd that each equation is linear in $\frac{dc}{\partial \theta },\&c\&\theta$ so that they are in fact a system of simultaneous linear differential equations in $c,c\mathrm{\&\#x2032;},c\mathrm{\&\#x2033;},\&c$ by whose integration (regarding $t$ as arbitrary & independent and treating $c,c\mathrm{\&\#x2032;},\&c$ and $\theta$ as the variables. These may be had moreover since $t$ is arbitrary any of these equations may be differentiated with regard to $t$ alone, & thus equations of greater simplicity obtained.

I think this is about the [illegible] view of Laplace’s method, but it is delicate & should be explained in full detail. It is extraordinary to me how I have forgot these things. I was once very familiar with this part of the Mec. Cel. — and yet I always found some thing catching in the reasoning & I know by experience how easy it is to make a slip in it.

I am much flattered by your & Sir J Macktintosh’s approbation of my book — one word of such praise is worth a whole volume of newspaper & magazine puffering with which it has been treated & which really had made me almost feel that I must have written something very foolish. But you reassure me. With comp[limen]ts to Dr S in which Mrs H joins, I am, dear Madam,

J F W Herschel