A study of Euler’s untranslated paper `De integratione aequationum differentialium’

UntitledThe following are some translations and annotations I made while studying a previously unstranslated paper in Latin by Leonhard Euler (1707-1783) entitled De integratione aequationum differentialium (On the integration of differential equations), 1763. (Eneström index number E269, obtained from the Euler archive). This contains, among other things, some interesting treatments of linear first-order equations as well as nonlinear first-order equations such as Bernoulli’s equation and Riccati’s equation. I was curious to see what Euler had to say about these, particularly Riccati’s equation. (Jacopo Francesco Riccati (1676-1754) was a Venetian mathematician who introduced an important class of nonlinear first-order differential equations which now bear his name, and which Euler investigated in greater depth).

To begin with, I will translate the first paragraph of the introduction to the paper as it gives a nice overview of the central idea as well as giving a flavour of Euler’s writing style:

1.

I consider herein differential equations of first order, involving only two variables, which one may represent in the general form M dx + N dy = 0, where M and N denote any functions of a pair of variables x and y. It is demonstrated that such equations always imply a certain relationship between the variables x and y, such that for any given value of one of them, the values for the other can be determined. Since moreover this definite relationship between the two variables is to be found through integration, the integral equation, regarded in its greatest generality, will accommodate a new constant quantity which provided that it depends completely on our choice enables as it were an infinite number of integral equations to be encompassed, which should all be equally compatible with the differential equation. 

Mathematical comments: What Euler is essentially saying in this first paragraph is that he is going to consider methods for integrating first-order differential equations of the form M(x, y) dx + N(x, y) dy = 0 in order to obtain corresponding relationships between the variables x and y which he calls `integral equations’ (aequationes integrales) of the form V(x, y) = Const. The term on the right-hand side is an arbitrary constant and no matter what value is assigned to this constant, the differential of this `integral equation’ will always be of the same form M dx + N dy = 0. \square

After some further introductory comments, the rest of Euler’s long paper consists of a multitude of problems, theorems, corollaries and examples concerning differential equations of this type. The first problem that Euler considers is important because it more or less sets the scene for the rest of the paper:

Problem 1.

7. Given that the differential equation M dx + N dy = 0 is such that \big(\frac{d M}{d y} \big) = \big(\frac{d N}{d x} \big), find its integral equation.   

Solution.

If it is to be that \big(\frac{d M}{d y} \big) = \big(\frac{d N}{d x} \big), then a function must exist defined over a pair of variables x and y, which when it is differentiated yields M dx + N dy. Let V be this function, and given that dV = M dx + N dy, M dx will be the differential of V when only the variable x is considered, and N dy will be its differential when only the variable y is considered. From here therefore V will be found, if either M dx is integrated with y regarded as a constant, or N dy is integrated with x regarded as a constant: and in this way the task is reduced to the integration of a differential formula involving a single variable, which might be achieved either by algebraic methods, or else quadrature of curves might be needed. But since it is the case that the quantity V can be found in two ways, and one integration treats as a constant any function whatever of y, while the other treats as a constant any function whatever of x, so it may then be that both V = \int M dx + Y and V = \int N dy + X hold, and it is always permissible to define these functions Y of y, and X of x, such that \int M dx + Y = \int N dy + X, and this is in all cases readily guaranteed. When this is done it is evident that since the quantity V is to be the integral of the formula M dx + N dy, the complete integral equation of the proposed equation M dx + N dy = 0 will then be V = Const., as this involves a constant which is dependent on our choice. 

Mathematical comments: The first thing to observe here is that Euler uses the notation \big(\frac{d M}{d y} \big) and \big(\frac{d N}{d x} \big) to represent the partial derivatives \frac{\partial M}{\partial y} and \frac{\partial N}{\partial x}. At the start of the solution he says that if \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} then a function V(x, y) must exist which yields M dx + N dy when it is totally differentiated. The reason for this immediately becomes clear if we totally differentiate V(x, y) to get dV = \frac{\partial V}{\partial x} dx + \frac{\partial V}{\partial y} dy. We see that M = \frac{\partial V}{\partial x} and N = \frac{\partial V}{\partial y} so the condition \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} simply amounts to saying that the cross-partials of V are equal, i.e., \frac{\partial^2 V}{\partial x \partial y} = \frac{\partial^2 V}{\partial y \partial x}. This must be true for any well-behaved function. So if the condition \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} holds it will always be possible to find a function V whose total differential is M dx + N dy. Conversely, if it is not true that \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}, there can be no function whose total differential is M dx + N dy. Euler deals with this latter case in a subsequent problem (see below). Since M = \frac{\partial V}{\partial x} we see that V = \int M dx + Y where Y is any function of y. Similarly, since N = \frac{\partial V}{\partial y} we have that V = \int N dy + X where X is any function of x. Equating these two expressions for V we can then work out what the functions X and Y must be in order to make \int M dx + Y and \int N dy + X mutually compatible. This is essentially the strategy that Euler is advocating here. \square

Euler provides a number of examples to illustrate the above procedure. The following is the first one:

Example 1.

12. Integrate this differential equation:

2axy dx + axx dy - y^3 dx - 3xyy dy = 0

The equation compared with the form M dx + N dy = 0 implies

M = 2axy - y^3 and N = axx - 3xyy.

Then the first thing to be considered is whether in this case the problem can be solved ? to which end we inquire as to the values of:

\big(\frac{d M}{d y}\big) = 2ax - 3yy, and \big(\frac{d N}{d y}\big) = 2ax - 3yy,

which since they are equal, the suggested method must necessarily succeed. Then it will be discovered, with y taken to be constant:

\int M dx = axxy - y^3 x + Y

which is of such a form that if the differential is taken, treating x as a constant, it will yield:

axx dy - 3yyx dy + dY = N dy,

and upon replacing N with its value axx - 3xyy, the result will be dY = 0, from which we get Y = 0, or Y = const. Hence we will have the required integral equation:

axxy - y^3 x = Const.

In the next extract, Euler introduces an integrating factor method to deal with situations in which the condition \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} does not hold:

Theorem.

16. If for the differential equation M dx + N dy = 0 it is not the case that \big(\frac{d M}{d y}\big) = \big(\frac{d N}{d x}\big), a multiplier is always available by means of which multiplication of the formula M dx + N dy will make it become integrable. 

Proof.

Since it is not the case that \big(\frac{d M}{d y}\big) = \big(\frac{d N}{d x}\big), also the formula M dx + N dy will not be integrable, or rather a function of x and y does not exist whose differential is M dx + N dy. In truth however it is not the formula M dx + N dy, but the equation M dx + N dy = 0, whose integral is being sought; and since this equation still stands if it is multiplied by any function L of x and y, so that it may be written L M dx + L N dy = 0, it is demonstrated that it is always possible to find a function L of this kind, such that the formula L M dx + L N dy = 0 will be integrable. For this to come about it will be necessary that:

\big(\frac{d LM}{d y}\big) = \big(\frac{d LN}{d x}\big)

or if it is supposed that dL = P dx + Q dy, then since \big(\frac{d L}{d y}\big) = Q, and \big(\frac{d L}{d x}\big) = P, the function L should be constructed so that:

L \big(\frac{d M}{d y}\big) + M Q = L \big(\frac{d N}{d x}\big) + NP

But it is apparent that this condition is sufficient to define a function L, which when it multiplies the formula M dx + N dy will make it integrable. 

Mathematical comments: What Euler is doing here is deriving a procedure for finding an integrating factor L(x, y) which will make LM dx + LN d y = 0 integrable. If this is integrable there will be a function V(x, y) whose total differential is LM dx + LN d y. But the total differential of V is d V = \frac{\partial V}{\partial x} dx + \frac{\partial V}{\partial y} dy from which we see that

\frac{\partial V}{\partial x} = LM and \frac{\partial V}{\partial y} = LN

Therefore Euler’s condition

\frac{\partial LM}{\partial y} = \frac{\partial LN}{\partial x}

again amounts to saying that the cross-partials of V must be equal. Expanding Euler’s condition we get

L \frac{\partial M}{\partial y} + M \frac{\partial L}{\partial y} = L \frac{\partial N}{\partial x} + N \frac{\partial L}{\partial x}

which when it is assumed that dL = P dx + Q dy we can write as

L \frac{\partial M}{\partial y} + MQ = L \frac{\partial N}{\partial x} + NP

This is the equation at the end of Euler’s proof and we can see that it is nothing more than a statement of the equality of the cross-partials of V. Euler does not explicitly take the final step, saying simply that it is evident from this that the required function L can be defined, but what he means is that L can now be defined in terms of the functions M, N, P and Q as

L = \frac{NP - MQ}{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}

When M dx + N dy is multiplied through by a function L defined in this way, the result will be a differential equation that is integrable. \square

I will now jump ahead in Euler’s paper to an application of the above procedure in the familiar case of a linear first-order differential equation:

Problem 4.

34. If the proposed differential equation is to be 

P dx + Q y dx + R dy = 0

where P, Q and R denote any functions whatever of x, and moreover the other variable y does not have more than one dimension, find the multiplier which makes it integrable. 

Solution.

Comparison of this equation with the form M dx + N dy implies M = P + Qy and N = R, and so we get 

\big(\frac{d M}{d y}\big) = Q and \big(\frac{d N}{d x}\big) = \frac{d R}{d x}

If now L is to be set up as the multiplier in question, let dL = p dx + q dy, and then the equation ought to satisfy

\frac{Np - Mq}{L} = Q - \frac{d R}{d x} = \frac{Rp - (P + Qy)q}{L}

Now since Q - \frac{d R}{d x} is to be a function only of x, L will likewise have to be accepted as a function only of x, so that q = 0 and dL = p dx; whence it will be the case that:

Q - \frac{d R}{d x} = \frac{Rp}{L}, or rather Q dx - dR = \frac{R dL}{L}

and therefore \frac{d L}{L} = \frac{Q dx}{R} - \frac{d R}{R}. Hence upon integration will be obtained lL = \int \frac{Q dx}{R} - lR, and taking e to be the number whose hyperbolic logarithm equals unity, it turns out that 

L = \frac{1}{R} e^{\int \frac{Q dx}{R}}

Then as a result of this discovered multiplier the integral equation will be:

\int \frac{P dx}{R} e^{\int \frac{Q dx}{R}} + y e^{\int \frac{Q dx}{R}} = Const.

Mathematical comments: In view of the previous theorem introducing the integrating factor approach using Euler’s idiosyncratic notation, Euler’s explanation of the solution to this problem is largely self-explanatory. Perhaps only the final step needs clarification. Having found the multiplier L we can multiply the original differential equation by it to get

L P dx + L Q y dx + L R dy = 0

or

\frac{P dx}{R} e^{\int \frac{Q dx}{R}} + \frac{Q dx}{R} y e^{\int \frac{Q dx}{R}} + e^{\int \frac{Q dx}{R}} dy = 0

or

\frac{P dx}{R} e^{\int \frac{Q dx}{R}} + d \big(y e^{\int \frac{Q dx}{R}}\big) = 0

This can now be integrated directly to get the final equation obtained by Euler. Although straightforward, it is interesting to compare Euler’s approach here with the type of argument usually given in modern treatments of linear first-order differential equations. In modern treatments the differential equation would usually be given as

\frac{d y}{d x} + Py = Q

We would then seek an integrating factor L(x) which would enable us to write

\frac{d}{dx}(y L) = QL

and this could then be integrated directly to find y. The unknown function L is found by first expanding the left-hand side to get

L \frac{d y}{d x} + y \frac{d L}{d x} = QL

Dividing through by L we get

\frac{d y}{d x} + \big(\frac{d L/d x}{L}\big)y = Q

Comparing this with the original equation we see that

\frac{1}{L} \frac{d L}{d x} = P

and so

L = e^{\int P dx}

Putting this into \frac{d}{dx}(y L) = QL and integrating directly we then get the required solution

y e^{\int P dx} = \int Q e^{\int P dx} dx + Const.

which is equivalent to the one obtained by Euler. \square

In the next problem Euler considers a nonlinear first-order differential equation which today is known as Bernoulli’s equation:

Problem 5.

37. If the proposed differential equation is to be: 

P y^n dx + Q y dx + R dy = 0

where P, Q and R denote any functions whatever of x, find the multiplier which makes it integrable. 

Solution.

Then it will be that M = P y^n + Qy and N = R, and hence

\big(\frac{d M}{d y}\big) = n P y^{n - 1} + Q, and \big(\frac{d N}{d x}\big) = \frac{d R}{d x}

Therefore assuming that the multiplier in question is L and dL = p dx + q dy, based on our previous findings it will be the case that:

 \frac{Rp - P y^n q - Q y q}{L} = n P y^{n - 1} + Q - \frac{d R}{d x}

If it is imagined that L = S y^m, where S is a function only of x, it will be the case that p = \frac{y^m d S}{d x}, and q = m S y^{m - 1}, so that substituting these values will yield:

\frac{R d S}{S d x} - m P y^{n - 1} - m Q = n P y^{n - 1} + Q - \frac{d R}{d x}

In order for this equation to hold it needs to take m = -n, so it will become

\frac{R d S}{S d x} = (1 - n) Q - \frac{d R}{d x} , or \frac{d S}{S} = \frac{(1 - n) Q dx}{R} - \frac{d R}{R}

Then since integration will yield S = \frac{1}{R} e^{(1 - n) \int \frac{Q dx}{R}}, the required multiplier, with m = -n, will be

L = \frac{y^{-n}}{R} e^{(1 - n) \int \frac{Q dx}{R}}

and the integral equation will be

\frac{y^{1 - n}}{1 - n} e^{(1 - n) \int \frac{Q dx}{R}} + \int \frac{P dx}{R} e^{(1 - n) \int \frac{Q dx}{R}} =  Const.

Mathematical comments: Again in view of the previous material, Euler’s solution to this problem is largely self-explanatory. To get the final equation we multiply the original differential equation by the multiplier to get

L P y^n dx + L Q y dx + L R dy = 0

or

\frac{P dx}{R} e^{(1 - n) \int \frac{Q dx}{R}} + \frac{Q dx}{R} y^{1 - n} e^{(1 - n) \int \frac{Q dx}{R}} + y^{-n} e^{(1 - n) \int \frac{Q dx}{R}} dy = 0

or

\frac{P dx}{R} e^{(1 - n) \int \frac{Q dx}{R}} + d \big(\frac{y^{1 - n}}{1 - n} e^{(1 - n) \int \frac{Q dx}{R}}\big) = 0

and this last equation can now be integrated directly. As with the previous problem, it is interesting to compare Euler’s approach to solving this problem with the way it would be done in modern treatments. Prior to Euler, the equation in this problem was also studied by two Bernoulli brothers, James Bernoulli (1654-1705) and John Bernoulli (1667-1748), as well as by Gottfried Wilhelm Leibniz (1646-1716). Ironically, the modern approach is essentially the one developed by John Bernoulli before Euler tackled it (John Bernoulli was Euler’s tutor). In modern times the equation would usually be written as

\frac{d y}{d x} + P y = Q y^n

and John Bernoulli’s approach was to make the change of variable z = y^{1 - n}. Then

\frac{d z}{d x} = \frac{1 - n}{y^n} \frac{d y}{d x}

and we can rewrite the original equation using this new variable as

\frac{d z}{d x} + (1 - n) P z = (1 - n) Q

This is now a first-order linear equation of the type solved in the previous problem. Following the approach discussed there, we seek a multiplier L which enables us to write the equation as

\frac{d}{dx} (z L) = (1 - n) Q L

Expanding the left-hand side and comparing with the original equation we find that

\frac{1}{L} \frac{d L}{d x} = (1 - n) P

so the multiplier is

L = e^{(1 - n) \int P dx}

Then we can write

\frac{d}{dx}\big(z e^{(1 - n) \int P dx}\big) = (1 - n) Q e^{(1 - n) \int P dx}

This can be integrated directly to get

z e^{(1 - n) \int P dx} = (1 - n) \int Q e^{(1 - n) \int P dx} dx + Const.

or

y^{1-n} e^{(1 - n) \int P dx} = (1 - n) \int Q e^{(1 - n) \int P dx} dx + Const.

which is equivalent to Euler’s solution. \square

The final extract from Euler’s paper that I want to include here is a problem dealing with a nonlinear differential equation now known as Riccati’s equation. The equation was apparently first referred to as `Riccati’s equation’ in 1770 (only a few years after Euler wrote this paper) by the mathematician Jean-Baptiste le Rond d’Alembert (1717-1783):

Problem 9. 

56. For this proposed differential equation:

dy + P y dx + Q y y dx + R dx = 0

where P, Q and R are functions only of x, if it is determined that this equation satisfies y = v where v is a function of x, find the multipliers which make this equation integrable. 

Solution.

Since this equation is satisfied by the value y = v, it will be the case that

dv + P v dx + Q v v dx + R dx = 0;

if one then puts y = v + \frac{1}{z}, one will obtain

-\frac{d z}{zz} + \frac{P d x}{z} + \frac{2Qv d x}{z} + \frac{Q d x}{zz} = 0

or:

dz - (P + 2Qv)z dx - Qdx = 0

which is rendered integrable by means of the multiplier

e^{-\int(P + 2Qv)dx}.

This multiplier will thus fit the proposed equation when it itself is multiplied by zz. When therefore it is the case that z = \frac{1}{y - v}, the multiplier that will render the proposed equation integrable will be

\frac{1}{(y - v)^2} \ e^{-\int(P + 2Qv)dx}

For the sake of brevity let e^{-\int(P + 2Qv)dx} = S. Since the integral of the equation dz - (P + 2Qv)z dx - Qdx = 0 is

Sz - \int QS dx = Const.

all the required multipliers will be encompassed by this form:

\frac{S}{(y - v)^2} funct. \big(\frac{S}{y - v} - \int QS dx \big)

where by hypothesis v is a known function of x, and also S = e^{-\int(P + 2Qv)dx}

Mathematical comments: In this problem Euler has developed a method for solving Riccati’s equation when a particular integral is known. The basic idea is that if the particular solution v(x) is known, then the change of variables y = v + \frac{1}{z} gives a simple linear first-order equation for z of the form dz - (P + 2Qv)z dx - Qdx = 0. This is because substituting y = v + \frac{1}{z} into Riccati’s equation we get

dy + Py dx + Qyy dx + Rdx =

\big(dv - \frac{1}{zz} dz\big) + \big(Pv dx + \frac{P dx}{z}\big) + \big(Qvv dx + \frac{2Qvdx}{z} + \frac{Qdx}{zz}\big) + Rdx =0

which reduces to

-\frac{d z}{zz} + \frac{P d x}{z} + \frac{2Qv d x}{z} + \frac{Q d x}{zz} = 0

\iff

dz - (P + 2Qv)z dx - Qdx = 0

The integrating factor which makes this simple linear first-order differential equation integrable can be found straightforwardly using the method of Problem 4, and is e^{-\int(P + 2Qv)dx}. Then the integrating factor that makes the equation

-\frac{d z}{zz} + \frac{P d x}{z} + \frac{2Qv d x}{z} + \frac{Q d x}{zz} = 0

integrable will be

zz e^{-\int(P + 2Qv)dx} = \frac{1}{(y - v)^2} \ e^{-\int(P + 2Qv)dx}

To clarify the last part of Euler’s solution, note that if we multiply through the equation dz - (P + 2Qv)z dx - Qdx = 0 by S = e^{-\int(P + 2Qv)dx} we get

Sdz - S(P + 2Qv)z dx - SQdx = 0

\iff

d(Sz) - QSdx = 0

This last equation can be integrated directly to give

Sz - \int QS dx = Const.

or

\frac{S}{y - v} - \int QS dx = Const.

Euler makes the point that in general the integrating factor for all cases of Riccati’s equation in which a particular solution v(x) is known will be of the form

\frac{S}{(y - v)^2} f\big(\frac{S}{y - v} - \int QS dx\big)

where f is some unspecified function. This works because the argument of the unspecified function f is an arbitrary constant, so df = 0. Thus, for example, starting from

-\frac{d z}{zz} + \frac{P d x}{z} + \frac{2Qv d x}{z} + \frac{Q d x}{zz} = 0

we can multiply through by -\frac{Sf}{(y - v)^2} instead of just -\frac{S}{(y - v)^2} to get

Sf dz - Sf (P + 2Qv) z dx - SfQ dx = 0

and since df = 0 we can add the term zS df to get

Sf dz - Sf (P + 2Qv) z dx + zSdf - SfQ dx = 0

\iff

d\big(Sfz\big) - QSf dx = 0

and this can be integrated directly to get

Sfz - \int QSf dx = Const. \square