1

Research into some Remarkable Integrations in the Analysis of Functions of Two Variables known as Partial Differential Equations

By Mr. Leonard Euler

Presented to the Academy 8 December 1777

Translated from French to English by Jasen Scaramazza of Rowan University

Translator’s Foreword:

In this paper, the Leonhard Euler finds the general solution to a set of homogeneous partial differential equations of two variables. He uses an ingenious and elegant method that reduces a daunting task to a clever application of three much simpler differential equation solutions that he finds in the “preliminary problem” sections.

There is a large number of small mistakes in the original printing from which I translated, and I hope to have corrected all of them. I put corrections and any helpful notes in italics where I believed necessary. The nature of the errors makes me think that perhaps they come from printing or even a scribe to whom Euler may have dictated this work. This paper was presented in 1777, a time in Euler’s life when he was almost completely blind, so there is a chance that he did not write the words himself. Nonetheless, all of the results are correct as far as I can tell.

Euler often uses interesting notation that is similar to that of modern day mathematicians, but has several subtleties. Because of this, I have tried to copy and paste as much of the mathematics as possible to maintain the authenticity of the translation. If the notation contained nothing special, I just used an equation editor to type it into the word processor. Sometimes, the copied and pasted work needed “touching up” from the original, which I did in MS Paint to make it more legible.

The topic may seem advanced, but it actually should be quite accessible to undergraduate students who have had at least a couple semesters of calculus. A firm grasp of Taylor series is important for understanding the finer points made about special cases. I do not even believe a course in partial differential equations is a necessary prerequisite because I was able to follow the work without one. The reader only needs to know that arbitrary functions in partial differential equations play a similar role that arbitrary constants do in ordinary differential equations. Furthermore, Euler is quite gentle in his presentation of the work. He gives examples of his general findings and explains very clearly why he makes the moves that he does.

Finally, Euler uses the term “différences partielles” to describe the analysis of functions of two variables. After much consideration, I decided on “partial differential equations” rather than “partial differences.”

Enjoy!

Jasen Scaramazza, 29 November 2011

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Taking z to be a function of two variables x and y, it is known that the first differentiation, depending on whether x or y is chosen to vary, produces two

differential formulas of the first degree:

z

x and

z

y. The second differentiation

results in three second order differential formulas:

z

x2,

z

xy,

z

y2. Differentiating a

third time leads to four third degree differential formulas:

3z

x3,

3z

x2y,

3z

xy2,

3z

y3.

The fourth differentiation produces these five fourth degree differential formulas:

4z

x4,

4z

x3y,

4z

x2y2,

4z

xy3,

4z

y4; etc. Here we omit the quotations marks in which

these formulas are normally written because there is no worry for ambiguity in the problems we are going to explore.

With that, I will consider the following expressions:

(correction:

Q x2 z

x2 2xy

z

xy y2

z

y2),

and so on and so forth. In general we have the following:

The expression should actually be:

Here I observe first that each of these expressions can be formed from the one immediately preceding it, and we will see that we can always write:

3

with the pattern continuing as indicated. I observe that if we designate O (not zero) as the formula that precedes the first P, we will have O = z; and we start with

To show that these equations are true, we start with the first that expresses

the value of Q. Since

P x*z

x y*

z

y, differentiation shows

From there, we derive this equation:

which easily reduces to this form:

x*P

x y*

P

y PQ .

Rearranging, we have

Q x*P

x y

P

y P

for the second of our equations because we have already defined

Q x2z

x2 2xy

z

xy y2

z

y2.

From here we derive

4

Now the combination of these two formulas produces:

This equation reduces to the following:

x*Q

x y*

Q

y 2Q R ,

from which we solve for R

R x*Q

x y*

Q

y 2Q

.

To show that the third of our equations is true, we have that

R x3 3z

x3 3xxy

3z

x2y 3xyy

3z

xy2 y3

3z

y3. From there we differentiate

Combining these equations we have

This equation reduces to this form:

xR

x y

R

y 3R S ,

which we rearrange to

S xR

x y

R

y 3R .

5

It would be superfluous to demonstrate the validity of the other equations using the same calculation because it is already clear that we will arrive at the formulas given earlier. Now we will find the integrals of these elegant relations between P, Q, R, etc., and even the complete integrals of the following differential equations:

1P 0;

2Q 0;

3R 0;

4S 0; and so on (Euler is just noting the degree of each

differential equation). To accomplish this task, we need only to solve the following three preliminary problems.

Preliminary Problem I

Find a function v of two variables x and y such that

xv

x y

v

y 0

.

Solution

Since v is a function of x and y, we suppose that we find in differentiating

while varying both x and y (though this is the full differential of v, a partial one is

written), such that

p v

x and

q v

y. Using our original equation, this must satisfy

from which we find

q xp

y. Substituting this value gives us

.

This formula must be integrable. We reduce it to the following form:

.

(This should actually be

)

We let

x

y p (Euler means

.) to have

pyt v .

Clearly, it is absolutely necessary that py be a function of the variable t in order for this formula to be integrable. The integral will then also be a function of the same value t.

Continuing, let us designate some function with the characters

so that , , or represent some function of

v px qy

6

t. Aside from that, we will use the generally accepted notation for differentials of a given order, specifically:

That stated, integrating our last equation gives . Equivalently, since

t x

y, we will have , so that we can take v as some function of

x

y. I

note here that all of these functions are categorized as homogeneous functions of x and y of null dimension.

Preliminary Problem II

Find a function of two variables x and y, called v, so that

nv xv

x y

v

y.

Solution

Like before, let us suppose that

v px qy . Since

p v

x and

q v

y, we will have

to satisfy

nv pxqy. We eliminate the variable q from both of these equations by

multiplying the first by y and the second by

y , and in subtracting the second

equation from the first, we find

.

We now must find the function p so that this equation becomes integrable.

To make the first part of this equation integrable, we need only divide it by

yn1. From there we see

.

Now since the expression is the differential of

x

y, we rewrite our

equation in the form

.

7

Here, it is evident that

p

yn1 must be a function of

x

y. Because the integral will also be

such a function, we will have after integrating . From there, we obtain the following value for the function v:

.

A function of

x

y multiplied by

x

y, or in general by

xn

yn, remains a function of

x

y.

Therefore, in place of we can write . Rearranging, the value for v could also be expressed by

,

or ,

or ,

and so on and so forth. Furthermore, we know that these are all homogeneous functions of dimension n.

Preliminary Problem III

Find a function of two variables x and y, called v, such that we have

.

Solution

Once again let

dv px qy to have

p v

x and

q v

y, so that we must

satisfy the condition . We now form from these two equations the following:

,

whose first part becomes integrable in dividing it by

yn1. We then have

8

.

To solve this equation, we let

x

y t, or just as well

x yt . In place of we write

T, such that T is a given function of t. Because

x ty yt , our equation will be

.

Now we integrate because it is allowable. Since

,

where we suppose , after integrating we will have:

.

Here we see that the formula must be some function of t that

we will designate as . Substituting this in, we will arrive at the following integrated equation:

.

(Euler does not indicate the difference between B:t and its integral.)

From there we see

.

We now replace t with its value

x

y and use in place of T, where it is

important to note that the character represents a specific function of

x

y because

it is already given in the differential equation. The character indicates some

arbitrary function of

x

y that is introduced in the ordinary integrations.

9

Consequently, we find the solution to our problem to be the following value of the function v:

.

Now we could probably ask what v would be in the case where the exponent equals n, since then the last piece of our equation would become infinite. To avoid this difficulty, we let

n , where is an infinitely small quantity. Then we will have

,

which will give us

.

Now since represents an arbitrary function, it is permissible to replace by this formula:

,

where marks some arbitrary function. Substituting in these values causes the infinite parts of the equation to disappear and the sought integral for the case

n

will be . We will now be able to solve the following problem.

I Problem

Find the complete integral of this differential equation:

xz

x y

z

z 0. In other

words, find the nature of the function z.

Solution

Here we have P = 0. The first preliminary problem will provide us with the sought integral, since we only have to write z in place of v. After substituting, our

complete integral will be . In other words we can take z as some

10

homogenous function of x and y of null dimension. For example, let

(actually should be

) and we will find

and ,

where it is easy to see that

xz

y y

z

y 0. In the same manner, in taking

, we will have

z

xyy xy

(xx yy)

3

2

and

z

yxx xy

(xx yy)

3

2

. From there

obviously follows

xz

x y

z

y 0.

II Problem

Find the complete integral of this second-degree differential equation:

xxz

x2 2xy

z

xy yy

z

y2 0.

Solution

We have that Q = 0. Since we found earlier that

Q xP

x y

P

y P , we must

solve this first-degree differential equation:

xP

x y

P

y P .

This integral is found using the second preliminary problem in replacing v

with P and letting n = 1. We find , where represents some function. We replace P with this value, and then we need to solve this first-degree differential equation:

.

Comparing this equation with the third preliminary problem gives us n = 0, = 1 and v = z, and the sought complete integral will be

11

.

Since the two functions are arbitrary, we can write

.

This equation contains two arbitrary functions as the nature of second order differential equations requires.

III Problem

Find the complete integral of this third-degree differential equation:

Solution

Here, it is clear to set R = 0. Putting in place of R its value found earlier in the paper, we have to solve

.

Comparing this equation with that of the second preliminary problem, we find v = Q and n = 2. The complete integral is therefore

.

Now, having

Q xP

x y

P

y P , we will solve the equation

.

Comparing this expression with that of the third preliminary problem gives v = P, n = 1, and = 2. Substituting these values produces the following integral:

,

or since the functions are arbitrary, we have

12

.

Finally, with

P xz

x y

z

z, we must solve the following equation:

.

This compared with the equation of the third preliminary problem will give us v = z and n = 0. For, we will have two different values: = 2 and = 1 because it is apparent that both can be treated in the same manner. Consequently, the complete integral of the proposed equation will be

,

or in changing the characters and signs of the arbitrary functions, we have

.

IV Problem

Find the complete integral of this fourth degree differential equation:

Typo in second term corrected:

Solution

Here we have S = 0. In other words

xR

x y

R

y 3R 0, which compared with

the second preliminary problem gives us v = R and n = 3. Therefore

(this should read ). Replacing R by its value, we are left to solve the equation

13

.

Comparing this to the third preliminary problem, since v = Q, n = 2 and = 3, gives

.

We can also write

.

Here we have v = P, n = 1 and = 2 or 3, and from this we find

or

.

By making a comparison, we get v = 2, n = 0 and = 1, 2 or 3. This gives

or

.

V General Problem

Find the complete integral of this nth degree differential equation:

xn nz

xnn

1xn1y

nz

xn1yn

1

n1

2xn2y2

nz

xn2y2 etc.

(Euler implicitly sets this equal to zero.)

14

Solution

Here it is easy to see that in doing the successive operations like the ones in the preceding problems, we will arrive at this complete integral:

The number of arbitrary functions is n, and is therefore equal to the degree of the proposed equation. We now see that the integral of any degree contains all of the integrals of each smaller degree, as well as one term that is exclusive to the degree proposed.

We have thus found the integrations of all of the differential equations, assigning to each of the letters the values that were given in the beginning of this paper:

For each case, the method that we use requires as many integrations as the degree of the differential. Now a young geometer, in doing the preceding calculations, has observed that all of these solutions can be found more easily in performing a single integration. This method even has another advantage over that which we used because it also applies to the integration of differential equations composed and contained in this general form:

Az BPCQDRES etc. 0. All the degrees of

the differentials are joined together, and the constant coefficients A, B, C, D, etc. can be chosen at will. The solution for all of these cases can be found from the second preliminary problem, where the differential equation

xv

x y

v

y nv 0

gives this complete integral:

.

To more clearly explain this new method, we will solve the following problems.

Problem I

Find the complete integral of this first degree differential equation

Az BP 0, or

Az B(xz

x y

z

y) 0.

15

Solution

For the solution, we use the (second) preliminary problem and set

v az to

have this equation:

a(xz

x y

z

y) naz 0. The integral is then

(should be ).

Now in place of

xz

x y

z

y, we use P, and the equation that we just integrated will be

aP naz 0. A comparison with the proposed

Az BP 0 gives

A na and

B a.

Consequently,

a B,

A nB, and therefore

AnB 0. Taking the value

n A

B

from this equation, the integral will be

.

This solution does not contain anything that could not have been done by the preceding method, but the following problem will truly bring to light the value of the new method.

Problem II

Find the complete integral of this second degree differential equation:

Az BPCQ 0.

Solution

To solve this equation, we suppose that in the (second) preliminary problem to have the integral

.

This is in agreement with the equation

a(xz

x y

z

y) naz b(x

P

x y

P

y) nbP 0 .

(To see why, review the second preliminary problem.)

16

We now replace

xP

x y

P

y (should be

xz

x y

z

y ) with its absolute value P that we

found in the formulas from the beginning. We also replace

xP

x y

P

y with its

absolute value

QP. We will then have the following:

aPbQbP naz nbP 0, or

naz(ab nb)P bQ 0.

This last result being compared to the given form

Az BPCQ 0 gives us the following values:

b C,

a BCnC, and

0 A n(n1)C,

(Typo alert: it should be )

from which we must find the value of n.

Now since this last equation is of the second degree it will have two roots. They are designated as and , and will provide us with particular values for a and b. These values are

n

n

a B(1)C

a B(1)C

b C

b C

From here we have two integral equations:

.

Now, we need only to eliminate the letter P from these two equations. This is done by subtracting one from the other, which gives:

Because these functions are absolutely arbitrary, the integral can be represented in this form:

17

.

Corollary

From here it is easy to find the integral of the equation

Q 0 that we already

dealt with earlier. We need only to suppose that

A 0,

B 0 and

C 1, and the equation for the number n becomes

n(n1) 0. The roots are

n 0 and

n 1. Therefore

0,

1, and the integral becomes

.

Problem II (should be III)

Find the complete integral of this third degree differential equation:

Az BPCQDR 0.

Solution

To find this solution, we suppose in the second preliminary problem that

v az bP cQ . The integration first gives us this equation:

This integral leads to the following differential equation:

Now in place of differential formulas, we put their finite values and come to this equation:

aPb(PQ) c(R2Q) naz nbP ncQ 0,

which reduces to the form

18

.

Comparing this form to the one proposed, we find the following relations

A na;

B a b(1 n);

c b(2 n)c;

D c.

Because the value of D is c, the third of these equations will give us

b C(n 2)D and the second of them will give us

a B(n1)C(n1)(n 2)D . Substituting these values into

A na gives us the following equation:

AnBn(n1)Cn(n1)(n 2)D 0.

This equation is of the third degree, and therefore has three roots denoted by

,

and

. Each of these roots will provide us with two distinct values for the letters a, b

and c. We will call them a, b and c for

; a’, b’ and c’ for

; and a’’, b’’ and c’’ for

. Each of these cases provides us with a particular integral, and the equations will be

From here it will be easy to find certain coefficients for each of these equations so that in adding the products together, the quantities P and Q will cancel. By this method, we arrive at the following final equation since these coefficients will not change the nature of the arbitrary functions:

.

This expresses the complete integral of our proposed differential equation.

Corollary

To find the integral of the equation

R 0 from this equation, we only need to set

A 0,

B 0,

C 0 and

D 1. The cubic equation to find the number n will then become

n(n1)(n 2) 0, so that

0,

1 and

2. We find the sought integral

for this case to be

.

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Problem III (should be IV)

Find the complete integral of this fourth degree differential equation:

Az BPCQDRES 0.

Solution

To solve this problem, we use the second preliminary problem and set

v az bP cQ dR . As a result, we will have this integral:

az bP cQ dR ,

which leads to the following differential equation:

(The “

” that seems out of place is really just a “d”.)

Let us write their finite values in place of the differential formulas, giving us

From here, arranging the terms will give

naz(ab(1 n))P(b c(2 n))Q(cd(3 n))RdS 0.

It remains only to make this form identical to the one proposed, which produces the following five equalities:

;

;

;

;

20

.

The last one gives us

d E; the fourth provides

c D(n 3)E; then the third gives

b C(n 2)D(n 2)(n 3)E; from the second we get

a B(n1)C(n1)(n 2)D(n1)(n 2)(n 3)E; and finally the first of these equations leads us to the following expression that will allow us to determine the value of n:

AnBn(n1)Cn(n1)(n 2)Dn(n1)(n 2)(n 3)E 0.

This last expression is of the fourth degree. Therefore, there will be four roots denoted by

,

,

and

that each have their own particular values for the

letters a, b, c and d. We then say that a, b, c and d correspond to the root

; a’, b’, c’ and d’ correspond to root

; a’’, b’’, c’’ and d’’ go to

and a’’’, b’’’, c’’’ and d’’’ go to

.

We will then have four different forms for the sought integral, which will be:

After having found these four equations, it is easy to eliminate the three quantities P, Q and R so that there will only remain the quantity z on the left side. Multiplying the arbitrary functions of the right by constants will not change them. We finally have the equation

.

It is noteworthy here that to find this equation, we had no need to find the values a, b, c and d or even the coefficients to eliminate P, Q and R.

Problem IV (V) General

Find the complete integral of this differential equation of any given degree:

Az BPCQDR etc. 0.

21

Solution

The entire solution to this problem reduces to the equation to determine every value of the number n; and from the previous problems it is clear that this equation will have the form

AnBn(n1)Cn(n1)(n 2)D etc. 0.

This expression will have the same degree as that of the differential equation proposed, giving n just as many values denoted by

,

,

,

, etc. The complete

integral of this equation will thus be

It is useful here to observe that since the two variables x and y enter equally into the calculation, in place of the powers

y ,

y ,

y , etc. we can also place similar

powers of x such as

x ,

x ,

x , etc. In fact, consider the formula . In lieu of

, we can put because

x

y is also a function of

x

y, and we

will have . Now taking

, we can put in place of

the expression . It is also clear that in general we could also write

given that the sum of the exponents

and

are equal to

, that is,

.

We will have no difficulty with this solution as long as the values

,

,

,

,

etc. are all real numbers and are not equal to each other. We will, however, encounter equations where some of these values are imaginary or equal. In the first case, we must make use of certain reductions to make the integral real. In the second case, we must keep the number of arbitrary functions the same in order to have a complete integral.

To resolve all of these difficulties, we start by considering the case where two values of n, called

and

, are imaginary. We know that these two values can

always be expressed by the forms

and ,

which leads to the terms of the integral being expressed by

22

and .

To reduce them to their real terms, suppose that

and .

Now the two terms in question will reduce to this form:

In place of y in the imaginary powers, we put

e ly , where is e the number

whose hyperbolic logarithm is equal to 1. The first formula

will become equal to , and the other

will become . Now we can see by known reduction formulas that

and

(here “fin” is meant to be “sin”, this is just the orthography of the day).

Since

ly , the form of our two terms will be

where we can omit both the real and imaginary constant coefficients. We then have the following in place of the two proposed terms:

.

These expressions hold when

and .

23

From this it is clear that when the number of imaginary values of n is 4, 6, 8, 10, etc., the reduction will always be done in the same matter since each couple reduces to the two expressions

and .

To give an example, we take the case where the equation to determine n is

1nn 0. This comes from the example where

AnB(n(n1)C 0, and we take

A B C 1. The differential equation to integrate becomes

zPQ 0.

Expanding this expression, we have

z xz

x y

x

y xx

z

x2 2xy

z

xy yy

z

y2 0

(there is a mistake in the third term in this equation, instead of

yx

y it should be

yz

y).

Since for n we will have the values and , we will have

0 and

1. We find first that

(the comma here does not signify anything).

To better clarify this case, we take and so that the specific integral will be

.

Here we see

and

.

Then

24

z

x2 0, ,

and

.

Substituting these values into the equation

zPQ 0 will give

(The expression for z is incorrect here, it should actually be ).

The sum here is

zPQ 0.

We now move on to the case where several values of n are equal to each other. We first suppose that

. In the integral form found, the first two terms

reduce to a single function, leaving the integral no longer complete. We propose that

to resolve this issue, taking

to be infinitely small. Because

y y y ,

we will have

y y y ly , and our original terms become

.

We can simply substitute for first two terms, and in place of

we just put . This substitution is such that instead of the original two terms, we have

.

25

As an example of this case, we suppose that the equation to determine the number n is

nn 0. This is a second-degree equation for which in general we have

AnBn(n1)C 0. We then get

A 0,

B 1 and

C 1, so that the differential equation to integrate will be

PQ 0, or

xz

x y

z

y xx

z

x2 2xy

z

xy yy

z

y2 0.

To solve this equation, we have

nn 0, and the two equal values of n will be

0 and

0. The sought integral will thus be

(Instead of just “y” in front of the second arbitrary function there should be an “ly” or as we write today, “lny”).

It is worth the trouble to see just how this solution satisfies, in general, the proposed equation. We will first differentiate these formulas according to the rule already established earlier in this paper:

and ,

and we find

(There is just one typo here, in the last derivative the

ought to be

)

From these we obtain the following formula:

26

or just . In the same way we find , and it becomes quite evident that . This development appears all the more important because it makes use of no specific rules for differentiating the (arbitrary) functions of two variables.

Now we consider the case where, in addition to the two equal roots , a third root called turns out to be equal as well. For the first two we reduced the

functions to (should be ).

We now must add , but this will join with the first term. We then suppose

, and since , we will have to go out to the third term. The second term of the expansion will join with the second term of the aforementioned expression. From there it is clear that these three terms will reduce to, after changing the arbitrary functions, the following:

To give an example, we consider the case where the equation for the number n takes the form

,

so that the three roots are all equal to each other: . This case corresponds to the third degree differential equation , for which we have already found

.

After expanding, we obtain

We must then make . Consequently, we find , so that our differential equation is

27

.

The complete integral, since , will be

.

To illustrate this with an example, we let and

. This way, a particular integral will be , and we calculate the following differentials:

(Two minor typos:

should be

and

should be

.)

We then find

and ,

This results in , which is perfectly correct.

Now it is already very apparent that if four values of n are all equal ( ), then instead of the four terms that would normally enter into the integral, we would have these:

Whatever the number of repeated roots, the reduction of the integral will present no difficulty. Finally, it is easy to understand that in all of these formulas, we can interchange x and y.

28

To prove this, I will show that in place of the terms

,

we can write

To this end, I observe that because both terms contain an arbitrary function of

, we

can multiply them by

. We then have .

Because

is also a function of

, in place of we can write

. We end up with , and it is clear to see that this permutation can always take place.

The integration of this quite general differential equation:

,

where P, Q, R, S, etc. designate the differential formulas described earlier, can be regarded as an excellent example of this Analysis that involves two variables. We must make the distinction between it and the ordinary analysis that only deals with functions of a single variable, for it is now quite clear that the two have important differences. These are differences not only in the functions examined, but also in the methods used to treat them. This is why the name “partial differentials”, used by many geometers to describe functions of two variables, does not seem to me to be adequate enough to illustrate their true character.

Their differences notwithstanding, we can often observe a beautiful harmony between these two types of analysis. Thus when we work with the differential

equation in ordinary analysis (when we look for z as a function of x), the ordinary method of integration leads to the following algebraic equation:

.

From here we must find all of the roots of n. The complete integral is expressed as

29

The letters represent arbitrary constants. This form exhibits an elegant relationship with the integral we found in this paper for the function z of two variables x and y.