One solution for many linear partial differential equations with terms of equal orders
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We disclose a simple and straightforward method of solving single-order linear partial differential equations. The advantage of the method is that it is applicable to any orders and the big disadvantage is that it is restricted to a single order at a time. As it is very easy compared to classical methods, it has didactic value.
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∑ (
(𝑛 − 𝑖)(𝑛 − 𝑖 − 1)
(2 + 𝑖)(1 + 𝑖)
𝑎𝑖,𝑛−𝑖 −
(𝑖 + 4)(𝑖 + 3)
(𝑖 + 6)(𝑖 + 5)
𝑎𝑖+4,𝑛−𝑖−4)
𝑛−4
𝑖=0
𝑥2+𝑖
𝑦𝑛−2−𝑖
+
2
𝑛(𝑛 − 1)
𝑎𝑛−2,2𝑥8
−
2
𝑛(𝑛 − 1)
𝑎2,𝑛−2𝑦8
+
3 × 2
(𝑛 − 1)(𝑛 − 2)
𝑎𝑛−3,3𝑥𝑛−1
𝑦
−
3 × 2
(𝑛 − 1)(𝑛 − 2)
𝑎3,𝑛−3𝑥𝑦𝑛−1
≡ 0. (22)
This identity provides
𝑎𝑖+4,𝑛−𝑖−4 =
(𝑖 + 6)(𝑖 + 5)(𝑛 − 𝑖)(𝑛 − 𝑖 − 1)
(1 + 4)(𝑖 + 3)(𝑖 + 2)(𝑖 + 1)
𝑎𝑖,𝑛−𝑖. (23)
The entire problem is conditioned to convergences for the coefficients
chosen with these restrictions. In particular, all coefficients 𝑎𝑖,𝑛−𝑖 can be
placed according to the coefficients of the types 𝑎0,𝑘, 𝑎1,𝑘; 𝑘 ∈ ℕ.
Note that on average, five-ninths of the coefficients 𝑎𝑖,𝑛−𝑖 are
obligatorily null. And those that are not necessarily null are all of types
𝑎4𝑖,4𝑘−4𝑖𝑥4𝑖
𝑦4𝑘−4𝑖
, 𝑎4𝑖,4𝑘+1−4𝑖𝑥4𝑖
𝑦4𝑘+1−4𝑖
, 𝑎4𝑖+1,4𝑘−4𝑖𝑥4𝑖+1
𝑦4𝑘−4𝑖
and
𝑎4𝑖+1,4𝑘+1−4𝑖𝑥4𝑖+1
𝑦4𝑘+1−4𝑖
, with 𝑛 = 4𝑘, 0 ≤ 𝑖 ≤ 𝑘 ∈ ℕ.
Conclusion
Classical methods such as those presented in [1] encompass the
method under study. However, this is explicit in many situations.
Furthermore, the mathematical tools needed to understand the method
are restricted to the most basic differential calculus. An approach such
as the one established would be of great encouragement for students in
their first contact with the theory of partial differential equations. And to
further facilitate learning, we could develop the results initially in
dimensions two and three. The linear transformation 𝐿(𝑓)(𝑥, 𝑦)
establishes an alternative method for the famous Laplacian operator.
Analogous situations must occur for many other operators. The analytic
functions that belong to the kernel of L seem to be simpler than the
harmonic functions.
References
[1]. GERALD B. FOLLAND. Introduction to Partial Differential Equations,
2nd ed. Princeton University Press, New Jersey, 1995.](https://image.slidesharecdn.com/endedpmethod-211107210423/85/One-solution-for-many-linear-partial-differential-equations-with-terms-of-equal-orders-6-320.jpg)
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One solution for many linear partial differential equations with terms of equal orders
- 1. 1 One solution for many linear partial differential equations with terms of equal orders 1Lohans de Oliveira Miranda; 2Lossian Barbosa Bacelar Miranda 1Universidad Europea del Atlântico, Spain, [email protected] 2IFPI, Brazil, [email protected] Abstract. We disclose a simple and straightforward method of solving single- order linear partial differential equations. The advantage of the method is that it is applicable to any orders and the big disadvantage is that it is restricted to a single order at a time. As it is very easy compared to classical methods, it has didactic value. 1. Basic concepts Consider: 1) 𝑥 ⃗ = (𝑥1, 𝑥2, … , 𝑥𝑛) ∈ 𝐴, 𝐴 open set of ℝ𝑛 ; 2) 𝑘 ∈ 𝐼𝑛 = {1, 2, 3, … , 𝑛}; 3) 𝑢: 𝐴 → ℝ, differentiable function of order 𝑘, with continuous derivatives (1) Let us consider the “𝑘-dimensional Hessian matrix” given by 𝐻 = ( 𝜕𝑘 𝑢(𝑥 ⃗) 𝜕𝑥𝑖1 𝜕𝑥𝑖2 … 𝜕𝑥𝑖𝑘 ) (2) From 𝐻, let us consider the following system, being 𝑏𝑖1𝑖2…𝑖𝑘 (𝑥 ⃗) and 𝑓𝑖1𝑖2…𝑖𝑘 (𝑥 ⃗) differentiable functions of order 𝑘, with continuous derivatives, and 𝑓𝑖1𝑖2…𝑖𝑘 (𝑥 ⃗) 𝑏𝑖1𝑖2…𝑖𝑘 (𝑥 ⃗) being well defined in 𝐴: (𝑏𝑖1𝑖2…𝑖𝑘 (𝑥 ⃗) 𝜕𝑘 𝑢(𝑥 ⃗) 𝜕𝑥𝑖1 𝜕𝑥𝑖2 … 𝜕𝑥𝑖𝑘 ) = (𝑓𝑖1𝑖2…𝑖𝑘 (𝑥 ⃗)) (3) Let us denote: 𝑔𝑖1𝑖2…𝑖𝑘 (𝑥 ⃗) = 𝑓𝑖1𝑖2…𝑖𝑘 (𝑥 ⃗) 𝑏𝑖1𝑖2…𝑖𝑘 (𝑥 ⃗) (4) So (3) will be written as ( 𝜕𝑘 𝑢(𝑥 ⃗) 𝜕𝑥𝑖1 𝜕𝑥𝑖2 … 𝜕𝑥𝑖𝑘 ) = (𝑔𝑖1…𝑖𝑘 (𝑥 ⃗)) (5)
- 2. 2 Observation 1. Repeated applications of the Fundamental Theorem of Calculus for each of the 𝑛𝑘 partial differential equations 𝜕𝑘 𝑢(𝑥 ⃗) 𝜕𝑥𝑖1 𝜕𝑥𝑖2 … 𝜕𝑥𝑖𝑘 = 𝑔𝑖1…𝑖𝑘 (𝑥 ⃗) (6) give us the 𝑛𝑘 solutions 𝑢𝑖1𝑖2…𝑖𝑘 (𝑥 ⃗) = ∭ … ∫ 𝑔𝑖1…𝑖𝑘 (𝑥 ⃗)𝜕𝑥𝑖1 𝜕𝑥𝑖2 … 𝜕𝑥𝑖𝑘−1 + ∑ 𝑐𝑠,𝑖1…𝑖𝑘 𝑘−1 𝑠=1 ∏ 𝑥𝑖𝜃 𝑘−1 𝜃=𝑠+1 + 𝑐𝑘−1,𝑖1𝑖2…𝑖𝑘 (7) Here, the c (under indexed) are real or complex numbers. Obviously, 𝜕𝑘 𝑢𝑖1𝑖2…𝑖𝑘 (𝑥 ⃗) 𝜕𝑥𝑗1 𝜕𝑥𝑗2 … 𝜕𝑥𝑗 = 𝑔𝑖1𝑖2…𝑖𝑘 (𝑥 ⃗) (8) if (𝑖1, 𝑖2, … , 𝑖𝑘) = (𝑗1, 𝑗2, … , 𝑗𝑘). Now, consider the function 𝑢 ̃(𝑥 ⃗) = ∑ 𝑢𝑖1𝑖2…𝑖𝑘 (𝑥 ⃗) 𝑖1,𝑖2,…,𝑖𝑘∈𝐼𝑛 (9) Now we can state the main result. 2. Main results Proposition 1. In the hypotheses established above, if for (𝑖1, 𝑖2, … , 𝑖𝑘) ≠ (𝑗1, 𝑗2, … , 𝑗𝑘) we have 𝜕𝑘 𝑢𝑖1𝑖2…𝑖𝑘 (𝑥 ⃗) 𝜕𝑥𝑗1 𝜕𝑥𝑗2 … 𝜕𝑥𝑗 = 0, (10) then 𝑢 ̃(𝑥 ⃗) = ∑ 𝑢𝑖1𝑖2…𝑖𝑘 (𝑥 ⃗) 𝑖1,𝑖2,…,𝑖𝑘∈𝐼𝑛 defined in (9) will be the solution of 𝑛𝑘 partial differential equations defined in (3), or alternatively in (6). In particular, 𝑢 ̃(𝑥 ⃗) will be a solution of the 2𝑛𝑘 − 1 differential equations defined by the sums of the elements of all non-empty subsets of the set 𝐵 = { 𝜕𝑘 𝑢(𝑥 ⃗) 𝜕𝑥𝑖1 𝜕𝑥𝑖2 … 𝜕𝑥𝑖𝑘 ; 𝑖1,𝑖2, … , 𝑖𝑘 ∈ 𝐼𝑛 }. (11) Demonstration. It is an immediate consequence of the construction of 𝑢 ̃(𝑥 ⃗) and of the hypothesis (𝑖1, 𝑖2, … , 𝑖𝑘) ≠ (𝑗1, 𝑗2, … , 𝑗𝑘). Observation 2. The thesis of Proposition 1 can still be obtained even if the assumptions established in (10) are not satisfied. To do so, it is enough
- 3. 3 to find the unknown functions involved that satisfy the required integral equations. Proposition 2. Let 𝑓: ℝ2 → ℝ, analytic function in your domain. So, if 𝑓 is affine function, then 𝐿(𝑓)(𝑥, 𝑦) ≝ ∬ 𝜕2 𝑓 𝜕𝑦2 (𝑥, 𝑦)𝜕𝑥𝜕𝑥 − ∬ 𝜕2 𝑓 𝜕𝑥2 (𝑥, 𝑦)𝜕𝑦𝜕𝑦 = 0. (12) Demonstration. If 𝑓(𝑥, 𝑦) = 𝑎𝑥 + 𝑏𝑦 + 𝑐, then 𝜕2𝑓 𝜕𝑦2 = 𝜕2𝑓 𝜕𝑥2 = 0 and from that follows (12). 3. Applications When considering order 2, the above method solves the main science equations such as the wave, Laplace and Poisson equations. In particular, they give the most obvious solutions in the case of the wave equation, such as simple vibrating string translations. In the case of the Poisson equation, they give very varied solutions. Let's look at a case of Laplace's equation. 3.1. Laplacian equation ( 𝜕2 𝑢 𝜕𝑥2 0 0 𝜕2 𝑢 𝜕𝑦2 ) = ( 𝑓(𝑥, 𝑦) 0 0 −𝑓(𝑥, 𝑦) ). (13) By the presented theory we have: 𝑢11(𝑥, 𝑦) = ∬ 𝑓(𝑥, 𝑦)𝜕𝑥𝜕𝑥 + 𝑘1(𝑦)𝑥 + 𝑘2(𝑦); 𝑢22(𝑥, 𝑦) = − ∬ 𝑓(𝑥, 𝑦)𝜕𝑦𝜕𝑦 + 𝑣1(𝑥)𝑦 + 𝑣2(𝑥); 𝑢 ̃(𝑥, 𝑦) = ∬ 𝑓(𝑥, 𝑦)𝜕𝑥𝜕𝑥 − ∬ 𝑓(𝑥, 𝑦)𝜕𝑦𝜕𝑦 + 𝑘1(𝑦)𝑥 + 𝑘2(𝑦) + 𝑣1(𝑥)𝑦 + 𝑣2(𝑥). (14) Note that the hypotheses established in (10) are not satisfied because, for example, 𝜕2𝑢11 𝜕𝑦2 (𝑥, 𝑦) = ∬ 𝜕2𝑓 𝜕𝑦2 (𝑥, 𝑦)𝜕𝑥𝜕𝑥 + 𝑘1′′(𝑦)𝑥 + 𝑘2′′(𝑦). But if we take 𝑘1(𝑦), 𝑘2(𝑦), 𝑣1(𝑥) and 𝑣2(𝑥) to be affine functions, all their second derivatives will be nullified and we will obtain from (14) the candidate for harmonic function, renamed with the same nomenclature, as follows: 𝑢 ̃(𝑥, 𝑦) = ∬ 𝑓(𝑥, 𝑦)𝜕𝑥𝜕𝑥 − ∬ 𝑓(𝑥, 𝑦)𝜕𝑦𝜕𝑦 + 𝑎𝑥𝑦 + 𝑏𝑥 + 𝑐𝑦 + 𝑑. (15)
- 4. 4 In (15), the parameters are any real or complex numbers. Derived (15) we will have 𝜕2𝑢 ̃ 𝜕𝑥2 (𝑥, 𝑦) + 𝜕2𝑢 ̃ 𝜕𝑥2 (𝑥, 𝑦) = (𝑓(𝑥, 𝑦) − ∬ 𝜕2𝑓 𝜕𝑥2 (𝑥, 𝑦) 𝜕𝑦𝜕𝑦) + (∬ 𝜕2𝑓 𝜕𝑦2 (𝑥, 𝑦) 𝜕𝑥𝜕𝑥 − 𝑓(𝑥, 𝑦)). This, equaled to zero gives us ∬ 𝜕2𝑓 𝜕𝑦2 (𝑥, 𝑦) 𝜕𝑦𝜕𝑦 − ∬ 𝜕2𝑓 𝜕𝑥2 (𝑥, 𝑦) 𝜕𝑥𝜕𝑥 = 0. By Proposition 2, 𝑢 ̃(𝑥, 𝑦) in (15) will be harmonic (solution of Laplace's equation) if 𝑓(𝑥, 𝑦) is affine function, that is, is of type 𝑓(𝑥, 𝑦) = 𝐴𝑥 + 𝐵𝑦 + 𝐷. (16) Substituting (16) into (15) gives the three-degree harmonic polynomial below: 𝑢 ̃(𝑥, 𝑦) = 𝐴 2 ( 𝑥3 3 − 𝑥𝑦2 ) + 𝐵 2 (𝑥2 𝑦 − 𝑦3 3 ) + 𝐷 2 (𝑥2 − 𝑦2) + 𝑎𝑥𝑦 + 𝑏𝑥 + 𝑐𝑦 + 𝑑 . (17) Proposition 3. The function 𝑢 ̃(𝑥, 𝑦) = ∬ 𝑓(𝑥, 𝑦)𝜕𝑥𝜕𝑥 − ∬ 𝑓(𝑥, 𝑦)𝜕𝑦𝜕𝑦 + 𝑎𝑥𝑦 + 𝑏𝑥 + 𝑐𝑦 + 𝑑 will be harmonic if and only if 𝑓(𝑥, 𝑦) belong to the kernel of the linear transformation 𝐿(𝑓)(𝑥, 𝑦) ≝ ∬ 𝜕2𝑓 𝜕𝑦2 (𝑥, 𝑦)𝜕𝑥𝜕𝑥 − ∬ 𝜕2𝑓 𝜕𝑥2 (𝑥, 𝑦)𝜕𝑦𝜕𝑦. Moreover, if 𝜕2𝑓 𝜕𝑦2 − 𝜕2𝑓 𝜕𝑥2 = 0, then 𝑓 will belong to that same kernel. Demonstration. (⟹). 0 = ∬ ( 𝜕2𝑓 𝜕𝑦2 + 𝜕2𝑓 𝜕𝑥2) 𝑑𝑥𝑑𝑥 − ∬ ( 𝜕2𝑓 𝜕𝑦2 + 𝜕2𝑓 𝜕𝑥2) 𝑑𝑦𝑑𝑦 = 𝐿(𝑓)(𝑥, 𝑦). (⟸). Conversely, we have ∆(𝑢 ̃)(𝑥, 𝑦) = 𝐿(𝑓)(𝑥, 𝑦. By hypothesis, the latter is the null function. From 𝜕2𝑓 𝜕𝑦2 = 𝜕2𝑓 𝜕𝑥2 follow, by simple derivation, ∬ 𝜕2𝑓 𝜕𝑦2 (𝑥, 𝑦)𝜕𝑥𝜕𝑥 = ∬ 𝜕2𝑓 𝜕𝑥2 (𝑥, 𝑦)𝜕𝑥𝜕𝑥 = 𝑓(𝑥, 𝑦) and ∬ 𝜕2𝑓 𝜕𝑦2 (𝑥, 𝑦)𝜕𝑦𝜕𝑦 = ∬ 𝜕2𝑓 𝜕𝑥2 (𝑥, 𝑦)𝜕𝑦𝜕𝑦 = 𝑓(𝑥, 𝑦). And from these results 𝐿(𝑓)(𝑥, 𝑦) = 0. Proposition 4. 𝐻(𝑥, 𝑦) = ∑ (∑ ℎ𝑖,𝑛−𝑖 𝑛 𝑖=0 𝑥𝑖 𝑦𝑛−𝑖 ) ∞ 𝑛=0 it is analytic harmonic function if and only if it exists analytic function 𝑓(𝑥, 𝑦) = ∑ (∑ 𝑎𝑖,𝑛−𝑖 𝑛 𝑖=0 𝑥𝑖 𝑦𝑛−𝑖 ) ∞ 𝑛=0 and numbers 𝑎, 𝑏, 𝑐, 𝑑 such that 𝐻(𝑥, 𝑦) = ∬ 𝑓(𝑥, 𝑦)𝑑𝑥𝑑𝑥 − ∬ 𝑓(𝑥, 𝑦)𝑑𝑦𝑑𝑦 + 𝑎𝑥𝑦 + 𝑏𝑥 + 𝑐𝑦 + 𝑑 and 𝑓(𝑥, 𝑦) belong to the kernel of the linear transformation 𝐿(𝑓)(𝑥, 𝑦) ≝ ∬ 𝜕2𝑓 𝜕𝑦2 (𝑥, 𝑦)𝜕𝑥𝜕𝑥 − ∬ 𝜕2𝑓 𝜕𝑥2 (𝑥, 𝑦)𝜕𝑦𝜕𝑦. Demonstration. (⟹) If 𝐻(𝑥, 𝑦) is an affine function, let's say 𝐻(𝑥, 𝑦) = 𝑎𝑥 + 𝑏𝑦 + 𝑐, let's take 𝑓(𝑥, 𝑦) = 0. Then 𝐻(𝑥, 𝑦) = 𝑎𝑥 + 𝑏𝑦 + 𝑐 = ∬ 0𝑑𝑥𝑑𝑥 − ∬ 0𝑑𝑦𝑑𝑦 + 𝑎𝑥 + 𝑏𝑦 + 𝑐. Let us consider the case where 𝐻(𝑥, 𝑦) is not an affine function. We have First, we're going to prove that if 𝑓 exists, then it belongs to 𝐾𝑒𝑟(𝐿) (kernel
- 5. 5 of L). Let 𝐻(𝑥, 𝑦) = ∬ 𝑓(𝑥, 𝑦)𝑑𝑥𝑑𝑥 − ∬ 𝑓(𝑥, 𝑦)𝑑𝑦𝑑𝑦 + 𝑎𝑥𝑦 + 𝑏𝑥 + 𝑐𝑦 + 𝑑. Then 0 = ∆𝐻(𝑥, 𝑦) = ∆(∬ 𝑓(𝑥, 𝑦)𝑑𝑥𝑑𝑥 − ∬ 𝑓(𝑥, 𝑦)𝑑𝑦𝑑𝑦 + 𝑎𝑥𝑦 + 𝑏𝑥 + 𝑐𝑦 + 𝑑) = 𝐿(𝑓)(𝑥, 𝑦). Now let's prove that 𝑓(𝑥, 𝑦) exists. We have 𝐻(𝑥, 𝑦) ≡ 𝐻(𝑥,𝑦) 2 + 𝐻(𝑥,𝑦) 2 ≡ 𝐻(𝑥,𝑦) 2 − 1 2 ∬ − 𝜕2𝐻 𝜕𝑦2 (𝑥, 𝑦)𝑑𝑦𝑑𝑦. Since, by hypothesis, − 𝜕2𝐻 𝜕𝑦2 (𝑥, 𝑦) ≡ 𝜕2𝐻 𝜕𝑥2 (𝑥, 𝑦), we have 𝐻(𝑥, 𝑦) = 1 2 ∬ 𝜕2𝐻 𝜕𝑥2 (𝑥, 𝑦) 𝑑𝑥𝑑𝑥 − 1 2 ∬ 𝜕2𝐻 𝜕𝑥2 (𝑥, 𝑦) 𝑑𝑦𝑑𝑦 ≡ ∬ 1 2 𝜕2𝐻 𝜕𝑥2 (𝑥, 𝑦)𝑑𝑥𝑑𝑥 − ∬ 1 2 𝜕2𝐻 𝜕𝑥2 (𝑥, 𝑦)𝑑𝑦𝑑𝑦. So, exists 𝑓(𝑥, 𝑦) ≡ 1 2 𝜕2𝐻 𝜕𝑥2 (𝑥, 𝑦). (⇐) ∆(𝐻) = ∆(∬ 𝑓(𝑥, 𝑦)𝑑𝑥𝑑𝑥 − ∬ 𝑓(𝑥, 𝑦)𝑑𝑦𝑑𝑦 + 𝑎𝑥𝑦 + 𝑏𝑥 + 𝑐𝑦 + 𝑑) = 𝐿(𝑓) = 0. Then 𝐻(𝑥, 𝑦) is harmonic analytic function. Observation 3. In relation to Proposition 4, note that: ∆(𝐻) = 2𝑎20 + 2𝑎02 + ∑ (∑ ℎ𝑖,𝑛−𝑖 𝑛 𝑖=0 (𝑖(𝑖 − 1)𝑥𝑖−2 𝑦𝑛−𝑖 ∞ 𝑛=3 + (𝑛 − 𝑖)(𝑛 − 𝑖 − 1)𝑥𝑖 𝑦𝑛−𝑖−2 )) ; (18) 𝐿(𝑓)(𝑥, 𝑦) ≡ 𝑎02𝑥2 − 𝑎20𝑦2 + ∑ (∑ 𝑎𝑖,𝑛−𝑖 𝑛 𝑖=0 ( (𝑛 − 𝑖)(𝑛 − 𝑖 − 1) (𝑖 + 2)(𝑖 + 1) 𝑥𝑖+2 𝑦𝑛−𝑖−2 ∞ 𝑛=3 − 𝑖(𝑖 − 1) (𝑛 − 𝑖 + 2)(𝑛 − 𝑖 + 1) 𝑥𝑖−2 𝑦𝑛−𝑖+2 )) ≡ 0. (19) Note that, for 𝑓(𝑥, 𝑦) = 𝑒−𝑥 𝑠𝑒𝑛𝑦, 𝐿(𝑓)(𝑥, 𝑦) ≡ 0 but 𝜕2𝑓 𝜕𝑦2 − 𝜕2𝑓 𝜕𝑥2 = −2𝑒−𝑥 𝑠𝑖𝑛𝑦 ≢ 0. See that ∬ 𝑓(𝑥, 𝑦)𝜕𝑥𝜕𝑥 = ∬ 𝑓(𝑥, 𝑦)𝜕𝑦𝜕𝑦 ⇔ 𝜕2𝑓 𝜕𝑦2 = 𝜕2𝑓 𝜕𝑥2 . Solving 𝐿(𝑓)(𝑥, 𝑦) = 0 considering 𝑓(𝑥, 𝑦) = 𝑔(𝑥)ℎ(𝑦) we get 𝑓(𝑥, 𝑦) = (𝑐1𝑒−𝑘𝑥 + 𝑐2𝑒𝑘𝑥 + 𝑐3𝑠𝑖𝑛𝑘𝑥 + 𝑐4𝑐𝑜𝑠𝑘𝑥)(𝑑1𝑒−𝑘𝑦 + 𝑑2𝑒𝑘𝑦 + 𝑑3𝑠𝑖𝑛𝑘𝑦 + 𝑑4𝑐𝑜𝑠𝑘𝑦). (20) The same equation can be solved by taking the series 𝑓(𝑥, 𝑦) = ∑ (∑ 𝑎𝑖,𝑛−𝑖 𝑛 𝑖=0 𝑥𝑖 𝑦𝑛−𝑖 ). ∞ 𝑛=0 (21) The null polynomial of degree 𝑛 obtained in 𝐿(𝑓)(𝑥, 𝑦) ≡ 0 is
- 6. 6 ∑ ( (𝑛 − 𝑖)(𝑛 − 𝑖 − 1) (2 + 𝑖)(1 + 𝑖) 𝑎𝑖,𝑛−𝑖 − (𝑖 + 4)(𝑖 + 3) (𝑖 + 6)(𝑖 + 5) 𝑎𝑖+4,𝑛−𝑖−4) 𝑛−4 𝑖=0 𝑥2+𝑖 𝑦𝑛−2−𝑖 + 2 𝑛(𝑛 − 1) 𝑎𝑛−2,2𝑥8 − 2 𝑛(𝑛 − 1) 𝑎2,𝑛−2𝑦8 + 3 × 2 (𝑛 − 1)(𝑛 − 2) 𝑎𝑛−3,3𝑥𝑛−1 𝑦 − 3 × 2 (𝑛 − 1)(𝑛 − 2) 𝑎3,𝑛−3𝑥𝑦𝑛−1 ≡ 0. (22) This identity provides 𝑎𝑖+4,𝑛−𝑖−4 = (𝑖 + 6)(𝑖 + 5)(𝑛 − 𝑖)(𝑛 − 𝑖 − 1) (1 + 4)(𝑖 + 3)(𝑖 + 2)(𝑖 + 1) 𝑎𝑖,𝑛−𝑖. (23) The entire problem is conditioned to convergences for the coefficients chosen with these restrictions. In particular, all coefficients 𝑎𝑖,𝑛−𝑖 can be placed according to the coefficients of the types 𝑎0,𝑘, 𝑎1,𝑘; 𝑘 ∈ ℕ. Note that on average, five-ninths of the coefficients 𝑎𝑖,𝑛−𝑖 are obligatorily null. And those that are not necessarily null are all of types 𝑎4𝑖,4𝑘−4𝑖𝑥4𝑖 𝑦4𝑘−4𝑖 , 𝑎4𝑖,4𝑘+1−4𝑖𝑥4𝑖 𝑦4𝑘+1−4𝑖 , 𝑎4𝑖+1,4𝑘−4𝑖𝑥4𝑖+1 𝑦4𝑘−4𝑖 and 𝑎4𝑖+1,4𝑘+1−4𝑖𝑥4𝑖+1 𝑦4𝑘+1−4𝑖 , with 𝑛 = 4𝑘, 0 ≤ 𝑖 ≤ 𝑘 ∈ ℕ. Conclusion Classical methods such as those presented in [1] encompass the method under study. However, this is explicit in many situations. Furthermore, the mathematical tools needed to understand the method are restricted to the most basic differential calculus. An approach such as the one established would be of great encouragement for students in their first contact with the theory of partial differential equations. And to further facilitate learning, we could develop the results initially in dimensions two and three. The linear transformation 𝐿(𝑓)(𝑥, 𝑦) establishes an alternative method for the famous Laplacian operator. Analogous situations must occur for many other operators. The analytic functions that belong to the kernel of L seem to be simpler than the harmonic functions. References [1]. GERALD B. FOLLAND. Introduction to Partial Differential Equations, 2nd ed. Princeton University Press, New Jersey, 1995.