3.6 Sign of the Green’s Function
3.6.3 The case (C3)
We study here the case (C3) for𝑎 = 𝑏. In this case, it is clear that 𝐺(𝑡, 𝑠) = [1 + 𝑎(𝑠 − 𝑡)]𝜒u�
0(𝑠) + 𝑎(𝑠 + 𝑡)𝜒0−u�(𝑠),
which we can rewrite as
𝐺(𝑡, 𝑠) = ⎧{ {{ { ⎨{ {{ {⎩ 1 + 𝑎(𝑠 − 𝑡), 0 ≤ 𝑠 ≤ 𝑡, −1 − 𝑎(𝑠 − 𝑡), 𝑡 ≤ 𝑠 ≤ 0, 𝑎(𝑠 + 𝑡), −𝑡 ≤ 𝑠 ≤ 0, −𝑎(𝑠 + 𝑡), 0 ≤ 𝑠 ≤ −𝑡, 0, otherwise.
Studying the expression of𝐺 we can obtain maximum and antimaximum principles. With this information, we can prove the following Lemma as we did with the analogous ones for cases (C1) and (C2).
Lemma 3.6.8. Assume𝑎 = 𝑏. Then, if 𝑎 > 0, the Green’s function of problem (3.5.5) is
• positive on{(𝑡, 𝑠), −𝑡 < 𝑠 < 0} and {(𝑡, 𝑠), 0 < 𝑠 < −𝑡}, • negative on{(𝑡, 𝑠), 𝑡 < 𝑠 < 0},
• positive on{(𝑡, 𝑠), 0 < 𝑠 < 𝑡} if and only if 𝑡 ∈ (0, 1/𝑎), and, if𝑎 < 0, the Green’s function of problem (3.5.5) is
• negative on{(𝑡, 𝑠), −𝑡 < 𝑠 < 0} and {(𝑡, 𝑠), 0 < 𝑠 < −𝑡}, • positive on{(𝑡, 𝑠), 0 < 𝑠 < 𝑡}.
• negative on{(𝑡, 𝑠), 𝑡 < 𝑠 < 0} if and only if 𝑡 ∈ (1/𝑎, 0).
As a corollary of the previous result we obtain the following one:
Lemma 3.6.9. Assume𝑎 = 𝑏. Then,
• if0 < 𝑎, the Green’s function of problem (3.5.5) is nonnegative on [0, 1/𝑎] × ℝ, • if𝑎 < 0, the Green’s function of problem (3.5.5) is nonpositive on [1/𝑎, 0] × ℝ,
• the Green’s function of problem (3.5.5) changes sign in any other strip not a subset of the aforementioned.
For this particular case we have another way of computing the solution to the problem.
Proposition 3.6.10. Let𝑎 = 𝑏 and assume 2𝑎𝑡0 ≠ 1. Let 𝐻(𝑡) ∶= ∫u�u�
0ℎ(𝑠) d 𝑠 and ℋ(𝑡) ∶= ∫u�
u�0𝐻(𝑠) d 𝑠. Then problem (3.5.5) has a unique solution given by 𝑢(𝑡) = 𝐻(𝑡) − 2𝑎ℋu�(𝑡) + 2𝑎 𝑡 − 12𝑎 𝑡
Proof. The equation is satisfied, since
𝑢′(𝑡) + 𝑎(𝑢(𝑡) + 𝑢(−𝑡)) = 𝑢′(𝑡) + 2𝑎𝑢 u�(𝑡)
=ℎ(𝑡) − 2 𝑎𝐻u�(𝑡) + 2𝑎 𝑐2𝑎 𝑡
0− 1 + 2 𝑎𝐻u�(𝑡) − 2𝑎 𝑐2𝑎 𝑡0− 1 = ℎ(𝑡).
The initial condition is also satisfied for, clearly, 𝑢(𝑡0) = 𝑐. The uniqueness of solution is
derived from the fact that2𝑎𝑡0≠ 1 and Lemma 3.5.6.
Example 3.6.11. Consider the problem𝑥′(𝑡)+𝜆(𝑥(𝑡)−𝑥(−𝑡)) = |𝑡|u�, 𝑥(0) = 1 for 𝜆, 𝑝 ∈
ℝ, 𝑝 > −1. For 𝑝 ∈ (−1, 0) we have a singularity at 0. We can arrive to the solution 𝑢(𝑡) = 1𝑝 + 1𝑡|𝑡|u�+ 1 − 2𝜆𝑡,
where ̄𝑢(𝑡) = 1
u�+1𝑡|𝑡|u�and ̃𝑢(𝑡) = 1 − 2𝜆𝑡.
̄𝑢 is positive in (0, +∞) and negative in (−∞, 0) independently of 𝜆, so the solution has better properties than the ones guaranteed by Lemma 3.6.9.
The next example shows that the estimate is sharp.
Example 3.6.12. Consider the problem
𝑢′
u�(𝑡) + 𝑢u�(𝑡) + 𝑢u�(−𝑡) = ℎu�(𝑡), 𝑡 ∈ ℝ; 𝑢u�(0) = 0, (3.6.4)
where𝜖 ∈ ℝ, ℎu�(𝑡) = 12𝑡(𝜖 − 𝑡)𝜒[0,u�](𝑡) and 𝜒[0,u�] is the characteristic function of the interval [0, 𝜖]. Observe that ℎu� is continuous. By means of the expression of the Green’s function for problem (3.6.4), we have that its unique solution is given by
𝑢u�(𝑡) = ⎧{ {{ ⎨{ {{ ⎩ −2𝜖3𝑡 − 𝜖4, if 𝑡 < −𝜖, −𝑡4− 2𝜖𝑡3, if − 𝜖 < 𝑡 < 0, 𝑡4− (4 + 2𝜖)𝑡3+ 6𝜖𝑡2, if 0 < 𝑡 < 𝜖, −2𝜖3𝑡 + 2𝜖3+ 𝜖4, if 𝑡 > 𝜖.
The a priory estimate on the solution tells us that𝑢u�is nonnegative at least in[0, 1]. Studying the function𝑢u�, it is easy to check that𝑢u�is zero at0 and 1+𝜖/2, positive in (−∞, 1+𝜖/2)\{0} and negative in(1 + 𝜖/2, +∞).
The case (C3.2) is very similar,
𝐺(𝑡, 𝑠) = ⎧{ {{ { ⎨{ {{ {⎩ 1 + 𝑎(𝑡 − 𝑠), 0 ≤ 𝑠 ≤ 𝑡, −1 − 𝑎(𝑡 − 𝑠), 𝑡 ≤ 𝑠 ≤ 0, 𝑎(𝑠 + 𝑡), −𝑡 ≤ 𝑠 ≤ 0, −𝑎(𝑠 + 𝑡), 0 ≤ 𝑠 ≤ −𝑡, 0, otherwise.
Figure 3.6.2: Graph of the function𝑢1andℎ1(dashed). Observe that𝑢 becomes zero at 𝑡 = 1 + 𝜖/2 = 3/2.
• positive on{(𝑡, 𝑠), −𝑡 < 𝑠 < 0}, {(𝑡, 𝑠), 0 < 𝑠 < 𝑡} and {(𝑡, 𝑠), 0 < 𝑠 < −𝑡}, • negative on{(𝑡, 𝑠), 𝑡 < 𝑠 < 0} if and only if 𝑡 ∈ (−1/𝑎, 0),
and, if𝑎 > 0, the Green’s function of problem (3.5.5) is
• negative on{(𝑡, 𝑠), −𝑡 < 𝑠 < 0}, {(𝑡, 𝑠), 𝑡 < 𝑠 < 0} and {(𝑡, 𝑠), 0 < 𝑠 < −𝑡}, • positive on{(𝑡, 𝑠), 0 < 𝑠 < 𝑡} if and only if 𝑡 ∈ (0, −1/𝑎).
As a corollary of the previous result we obtain the following one:
Lemma 3.6.14. Assume𝑎 = −𝑏. Then,
• if𝑎 > 0,the Green’s function of problem (3.5.5) is nonnegative on [0, +∞) × ℝ, • if𝑎 < 0 the Green’s function of problem (3.5.5) is nonpositive on (−∞, 0] × ℝ,
• the Green’s function of problem (3.5.5) changes sign in any other strip not a subset of the aforementioned.
Again, for this particular case we have another way of computing the solution to the prob- lem.
Proposition 3.6.15. Let𝑎 = −𝑏, 𝐻(𝑡) ∶= ∫0u�ℎ(𝑠) d 𝑠 and ℋ(𝑡) ∶= ∫0u�𝐻(𝑠) d 𝑠. Then prob-
lem (3.5.5) has a unique solution given by
𝑢(𝑡) = 𝐻(𝑡) − 𝐻(𝑡0) − 2𝑎(ℋu�(𝑡) − ℋu�(𝑡0)) + 𝑐.
Proof. The equation is satisfied, since
𝑢′(𝑡) + 𝑎(𝑢(𝑡) − 𝑢(−𝑡)) = 𝑢′(𝑡) + 2 𝑎𝑢
u�(𝑡) = ℎ(𝑡) − 2 𝑎𝐻u�(𝑡) + 2 𝑎𝐻u�(𝑡) = ℎ(𝑡).
Example 3.6.16. Consider the problem
𝑥′(𝑡) + 𝜆(𝑥(−𝑡) − 𝑥(𝑡)) = 𝜆𝑡2− 2𝑡 + 𝜆
(1 + 𝑡2)2 , 𝑥(0) = 𝜆
for𝜆 ∈ ℝ. We can apply the theory in order to get the solution
𝑢(𝑡) = 11 + 𝑡2 + 𝜆(1 + 2𝜆𝑡) arctan 𝑡 − 𝜆2ln(1 + 𝑡2) + 𝜆 − 1
where ̄𝑢(𝑡) = 1
1+u�2 + 𝜆(1 + 2𝜆𝑡) arctan 𝑡 − 𝜆2ln(1 + 𝑡2) − 1. Observe that the real function
ℎ(𝑡) ∶= 𝜆𝑡(1 + 𝑡2− 2𝑡 + 𝜆2)2
is positive onℝ if 𝜆 > 1 and negative on ℝ for all 𝜆 < −1. Therefore, Lemma 3.6.14 guaran- tees that ̄𝑢 will be positive on (0, ∞) for 𝜆 > 1 and in (−∞, 0) when 𝜆 < −1.
In the previous chapter we dealt with order one differential equations with reflection, constant coefficients and different boundary conditions. Now, following [41] we reduce a new, more general problem containing nonconstant coefficients and arbitrary differentiable involutions, to the one studied in Chapter 3. As we will see, we will do this in three steps. First we add a term depending on𝑥(𝑡) which does not change much with respect to the previous situations. Then, moving from the reflection to a general involution is fairly simple using some of the knowledge gathered in Chapter 1.
The last step, changing from constant to nonconstant coefficients, is another matter. In the nonconstant case computing the Green’s function gets trickier and it is only possible in some situations. We use a special change of variable (only valid in some cases) that allows the obtaining the Green’s function of problems with nonconstant coefficients from the Green’s functions of constant-coefficient analogs.
4.1
Order one linear problems with involutions
Assume𝜑 is a differentiable involution on a compact interval 𝐽1⊂ ℝ. Let 𝑎, 𝑏, 𝑐, 𝑑 ∈ L1(𝐽1) and consider the following problem
𝑑(𝑡)𝑥′(𝑡) + 𝑐(𝑡)𝑥′(𝜑(𝑡)) + 𝑏(𝑡)𝑥(𝑡) + 𝑎(𝑡)𝑥(𝜑(𝑡)) = ℎ(𝑡), 𝑥(inf 𝐽
1) = 𝑥(sup 𝐽1).
(4.1.1) It would be interesting to know under what circumstances problem (4.1.1) is equivalent to another problem of the same kind but with a different involution, in particular the reflection. The following corollary of Lemma 1.2.14 will help us to clarify this situation.
Corollary 4.1.1 (CHANGE OF INVOLUTION). Under the hypothesis of Lemma 1.2.14, problem (4.1.1)
is equivalent to
𝑑(𝑓 (𝑠))
𝑓′(𝑠) 𝑦′(𝑠) + 𝑐(𝑓 (𝑠))𝑓′(𝜓(𝑠))𝑦′(𝜓(𝑠)) + 𝑏(𝑓 (𝑠))𝑦(𝑠) + 𝑎(𝑓 (𝑠))𝑦(𝜓(𝑠)) = ℎ(𝑓 (𝑠)),
𝑦(inf 𝐽2) = 𝑦(sup 𝐽2).
(4.1.2)
Proof. Consider the change of variable𝑡 = 𝑓 (𝑠) and 𝑦(𝑠) ∶= 𝑥(𝑡) = 𝑥(𝑓 (𝑠)). Then, using
Lemma 1.2.14, it is clear that d 𝑦
d 𝑠(𝑠) = d 𝑥d 𝑡 (𝑓 (𝑠))d 𝑓d 𝑠(𝑠) and d 𝑦d 𝑠(𝜓(𝑠)) = d 𝑥d 𝑡 (𝜑(𝑓 (𝑠)))d 𝑓d 𝑠(𝜓(𝑠)).
This last results allows us to restrict our study of problem (4.1.1) to the case where𝜑 is the reflection𝜑(𝑡) = −𝑡.
Now, take 𝑇 ∈ ℝ+, 𝐼 ∶= [−𝑇, 𝑇]. Equation (4.1.1), for the case 𝜑(𝑡) = −𝑡, can be reduced to the following system
Λ (𝑥′u�
𝑥′
u�) = (𝑎
u�− 𝑏u� −𝑎u�− 𝑏u�
𝑎u�− 𝑏u� −𝑎u�− 𝑏u�) (𝑥 u�
𝑥u�) + (ℎ u�
ℎu�) ,
where
Λ = (𝑐u�+ 𝑑u� 𝑑u�− 𝑐u�
𝑐u�+ 𝑑u� 𝑑u�− 𝑐u�) .
To see this, just compute the even and odd parts of both sides of the equation taking into account Corollary 1.1.7.
Now, ifdet(Λ(𝑡)) = 𝑑(𝑡)𝑑(−𝑡) − 𝑐(𝑡)𝑐(−𝑡) ≠ 0 for a. e. 𝑡 ∈ 𝐼, Λ(𝑡) is invertible a. e. and
(𝑥′u�
𝑥′
u�) = Λ
−1(𝑎u�− 𝑏u� −𝑎u�− 𝑏u�
𝑎u�− 𝑏u� −𝑎u�− 𝑏u�) (𝑥 u�
𝑥u�) + Λ
−1(ℎu�
ℎu�) .
So the general case where𝑐≡0 is reduced to the case 𝑐 = 0, taking Λ−1(𝑎u�− 𝑏u� −𝑎u�− 𝑏u�
𝑎u�− 𝑏u� −𝑎u�− 𝑏u�)
as coefficient matrix.
Hence, in the following section we will further restrict our assumptions to the case where 𝑐 ≡ 0 in problem (4.1.1).