2.8 COMPOSTAJE EN PILAS POR VOLTEO MANUAL
2.8.3. PARÁMETROS DE CONTROL DURANTE EL PROCESO
For basic Fourier series theory we will need the following three eigenvalue problems. We will consider more general equations and boundary conditions, but we will postpone this
until . 𝑥00+ 𝜆𝑥 =0, 𝑥(𝑎)=0, 𝑥(𝑏)=0, (4.1) 𝑥00+ 𝜆𝑥 =0, 𝑥0(𝑎) =0, 𝑥0(𝑏) =0, (4.2) and 𝑥00+ 𝜆𝑥 =0, 𝑥(𝑎)= 𝑥(𝑏), 𝑥0(𝑎)= 𝑥0(𝑏). (4.3)
A number𝜆is called aneigenvalueof ( ) (resp. ( ) or ( )) if and only if there exists a nonzero (not identically zero) solution to ( ) (resp. ( ) or ( )) given that specific𝜆. A nonzero solution is called a correspondingeigenfunction.
Note the similarity to eigenvalues and eigenvectors of matrices. The similarity is not just coincidental. If we think of the equations as differential operators, then we are doing the same exact thing. Think of a function𝑥(𝑡)as a vector with infinitely many components (one for each𝑡). Let𝐿 =−𝑑2
𝑑𝑡2 be the linear operator. Then the eigenvalue/eigenfunction
pair should be 𝜆and nonzero 𝑥 such that𝐿𝑥 = 𝜆𝑥. In other words, we are looking for nonzero functions𝑥satisfying certain endpoint conditions that solve(𝐿−𝜆)𝑥 =0. A lot of
the formalism from linear algebra still applies here, though we will not pursue this line of reasoning too far.
Example 4.1.3: Let us find the eigenvalues and eigenfunctions of 𝑥00+
𝜆𝑥 =0, 𝑥(0)=0, 𝑥(𝜋)=0.
We have to handle the cases𝜆 > 0,𝜆 = 0,𝜆 < 0 separately. First suppose that𝜆 > 0. Then the general solution to𝑥00+𝜆𝑥 =0 is
𝑥 =𝐴cos( √
𝜆𝑡) +𝐵sin(
√
𝜆𝑡). The condition 𝑥(0)=0 implies immediately𝐴 =0. Next
0=𝑥(𝜋)= 𝐵sin( √
𝜆 𝜋).
If 𝐵 is zero, then 𝑥 is not a nonzero solution. So to get a nonzero solution we must have that sin(
√
𝜆 𝜋) = 0. Hence, √
√
𝜆 = 𝑘 for a positive integer 𝑘. Hence the positive eigenvalues are 𝑘2 for all integers 𝑘 ≥ 1. Corresponding eigenfunctions can be taken as𝑥=sin(𝑘𝑡). Just like for eigenvectors,
constant multiples of an eigenfunction are also eigenfunctions, so we only need to pick one. Now suppose that𝜆= 0. In this case the equation is𝑥00 =0, and its general solution is𝑥 =𝐴𝑡 +𝐵. The condition 𝑥(0)=0 implies that𝐵 =0, and𝑥(𝜋)=0 implies that𝐴=0. This means that𝜆=0 isnotan eigenvalue.
Finally, suppose that𝜆 <0. In this case we have the general solution 𝑥= 𝐴cosh(
√
−𝜆𝑡) +𝐵sinh(
√
−𝜆𝑡).
Letting𝑥(0) =0 implies that𝐴=0 (recall cosh 0=1 and sinh 0=0). So our solution must be𝑥 =𝐵sinh(
√
−𝜆𝑡)and satisfy𝑥(𝜋)=0. This is only possible if𝐵is zero. Why? Because sinh𝜉is only zero when𝜉=0. You should plot sinh to see this fact. We can also see this from the definition of sinh. We get 0 = sinh𝜉 = 𝑒𝜉−2𝑒−𝜉. Hence 𝑒𝜉 = 𝑒−𝜉, which implies
𝜉=−𝜉and that is only true if𝜉 =0. So there are no negative eigenvalues. In summary, the eigenvalues and corresponding eigenfunctions are
𝜆𝑘 = 𝑘2 with an eigenfunction 𝑥𝑘 =sin(𝑘𝑡) for all integers𝑘 ≥ 1.
Example 4.1.4: Let us compute the eigenvalues and eigenfunctions of 𝑥00+
𝜆𝑥 =0, 𝑥0(0)=0, 𝑥0(𝜋)=0.
Again we have to handle the cases𝜆 > 0,𝜆 =0,𝜆 < 0 separately. First suppose that
𝜆 >0. The general solution to𝑥00+𝜆𝑥 =0 is𝑥 =𝐴cos(
√ 𝜆𝑡) +𝐵sin( √ 𝜆𝑡). So 𝑥0 =−𝐴 √ 𝜆 sin( √ 𝜆𝑡) +𝐵√𝜆 cos( √ 𝜆𝑡). The condition 𝑥0(0)=0 implies immediately𝐵 =0. Next
0=𝑥0(𝜋)=−𝐴 √
𝜆 sin(
√
𝜆 𝜋). Again𝐴cannot be zero if𝜆is to be an eigenvalue, and sin(
√
𝜆 𝜋)is only zero if √
𝜆 = 𝑘for
a positive integer𝑘. Hence the positive eigenvalues are again 𝑘2for all integers𝑘 ≥ 1. And the corresponding eigenfunctions can be taken as𝑥 =cos(𝑘𝑡).
Now suppose that𝜆 =0. In this case the equation is𝑥00=0 and the general solution is
𝑥= 𝐴𝑡+𝐵so𝑥0= 𝐴. The condition𝑥0(0)=0 implies that 𝐴=0. The condition𝑥0(𝜋)=0 also implies 𝐴 = 0. Hence 𝐵 could be anything (let us take it to be 1). So𝜆 = 0 is an eigenvalue and𝑥 =1 is a corresponding eigenfunction.
Finally, let𝜆 <0. In this case the general solution is 𝑥= 𝐴cosh( √ −𝜆𝑡) +𝐵sinh( √ −𝜆𝑡) and 𝑥0 = 𝐴 √ −𝜆 sinh( √ −𝜆𝑡) +𝐵 √ −𝜆 cosh( √ −𝜆𝑡).
∗Recall that cosh𝑠 = 1 2(𝑒
𝑠+𝑒−𝑠)
and sinh𝑠 = 12(𝑒𝑠−𝑒−𝑠)
. As an exercise try the computation with the general solution written as𝑥=𝐴𝑒
√
−𝜆𝑡+𝐵𝑒− √
−𝜆𝑡
We have already seen (with roles of𝐴and𝐵switched) that for this expression to be zero at 𝑡 =0 and𝑡 =𝜋, we must have𝐴=𝐵 =0. Hence there are no negative eigenvalues.
In summary, the eigenvalues and corresponding eigenfunctions are
𝜆𝑘 = 𝑘2 with an eigenfunction 𝑥𝑘 =cos(𝑘𝑡) for all integers 𝑘 ≥ 1, and there is another eigenvalue
𝜆0=0 with an eigenfunction 𝑥0=1.
The following problem is the one that leads to the general Fourier series. Example 4.1.5: Let us compute the eigenvalues and eigenfunctions of
𝑥00+
𝜆𝑥 =0, 𝑥(−𝜋)=𝑥(𝜋), 𝑥0(−𝜋)= 𝑥0(𝜋).
We have not specified the values or the derivatives at the endpoints, but rather that they are the same at the beginning and at the end of the interval.
Let us skip𝜆 < 0. The computations are the same as before, and again we find that there are no negative eigenvalues.
For 𝜆 = 0, the general solution is 𝑥 = 𝐴𝑡 +𝐵. The condition 𝑥(−𝜋) = 𝑥(𝜋) implies that𝐴=0 (𝐴𝜋+𝐵 =−𝐴𝜋+𝐵implies 𝐴=0). The second condition𝑥0(−𝜋) =𝑥0(𝜋)says nothing about𝐵 and hence𝜆 = 0 is an eigenvalue with a corresponding eigenfunction 𝑥=1.
For𝜆 >0 we get that𝑥 =𝐴cos(
√ 𝜆𝑡) +𝐵sin( √ 𝜆𝑡). Now 𝐴cos(− √ 𝜆 𝜋) +𝐵sin(− √ 𝜆 𝜋) | {z } 𝑥(−𝜋) =𝐴cos( √ 𝜆 𝜋) +𝐵sin( √ 𝜆 𝜋) | {z } 𝑥(𝜋) .
We remember that cos(−𝜃)=cos(𝜃)and sin(−𝜃)=−sin(𝜃). Therefore, 𝐴cos( √ 𝜆 𝜋) −𝐵sin( √ 𝜆 𝜋)= 𝐴cos( √ 𝜆 𝜋) +𝐵sin( √ 𝜆 𝜋). Hence either𝐵=0 or sin(
√
𝜆 𝜋)=0. Similarly (exercise) if we differentiate 𝑥and plug in the second condition we find that 𝐴= 0 or sin(
√
𝜆 𝜋) = 0. Therefore, unless we want𝐴
and 𝐵to both be zero (which we do not) we must have sin( √
𝜆 𝜋) = 0. Hence, √
𝜆 is an integer and the eigenvalues are yet again𝜆= 𝑘2for an integer𝑘 ≥1. In this case, however,
𝑥= 𝐴cos(𝑘𝑡) +𝐵sin(𝑘𝑡)is an eigenfunction for any 𝐴and any𝐵. So we have two linearly
independent eigenfunctions sin(𝑘𝑡)and cos(𝑘𝑡). Remember that for a matrix we can also have two eigenvectors corresponding to a single eigenvalue if the eigenvalue is repeated.
In summary, the eigenvalues and corresponding eigenfunctions are
𝜆𝑘 = 𝑘2 with eigenfunctions cos(𝑘𝑡) and sin(𝑘𝑡) for all integers 𝑘 ≥ 1,