Per il teorema di Rouch\u00e9-Capelli, poich\u00e9 det(A) \u2260 0, ossia A non \u00e8 singolare, esiste ed \u00e8 unica la soluzione v \u03f5 \u211cn<\/sup> data da:<\/p>\n\n\n\n v = A-1<\/sup>b<\/p>\n\n\n\n Tuttavia, poich\u00e9 il calcolo della matrice inversa A-1<\/sup> \u00e8 costoso e in generale malcondizionato, si cercano altri metodi per determinare la soluzione, i quali non richiedono il calcolo esplicito della matrice inversa A-1<\/sup>.<\/p>\n\n\n\n Alla base dei metodi iterativi c\u2019\u00e8 l\u2019idea do costruire un\u2019opportuna successione di vettori x(k)<\/sup> tale che:<\/p>\n\n\n\n Per semplicit\u00e0, sia con A \u03f5 \u211c3 x 3<\/sup> con aii<\/sub> \u2260 0 per ogni i = 1,\u2026,3 cio\u00e8 il sistema lineare:<\/p>\n\n\n\n Tale sistema pu\u00f2 essere riscritto come:<\/p>\n\n\n\n Tale riscrittura suggerisce direttamente la formulazione di un metodo iterativo nel modo seguente:<\/p>\n\n\n\n Ove si assume che venga fornito un vettore x(0)<\/sup>, detto vettore d\u2019innesco, a partire dal quale viene generata la successione. Per l\u2019i-esima equazione si pu\u00f2 formalizzare in:<\/p>\n\n\n\n Per i = 1, \u2026, n.<\/p>\n\n\n\n Il costo in termini di operazioni per calcolare una nuova iterazione \u00e8 n2<\/sup>. Tale costo dovr\u00e0 essere a sua volta moltiplicato per il numero di iterazioni effettuate.<\/p>\n\n\n\n Infine, si tenga presente che se l\u2019ipotesi aii<\/sub> \u2260 0 per ogni i non \u00e8 verificata, si possono scambiare fra loro equazioni del sistema cos\u00ec che risulti soddisfatta.<\/p>\n\n\n\n Una variante del metodo di Jacobi pu\u00f2 essere semplicemente ottenuta utilizzando le componenti della soluzione gi\u00e0 aggiornate. Pi\u00f9 precisamente, nel caso n = 3 il metodo Gauss-Seidel \u00e8 dato da:<\/p>\n\n\n\n Infatti, nella seconda equazione si utilizza il valore x1<\/sub>(k+1)<\/sup> appena calcolato al posto del valore x1<\/sub>(k)<\/sup> e nella terza equazione si utilizzano i valori x1<\/sub>(k+1)<\/sup> e x2<\/sub>(k+1)<\/sup> appena calcolati al posto di x1<\/sub>(k)<\/sup> e x2<\/sub>(k)<\/sup>.<\/p>\n\n\n\n Per un sistema di generica dimensione n il metodo di Gauss-Seidel ha la seguente formulazione:<\/p>\n\n\n\n Per i = 1, \u2026, n.<\/p>\n\n\n\n Come nel caso del metodo di Jacobi, il costo in termini di operazioni di tipo moltiplicativo per calcolare una nuova iterazione \u00e8 n2<\/sup>. Tale costo dovr\u00e0 essere moltiplicato per il numero di iterazioni.<\/p>\n\n\n\n Se la matrice A \u00e8 diagonalmente dominante in senso stretto, ossia se per ogni i = 1,\u2026,n si ha:<\/p>\n\n\n\n Allora i metodi di Jacobi e Gauss-Seidel convergono.<\/p>\n\n\n\n Esempio:<\/p>\n\n\n\n Quindi:<\/p>\n\n\n\n La soluzione analitica \u00e8 semplice e fornisce x = [1,2,3]T<\/sup>.<\/p>\n\n\n\n La matrice \u00e8 diagonalmente dominante, poich\u00e9 per ogni elemento della diagonale, la somma degli altri due elementi di riga \u00e8 minore.<\/p>\n\n\n\n Il seguente codice in C# trova la soluzione con il metodo iterativo di Gauss-Seidel:<\/p>\n\n\n Facciamo un esempio concreto.<\/p>\n\n\n\n La legge di Darcy-Weisbach<\/strong> stabilisce che la differenza di pressione ai capi di un tubo lungo L metri, di diametro D vale:<\/p>\n\n\n\n Dove \u03bb<\/em> \u00e8 il coefficiente d\u2019attrito<\/strong>, \u03c1<\/em> e w<\/em> sono rispettivamente la densit\u00e0 e la velocit\u00e0 del fluido<\/strong>.<\/p>\n\n\n\n In un tubo liscio, il coefficiente d\u2019attrito vale \u03bb = 64 \/ Re se Re < 2000, mentre vale \u03bb \u2248 0.316\u2219Re-0.25<\/sup> altrimenti (per Re < 105<\/sup>). Dove introduciamo la viscosit\u00e0 dinamica \u03bc<\/strong>.<\/p>\n\n\n\n Se Re \u00e8 minore o uguale a 2000 il moto \u00e8 laminare, cio\u00e8 dominato dalle forze viscose. Se invece \u00e8 compreso tra 2000 e 4000 c\u2019\u00e8 una zona di transizione, dove il moto tende a mantenere le proprie caratteristiche, al di sopra \u00e8 turbolento.<\/p>\n\n\n\n Se conosco la portata volumetrica Q di un fluido noto (\u03c1 e \u03bc noti) che attraversa il tubo, \u00e8 dunque facile ricavare \u0394p, perch\u00e9 la velocit\u00e0 media \u00e8 w = Q\/A, quindi calcolo Re. A seconda che sia maggiore di 2000 oppure no, calcolo il coefficiente d\u2019attrito \u03bb appropriato e infine applico la legge di Darcy-Weisbach.<\/p>\n\n\n\n Un po\u2019 meno ovvio \u00e8 se conosco il \u0394p e mi chiedono la portata.<\/p>\n\n\n\n Il problema \u00e8 che questa volta devo partire dalla legge di Darcy-Weisbach, ma non conosco ancora il coefficiente d\u2019attrito \u03bb, perch\u00e9 ancora non so dire se il moto \u00e8 laminare oppure turbolento.<\/p>\n\n\n\n Anche in questo caso, un algoritmo iterativo ci pu\u00f2 venire in soccorso. Ed ecco il codice, sempre in C#:<\/p>\n\n\n using System;<\/span> \u00c8 dato il sistema lineare: Ax = b con A \u03f5 \u211cn x n e x,b \u03f5 \u211cn con det(A) \u2260 0. Per il teorema di Rouch\u00e9-Capelli, poich\u00e9 det(A) \u2260 0, ossia A non \u00e8 singolare, esiste ed \u00e8 unica la soluzione v \u03f5 \u211cn data da: v = A-1b Tuttavia, poich\u00e9 il calcolo della […]<\/p>\n","protected":false},"author":1,"featured_media":122,"comment_status":"open","ping_status":"open","sticky":true,"template":"","format":"standard","meta":{"footnotes":""},"categories":[24],"tags":[11,13,7,8,10,9,12],"class_list":["post-112","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-metodi-numerici","tag-darcy","tag-darcy-weisbach","tag-gauss","tag-gauss-seidel","tag-jacobi","tag-metodi-iterativi","tag-weisbach"],"_links":{"self":[{"href":"https:\/\/www.paoloferraresi.it\/index.php?rest_route=\/wp\/v2\/posts\/112","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.paoloferraresi.it\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.paoloferraresi.it\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.paoloferraresi.it\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.paoloferraresi.it\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=112"}],"version-history":[{"count":12,"href":"https:\/\/www.paoloferraresi.it\/index.php?rest_route=\/wp\/v2\/posts\/112\/revisions"}],"predecessor-version":[{"id":170,"href":"https:\/\/www.paoloferraresi.it\/index.php?rest_route=\/wp\/v2\/posts\/112\/revisions\/170"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.paoloferraresi.it\/index.php?rest_route=\/wp\/v2\/media\/122"}],"wp:attachment":[{"href":"https:\/\/www.paoloferraresi.it\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=112"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.paoloferraresi.it\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=112"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.paoloferraresi.it\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=112"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}![\"\"](\"http:\/\/www.paoloferraresi.it\/wp-content\/uploads\/2020\/07\/immagine.jpg\")
<\/figure><\/div>\n\n\n\nMetodo di Jacobi<\/h2>\n\n\n\n
![\"\"](\"http:\/\/www.paoloferraresi.it\/wp-content\/uploads\/2020\/07\/immagine-2.jpg\")
<\/figure><\/div>\n\n\n\n![\"\"](\"http:\/\/www.paoloferraresi.it\/wp-content\/uploads\/2020\/07\/immagine-3.jpg\")
<\/figure><\/div>\n\n\n\n![\"\"](\"http:\/\/www.paoloferraresi.it\/wp-content\/uploads\/2020\/07\/immagine-4.jpg\")
<\/figure>\n\n\n\n![\"\"](\"http:\/\/www.paoloferraresi.it\/wp-content\/uploads\/2020\/07\/immagine-5.jpg\")
<\/figure><\/div>\n\n\n\nMetodo di Gauss-Seidel<\/h2>\n\n\n\n
![\"\"](\"http:\/\/www.paoloferraresi.it\/wp-content\/uploads\/2020\/07\/immagine-6.jpg\")
<\/figure><\/div>\n\n\n\n![\"\"](\"http:\/\/www.paoloferraresi.it\/wp-content\/uploads\/2020\/07\/immagine-7.jpg\")
<\/figure><\/div>\n\n\n\n![\"\"](\"http:\/\/www.paoloferraresi.it\/wp-content\/uploads\/2020\/07\/immagine-8.jpg\")
<\/figure><\/div>\n\n\n\n![\"\"](\"http:\/\/www.paoloferraresi.it\/wp-content\/uploads\/2020\/07\/immagine-9.jpg\")
<\/figure><\/div>\n\n\n\n![\"\"](\"http:\/\/www.paoloferraresi.it\/wp-content\/uploads\/2020\/07\/immagine-10.jpg\")
<\/figure><\/div>\n\n\n\nusing System;<\/span>
namespace GaussSeidel<\/span>
{<\/span>
class Program<\/span>
{<\/span>
\/\/\/ <summary> Solve Ax=b, with the Gauss-Seildel iterative method<\/span>
\/\/\/ static bool GaussSeidel(A, b, x, MaxIter=20, AbsError = 1e-15)<\/summary><\/span>
\/\/\/ <param name=\"double[,] A\">A matrix<\/param><\/span>
\/\/\/ <param name=\"double[] x\">b vector<\/param><\/span>
\/\/\/ <param name=\"double[] b\">b vector<\/param><\/span>
\/\/\/ <param name=\"int MaxIter=20\">Maximum number or iterations<\/param><\/span>
\/\/\/ <param name=\"double AbsError = 1e-15\">Absolute error for convergence<\/param><\/span>
\/\/\/ <returns>bool convergence. True if the solution converges.<\/returns><\/span>
static bool GaussSeidel(double[,] A, double[] x, double[] b, int MaxIter=20, double AbsError = 1e-15)<\/span>
{<\/span>
int n = b.Length;<\/span>
double[] y = new double[n], last = (double[])x.Clone();<\/span>
bool convergence;<\/span>
do<\/span>
{<\/span>
\/\/ Gauss-Seidel<\/span>
for(int i=0;i<n;++i)<\/span>
{<\/span>
x[i] = (b[i] \/ A[i, i]);<\/span>
for (int j = 0; j < n; ++j)<\/span>
{<\/span>
if (j != i)<\/span>
{<\/span>
x[i] -= ((A[i, j] \/ A[i, i]) * y[j]);<\/span>
y[i] = x[i];<\/span>
}<\/span>
}<\/span>
}<\/span>
\/\/ check for convergence<\/span>
convergence = true;<\/span>
for (int i = 0; i < n; ++i)<\/span>
{<\/span>
if (Math.Abs(x[i] - last[i]) > AbsError)<\/span>
{<\/span>
convergence = false;<\/span>
break;<\/span>
}<\/span>
}<\/span>
Array.Copy(x, last, n);<\/span>
} while (!convergence && --MaxIter > 0);<\/span>
return convergence;<\/span>
}<\/span>
static double[,] A = new double[,]<\/span>
{<\/span>
{4,-1,0},<\/span>
{-1,4,-1},<\/span>
{0,-1,4}<\/span>
};<\/span>
static double[] b = new double[] { 2, 4, 10 };<\/span>
static void Main(string[] args)<\/span>
{<\/span>
double[] x = { 1, 1, 1 }; \/\/ innesco<\/span>
bool cnv = GaussSeidel(A, x, b);<\/span>
if (cnv)<\/span>
{<\/span>
string s = string.Join(\",\", x);<\/span>
Console.WriteLine(\"La soluzione \u00e8: {0}\", s);<\/span>
} else<\/span>
{<\/span>
Console.WriteLine(\"La soluzione non converge!\");<\/span>
}<\/span>
}<\/span>
}<\/span>
}<\/span><\/pre>\n\n\nApplicazione ai sistemi non lineari<\/h2>\n\n\n\n
![\"\"](\"http:\/\/www.paoloferraresi.it\/wp-content\/uploads\/2020\/07\/immagine-11.jpg\")
<\/figure><\/div>\n\n\n\n
Re \u00e8 il numero di Reynolds<\/strong>:<\/p>\n\n\n\n![\"\"](\"http:\/\/www.paoloferraresi.it\/wp-content\/uploads\/2020\/07\/immagine-12.jpg\")
<\/figure><\/div>\n\n\n\n![\"\"](\"http:\/\/www.paoloferraresi.it\/wp-content\/uploads\/2020\/07\/immagine-13.jpg\")
<\/figure>\n\n\n\n
Si tratta di ricalcolare iterativamente le tre equazioni, fino a quando si trova una convergenza o al massimo numero di tentativi.<\/p>\n\n\n\n
namespace Darcy<\/span>
{<\/span>
\u00a0\u00a0\u00a0 class Program<\/span>
\u00a0\u00a0\u00a0 {<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 static double FlowRateFromdP(double dP, double L, double D,<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 double rho, double mu, double lambda, int MaxIter = 100,<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 double AbsError = 1e-15)<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 {<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 if (Math.Abs(dP) < AbsError) return 0.0;<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 double sgn;<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 if (dP < 0.0)<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 {<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 dP = -dP;<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 sgn = -1.0;<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 }<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 else sgn = 1.0;<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 double w = 0; \/\/ flow velocity<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 double Re = 0; \/\/ Reynolds number<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 double wLast = w, ReLast = Re,<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 lambdaLast = lambda;<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 bool convergence;<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 do<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 {<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 if (lambda > 0.0)<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 {<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 w = Math.Sqrt(2 * D * dP \/ (rho * L * lambda));<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 }<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 else w = 0.0;<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Re = rho * w * D \/ mu;<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 if (Re<=64) lambda = 1;<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 else if (Re <= 2000) lambda = 64 \/ Re;<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 else lambda = 0.3164 \/ Math.Pow(Re, 0.25);<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 convergence = (Math.Abs(w – wLast) < AbsError) &&<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (Math.Abs(Re – ReLast) < AbsError) &&<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (Math.Abs(lambda – lambdaLast) < AbsError);<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 wLast = w;<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 ReLast = Re;<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 lambdaLast = lambda;<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 } while (!convergence && –MaxIter > 0);<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 return sgn*w * (0.25 * Math.PI * D * D);<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 }<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 static void Main(string[] args)<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 {<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 double dP = 7.85e+5;<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 double L = 1.0;<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 double D = 1.0e-2;<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 double rho = 850.0;<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 double mu = 0.051;<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 double lambda = 0;<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 double Q = FlowRateFromdP(dP, L, D, rho,<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 mu, lambda, AbsError: 1e-6);<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Console.WriteLine(“Q = {0} m3\/s”,Q);<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Console.ReadKey();<\/span>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 }<\/span>
\u00a0\u00a0\u00a0 }<\/span>
}<\/span><\/p>","protected":false},"excerpt":{"rendered":"