initial commit / add all books

5 files changed, 201 insertions, 0 deletions

diff --git a/260/CH15/EX15.1/15_1.sce b/260/CH15/EX15.1/15_1.sce
new file mode 100644
index 000000000..fae9efee4
--- /dev/null
+++ b/260/CH15/EX15.1/15_1.sce
@@ -0,0 +1,47 @@
+//Eg-15.1

+//pg-605

+

+clear

+clc

+

+//using d/dx to denote the partial derivatve w.r.t x

+

+//Consider the general equation defined by the equation [2] on pg-605

+

+// A * d^2(F)/dx^2 + B * d^2(F)/dxdy + C * d^2(F)/dy^2 + D = 0;

+//for part (i) 

+

+A(1) = 1;

+B(1) = 0;

+C(1) = 1;

+D(1) = 0;

+

+//for part (ii)

+

+A(2) = 1;

+B(2) = 0;

+C(2) = -1;

+D(2) = 0;

+

+//for part(iii)

+

+A(3) = 1;

+B(3) = 0;

+C(3) = 0;

+D(3) = 0;

+

+for(i = 1:3)

+    dt(i) = B(i)^2 - 4*A(i)*C(i);

+    

+    if(dt(i) > 0)

+        printf('The discriminant of the PDE in part %d is  %f ,so it is Hyperbolic\n',i,dt(i))

+    

+    elseif(dt(i) < 0)

+        printf('The discriminant of the PDE in part %d is  %f ,so it is Elliptic\n',i,dt(i)')

+    

+    else(dt(i) == 0)

+        printf('The discriminant of the PDE in part %d is  %f ,so it is parabolic\n',i,dt(i)')

+    end

+    

+end

+

diff --git a/260/CH15/EX15.2/15_2.sce b/260/CH15/EX15.2/15_2.sce
new file mode 100644
index 000000000..88e655b83
--- /dev/null
+++ b/260/CH15/EX15.2/15_2.sce
@@ -0,0 +1,33 @@
+//Eg-15.2

+//pg-608

+

+clear

+clc

+

+// Using D in the place of greek alphabet delta

+

+// Taking square grid => Dx = Dy = 0.25

+

+// Applying central difference approximation to the second derivatives, we obtain

+

+// [T(i+1,j) - 2*T(i,j) + T(i-1,j)]/(Dx)^2 + [T(i,j+1) - 2*T(i,j) + T(i,j-1)]/(Dy)^2 = 0.    i = 1,2,3; j = 1,2,3.....this can be simplified as

+

+// T(i+1,j) + T(i-1,j) + T(i,j+1) + T(i,j-1) - 4*T(i,j) = 0

+

+//Applying the above equation to the 9 points analytically leaves us with 9 equations and 9 variables T11 to T33. This can be written in the matrix equation form Ax = B.

+

+A = [4 -1 0 -1 0 0 0 0 0;-1 4 -1 0 -1 0 0 0 0;0 -1 4 0 0 -1 0 0 0;-1 0 0 4 -1 0 -1 0 0;0 -1 0 -1 4 -1 0 -1 0;0 0 -1 0 -1 4 0 0 -1;0 0 0 -1 0 0 4 -1 0;0 0 0 0 -1 0 -1 4 -1;0 0 0 0 0 -1 0 -1 4];

+

+B = [65;25;125;40;0;100;90;50;150];

+printf('Solving the Equation Ax = B will give the values of Temperatures, where A = \n')

+disp(A)

+

+printf('\nand B = ')

+

+disp(B)

+

+printf('\nTherefore the matrix representing T11 to T33 is \n')

+

+x = inv(A)*B;

+

+disp(x)

diff --git a/260/CH15/EX15.3/15_3.sce b/260/CH15/EX15.3/15_3.sce
new file mode 100644
index 000000000..48da6eae9
--- /dev/null
+++ b/260/CH15/EX15.3/15_3.sce
@@ -0,0 +1,39 @@
+//Eg-15.3

+//pg-611

+

+clear

+clc

+

+// The value of y can be computed using the equation : 

+// y(i,j+1) = 0.68*y(i,j) + 0.16*y(i+1,j) + 0.16*y(i-1,j)

+y = zeros(5,5);

+y(1,1:5) = 25;

+y(1:5,1) = 100;

+y(1:5,5) = 100;

+

+//disp(y)

+

+for(j = 2:4)

+    for(i = 1:4)

+        y(i+1,j) = 0.68*y(i,j) + 0.16*y(i,j+1) + 0.16*y(i,j-1);

+    end

+

+end

+

+printf('The values of y for different t are shown in the table(Rows represents different time level and columns represent x0,x1,x2,x3,x4) below and the plot corresponding to this value is shown in the figure generated\n')

+disp(y)

+

+//There is a mistake in the textbook

+

+x = 0:0.25:1;

+

+y1 = y(1,1:5);

+y2 = y(2,1:5);

+y3 = y(3,1:5);

+y4 = y(4,1:5);

+y5 = y(5,1:5);

+

+plot(x,y1,x,y2,x,y3,x,y4,x,y5)

+legend('t0','t1','t2','t3','t4')

+xlabel('x')

+ylabel('y')

diff --git a/260/CH15/EX15.4/15_4.sce b/260/CH15/EX15.4/15_4.sce
new file mode 100644
index 000000000..2f31d7315
--- /dev/null
+++ b/260/CH15/EX15.4/15_4.sce
@@ -0,0 +1,58 @@
+//Eg-15.4

+//pg-617

+

+clear

+clc

+

+close()

+

+//Using equation [21] in this problem

+

+//Solving the equations analytically gives 3 equatinos in y11,y21,y31 which can be solved using matrix inversion.

+

+printf('Using the equation [21] we get :\n')

+x1 = inv([1.16 -0.08 0;-0.08 1.16 -0.08;0 -0.08 1.16])*[33;25;33];

+

+printf(' y11 = %f\n y21 = %f\n y31 = %f\n',x1(1,1),x1(2,1),x1(3,1))

+

+//Similarly solving for the elements in the column 2,3&4

+

+x2 = inv([1.16 -0.08 0;-0.08 1.16 -0.08;0 -0.08 1.16])*[43.442;26.4406;43.4442];

+

+printf(' y12 = %f\n y22 = %f\n y32 = %f\n',x2(1,1),x2(2,1),x2(3,1))

+

+x3 = inv([1.16 -0.08 0;-0.08 1.16 -0.08;0 -0.08 1.16])*[51.3531;30.0152;51.3531];

+

+printf(' y13 = %f\n y23 = %f\n y33 = %f\n',x3(1,1),x3(2,1),x3(3,1))

+

+x4 = inv([1.16 -0.08 0;-0.08 1.16 -0.08;0 -0.08 1.16])*[57.6403;34.5618;57.6403];

+

+printf(' y14 = %f\n y24 = %f\n y34 = %f\n',x4(1,1),x4(2,1),x4(3,1))

+

+y = zeros(4,5)

+

+y(1:4,1) = 100;

+y(1:4,5) = 100;

+

+y(1,2:4) = x1';

+y(2,2:4) = x2';

+y(3,2:4) = x3';

+y(4,2:4) = x4';

+

+printf('The values of y at the grid points are shown in the following table :\n')

+disp(y)

+

+printf('\nThe profiles of y by Crank-Nicolson method  is shown in the figure\n')

+

+x = 0:0.25:1;

+

+y1 = y(1,1:5);

+y2 = y(2,1:5);

+y3 = y(3,1:5);

+y4 = y(4,1:5);

+

+

+plot(x,y1,x,y2,x,y3,x,y4)

+legend('t1','t2','t3','t4')

+xlabel('x')

+ylabel('y')

diff --git a/260/CH15/EX15.5/15_5.sce b/260/CH15/EX15.5/15_5.sce
new file mode 100644
index 000000000..f61ac0617
--- /dev/null
+++ b/260/CH15/EX15.5/15_5.sce
@@ -0,0 +1,24 @@
+//Eg-15.5

+//pg-619

+

+clear

+clc

+close()

+

+y = zeros(5,5);

+

+y(1:5,1) = 10;

+y(1:5,5) = 10;

+y(1,5:4) = 0;

+//y(1,2:4) = 10;

+

+for(i = 3:5)

+    for(j = 2:4)

+        y(i,j) = y(i-1,j+1) + y(i-1,j-1) - y(i-2,j);

+    end

+end

+

+printf('The table of the y values with different rows as different timelines :\n')

+

+Y = y(2:5,:)

+disp(Y)
\ No newline at end of file