Consider the first Galerkin Example (video:
GalerkinMethod_Example). Solve this example using three trial
functions, 1(x) = x, 2(x) = x2 , and 3(x) = x3 .
Homework Answers
In this Galerkin method the Residual function R is multiplied with the weight function Wi and it is forced to be zero.
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Consider the first Galerkin Example (video:
GalerkinMethod_Example). Solve this example using three trial
functions, 1(x) =...
Consider the second Galerkin Example (videos: GalerkinDiscrete-Example_1 to 3). Solve this example if u(0) = 0,...
Consider the second Galerkin Example (videos:
GalerkinDiscrete-Example_1 to 3). Solve this example if u(0) = 0,
du(2)/dx =0, and 0 ≤ x ≤2. Every single step must be
shown.
EXAMPLE Solve ODE using Galerkin method for two equal-length elements du u(0) = 0 +1 = 0, 0 < x < 1 dx2 du Boundary conditions (1) dx We know for three nodes: X2 = 0, X2=0.5, X3=1.0; displacement at nodes = Uy, U2, U3; length of elements L1=0.5, L2=0.5 -...
Problem 1 (3 points) Solve the following boundary value problem with the Galerkin method in which...
Problem 1 (3 points) Solve the following boundary value problem with the Galerkin method in which a three-term approximation, u(x) = 60, () + 0,02(x) + C303(x) with 0,(x)=(1-x), 02(x)=x(1-x), and 03(x)=x (1-x), is used. du + x2 = 0, 0<x<1 dx2 f(0) = 1 Boundary Conditions: lu(1) = 0
solve problem #1 depending on the given information Consider the following 1D second order elliptic equation with Dirichlet boundary conditions du(x) (c(x)du ) = f(x) (a $15 b), u(a) = ga, u(b) = gb...
solve problem #1 depending on the given information
Consider the following 1D second order elliptic equation with Dirichlet boundary conditions du(x) (c(x)du ) = f(x) (a $15 b), u(a) = ga, u(b) = gb dr: where u(x) is the unknown function, ga and gb are the Dirichlet boundary values, c(x) is a given coefficient function and f(x) is a given source function. See the theorem 10.1 in the textbook for the existence and uniqueness of the solution. 1.1 Weak Formulation...
3.24 Solve the differential equation in Example 3.4.1 for the mixed boundary conditions u(0) = 0,...
3.24 Solve the differential equation in Example 3.4.1 for the mixed boundary conditions u(0) = 0, (d) = 1 dx/x=1 Use the uniform mesh of three linear elements. The exact solution is mm)_ 2 cos(1 – 2) - sin 2 - + x2 – 2 cos(1) Answer: U2 = 0.4134, Uz = 0.7958, U4 = 1.1420, (Q1)def = -1.2402. Example 3.4.1 Use the finite element method to solve the problem described by the following differential equation and boundary conditions (see...
Question Question 1 (1 mark) Attempt 1 Consider the boundary value problem: du+ U=1, 2<c<13 with...
Question Question 1 (1 mark) Attempt 1 Consider the boundary value problem: du+ U=1, 2<c<13 with u(2) = 4 and u(13) = 5 Find functions g and , such that u=gta, is a quadratic approximation that satisfies the boundary conditions. Your answer should consist of two expressions, the first representing the term g and the second representing the term ,. Both should be expressed in terms of the independent variable x. Your answers should be expressed as a function of...
do 11.3 please Example 11.2b Let us reconsider Example 11.2a, where we have 5 to invest among three projects whose return functions are f(x) = 2x . 1+x f(x) = 10( I-e-x). Let xi (j) denote the optimal...
do 11.3 please
Example 11.2b Let us reconsider Example 11.2a, where we have 5 to invest among three projects whose return functions are f(x) = 2x . 1+x f(x) = 10( I-e-x). Let xi (j) denote the optimal amount to invest in project 1 when we have maxlfi(l), f2(1), f3(1))-max(5, 1632 6.32, a total of j to invest. Because we see that Xi(1)=0, X2(I) = 0, x3(1)=1. Since max(f(xdl) + I)-f(xdl)) = max(5, I, 8.65-6.32) = 5. we have X1(2)...
For : U(x,0) = Sin(ax) a= 2.6 using the Explicit Forward Euler and Crank-Nicholson methods. Example 92. One-Dimensional...
For : U(x,0) = Sin(ax) a=
2.6
using the Explicit Forward Euler and Crank-Nicholson
methods.
Example 92. One-Dimensional Parabolic PDE: Heat Flow Equation. Consider the parabolic PDE d-u(x, t) du(x, t) 0t with the initial condition and the boundary conditions (E9.2.2) We were unable to transcribe this image
Example 92. One-Dimensional Parabolic PDE: Heat Flow Equation. Consider the parabolic PDE d-u(x, t) du(x, t) 0t with the initial condition and the boundary conditions (E9.2.2)
(1 point) Solve the heat problem with non-homogeneous boundary conditions du (x, 1) = ot (x,1),...
(1 point) Solve the heat problem with non-homogeneous boundary conditions du (x, 1) = ot (x,1), 0<x<2, t> 0 dx (0,t) = 0, (2, 1) = 2, t> 0, u(x,0) = 0<x<2. Recall that we find h(x), set u(x, t) = u(x, t)-h(x), solve a heat problem for u(x, t) and write u(x, t) = u(x, t) + h(x). Find h(x) h(x) = The solution u(x, t) can be written as u(x, t) = h(x) + u(x, t), where u(x,...
By using Karush–Kuhn–Tucker (KKT) Conditions and condition for lamda solve the example: We were unable to...
By using Karush–Kuhn–Tucker (KKT) Conditions and condition for
lamda solve the example:
We were unable to transcribe this imageExample: Consider the constrained minimization problem: 2 4 3 min xi + X2 VER? 2 8 subject to 1- Xı – x2 > 0 1- xy + x, 20 1+ x - x2 > 0 1+x+x, 20.
Consider the boundary value problem (12")=-45+4 with (2)=5 and u(6)=-1 Find the functions g., and ,...
Consider the boundary value problem (12")=-45+4 with (2)=5 and u(6)=-1 Find the functions g., and , so that y = 9+Q81+ay, is the approximate quadratic solution that satisfies the essential boundary condition. Your answer should consist of three expressions, the first representing the term g, the second representing the term , and the third representing the term og. All three expressions should be expressed in terms of the independent variable x. Your answers should be expressed as a function of...
Consider the second Galerkin Example (videos:
GalerkinDiscrete-Example_1 to 3). Solve this example if u(0) = 0,
du(2)/dx =0, and 0 ≤ x ≤2. Every single step must be
shown.
EXAMPLE Solve ODE using Galerkin method for two equal-length elements du u(0) = 0 +1 = 0, 0 < x < 1 dx2 du Boundary conditions (1) dx We know for three nodes: X2 = 0, X2=0.5, X3=1.0; displacement at nodes = Uy, U2, U3; length of elements L1=0.5, L2=0.5 -...
Problem 1 (3 points) Solve the following boundary value problem with the Galerkin method in which a three-term approximation, u(x) = 60, () + 0,02(x) + C303(x) with 0,(x)=(1-x), 02(x)=x(1-x), and 03(x)=x (1-x), is used. du + x2 = 0, 0<x<1 dx2 f(0) = 1 Boundary Conditions: lu(1) = 0
solve problem #1 depending on the given information
Consider the following 1D second order elliptic equation with Dirichlet boundary conditions du(x) (c(x)du ) = f(x) (a $15 b), u(a) = ga, u(b) = gb dr: where u(x) is the unknown function, ga and gb are the Dirichlet boundary values, c(x) is a given coefficient function and f(x) is a given source function. See the theorem 10.1 in the textbook for the existence and uniqueness of the solution. 1.1 Weak Formulation...
3.24 Solve the differential equation in Example 3.4.1 for the mixed boundary conditions u(0) = 0, (d) = 1 dx/x=1 Use the uniform mesh of three linear elements. The exact solution is mm)_ 2 cos(1 – 2) - sin 2 - + x2 – 2 cos(1) Answer: U2 = 0.4134, Uz = 0.7958, U4 = 1.1420, (Q1)def = -1.2402. Example 3.4.1 Use the finite element method to solve the problem described by the following differential equation and boundary conditions (see...
Question Question 1 (1 mark) Attempt 1 Consider the boundary value problem: du+ U=1, 2<c<13 with u(2) = 4 and u(13) = 5 Find functions g and , such that u=gta, is a quadratic approximation that satisfies the boundary conditions. Your answer should consist of two expressions, the first representing the term g and the second representing the term ,. Both should be expressed in terms of the independent variable x. Your answers should be expressed as a function of...
do 11.3 please
Example 11.2b Let us reconsider Example 11.2a, where we have 5 to invest among three projects whose return functions are f(x) = 2x . 1+x f(x) = 10( I-e-x). Let xi (j) denote the optimal amount to invest in project 1 when we have maxlfi(l), f2(1), f3(1))-max(5, 1632 6.32, a total of j to invest. Because we see that Xi(1)=0, X2(I) = 0, x3(1)=1. Since max(f(xdl) + I)-f(xdl)) = max(5, I, 8.65-6.32) = 5. we have X1(2)...
For : U(x,0) = Sin(ax) a=
2.6
using the Explicit Forward Euler and Crank-Nicholson
methods.
Example 92. One-Dimensional Parabolic PDE: Heat Flow Equation. Consider the parabolic PDE d-u(x, t) du(x, t) 0t with the initial condition and the boundary conditions (E9.2.2) We were unable to transcribe this image
Example 92. One-Dimensional Parabolic PDE: Heat Flow Equation. Consider the parabolic PDE d-u(x, t) du(x, t) 0t with the initial condition and the boundary conditions (E9.2.2)
(1 point) Solve the heat problem with non-homogeneous boundary conditions du (x, 1) = ot (x,1), 0<x<2, t> 0 dx (0,t) = 0, (2, 1) = 2, t> 0, u(x,0) = 0<x<2. Recall that we find h(x), set u(x, t) = u(x, t)-h(x), solve a heat problem for u(x, t) and write u(x, t) = u(x, t) + h(x). Find h(x) h(x) = The solution u(x, t) can be written as u(x, t) = h(x) + u(x, t), where u(x,...
By using Karush–Kuhn–Tucker (KKT) Conditions and condition for
lamda solve the example:
We were unable to transcribe this imageExample: Consider the constrained minimization problem: 2 4 3 min xi + X2 VER? 2 8 subject to 1- Xı – x2 > 0 1- xy + x, 20 1+ x - x2 > 0 1+x+x, 20.
Consider the boundary value problem (12")=-45+4 with (2)=5 and u(6)=-1 Find the functions g., and , so that y = 9+Q81+ay, is the approximate quadratic solution that satisfies the essential boundary condition. Your answer should consist of three expressions, the first representing the term g, the second representing the term , and the third representing the term og. All three expressions should be expressed in terms of the independent variable x. Your answers should be expressed as a function of...
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