Scientific Computing Partial Differential Equations Implicit Solution of Heat Equation.

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Scientific Computing Partial Differential Equations Implicit Solution of Heat Equation

Transcript of Scientific Computing Partial Differential Equations Implicit Solution of Heat Equation.

Scientific Computing

Partial Differential EquationsImplicit Solution of

Heat Equation

Explicit vs Implicit Methods

• Explicit methods have problems relating to stability

• Implicit methods overcome this but at the expense of introducing a more complicated algorithm

• In the implicit algorithm, we develop simultaneous equations for u at the j-th and (j+1)-st time steps

Explicit Method

We use the centered-difference approximation for uxx at time step j:

ui−1, j − 2ui, j + ui+1, j

h2

grid point involved with space difference grid point involved with time difference

Implicit Method

We use the centered-difference approximation for uxx at step (j+1) :

ui−1, j+1 − 2ui, j+1 + ui+1, j+1

h2

grid point involved with space difference grid point involved with time difference

• We use the forward-difference formula for the time derivative and the centered-difference formula for the space derivative at time step (j+1):

Implicit Method

ut (x i, t j ) ≅ui, j+1 − ui, j

k

cuxx (x i, t j ) ≅ cui−1, j+1 − 2ui, j+1 + ui+1, j+1

h2

• Then the heat equation (ut =cuxx ) can be approximated as

Or,

Let r = (ck/h2) Solving for ui,j we get:

Implicit Method

ui, j+1 − ui, jk

= cui−1, j+1 − 2ui, j+1 + ui+1, j+1

h2

ui, j+1 − uij =ck

h2 ui−1, j+1 − 2ui, j+1 + ui+1, j+1( )

ui, j = −rui−1, j+1 + (1+ 2r)ui, j+1 − rui+1, j+1( )

• Putting in the boundary conditions, we get the following equations for the (implicit) solution to the heat equation is:

Implicit Method

ui, j = −rui−1, j+1 + (1+ 2r)ui, j+1 − rui+1, j+1( ) j ≥ 0 i = 2,K n − 2

u1, j = −rg0, j+1 + (1+ 2r)u1, j+1 − ru2, j+1( ) j ≥ 0

un−1, j = −run−2, j+1 + (1+ 2r)un−1, j+1 − rg1, j+1( ) j ≥ 0

ui,0 = f i i = 0K n

Matrix Form of Solution

1+ 2r −r

−r 1+ 2r −r

O O O

−r 1+ 2r −r

−r 1+ 2r

⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎥ ⎥ ⎥ ⎥ ⎥ ⎥

u1, j+1

u2, j+1

M

un−2, j+1

un−1, j+1

⎜ ⎜ ⎜ ⎜ ⎜ ⎜

⎟ ⎟ ⎟ ⎟ ⎟ ⎟

+

−rg0, j+1

0

M

0

−rg1, j+1

⎜ ⎜ ⎜ ⎜ ⎜ ⎜

⎟ ⎟ ⎟ ⎟ ⎟ ⎟

=

u1, j

u2, j

M

un−2, j

un−1, j

⎜ ⎜ ⎜ ⎜ ⎜ ⎜

⎟ ⎟ ⎟ ⎟ ⎟ ⎟

ui, j = −rui−1, j+1 + (1+ 2r)ui, j+1 − rui+1, j+1( ) j ≥ 0 i = 2,K n − 2

u1, j = −rg0, j+1 + (1+ 2r)u1, j+1 − ru2, j+1( ) j ≥ 0

un−1, j = −run−2, j+1 + (1+ 2r)un−1, j+1 − rg1, j+1( ) j ≥ 0

ui,0 = f i i = 0K n

This is of the formTo solve for u(:,j) we need to use a linear systems method.Since we have to solve this system repeatedly, a good choice

is the LU decomposition method.

Matrix Form of Solution

Au(:, j +1) +b = u(:, j)

1+ 2r −r

−r 1+ 2r −r

O O O

−r 1+ 2r −r

−r 1+ 2r

⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎥ ⎥ ⎥ ⎥ ⎥ ⎥

u1, j+1

u2, j+1

M

un−2, j+1

un−1, j+1

⎜ ⎜ ⎜ ⎜ ⎜ ⎜

⎟ ⎟ ⎟ ⎟ ⎟ ⎟

+

−rg0, j+1

0

M

0

−rg1, j+1

⎜ ⎜ ⎜ ⎜ ⎜ ⎜

⎟ ⎟ ⎟ ⎟ ⎟ ⎟

=

u1, j

u2, j

M

un−2, j

un−1, j

⎜ ⎜ ⎜ ⎜ ⎜ ⎜

⎟ ⎟ ⎟ ⎟ ⎟ ⎟

function z = implicitHeat(f, g0, g1, T, n, m, c)%Simple Implicit solution of heat equation % Constants h = 1/n; k = T/m; r = c*k/h^2; % x and t vectors x = 0:h:1; t = 0:k:T; % Boundary conditions u(1:n+1, 1) = f(x)'; u(1, 1:m+1) = g0(t); u(n+1, 1:m+1) = g1(t);

Matlab Implementation

% Set up tri-diagonal matrix for interior points of the grid A = zeros(n-1, n-1); % 2 less than x grid size for i= 1: n-1 A(i,i)= (1+2*r); if i ~= n-1 A(i, i+1) = -r; end if i ~= 1 A(i, i-1) = -r; end end % Find LU decomposition of A [LL UU] = lu_gauss(A); % function included below

Matlab Implementation

% Solve for u(:, j+1); for j = 1:m % Set up vector b to solve for in Ax=b b = zeros(n-1,1); b = u(2:n, j)'; % Make a column vector for LU solver b(1) = b(1) + r*u(1,j+1); b(n-1) = b(n-1) + r*u(n+1,j+1); u(2:n, j+1) = luSolve(LL, UU, b); end z=u'; % plot solution in 3-d mesh(x,t,z);end

Matlab Implementation

Usage: f = inline(‘x.^4’); g0 = inline(‘0*t’); g1 = inline(‘t.^0’); n=5; m=5; c=1; T=0.5; z = implicitHeat(f, g0, g1, T, n, m, c);

Matlab Implementation

Calculated solution appears stable:

Matlab Implementation

To analyze the stability of the method, we again have to consider the eigenvalues of A for the equation

We have

Using the Gershgorin Theorem, we see that eigenvalues are contained in circles centered at (1+2r) with max radius of 2r

Stability

Au(:, j +1) +b = u(:, j)

A =

1+ 2r −r

−r 1+ 2r −r

O O O

−r 1+ 2r −r

−r 1+ 2r

⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎥ ⎥ ⎥ ⎥ ⎥ ⎥

Thus, if λ is an eigenvalue of A, we have

Thus, all eigenvalues are at least 1. In solving for u(:,j+1) in we haveSince the eigenvalues of A-1 are the reciprocal of the

eigenvalues of A, we have that the eigenvalues of A-1 areall <= 1. Thus, the implicit algorithm yields stable iterates – they do

not grow without bound.

Stability

1+ 2r − 2r ≤ λ ≤1+ 2r + 2r

⇒ 1≤ λ ≤1+ 4r

Au(:, j +1) +b = u(:, j)

u(:, j +1) = A−1(u(:, j) −b)

• Convergence means that as x and t approach zero, the results of the numerical technique approach the true solution

• Stability means that the errors at any stage of the computation are attenuated, not amplified, as the computation progresses

• Truncation Error refers to the error generated in the solution by using the finite difference formulas for the derivatives.

Convergence and Stability

Example: For the explicit method, it will be stable if r<= 0.5 and will have truncation error of O(k+h2) .

Implicit Method is stable for any choice of r and will again have truncation error of O(k+h2) .

It would be nice to have a method that is O(k2+h2). The Crank-Nicholson method has this nice feature.

Convergence and Stability

Crank-Nicholson Method

We average the centered-difference approximation

for uxx at time steps j and j+1:

ui−1, j − 2ui, j + ui+1, j

h2

grid point involved with space difference grid point involved with time difference

ui−1, j+1 − 2ui, j+1 + ui+1, j+1

h2

Crank-Nicholson Method

We get the following

ui, j+1 − uij =ck

2h2 ui−1, j − 2ui, j + ui+1, j( )

+ck

2h2 ui−1, j+1 − 2ui, j+1 + ui+1, j+1( )

• Putting in the boundary conditions, we get the following equations for the (C-N) solution to the heat equation is:

Crank-Nicholson Method

−r

2ui−1, j+1 + (1+ r)ui, j+1 −

r

2ui+1, j+1

⎝ ⎜

⎠ ⎟=r

2ui−1, j + (1− r)ui, j +

r

2ui+1, j

⎝ ⎜

⎠ ⎟

−r

2g0, j+1 + (1+ r)u1, j+1 −

r

2u2, j+1

⎝ ⎜

⎠ ⎟=r

2g0, j + (1− r)u1, j +

r

2u2, j

⎝ ⎜

⎠ ⎟

−r

2un−2, j+1 + (1+ r)un−1, j+1 −

r

2g1, j+1

⎝ ⎜

⎠ ⎟=r

2un−2, j + (1− r)un−1, j +

r

2g1, j

⎝ ⎜

⎠ ⎟

ui,0 = f i

• Class Project: Determine the matrix for the C-N method and revise the Matlab implicitHeat function to implement C-N.

Crank-Nicholson Method

−r

2ui−1, j+1 + (1+ r)ui, j+1 −

r

2ui+1, j+1

⎝ ⎜

⎠ ⎟=r

2ui−1, j + (1− r)ui, j +

r

2ui+1, j

⎝ ⎜

⎠ ⎟

−r

2g0, j+1 + (1+ r)u1, j+1 −

r

2u2, j+1

⎝ ⎜

⎠ ⎟=r

2g0, j + (1− r)u1, j +

r

2u2, j

⎝ ⎜

⎠ ⎟

−r

2un−2, j+1 + (1+ r)un−1, j+1 −

r

2g1, j+1

⎝ ⎜

⎠ ⎟=r

2un−2, j + (1− r)un−1, j +

r

2g1, j

⎝ ⎜

⎠ ⎟

ui,0 = f i