KTH – ROYAL INSTITUTE OF TECHNOLOGY

School of Engineering Sciences

Course: Applied Numerical Methods

Course code: SF2520

Authors: 1) Andreas Angelou

2) Vasileios Papadimitriou

3) Dionysios Zelios

Lab name: Partial Differential Equation of Parabolic Type

Date of submission: 5/12/2013

CONTENTS

Part A ……………………………………………………………………………………………………………….3

Part B………………………………………………………………………………………………………………..4 Part C………………………………………………………………………………………………………………..6 Part D………………………………………………………………………………………………………………..8 Part E………………………………………………………………………………………………………………...8 Part F……………………………………………………………………………………………………………….…9 Appendix……………………………………………………………………………………………………………13

References…………………………………………………………………………………………………………15

Partial Differential Equation of Parabolic Type

Problem description: We have a metallic rod that is isolated on the right end. At time t=0 we

heat the left end. After some time, the right end will be warmer and then it cools off again.

The density of the rod is given as ρ[kg/m3]. The heat capacity is Cp[J/(kg*C)] and

k[J/(m*s*C)] is the thermal conductivity. The problem can be described by the following:

Length of rod is L[m]. At x=0[m] we have the left end and at x=L[m] we have the

right end.

Initially (t<0) the temperature of the rod is T=0[C].

At time t=0 we heat the left end of the rod with a pulse of temperature T=T0.

The heat pulse lasts for a duration of t=tp[s].

The heat diffusion process can be described by the following partial differential

equation: ρ*Cp*(θT/θt)=k*(θ2/θx2), t>0, 0<=x<=L

The Boundary Conditions are: 1A) T(0,t)=T0, if 0<=t<=tp

1B) T(0,t)=0, if t>tp

2) θΤ/θt(L,t)=0

Initial Conditions: 1) T(x,0)=T0, if x=0

2) T(x,0)=0, if 0<x<=L

Part A

Show that with the new values u, ξ and τ defined by: T=T0*u, x=L*ξ and t=tp*τ, the problem

can be transformed(scaled) into the following dimensionless form: θu/θτ=α*(θ2u/θξ2), τ>0,

0<ξ<1, with Boundary Conditions:

1a) u(0,τ)=1, if 0<=τ<=1

1b) u(0,τ)=0, if τ>1

2)θu/θξ(1,τ)=0 and Initial Conditions: 1) u(ξ,0)=1, if ξ=0, 2) u(ξ,0)=0, if 0<ξ<=1.

Show that the only remaining parameter α is dimensionless.

Solution

The differential equation of the problem is ρ*Cp*(θT/θt)=k*(θ2/θx2), t>0, 0<=x<=L.

So we can say that:

t>0 => tp*τ>0 => τ>0 (since tp>0)

0<=x<=L => 0<=L*ξ<=L =>(we divide by L) 0<=ξ<=1

(1) Τ=Τ0*u => θΤ=Τ0θu, (2) x=L*ξ => θx=L*θξ and (3) t=tp*τ => θt=tp*θτ

θ2Τ/θx2 = θ/θx(θΤ/θx )= θ/(L*θξ)[(Τ0*θu)/(L*θξ)] = (Τ0/L2)*( θ2u/θξ2).

Thus, the partial differential equation of the problem can be rewritten by using the

relations (1), (2) and (3) as ρ*Cp*(θT/θt)=k*(θ2/θx2) =>

ρ*Cp*[(T0*θu)/(tp*θt)]=k*T0/L2*(θ2u/θξ2) =>(we transfer all constants to the right hand

side) θu/θt=(k*tp)/(ρ*Cp*L2)*(θ2u/θξ2). We “name” α=(k*tp)/(ρ*Cp*L2) and we have

θu/θt=α*(θ2u/θξ2). Also, we can easily see that α is dimensionless:

[α]=[((J/m*s*C)*s)/((kg/m3)*(J/kg*s)*m2)] => [α]=[(J*s*m3*kg*C)/(m*s*C*kg*J*m2)] =>

[α]=[] (dimensionless).

The Boundary Conditions are: 1A) T(0,t)=T0, if 0<=t<=tp , 1B) T(0,t)=0, if t>tp

For 0<=t<=tp => 0<=tp*τ<=tp => 0<=τ<=1 we have T(0,t)=T0 => T0*u(0,τ)=Τ0 => u(0,τ)=1

For t>tp => tp*τ>tp => τ>1 we have T(0,t)=0 => T0*u(0,τ)=0 => u(0,τ)=0

For x=L we have x=L*ξ => L=L*ξ =>ξ=1 and t=tp*τ, so θΤ/θx(L,t)=0 => (T0*θu)/(L*θξ)(1,τ)=0 =>

θu/θξ(1,τ)=0

Hence, the Boundary Conditions can be written as: 1A) u(0,τ)=1, if 0<=τ<=1 , 1b) u(0,τ)=0, if

τ>1, 2) θu/θξ(1,τ)=0

The Initial Conditions are: 1) T(x,0)=T0, if x=0 , 2) T(x,0)=0, if 0<x<=L

We have that x=L*ξ, so for x=0 => L*ξ=0 => ξ=0 and T(x,0)=T0 => T0*u(ξ,0)=T0 => u(ξ,0)=1

For x=L => L*ξ=L => ξ=1 and T(x,0)=0 => T0*u(ξ,0)=0 => u(ξ,0)=0.

Hence, the Initial Conditions can be written as 1) u(ξ,0)=1, if ξ=0, 2) u(ξ,0)=0, if 0<ξ<=1

We can consider a=1, for the following parts.

Part B

Discretize the ξ variable using constant stepsize h and central differences to obtain an ODE

system du/dt=A*u+b(τ), u(0)=u0, where u0 is the zero vector. Show the dimensions and

structures of A, b(τ) and u0. Show also the discretized grid of the ξ-axis you have used and

how the gridpoints are numbered.

Solution

Our model problem is θu/θτ=α*(θ2u/θξ2) =>(α=1) θu/θτ=(θ2u/θξ2) with Boundary

Conditions u(0,τ)=1, if 0<=τ<=1, u(0,τ)=0, if τ>1 and θu/θξ(1,τ)=0 and with Initial

Condition u(ξ,0)=1, if ξ=0 and u(ξ,0)=0, if 0<ξ<=1.

We make a discretization with the MoL by introducing an extra gridpoint after the

right boundary point.

Hence ξi=i*hξ, where i=0,1,2,…,N+1 and hξ=1/(Ν+1). The ODE system is

θ²ui,k(t,ξ)/θτ²=(ui+1,k(τ,ξ)-2* ui,k(τ,ξ)+ ui-1,k(τ,ξ))/ hξ2, where ui(τ,ξ) is a time dependent solution

function associated to the space point ξi. We also know that θu(τ,ξ)= (ui,k+1(τ,ξ)- ui,k(τ,ξ))/2 ht

Hence, substituting those partial differential equations to our initial equation and solving

with respect to ui+1,k :

ui+1,k =((2* ht )/ hξ2

)*(ui+1,k(τ,ξ)-2* ui,k(τ,ξ)+ ui-1,k(τ,ξ)) + ui,k(τ,ξ)

The Boundary condition can now be transformed:

θu(1,τ)/ θτ =0 => (UN+1- UN-1)/ hξ2 =0 => UN+1= UN-1

Initial Condition is given u(ξ,0)=1 for ξ=0, and u(ξ,0)=0 for 0<ξ<=1 . The above ODEs can be

written in the matrix form θu/θτ=A*u+b(τ) (stiff system of ODEs), u(0)=u0, where u0 is the

zero vector.

A=(1/hξ2)*

2 01 0

1 2 1

0 0

11 2

0 20 1

b(τ)= =(1/hξ2)*

1

1

1

1

0

and u(0)=

1

0

0

0

0

Part C

We discretize the ODE with the Euler’s explicit method using constant stepsize Δt. In order to

make our calculations efficient we write our code so that the vector Au+b is formed directly.

We store the whole approximate solution (including initial and boundary condition) in a

large matrix U, where the first column is the initial condition while the first and the last row is

the boundary condition.

We use ‘surf’ to draw a 3D plot of the solution. We make a graph showing a stable solution

and one graph with an unstable solution.

Stable solution

h Δτ Δτ/h²

0.2 0.004 0.1

Unstable solution

h Δτ Δτ/h²

0.2 0.013 0.325

Part D

In this part of the lab we compare the explicit method in ode23 and the implicit method in

ode23s, suitable for stiff problems, in the Matlab library. The two functions run under the

similar conditions (same problem, same tolerance) and for three values of the stepsize h

corresponding to N=10, N=20, N=40 grid points on the ξ-axis.

The comparison comprises:

i. The number of time steps needed to reach τ=2

ii. The cpu-time needed to do each computation

iii. The maximal stepsize that each method could take

The results are presented in the following table:

N Time steps Cpu-time (sec) hmax

Ode23 | Ode23s Ode23 | Ode23s Ode23 | Ode23s

10 458 | 252 0.562 | 0.807 0.0053 | 0.0456

20 1477 | 298 0.754 | 1.055 0.0006 | 0.0456

40 5424 | 347 1.440 | 1.610 0.0004 | 0.0434

We conclude that the number of time steps is considerably smaller when using a stiff

method.

Part E

With the Malab function odeset , options can be set up by the user in order to make the

computation more efficient. By activating the options ‘Jacobian’, ‘Jconstant’ and ‘Jpattern’,

we see that we can smooth out the graph for ode-23s and we have the following table:

N Time steps Cpu-time (sec) hmax

Ode23 | Ode23s Ode23 | Ode23s Ode23 | Ode23s

10 458 | 876 0.562 | 0.710 0.0053 | 0.0046

20 1477 | 2099 0.754 | 1.019 0.0006 | 0.0012

40 5424 | 6892 1.440 | 2.396 0.0004 | 0.0003

Part F

We visualize the result of a successful computation from part D in graphs.

The following graphs show two-dimensional plots of u(t,ξ), 0<ξ,=1 at four timepoints:

τ=0.5, 1, 1.5, 2.

Ode23:

Ode23s

The following graphs show three-dimensional plots of u as a function of τ and ξ.

Ode23:

Ode23s:

We conclude that if we want efficient numerical calculations for parabolic problems we

should use non-stiff solver (ode-23) for time efficiency. However, after using the Jacobian

matrix properties in the odeset the stiff solver (ode-23s)becomes equally efficient but with

less time efficiency. The structure of the Jacobian is sparse. The type of the linear equation

solver that we used in the implicit ODE method was solver for full matrices.

Appendix

clear all close all clc

%========= Solution of parabolic model problem with the Mol:

===========

%Solution of u_t=u_xx, u(x,0)=0 , u(0,t)=1, u_x(1,t)=0 using the

MoL.

%================ discretization parameters

============================= dx=0.2; %stepsize in x direction x=0:dx:1; %inner boundary points N=1/dx; %total number of stepsizes dt=0.013; % timestep T=10; %Total number of timesteps c=(2*dt)/((dx)^2); %constant given that a=1

%=============== initialize arrays of boundary and initial conditions

==== u0=[]; b=[]; %initiating initial and boundary

values u0(1)=1; %setting the initial value for

ξ=0 for i=2:N+1 %setting the initial values for

ξ>0 u0(i)=0; end for j=1:T %setting the boundary values for

t<1 b0(j)=1; end

u=u0'; % initial value matrix

b=b0; %upper boundary condition matrix

upast = zeros(1,N+1); %initiating the backwards time line U=zeros(N+1,T); %initiating the solution matrix U

U(:,1)=u; %adding initial condition to the

solution matrix U(1,:)=b; %adding boundary condition to the

solution matrix

%==================== Creating the rest matrix

========================== for i=2:T upast=u; %condition for saving the backwards

time line if (dt*i)>1 %setting the boundary values for

t>1 u(1)=0;

end for j=2:N %creating the rest of the solution

matrix u(j)=c*upast(j+1)-(2*c-1)*upast(j)+c*upast(j-1); U(:,i)=u; end u(N+1)=u(N-1); %setting the lower boundary condition end

%3-D plot surf(U) title('solution of u_t=u_xx, u(x,0)=[1,0,....,0],

u(0,t<1)=[1,1,....,1] u(0,t=1)=0 using the MoL.') xlabel('space variable x') ylabel('time variable y') zlabel('solution variable u(x,t)')

%========= Solution of parabolic model problem with the Mol:

===========

%Solution of u_t=u_xx, u(x,0)=[1,0,....,0], u(0,t<1)=[1,1,....,1]

u(0,t=1)=0 using the MoL. Discretized system store in heatflow.m

clear all clc

global N hx %discretization parameters N=40; %number of inner x points hx=1/(N+1); %stepsize in x direction x=hx:hx:1-hx; %inner points u0(1)=1; for i=2:N u0(i)=0; % initial values end t0=0; tend=2; %time interval options = odeset('RelTol',1e-6, 'AbsTol',1e-

6,'Jacobian',1,'Jconstant',1,'Jpattern',1); [time,result]=ode23s(@heatflow,[t0,tend],u0,options); %non-

stiff ode-solver resultp=[zeros(size(time)) result zeros(size(time))]; %add

BCs xp=0:hx:1;

%add boundary points mesh(xp,time,resultp) title('solution of u_t=u_xx, u(x,0)=[1,0,....,0],

u(0,t<1)=[1,1,....,1] u(0,t=1)=0 using the MoL.') xlabel('space variable x') ylabel('time variable y') zlabel('solution variable u(x,t)')

%Function used in the parabolic solver for creating the solution

matrix function rhs=heatflow(t,u) global N hx %discretization parameters rhs=zeros(N,1); rhs(1)=(-2*u(1)+u(2)+1)/(hx*hx); rhs(2:N-1)=(u(1:N-2)-2*u(2:N-1)+u(3:N))/(hx*hx); rhs(N)=(u(N-1)-u(N))/(hx*hx); end

References

Introduction to computation and modeling for differential equations,

Lennart Edsberg