INF-MAT Mandatory Exercise 1

Transcription

1 INF-MAT336 - Mandatory Exercise 1 Vegard Kjelseth University of Oslo vokjelse@ifi.uio.no March 8, 213 Exercise 1 a) The solution of the ODE u (t) = au (t), t >, u () = 1 is u (t) = Ce at. The value of C is found using the initial conditions. u () = C = 1. The full solution is u (t) = e at. b) Given the following permutation of the initial conditions, ũ () = 1 + ɛ, the solution is given by: ũ (t) = (1 + ɛ) e at The error is then given by: e (t) = ũ (t) u (t) = ɛe at The error increases exponentially with t for a >, while for a < the error decreases exponentially. It follows that the problem is stable for a < and unstable for a >. The code following below plots the difference between the solution and the perturbed solution. As expected, the difference decreases exponentially for a <, while for a > it increases exponentially. This confirms the analysis above. 1 % I n i t i a l i z e t 2 t = linspace (, 1, 1 ) ; 3 4 % I n i t i a l i z e a 5 a = 1 ; 6 7 % I n i t i a l i z e u 8 u = exp ( a * t ) ; 9 1 % I n i t i a l i z e u _ p e r t u r b e d 11 epsilon =. 1 ; 12 u_perturbed = (1+ epsilon ) * exp ( a * t ) ; % P l o t d i f f e r e n c e between s o l u t i o n s 15 plot ( t, abs ( u u_perturbed ), b ) ; 16 legend ( D i f f e r e n c e ) ; 1

2 c) The code for the forward Euler method follows below. The elements are calculated using the formula: v m+1 = v m + hav m = v m (1 + ha) Solving this yields the following general formula for the different elements: v m = (1 + ha) n This shows that there is a clear danger with this approach in cases where ha < 1 since this will result in an oscillating solution and an unstable scheme. Running the script with the given values confirm this since for k = 1 and k = 1 we get increasingly higher oscillations that don t correspond to the actual solution. Since h = 1/3 this is explained by the fact that ha < 1 for these values of k. 1 % S c r i p t f o r s o l v i n g u ( t )= a * u ( t ) using f o r w a r d E u l e r 2 3 % I n i t i a l i z e time s t e p 4 n = 3 ; % Number o f s t e p s 5 h = 1/n ; 6 7 % I n i t i a l i z e u v e c t o r 8 u = zeros ( 1, n + 1 ) ; 9 u ( 1 ) = 1 ; % I n i t a l i z e u ( t =) 1 11 % I n i t i a l i z e a 12 a = 1; % Perform f o r w a r d e u l e r 15 for i =1:n 16 der = a * u ( i ) ; 17 u ( i +1) = u ( i )+h * der ; 18 end 19 2 % P l o t s o l u t i o n 21 t = linspace (, 1, n + 1 ) ; 22 plot ( t, u, b ) ; 23 hold on ; 24 plot ( t, exp ( a * t ), g ) 25 legend ( s t r c a t ( Forward Euler s o l u t i o n to u, char ( 3 9 ), ( t )= a * u ( t ) ), Exact Solution : exp ( a * t ) ) ; The code for the backwards Euler method follows below. The backwards Euler solution is calculated using v m+1 h f (v m+1 ) = v m which yields: v m+1 h a v m+1 = v m+1 (1 ha) = v m v m+1 = v m / (1 ha) Solving this yields the following general formula: v m = 1 (1 ha) n The problem with this approach is that for ha = 1 it will divide by zero, in other words for a = 3. For 3 < a < 6 we will get an oscillating behaviour with 2

3 increasingly volatile oscillations, while for a > 6 we will get an oscillating behaviour converging towards zero. The solution is therefore not stable for any values of a resulting in ha 1, something that is confirmed by running the script for these values. 1 % S c r i p t f o r s o l v i n g u ( t )= a * u ( t ) using backward E u l e r 2 3 % I n i t i a l i z e time s t e p 4 n = 3 ; % Number o f s t e p s 5 h = 1/n ; 6 7 % I n i t i a l i z e u v e c t o r 8 u = zeros ( 1, n + 1 ) ; 9 u ( 1 ) = 1 ; % I n i t a l i z e u ( t =) 1 11 % I n i t i a l i z e a 12 a = 1 ; % Perform f o r w a r d e u l e r 15 for i =1:n 16 u ( i +1) = u ( i )/(1 h * a ) ; 17 end % P l o t s o l u t i o n 2 t = linspace (, 1, n + 1 ) ; 21 plot ( t, u, b ) ; 22 hold on ; 23 plot ( t, exp ( a * t ), g ) 24 legend ( s t r c a t ( Backward Euler s o l u t i o n to u, char ( 3 9 ), ( t )= a * u ( t ) ), Exact Solution : exp ( a * t ) ) ; Exercise 2 a The code for solving the equations follows below. The equations are solved using the forward Euler method. The plot for ζ = can be found in figure 1 on page 5, while the plot for ζ =.1 is in figure 2 on page 6. The results are similar in both cases, but clearly show that if the model is accurate there won t be a zombie apocalypse even if there is an outbreak. Both cases show that the number of infected and the number of zombies never rise significantly above. This can be explained by the fact that α > β which means that the zombies are killed faster than the susceptible are being infected. As for the difference between ζ = and ζ =.1, the plot for ζ =.1 shows a slightly higher amount of deceased people than for ζ =. Using a value of ζ = means that the number of zombies increases not just because of people being infected, but that dead susceptible people can become zombies as well. Given the model we re using this is not realistic since we are assuming that people are infected through contact with zombies, and not through black magic. Running the simulation for longer than t = 1 shows the number of suscep- 3

4 tible continuing to decrease and the the dead increasing, but with the number of zombies and infected staying close to. One could therefore argue that even though the zombies don t take over the planet, an outbreak would still cause the death of all of humandkind (assuming all people are susceptible). The reason this won t happen is that the continued decrease in the number of susceptible in the simulation is due to the fact that the number of zombies is >, but it is still << 1. Since it s not possible to have a fraction of a zombie, the reality is that given the parameters used the zombie outbreak would be put down quickly and we would all be able to continue with our lives. 1 % I n i t i a l i z e time s t e p 2 h =. 1 ; 3 n = 1/h ; 4 5 % C r e a t e v e c t o r s 6 S = zeros ( 1, n + 1 ) ; % S u s c e p t i b l e 7 I = zeros ( 1, n + 1 ) ; % I n f e c t e d 8 Z = zeros ( 1, n + 1 ) ; % Zombies 9 R = zeros ( 1, n + 1 ) ; % Removed / Dead 1 11 % I n i t i a l i z e v e c t o r s 12 S ( 1 ) = 1 ; 13 I ( 1 ) = ; 14 Z ( 1 ) = 2 ; 15 R ( 1 ) = ; % S e t c o n s t a n t s 18 Sigma = ; 19 alpha =. 3 ; 2 beta_var =. 6 ; 21 d e l t a =. 1 ; 22 sigma =. 1 ; 23 rho =. 1 ; 24 zeta = ; % Constant t o vary c o n v e r s i o n from dead t o zombie % Perform f o r w a r d e u l e r 27 for i =1:n 28 % S u s c e p t i b l e 29 der = Sigma beta_var * S ( i ) * Z( i ) d e l t a * S ( i ) ; 3 S ( i +1) = S ( i )+h * der ; % I n f e c t e d 33 der = beta_var * S ( i ) * Z( i ) rho * I ( i ) d e l t a * I ( i ) ; 34 I ( i +1) = I ( i )+h * der ; % Zombies 37 der = rho * I ( i ) alpha * S ( i ) * Z( i )+ zeta * R( i ) sigma * Z( i ) ; 38 Z( i +1) = Z( i )+h * der ; 39 4 % Removed / Dead 41 der = d e l t a * S ( i )+ d e l t a * I ( i ) zeta * R( i )+ alpha * S ( i ) * Z( i ) ; 42 R( i +1) = R( i )+h * der ; 43 end % P l o t s o l u t i o n 46 t = linspace (, 1, n + 1 ) ; 47 plot ( t, S, b ) ; 48 hold on ; 49 plot ( t, I, r ) ; 5 hold on ; 51 plot ( t, Z, k ) ; 4

5 Figure 1: Plot for ζ = 52 hold on ; 53 plot ( t, R, g ) ; 54 legend ( S u s c e p t i b l e, I n f e c t e d, Zombies, Removed / Dead ) ; b In this exercise I have used the same code as listed above, but switching the values of β and α so β =.6 and α =.3. The plot showing this scenario can be found in figure 3 on the following page. As noted briefly above using these values means that people are infected faster than zombies are being put down, and the result is a zombie apocalypse which can be seen in the figure. Exercise 3 Positive Definiteness For the operator L to be positive definite it has to satisfy Lu, u with equality only if u =. Lu, u = (u (x)) 2 u (x) u (x) dx Using integration by parts on the second expression yields: u (x) u (x) dx = u (x) u (x) 1 1 ( u (x) ) 2 dx 5

6 Figure 2: Plot for ζ =.1 Figure 3: Plot for β =.6, α =.3 6

7 u (x) u (x) 1 = since u () = u (1) =. Plugging this into the initial equation yields: Lu, u = (u (x)) 2 + ( u (x) ) 2 dx This is clearly non-negative, and for the expression to be equal to zero both u and u has to be equal to zero. This proves that the operator L is positive definite. Symmetry For the operator L to be symmetric it has to satisfy the following equality: Lu, v = u, Lv Where u, v are solutions to the system described by L. Lu, v = 1 u (x) v (x) u (x) v (x) dx u, Lv = 1 u (x) v (x) u (x) v (x) dx From this it is clear to see that for the equality to hold be equal to yields: u (x) v (x) dx must u (x) v (x) dx. Using integration by parts on the first expression u (x) v (x) dx = u (x) v (x) 1 u (x) v (x) u (x) v (x) dx Since u and v both are solutions to L this means that u () = v () = u (1) = v (1) =. It follows that u (x) v (x) 1 = u (x) v (x) 1 =, which yields : u (x) v (x) dx = So Lu, v = u, Lv and L is symmetric. u (x) v (x) dx Uniqueness Assume that u 1 and u 2 both are solutions to L, so Lu 1 = f and Lu 2 = f. I let e = u 1 u 2. This yields: Le = Lu 1 Lu 2 = f f = From this it follows that Le, e = due to the fact that Le =. Since L is positive definite it also follows from Lemma 2.4 that e = and u 1 = u 2. It is therefore clear that problem has a unique solution. 7

8 What is the solution? Since f (x) = sin (kπx) an initial guess would be that the solution is of the form u (x) = A sin (kπx). This covers the boundary conditions since u () = u (1) =. I plug this into the differential equation: u (x) u (x) = A sin ( (kπx) + A (kπ) 2 sin (kπx) = A 1 + (kπ) 2) sin (kπx) = f (x) = sin (kπx) Clearly ( the solution u (x) = A sin (kπx) solves the differential equation as long as A 1 + (kπ) 2) = 1. This gives us the following value for A: A = (kπ) 2 The final solution is then: u (x) = sin (kπx) 1 + (kπ) 2 Implement a numerical scheme to solve the PDE The differential equation can be written as: u (x) = u (x) + f (x) I will use the result of a Taylor series expansion given by: u (x + h) 2u (x) + u (x h) h 2 = u (x) + E h (x) u (x) Where the error term E h satisfies E h (x) M uh 2 12, and where the constant M u is given by : M u = sup u (4) (x) x This yields: u (x) = u (x + h) 2u (x) + u (x h) h 2 + f (x) Letting v j denote the discrete solution for u ( x j ) and rearranging the expression gives us: v j+1 + ( 2 + h 2) v j v j 1 h 2 = f ( ) x j vj+1 + (2 + h 2) v j v j 1 = h 2 f ( ) x j I let b denote the vector where b i = h 2 f (x i ) and v denote the vector where v i = v i for i = 1,..., n. v and v n+1 are not included since v = v n+1 =. The 8

9 solution can be found by solving the equation Av = b where A is given by: 2 + h h A = h h 2 We know that a solution exists if the matrix A is diagonally dominant. Here α = 2 + h 2, γ = β = 1 and the matrix is diagonally dominant if α > γ and α β + γ (I dropped subscripts since they re the same regardless of index). This is clearly true for A, so it is diagonally dominant and has a unique solution. The MATLAB code for solving this follows below: 1 % I n i t i a l i z e h and n 2 n = 1 ; % Number o f p o i n t s between and 1 3 h = 1/(n + 1 ) ; 4 5 % I n i t i a l i z e v 6 v = zeros ( 1, n + 2 ) ; 7 8 % I n i t i a l i z e x v a l u e s 9 x = linspace (, 1, n + 2 ) ; 1 11 % I n i t i a l i z e k 12 k = 1 ; % I n i t i a l i z e f u n c t i o n f 15 f x, k ) ( sin ( k * pi * x ) ) ; % I n i t i a l i z e b 18 b = h^2 * f ( x ( 2 : n +1), k ) ; 19 2 % I n i t i a l i z e A 21 A = g a l l e r y ( t r i d i a g, ones ( 1, n 1),(2+h^2) * ones ( 1, n), ones ( 1, n 1 ) ) ; % C a l c u l a t e v 24 v ( 2 : n+1) = A\b ; % P l o t s o l u t i o n 27 plot ( x, v, b ) ; 28 hold on ; 29 exact = sin ( k * pi * x ) / ( 1 + ( k * pi ) ^ 2 ) ; 3 plot ( x, exact, r ) ; 31 legend ( Numerical Solution, Exact Solution ) ; % L a r g e s t e r r o r 34 e r r o r = max( abs ( v exact ) ) ; 35 disp ( s t r c a t ( The l a r g e s t e r r o r i s :, num2str ( e r r o r ) ) ) ; Analysis Using h = 1/1 was not a problem on my computer. 9

10 k=1 For h = 1/1 to 1/1 the maximum errors are e 6, 6.857e 8 and e 1. k=1 For h = 1/1 to 1/1 the maximum errors are e 6, 8.33e 8 and e 1. k=1 For h = 1/1 to 1/1 the maximum errors are e 5, e 8 and e 1. CPU time and complexity The script I used for timing this exercise and the next follows below. The timing was done with the plot part of the exercise commented out. For a given n, h is defined as h = 1/ (n + 1). Running this exercise with n = 1 led to an average time of t.95s. Using n = 1 led to an average time of t.135s, while using n = 1 had an average time of t =.7s. Going from n = 1 to n = 1 leads to an increase in time by a factor of around 1.4, while increasing from n = 1 to n = 1 only increases the time by a factor of approximately 5. A reasonable explanation for this is that at lower n the initialization cost of variables accounts for such a large part of the total operations that the increase in the time doesn t properly reflect the complexity of the core computations in the script. I therefore increased n and performed the timing for n = 2 and n = 3 as well, yielding the times t =.14s and t =.21s respectively. This amounts to an increase in time of 2 and 3 for the same increase in n, indicating that the complexity is linear. I therefore conclude that the complexity is O (n). This complexity seemed strange to me initially since we re dealing with an n-by-n matrix A. On the other hand this matrix is initialized using the gallery function which does not save elements that are equal to, which results in the number of elements being n + 2 (n 1) = 3n 2. Additionally the script computes A\b, which is the same as inv (A) b to find the values of the elements of v. Finding the inverse of a matrix generally has a complexity larger than O ( n 2), so this requires an explanation. An alternative to this would be to use Algorithm 2.1 from the textbook to solve this problem. We could have used this since the matrix A is diagonally dominant, and this algorithm is linear in time. So it is clearly possible to solve this in linear time. My explanation for the fact that the operation A\b only is linear in time is that MATLAB exploits the fact that A is tridiagonal and diagonally dominant and that this is the reason it is able to find the solution in linear time. 1 % S c r i p t f o r t i m i n g e x e r c i s e 3 and 4 1

11 2 i t e r = 5. ; 3 t = cputime ; 4 5 for i =1: i t e r 6 run ( e x e r c i s e 4 ) ; 7 end 8 9 % Get a v e r a g e time 1 avg = ( cputime t )/ i t e r ; % P r i n t a v e r a g e time 13 disp ( s t r c a t ( Average time :, num2str ( avg ) ) ) ; Exercise 4 Positive Definiteness For the operator L to be positive definite it has to satisfy Lu, u with equality only if u =. Lu, u = u (x) u (x) cu (x) u (x) dx I use integration by parts on the first expression which yields: 1 u (x) u (x) dx = u (x) u (x) 1 + ( u (x) ) 1 2 dx = ( u (x) ) 2 dx Where I used the fact that u () = u (1) = to cancel out u (x) u (x) 1. I proceed by using integration by parts on the second expression. 1 u (x) u (x) dx = u (x) u (x) 1 + u (x) u (x) = u (x) u (x) Where I used the fact that u () = u (1) = to cancel out u (x) u (x) 1. Moving the left hand-side over yields: 2 u (x) u (x) = u (x) u (x) = Plugging these expressions into the initial equation yields: Lu, u = ( u (x) ) 2 dx This expression is clearly non-negative. For the expression to equal zero we must have u (x) =. This means that u (x) must be a constant. Since we know that u () = u (1) =, we must have u (x) =, thus Lu, u is always positive except when u =, in which case Lu, u =. It follows that L is positive definite. 11

12 Symmetry For the operator L to be symmetric it has to satisfy the following equality: Lu, v = u, Lv Where u, v are solutions to the system described by L. I showed that Lu, v = 1 u (x) v (x) cu (x) v (x) dx u, Lv = 1 u (x) v (x) cu (x) v (x) dx u (x) v (x) dx = 1 prove equality I need to show that by using integration by parts: 1 u (x) v (x) dx = u (x) v (x) 1 u (x) v (x) dx in the previous exercise. To u (x) v (x) dx = 1 u (x) v (x) dx = u (x) v (x) dx. I do this u (x) v (x) dx Where I used the fact that u () = u (1) = v () = v (1) = to cancel out u (x) v (x) 1 1. Since u (x) v (x) dx = 1 u (x) v (x) 1 dx it is clear that u (x) v (x) dx = u (x) v (x) dx, and it follows that Lu, v = u, Lv, so L is not symmetric. Uniqueness The proof for uniqueness in the following exercise was based on the fact that L was positive definite. The same proof can be applied here, since L is positive definite in this case as well. Assume that u 1 and u 2 both are solutions to L, so Lu 1 = f and Lu 2 = f. I let e = u 1 u 2. This yields: Le = Lu 1 Lu 2 = f f = From this it follows that Le, e = due to the fact that Le =. Since L is positive definite it also follows from Lemma 2.4 that e = and u 1 = u 2. It is therefore clear that problem has a unique solution. What is f? Since u (x) = sin (kπx) I simply plug this into the differential equation to find u. f (x) = u (x) cu (x) = (kπ) 2 sin (kπx) cπk cos (kπx) 12

13 Implement a numerical scheme to solve the PDE The differential equation can be written as: u (x) cu (x) = f (x) To find a numerical solution I will use the following results of Taylor series expansions: u(x+h) u(x h) 2h u(x+h) 2u(x)+u(x h) h 2 = u (x) + O ( h 2) u (x) = u (x) + E h (x) u (x) The second expression is the same as the one used in the previous exercise, while the expression O ( h 2) satisfies : ( O h 2) h 2 sup u (3) (x) x Plugging this into the differential equation gives us: u (x + h) 2u (x) + u (x h) u (x + h) u (x h) h 2 c = f (x) 2h Multiplying through by 2h 2, rearranging and letting v j denote the discrete solution for u ( x j ) yields: v j+1 ( 2 ch) + 4v j + v j 1 ( 2 + ch) = 2h 2 f (x) I let b denote the vector where b i = 2h 2 f (x i ) and v denote the vector where v i = v i for i = 1,..., n. v and v n+1 are not included since v = v n+1 =. The solution can be found by solving the equation Av = b where A is given by: ch ch ch... A = ch ch... 2 ch 4 We know that a solution exists if the matrix A is diagonally dominant. Here α = 4, γ = 2 + ch and β = 2 ch. (Like the previous exercise I ve dropped the subscripts since they re the same regardless of index.) The first requirement is that α > γ. γ = 2 ch for ch 2 which is clearly less than α = 4. For ch > 2 we have γ = ch 2 which is less than α = 4 for ch < 6. The second requirement is that α β + γ. For ch 2 we get β + γ = 2 + ch + 2 ch = 4, which satisifies the requirement since it s equal to α. For ch > 2 we get β + γ = 2 + ch 2 + ch = 2ch > 4, which does not satisfy this requirement. 13

14 In conclusion, the matrix A is diagonally dominant as long as ch is 2. Since we re only operating with values where c/h < 2 c < 2h we get ch < 2h 2 < 2 since h will always be far less than 1. The matrix A is therefore diagonally dominant for all values we operate with. The MATLAB code for solving this follows below: 1 % I n i t i a l i z e h, n, c 2 n = 3; % Number o f p o i n t s between and 1 3 h = 1/(n + 1 ) ; 4 c = h ; 5 6 % I n i t i a l i z e v 7 v = zeros ( 1, n + 2 ) ; 8 9 % I n i t i a l i z e x v a l u e s 1 x = linspace (, 1, n + 2 ) ; % I n i t i a l i z e k 13 k = 1 ; % I n i t i a l i z e f u n c t i o n f 16 f x, k ) ( ( sin ( k * pi * x ) * ( k * pi )^2) c * k * pi * cos ( k * pi * x ) ) ; % I n i t i a l i z e b 19 b = 2 * h^2 * f ( x ( 2 : n +1), k ) ; 2 21 % I n i t i a l i z e A 22 A = g a l l e r y ( t r i d i a g,( 2 c * h ) * ones ( 1, n 1),4 * ones ( 1, n),( 2+ c * h ) * ones ( 1, n 1 ) ) ; % C a l c u l a t e v 25 v ( 2 : n+1) = A\b ; % P l o t s o l u t i o n 28 %p l o t ( x, v, b ) ; 29 %h o l d on ; 3 %e x a c t = s i n ( k * p i * x ) ; 31 %p l o t ( x, e x a c t, r ) ; 32 %l e g e n d ( Numerical S o l u t i o n, Exact S o l u t i o n ) ; % L a r g e s t e r r o r 35 %e r r o r = max ( a b s ( v e x a c t ) ) ; 36 %d i s p ( s t r c a t ( The l a r g e s t e r r o r i s :, num2str ( e r r o r ) ) ) ; Analysis Using h = 1/1 was not a problem on my computer. The following errors are calculated using c = h. k=1 For h = 1/1 to 1/1 the maximum errors are :.13762, and 1.345e 5. 14

15 k=1 For h = 1/1 to 1/1 the maximum errors are.87552, and e 5. k=1 For h = 1/1 to 1/1 the maximum errors are , and e 5. CPU time and complexity I used the same script for timing the exercise as I did for exercise 3. Running this exercise with n = 1 led to an average time of t.1s. Using n = 1 led to an average time of t.14s, while using n = 1 had an average time of t =.71s. This amounts to an increase by a factor of approximately 1.4 and 5 respectively and paints the same picture as what I encountered in exercise 3. Assuming the cost of the initialization of variables takes a larger share of the computational cost for lower values of n I performed this using n = 2 and n = 3 like I did in the previous exercise. The results were average times of t =.1317s and t =.2873s respectively, corresponding to approximately an increase in time of a factor 2 for an increase in n by 2, and and increase in time by 3 for an increase in n by 3. This indicates that it runs in linear time like the script in the previous exercise. The complexity is O (n). The script for this exercise is basically the same as in exercise 3, but tailored for this specific problem, so for an explanation of this I refer to the explanation in the last exercise. 15