Numercal Algorthms for Vsual Computng 008/09 Example Solutons for Assgnment 4 Problem (Shft nvarance of the Laplace operator The Laplace equaton s shft nvarant,.e., nvarant under translatons x x + a, y y + b, a, b R. The shft nvarance can then be wrtten as u xx + u yy u x x + u y y To see ths explctly, we consder x, y as mappngs dependng on x and y, respectvely, and we compute y(y y b, x(x x a y(y y, x(x x. It s not wrong to consder x x(x, y, y y(x, y, so that y y, x x. It follows u x x u(x, y x x ( u(x, y x x x x x x u x(x, y x x x x u(x, y u x (x, y u xx (x, y. u y y u yy follows analogously. Problem (What s the matrx, what s the matrx?
. For the orderng (u u u 3 u 4 u 5 u 6 u 7 u 8 u 9 u 0 u u u 3 u 4 u 5 u 6 ( and the underlyng process [ u+,j u j + u,j x + u ],j+ u j + u,j y f j ( we get the followng matrx system 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 (3. For the orderng (u u 3 u 6 u 0 u u 5 u 9 u 3 u 4 u 8 u u 5 u 7 u u 4 u 6 (4
we get the followng matrx system 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 (5 Problem 3 (Crossng dervatves. We have been gven the followng cross dervatve dscretsaton by use of central dfference methods x y ( u,j+ u,j x h ( u,j+ x h x h u +,j+ u,j+ 4h ( u,j u +,j u,j 4h 4h (u +,j+ u,j+ u +,j + u,j wth h x y. For ths 4 ponts we can compute the D- 3
Taylorexpanson ( ( u(x 0, y 0 f(x, y + 0 x f(x, y(x x 0 + y f(x, y(y y 0 + (( ( 0 x f(x, y(x x 0 + x y f(x, y(x x 0(y y 0 ( + f(x, y y (y y 0 Ths gves for our smple ponts the followng approxmaton: u(x ± h, y + h u ± hu x + u y + h (u xx ± u xy + u yy + 6 h3 (±u xxx + 3u xxy ± u xyy + 3u yyy + 4 h4 (u xxxx ± 4u xxxy + 6u xxyy ± 4u xyyy + u yyyy + O(h 5 u(x + h, y h u + hu x u y + h (u xx u xy + u yy + 6 h3 (u xxx 3u xxy + u xyy 3u yyy + 4 h4 (u xxxx 4u xxxy + 6u xxyy 4u xyyy + u yyyy + O(h 5 u(x h, y h u hu x u y + h (u xx + u xy + u yy 6 h3 (u xxx + 3u xxy + u xyy + 3u yyy + 4 h4 (u xxxx + 4u xxxy + 6u xxyy + 4u xyyy + u yyyy + O(h 5 4
Now we can nput ths approxmaton nto our four pxel scheme: u xy 4h (u +,j+ u,j+ u +,j + u,j 4h (u + hu x + u y + h (u xx + u xy + u yy + 6 h3 (u xxx + 3u xxy + u xyy + 3u yyy + 4 h4 (u xxxx + 4u xxxy + 6u xxyy + 4u xyyy + u yyyy + O(h 5 (u hu x + u y + h (u xx u xy + u yy + 6 h3 ( u xxx + 3u xxy u xyy + 3u yyy + 4 h4 (u xxxx 4u xxxy + 6u xxyy 4u xyyy + u yyyy + O(h 5 (u + hu x u y + h (u xx u xy + u yy + 6 h3 (u xxx 3u xxy + u xyy 3u yyy + 4 h4 (u xxxx 4u xxxy + 6u xxyy 4u xyyy + u yyyy + O(h 5 +u hu x u y + h (u xx + u xy + u yy 6 h3 (u xxx + 3u xxy + u xyy + 3u yyy + 4 h4 (u xxxx + 4u xxxy + 6u xxyy + 4u xyyy + u yyyy + O(h 5 5
If we combne ths terms together, we wll get: (u ( + +hu 4h x ( + +hu y ( + + h (u xx + u xy + u yy u xx + u xy u yy u xx + u xy u yy + u xx + u xy + u yy 8u xy + 6 h3 (u xxx ( + +3u xxy ( + + 3u xyy ( + +u yyy ( + + 4 h4 (u xxxx ( + +6u xxyy ( + Ths sums up to +4u xxxy ( + + + 4 +4u xyyy ( + + + 4 4h (4h u xy + 6 4 (u xxxy + u xyyy + O(h 5 u xy + 6 h (u xxxy + u xyyy + O(h 3 u xy + 6 h u xy + O(h 3 ( + 6 h u xy + O(h 3 O(h Overall we get a O(h error term for the cross dervatve approxmaton.. Ths dscretsaton s sotropc, as the error term ncorporates an addtonal sotropc Laplace operator onto u xy, whch we wanted to approxmante n the frst place. +u yyyy ( + + O(h 5 Problem 4 (Cookng norms. We want to prove the followng statement: n x x x 3 n x 6
We wll do ths step by step. So at frst we prove the frst nequalty x n x n n n n x max x max x. max x Now we wll have a closer look at the second nequalty x x. For ths, however we consder now the squared norms, as ths does not volate the monotoncty of the normng functon: x max,...,n x x x. Now we only need to prove the last nequalty, so we compute whch concludes the proof. x ( x n max x n max x n max x n x. We want to prove the followng statement: n x x x 3 n x 7
At frst, we wll prove the thrd nequalty by use of the Cauchy-Schwarz nequalty ( ( ( x y, so we can compute x x ( ( x x n x. From ths, the frst nequalty s easly dervable. The bggest problem s now the second nequalty. For ths we wll now consder the vector y x x x x k k We wll now show that y whch wll help us later. y We use ths result now n: what we wanted to show. P n k x k x x ( x k k n k x y x n k x k x x x x x y x x,. Problem 5 (Provng Banach Let an arbtrary x 0 D be gven. As F : D D, the sequence (x k k N0 s unquely determned by x k+ F (x k and for k N t holds: x k+ x k F (x k F (x k L x k x k L x k x k... (. 8
Also, by teraton of the frst approxmaton t follows for k n: and from that also for m n: x k+ x k L k+ n x n x n (. x m x n (x m x m + (x m x m +... (x n+ x n m x k+ x k Trangle-Inequalty kn ( m L k+ m x n x n kn ( L j x n x n j ( L L j x n x n j Therefore t holds for m n: L L x n x n Ln L x x 0 see ( or (. x m x n L L x n x n Ln L x x 0 ( Due to L < t holds that L n 0 (n and therefore (x n n N0 s a Cauchy sequence that s convergent wth the lmt x. As D s compact, wth, x D. F s contnuousand therefore the lmt x s a fxed pont of F. x s the only fxed pont n D. If x x were another fxed pont, t would follow: x x F (x F (x L x x < x x whch s a contradcton. The proposed error approxmaton follows drectly from (, f m. 9