Lecture 8: Contents. Boundary Conditions. Example Consider the wave equation

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1 Lecture 8: Contents Boundary Conditions Wave Propagation Reflecting boundary conditions. Free boundary or Absorbing boundary conditions. Transport or Wave-Propagation in two dimensions. Fourier Methods for Periodic Problems Discrete Fourier Series. Periodic Solutions. Example: A Periodic solution to the wave equation. Example Consider the wave equation u tt = c 2 u xx in(, ) [, T). with initial conditions u(x, ) = u t (x, ) =. Question What kind of boundary conditions can we impose at x = and x =? May 8, 27 Sida / 29 May 8, 27 Sida 2 / 29 Method Set Introduce a grid {(x i, t j )}, i =,,..., n+ and j =,,..., m. Stepsizes are h and k. Discretize at interior points u i,j+ = u i,j +2u i,j +( ck h )2 (u i+,j 2u i,j +u i,j )+O(h 2 +k 2 ). This is a two step method in time. It is consistent. The initial condition u t (x, ) = is approximated by (u t ) i, = 2k (u i, u i, )+O(k 2 ). Eliminate u i, from the difference formula for u i,. Matlab Set the grid and boundary function first c=.5;n=2;t=2; x=linspace(,,n+2);h=x(2)-x(); k=h/c;m=round(t/k);t=m*k; t=linspace(,t,m+);k=t(2)-t(); r=(c*k/h)^2 f=*t;ind=find(t <); f(ind)=sin(pi*t(ind)); u=zeros(n+2,m+); Remark This makes sure r= and the numerical method is exact. May 8, 27 Sida 3 / 29 May 8, 27 Sida 4 / 29

2 Dirichlet Conditions Matlab The time stepping is different for the first step and the rest of them. For regular time steps we use end for j=2:m Example Let f(t) = sin(πt), for < t <, and zero otherwise. Set for i=2:n+, % Interior points boundary conditions: u(i,j+)=-u(i,j-)+2*u(i,j)+r*(u(i-,j)-2*u(i,j)+u(i+,j))/2; end u(, t) = f(t) and u(, t) =. % Use boundary conditions to update % u(,j+) and u(n+2,j+); Remark The method is explicit. We can update all interior points first. Then use the boudnary conditions to get u(, t j+ ) and u(, t j+ ). Numerical Method Simply set u,j+ = f(t j+ ) and u n+.j+ = at each step. May 8, 27 Sida 5 / 29 May 8, 27 Sida 6 / 29 Neumann Conditions Example Let f(t) = sin(πt), for < t <, and zero otherwise. Set boundary conditions: u(, t) = f(t) and u x (, t) =. Numerical Method Simply set u,j+ = f(t j+ ) and Results A sine-wave moves to the right and an mirror image is sent back. at each step. u n+,j+ = u n+,j + 2u n+,j + 2( ck h )2 (u n,j u n+,j ) Lemma Setting u = is a mirror wave condition. May 8, 27 Sida 7 / 29 May 8, 27 Sida 8 / 29

3 Robin Conditions Example Let f(t) = sin(πt), for < t <, and zero otherwise. Set boundary conditions: u(, t) = f(t) and u t (, t)+cu x (, t) =. Numerical Method Simply set u,j+ = f(t j+ ) and Results A sine-wave moves to the right and an is reflected back. Lemma Setting n u = is a reflecting wave condition. at each step. u n+,j+ = u n+,j ( ck h )(u n,j u n+,j ) May 8, 27 Sida 9 / 29 May 8, 27 Sida / 29 Animations Example Suppose the solution u ij is stored in a matrixu. We can plot the solution for a given time t j. Put together a movie from individual frames. for j=:m+ plot(x,u(:,j), LineWidth,.5) axis([ -2 2]) mov(j)=getframe; end movie(mov,) Results A sine-wave moves to the right and is mostly gone. Lemma Setting u(, t)+cu x (, t) = is an open boundary or an absorbing boundary. Also works withcontour,image, etc. Write a file with VideoWriter. Remark It is difficult to understand wave propagation without animations. May 8, 27 Sida / 29 May 8, 27 Sida 2 / 29

4 Boundary conditions in 2D Example In acoustics we generate a wave front that is reflected off objects in its path Example Consider the wave equation u tt = c 2 (u xx + u yy ), x >. A family of solutions is given by u(x, y, t) = e i( ξ 2 ξ2 2 x+cξ t+ξ 2 y), where ξ = (ξ,ξ 2 ) is a fixed frequency. Design a boundary condition at x= that keeps the wave undisturbed. Remark On the inclusion we can use n u = but how to set conditions on the surrounding box? Reference B. Engquist and A. Majda. Absorbing Boundary Conditions for the Numerical Simulation of Waves. Mathematics of Computation, Vol. 3, No. 39, pp , 977. This is an artificial boundary! May 8, 27 Sida 3 / 29 May 8, 27 Sida 4 / 29 Approximate Absorbing Conditions Lemma Consider the wave equation u tt = c 2 (u xx + u yy ), x >. The perfectly absorbing boundary conditon, ( x c t 2 y 2 )u =, on x =. does not have any effect on the solution for x >. Lemma A first order accurate absorbing boundary condition is, for the line x =, is u x cu t =, on x =. Remark The solution can be written as a linear combination of plane waves. The makes it impossible to implement this exactly. Alternative? Remark Based on +x = +O( x ). Easy to generalize to waves hitting a boundary descrbed by a straight line. May 8, 27 Sida 5 / 29 May 8, 27 Sida 6 / 29

5 Similarily we use +x = + x 2 +O( x 2 ) to obtain u xt cu tt + c 2 u yy =, on x =. There is also an approximation based on +x = + x 2+(x/2) +O( x 3 ). Remark This allows very accurate implementations of open boundaries involving straight lines. Lemma In cylindrical coordinates the wave equation takes the form, u tt = c 2 (u rr + r u r + r 2 u θθ) =, and an absorbing boundary condition, for r = a, is u r + 2a u cu t = on r = a. Example A sonar pulse is sent out and any sound reflected back is measured. Simulate the response given a specific object. May 8, 27 Sida 7 / 29 May 8, 27 Sida 8 / 29 Example Consider a waveguide with is feed a wave at one end. Γ Γ 2 Γ 3 Γ 4 The boundaries Γ 2 and Γ 3 depends on the physics. The feeding boundary Γ has to generate the correct type of wave, i.e. a Dirichlet Condition. The outlet Γ 4 should not influence the solution, i.e. an absorbing boundary. Periodic Solutions Example Consider the equation a(x)u xx = u tt, in(, ) R. with boundary conditions u(, t) = f(t), and u x (, t) =. If f(t) is periodic, with period T, then u(x, t) is also periodic. Remark Also works for, e.g., a(x, y)u xx + b(x, y)u yy + u tt =, in(, ) 2 R. May 8, 27 Sida 9 / 29 May 8, 27 Sida 2 / 29

6 Lemma A periodic function, with period T, can be written as a Fourier series, f(t) = T k= f(ξ k )e iξ kt dt, ξ k = 2πk T, Lemma Let = t < t < < t n = T and f = (f, f,..., f n ) T. The discrete Fourier series is f j = n f k e iξ kt j, T k= ξ k = 2πk T, t j = Tj n. Remark For general functions we have the Fourier Transform f(ξ) = 2π f(t)e iξt dt. Remark Periodic means f(t ) = f(t n ). Very accurate representation of smooth functions. We have both positive and negative frequencies. Compare with Von Neumann analysis. Interval and boundary conditions gives discrete frequencies. May 8, 27 Sida 2 / 29 May 8, 27 Sida 22 / 29 Example Suppose we have a n = 5 samples {f(t k )} from a function f(t). How to approximate f(t) using trigonometric interpolation? Matlab The function fft computes the discrete Fourier coefficients..9.7 ii=sqrt(-);n=length(f); F=fft(f)/n; xi=2*pi*[:ceil(n/2)- -(floor(n/2):-:)] ; N=5; t=(:n-)/(n-); T=zeros([n ]); for k=:length(t), T(k)=real(sum(F.*exp(ii*xi*t(k)))); end; Remark The points t = and t = and the same because of periodicity. Remark The derivative of f(t) can be computed using iξ k f(ξ k ). May 8, 27 Sida 23 / 29 May 8, 27 Sida 24 / 29

7 Example Consider the equation a(x)u xx + u tt =, in(, ) (, T). with boundary conditions u(, t) = f(t), and u x (, t) =. The trigonometric interpolant, for n = 5 and n = 25, evaluates on a grid tt = ( : N )/(N ), N = 52. Note the Gibbs phenomena around the discontinuity. and a periodic solution in the time variable. Solve the problem using the discrete Fourier transform. Remark The Matlab functionfft computes the discrete Fourier transform of a vector v R n. May 8, 27 Sida 25 / 29 May 8, 27 Sida 26 / 29 Matlab Create a grid and frequencies. Solve one frequency at a time. x=(:n) /n;dx=x(2)-x(); t=t*(:m-) /m;dt=t(2)-t(); xi=(:m-) ; xi(ceil(m/2)+:m)=-(m-ceil(m/2):-:); xi=xi*2*pi/t; u=zeros(n+,m); u(,:)=fft(f); for k=:m e=ones(n,);ah=spdiags([e -2*e e],[:2],n,n+2); Ah=Ah(:,2:n+);Ah(n,n-)=2; Ah=spdiags(a(2:n+),,n,n)*Ah/dx^2; Ah=Ah+xi(k)^2*speye(n); b=zeros(n,);b()=-a(2)*u(,k)/dx^2; u(2:n+,k)=ah\b; end Matlab Transform back and check for unwanted imaginary parts. for k=:n+, u(k,:)=ifft(u(k,:)); end; err=sum(sum(abs(imag(u)))); if err > ^- fprintf(, Warning: Large imaginary part in Press end; u=real(u); Question What happens with a stability analysis? Can this fail? May 8, 27 Sida 27 / 29 May 8, 27 Sida 28 / 29

8 Pressure u(,t) Time t The boundary data f(t) = u(, t) and the solution u(x, t). The preiodicity condition replaces u(x, o) and u t (x, t). Remark The absorbing boundary condition transforms into û x (,ξ k )+a()iξ k û(,ξ k ) and can be easily implemented. May 8, 27 Sida 29 / 29