solving the Korteweg-de Vries equation in Matlab

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Yoav Eshel 2020-12-12

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评论： Rahul 2025-1-17

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Hey,

I'm trying to solve the Korteweg-de Vries -

in matlab using ode45 and fft, but my code is very inefficient and I figure out why. The odefunc is

function dydt = kdv(y,k)
        % du/dt = -d3u/dx3 - 6u du/dx
        % numerically approximates the derivative in the fourier domain
        yhat = fft(y);
        dyhat = 1i*k.*yhat;
        d3yhat = -1i*(k.^3).*yhat;
        dy = ifft(dyhat);
        d3y = ifft(d3yhat);
        dydt = - d3y - 6*(y.*dy);
end

and then I call it using

[t,U] = ode45(@(t,y)kdv(y,k),tspan, y0);

but even for very small arrays it takes forever to run. I tried timing each step becuase I suspected that evaluating the fourier transform every iteration is the reason for the inefficiency, but it turns out that this part of the code runs quite fast and the bulk of the computtion time is in ode45 computations between steps.

Any suggestions are appriciated!

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Yoav Eshel 2020-12-20

I'm trying to solve it without the partial differential equation toolbox. Doesn't the use of fft turns it into an ode?

Rahul 2025-1-17

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You are right. The FFT turns it into a ordinary differentuial equation..

the code below does not work , i suspect the problem is in the conv() part

L = 100;
N = 1000;
dx = L/N;
x = -L/2:dx:L/2-dx;
kappa = (2*pi/L)*(-N/2 : N/2-1);
kappa = fftshift(kappa);
u0 = sech(x);
t = 0:0.1:10;
[t,uhat] = ode45(@(t,uhat)kortewegdevries(t,uhat,kappa),t,fft(u0));
for k = 1:length(t)
    u(k,:)= ifft(uhat(k,:));
end
figure, waterfall( (u(1:10:end,:)));
figure, imagesc( flipud(u));
function duhatdt = kortewegdevries(t,uhat,kappa)
duhatdt = -1i*(kappa.^3)'.*uhat + conv(uhat,1i*kappa.*uhat); % -(kappa.^2)'.*uhat;%
end