Suppose, v(x) is a user-defined function. How, to get v(-x) which is going to be used in the same code?

Is there any in-built MATLAB function to do it. Should we define user-defined function, say, a local function. If so, how that can be done?

Answer 1

[Titus Edelhofer]

Hi,

sorry, I don't understand the question. If you have the function, why don't you simply call v(-x)?

Or are you looking for defining this function? In this case you could do e.g.

minusv = @(x) v(-x);

Titus

[SAMIT KUMAR GUPTA]

Let me give you a part of the code I used:

L=80; N=4096; dt=0.0001; tmax=20; nmax=round(tmax/dt); dx=L/N; x=[-L/2:dx:L/2-dx]'; k=[0:N/2-1 -N/2:-1]'*2*pi/L; k2=k.^2;

a=0.0;
y=-x-a;
u=sech(x);
ucon=sech(y);
udata=u; tdata=0;
for nn=1:nmax                                       % Integration begins
  du1=1i*(-ifft(0.5*k2.*fft(u))+u.*u.*conj(ucon));  v=u+0.5*du1*dt;
  du2=1i*(-ifft(0.5*k2.*fft(v))+v.*v.*conj(vcon));  v=u+0.5*du2*dt;
  du3=1i*(-ifft(0.5*k2.*fft(v))+v.*v.*conj(vcon));  v=u+    du3*dt;
  du4=1i*(-ifft(0.5*k2.*fft(v))+v.*v.*conj(vcon));
  u=u+(du1+2*du2+2*du3+du4)*dt/6;

    if mod(nn,round(nmax/25)) == 0
       udata=[udata u]; tdata=[tdata nn*dt];
    %end
    end
    ucon=u;
  end       
subplot(2,2,1)
  imagesc(x,tdata,abs(udata'));
  colorbar
  title('Evolution of the rogue wave')
  ylabel('Time','fontsize',12)
  xlabel('Transverse co-ordinate','fontsize',12)

My question is as follows: in the integration part by Runge-Kutta method, ucon is being calculated by the usual way as it has been defined ucon=conj(u(-x)). Since, my u(x)=sech(x), there is no as such problem for matlab to calculate ucon. Then using du1, it finds v, using that when it calculates du2, it needs information of vcon=v(-x). Now, the form of v(x) is not fully known. So, my question is how to define and v(-x) so that it can be called and used in the code to get expected result.

Thanks and regards,

Samit

Answer 2

[Stephen23]

You can't define v(x) as a function , only v could be defined as a function.

Once you define a function v, you can call it with either v(x) or v(-x) or any other suitable values that you want to use. Calling a function is independent of how it is defined:

>> A = @(x)disp(x); % define a function A
>> A(3)
     3

>> A(-3)
    -3

[SAMIT KUMAR GUPTA]

Let me give you a part of the code I used:

L=80; N=4096; dt=0.0001; tmax=20; nmax=round(tmax/dt); dx=L/N; x=[-L/2:dx:L/2-dx]'; k=[0:N/2-1 -N/2:-1]'*2*pi/L; k2=k.^2;

a=0.0;
y=-x-a;
u=sech(x);
ucon=sech(y);
udata=u; tdata=0;
for nn=1:nmax                                       % Integration begins
  du1=1i*(-ifft(0.5*k2.*fft(u))+u.*u.*conj(ucon));  v=u+0.5*du1*dt;
  du2=1i*(-ifft(0.5*k2.*fft(v))+v.*v.*conj(vcon));  v=u+0.5*du2*dt;
  du3=1i*(-ifft(0.5*k2.*fft(v))+v.*v.*conj(vcon));  v=u+    du3*dt;
  du4=1i*(-ifft(0.5*k2.*fft(v))+v.*v.*conj(vcon));
  u=u+(du1+2*du2+2*du3+du4)*dt/6;

    if mod(nn,round(nmax/25)) == 0
       udata=[udata u]; tdata=[tdata nn*dt];
    %end
    end
    ucon=u;
  end       
subplot(2,2,1)
  imagesc(x,tdata,abs(udata'));
  colorbar
  title('Evolution of the rogue wave')
  ylabel('Time','fontsize',12)
  xlabel('Transverse co-ordinate','fontsize',12)

My question is as follows: in the integration part by Runge-Kutta method, ucon is being calculated by the usual way as it has been defined ucon=conj(u(-x)). Since, my u(x)=sech(x), there is no as such problem for matlab to calculate ucon. Then using du1, it finds v, using that when it calculates du2, it needs information of vcon=v(-x). Now, the form of v(x) is not fully known. So, my question is how to define and v(-x) so that it can be called and used in the code to get expected result.

Thanks and regards,

Samit