I Numerically solving a non-local PDE

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I have a PDE to solve numerically on the region ##x \in [0,1]## and ##t \in (0, \infty)##. It is of the form:$$\frac{\partial f(x,t)}{\partial t} = g(x,t) + \int_0^1 h(x, x') f(x', t) dx'$$The second term is the tricky part. The change in ##f(x,t)## at ##x## depends on the value ##f(x',t)## of every other point ##x'## in the space.

In discretized form, it is something like$$f(x_i,t_{j+1}) = f(x_i,t_j) + \Delta t \left[ g(x_i,t_i) + \sum_{j \neq i} h(x_i, x_j) f(x_j, t_i) \right]$$How can I apply an accurate scheme like Runge-Kutta to this system?
 
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Why do you exclude j = i from the sum? And what are the boundary conditions?

After discretizing in space, you must end up with <br /> \dot{\mathbf{f}} = M\mathbf{f} + \mathbf{g} where f_i(t)= f(x_i, t) etc. with the matrix M depending on how you do the numerical integration. For the trapezoid rule with x_n = \frac{n}{N - 1} = n \Delta x, <br /> \int_0^1 h(x,x&#039;)f(x&#039;,t)\,dx&#039; \approx \frac12 \Delta x (h(x,x_0)f_0 + 2h(x,x_1)f_1 + \dots + h(x,x_{N-1})f_{N-1}) so that <br /> M_{ij} = \begin{cases} \frac12 \Delta x h(x_i,x_j) &amp; j = 0, N - 1 \\<br /> \Delta x h(x_i, x_j) &amp; j = 1 , \dots, N - 2 \end{cases} You may need to adjust rows 0 and N - 1 in order to enforce a boundary condition.
 
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Thanks, I'm making progress...
 
Another alternative is to use Gauss-Legendre quadrature, <br /> \int_0^1 h(x,x&#039;)f(x&#039;,t)\,dx = \frac12\int_{-1}^1 h(x,\tfrac12(z + 1))f(\tfrac12(z + 1),t)\,dz<br /> \approx \sum_{j=0}^{N-1} \tfrac12 w_j h(x,\tfrac12(z_j + 1))f_j which is exact for polynomials of order up to 2N - 1 in x&#039;. The z_i \in [-1,1] are then the Gauss-Legendre points with x_i = \frac12(z_i + 1) and <br /> M_{ij} = \frac12 w_jh(x_i,x_j).
 
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I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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