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This number, is it always equal to the order of the differential equation?

in which case, one can reduce this number?

thanks in advance

1 Answers 1

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If you are given $\frac{{{d^n}y}}{{d{x^n}}} = f(x,y,y',y'', \ldots ,{y^{(n - 1)}})$ then you need the values of $y({x_0}),y'({x_0}),y''({x_0}), \ldots ,{y^{(n - 1)}}({x_0})$ and the function itself. So you are right -- the number of initial conditions corresponds to the order of the differential equation. But this differential equation can, of course, be reduced to $\frac{{dy}}{{dx}} = {y_1},\frac{{d{y_1}}}{{dx}} = {y_2}, \ldots ,\frac{{d{y_{n - 2}}}}{{dx}} = {y_{n - 1}},\frac{{d{y_{n - 1}}}}{{dx}} = f(x,{y_1},{y_2}, \ldots ,{y_{n - 1}})$ using substitutions ${y_1} = y',{y_2} = y'', \ldots ,{y_{n - 1}} = {y^{(n - 1)}}.$

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    If the initial surface is space-like then the Cauchy problem can be posed, but when the initial surface is a characteristic, solution are not guaranteed to be unique or may not exist if I remember correctly. I am not sure about the number of initial conditions. I think it depends on the PDE and a given surface.2012-10-20