A Is a solution of a differential equation a function of its parameters?

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Hi everyone,

Imagine I have a system of linear differential equations, e.g. the Maxwell equations.

Imagine my input variables are the conductivity $\sigma$. Is it correct from the mathematical point of view to say that the electric field solution, $E$, is a function of sigma in general, E(r,t,sigma)?

Thank you
 
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Yes. Unless something weird is going on, the value of the solution at a particular point and time is determined by the parameters and the initial and boundary conditions.
 
pasmith said:
Yes. Unless something weird is going on, the value of the solution at a particular point and time is determined by the parameters and the initial and boundary conditions.

Thank you
 
Basically, any differential equation describes a family of curves, surfaces, volumes...

It's the boundary conditions and initial parameters that fix it to one curve, surface or volume...
 

Thread 'A question about variables of integration'

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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