Chorux

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6 years, 116 days

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These are questions asked by Chorux

Given a set (list) of PDE, is there a way to search all possible solution sets? For instance, pdsolve will output the solution

{_eta[0](t, x) = 0, _xi[t](t, x, u) = _C1, _xi[x](t, x, u) = _C2, eta[1](t, x) = 0}

for the list of PDEs below. But I am aware that there is another solution different from the above, is there way to seek these other solutions?

 

[alpha*u^2*(diff(eta[1](t, x), t))+alpha*u*(diff(_eta[0](t, x), t))-u*(diff(eta[1](t, x), t))-u*(diff(eta[1](t, x), x))+u*(diff(eta[1](t, x), x, x, t))-(diff(_eta[0](t, x), x))-(diff(_eta[0](t, x), t))+diff(_eta[0](t, x), x, x, t), -(diff(_xi[x](t, x, u), u, u, u)), -(diff(_xi[t](t, x, u), x))-(diff(_xi[t](t, x, u), t)), -(diff(_xi[t](t, x, u), x, x)), -2*(diff(_xi[t](t, x, u), x, x))-(diff(_xi[x](t, x, u), x, x))+2*(diff(eta[1](t, x), x)), diff(eta[1](t, x), t)-2*(diff(_xi[x](t, x, u), x, x)), -(diff(_xi[x](t, x, u), t))*alpha*u-(diff(_xi[x](t, x, u), x, x, x))-(diff(_xi[t](t, x, u), x))+2*(diff(eta[1](t, x), x, t)), -(diff(_xi[x](t, x, u), x)), -(diff(_xi[t](t, x, u), t))*alpha*u+(diff(_xi[t](t, x, u), x))*alpha*u+(diff(_xi[x](t, x, u), x))*alpha*u+(diff(_xi[x](t, x, u), t))*alpha*u+eta[1](t, x)*alpha*u+alpha*_eta[0](t, x)+diff(_xi[t](t, x, u), t)-(diff(_xi[t](t, x, u), x, x, x))-(diff(_xi[x](t, x, u), x))-(diff(_xi[x](t, x, u), t))+diff(eta[1](t, x), x, x), -2*(diff(_xi[x](t, x, u), u)), -2*(diff(_xi[t](t, x, u), x, u)), -2*(diff(_xi[t](t, x, u), u))-(diff(_xi[x](t, x, u), u)), -(diff(_xi[t](t, x, u), u)), -(diff(_xi[t](t, x, u), u)), -5*(diff(_xi[x](t, x, u), u, u)), -(diff(_xi[t](t, x, u), u, u)), -3*(diff(_xi[t](t, x, u), x, u)), -2*(diff(_xi[x](t, x, u), u)), -(diff(_xi[t](t, x, u), u)), -(diff(_xi[t](t, x, u), u)), -2*(diff(_xi[t](t, x, u), u, u))-(diff(_xi[x](t, x, u), u, u)), -(diff(_xi[t](t, x, u), u, u, u)), -3*(diff(_xi[t](t, x, u), u, u)), -3*(diff(_xi[t](t, x, u), x, u, u)), (diff(_xi[t](t, x, u), u))*alpha*u-(diff(_xi[x](t, x, u), u))-3*(diff(_xi[t](t, x, u), x, x, u)), -2*(diff(_xi[x](t, x, u), x, u))-4*(diff(_xi[t](t, x, u), x, u)), -(diff(_xi[t](t, x, u), u))*alpha*u+(diff(_xi[x](t, x, u), u))*alpha*u+diff(_xi[t](t, x, u), u)-(diff(_xi[x](t, x, u), u)), -3*(diff(_xi[x](t, x, u), x, x, u))-(diff(_xi[t](t, x, u), u)), -7*(diff(_xi[x](t, x, u), x, u)), -3*(diff(_xi[x](t, x, u), x, u, u))]

Please I will like to know if there exists any command to isolate the known regular mathematical functions from an expression. For example, given the equation 

I need a means of isolating (ex & sin(v)) from the above expression

Thanks

I want to reverse a catenated expression (eg. xxtt ->x,x,t,t). I don't know the command to achieve this. Anyone with an idea please help

Thanks

Hello,

I am writing an arrow procedure and will like to know if there is a way to implement the following 
 

bj := (G, y) ->  (`@`(seq((t -> u -> v -> G(u, t) - v)(args[1 + nargs - i]), i = 1 .. nargs - 2)))(y)
bj(F, y, a, b, c, d);
v:=[p, q, r, s]
the output

My question is, how can I replace the v with each element of the list to get the following as output
F(F(F(F(y,a)-p),b)-q,c)-r,d)-s

Aany suggestion will be highly appreciated

 

I am trying to develop a recursive code to the above equations.  Note, U,X&Y are multivariate functions (in this case bivariate functions of (x,y)) i.e. U=U(x,y), X=X(x,y) etc.

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