Question: How can I declare a the square root of a certain function is positive?

Hi,

 

I have a problem of having a problem to solve the following equation with the this error. I found out that it might be due to that I have a differential that is squared and maple could not calculate it as after square root it will have positive and negative. May I know how to overcome this other than changing my equations?

ODE_summarize.mw
 

restart; with(DEtools)

````

sol1 := alpha*(diff(f(y), y, y, y)) = -(1/4)*f(y)*(diff(f(y), y, y))*n+(1/2)*(diff(f(y), y))^2*n-(3/4)*f(y)*(diff(f(y), y, y))+(1/2)*(diff(f(y), y))^2-g(y), diff(g(y), y, y) = -(1/4)*P[r](f(y)*(diff(g(y), y))*n-4*n*(diff(f(y), y))*g(y)+3*f(y)*(diff(g(y), y)))

alpha*(diff(diff(diff(f(y), y), y), y)) = -(1/4)*f(y)*(diff(diff(f(y), y), y))*n+(1/2)*(diff(f(y), y))^2*n-(3/4)*f(y)*(diff(diff(f(y), y), y))+(1/2)*(diff(f(y), y))^2-g(y), diff(diff(g(y), y), y) = -(1/4)*P[r](f(y)*(diff(g(y), y))*n-4*n*(diff(f(y), y))*g(y)+3*f(y)*(diff(g(y), y)))

(1)

bc := (D(f))(y)

(D(f))(y)

(2)

gg := convert(bc, diff)

diff(f(y), y)

(3)

gg1 := limit(gg, y = infinity)

limit(diff(f(y), y), y = infinity)

(4)

gg2 := convert(gg1, D) = 0

limit((D(f))(y), y = infinity) = 0

(5)

gg2

(D(f))(y) = 0

(6)

bc1 := eval(f(y), y = 0) = 0, eval((D(f))(y), y = 0) = 0, convert(gg1, D) = 0, limit(g(y), y = infinity) = 0, eval(g(y), y = 0) = 1

f(0) = 0, (D(f))(0) = 0, limit((D(f))(y), y = infinity) = 0, limit(g(y), y = infinity) = 0, g(0) = 1

(7)

``

odesub1 := eval(eval(sol1, n = .6), P[r] = 0)

alpha*(diff(diff(diff(f(y), y), y), y)) = -.9000000000*f(y)*(diff(diff(f(y), y), y))+.8000000000*(diff(f(y), y))^2-g(y), diff(diff(g(y), y), y) = 0

(8)

sol2 := dsolve({bc1, odesub1}, numeric, [f(y), g(y)], output = operator, continuation = alpha, initmesh = 2400, method = bvp[midrich])

Error, (in dsolve/numeric/bvp/convertsys) unable to convert to an explicit first-order system

 

``


 

 

 

 

 

 

 

 

 

 

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