Possible to solve for V in MATLAB r2011a?

(V^1.4*A^.4*C1-1/V)dV=dA/A

where

C1=constant, 
A=.25*pi*(.022^2-x^2), and 
V=f(x).

I'm trying to derive a symbolic expression for V. Is it possible to change this eq. into a form with no derivatives or integrals? Or could I just shove this into some "solve" function? I have r2011a so the solve functions are less powerful.

[Walter Roberson]

Could you confirm that you mean

(V(x)^(7/5)*A(x)^(2/5)*C1-1/V(x))*(diff(V(x), x)) = (diff(A(x), x)) / A(x)
A(x) = (1/4)*Pi*((22/1000)^2-x^2)

Answer 1

[Walter Roberson]

If you do mean

(V(x)^(7/5)*A(x)^(2/5)*C1-1/V(x))*(diff(V(x), x)) = (diff(A(x), x)) / A(x)
A(x) = (1/4)*Pi*((22/1000)^2-x^2)

then No. If there is a closed form solution for that it is difficult to solve.

[Yang Shi]

Is there any way to approximate V(x)? I have one initial condition for V(x) and A(x).

[Walter Roberson]

Are you looking for real values or complex values?

If you are looking for real values than for any given x, the easiest way I found so far was to split real and imaginary parts and examine the places where the imaginary part crosses 0. With the x I tested on (only a small number) there were two solutions for the imaginary part being 0, one positive and one negative, roughly -Pi and + 3*Pi/2 (roughly!)

The general form is No Real Help:

V(x) = RootOf(-ln(250000 * x^2 - 121) + 500000 * Intat( (3/250000) *(C1 * Pi * TEMP^4 + 100 * (-1)^(3/5) * 100^(1/5) * (Pi^3 * TEMP^8)^(1/5)) / (TEMP * (C1 * Pi * TEMP^4 - 500 * (-1)^(3/5) * 100^(1/5) * (Pi^3 * TEMP^8)^(1/5))), TEMP = Z) + 500000 * C2, Z) / (250000 * x^2 - 121)^(1/6)

In the above, C2 is an arbitrary constant that has to do with the boundary condition on V(x)

RootOf(F(Z), Z) for some expression F in Z, represents the set of values, Z, such that F(Z) = 0 -- the roots of the expression.

IntAt(F(TEMP),TEMP=VALUE) was something I had not encountered before. It means to take the indefinite integral of the expression F(TEMP) in the variable TEMP, and then evaluate it at the point TEMP = VALUE. For example, IntAt(x^2, x=3) would be int(x^2,x) evaluated at x=3, so 1/3*x^3, evaluated at x=3, so 1/3*3^3, so 9. In the place that it occurs in the expression, it is in the middle of RootOf(something + IntAt(F(TEMP), TEMP=Z),Z) . Basically that means that you have to find the value Z such that the something plus the integral to Z, gives 0. Which is to say that you have to solve for the Z such that the indefinite integral, when evaluated at Z, gives the negative of the "something". Not so fun. But more on that in a moment.

[Walter Roberson]

Let us look again at the RootOf(), and let us restrict to wanting real valued solutions. The RootOf() has the components

500000*C2 - ln(250000*x^2-121) + 500000 * TheIntegral

This splits into two cases. ln(25000*x^2-121) is real-valued if 25000*x^2-121 is >= 0 but the ln() is imaginary if 250000*x^2-121 < 0. Let us look first at the real-valued case, 25000*x^2-121 >= 0 and corresponding real-valued ln(). Then provided that the boundary condition C2 is real-valued, we can see that the overall expression would be real-valued if and only if 50000 * TheIntegral is real-valued, which occurs if TheIntegral is real-valued. So we study when the imaginary component is 0.

It is not immediately obvious from the expression what imaginary part will be, considering that there are fractional powers of -1 on both the numerator and denominator. Maple tells me that the imaginary component is

(9/1250) * TEMP^3 * 10^(2/5) * Pi^(3/5) *(TEMP^8)^(1/5) * sin((2/5)*Pi) / (1000 * cos((2/5)*Pi) *Pi^(3/5) * (TEMP^8)^(1/5) * 10^(2/5) * TEMP^4 + Pi * TEMP^8 + 250000 * 10^(4/5) * Pi^(1/5) * (TEMP^8)^(2/5))

Making the assumption that we are not going to push TEMP so that we are dividing by infinity, we can solve for that being 0 by examining the numerator

9 * TEMP^3 * 10^(2/5) * Pi^(3/5) * (TEMP^8)^(1/5) * sin((2/5)*Pi)

and simple inspection shows that is only true if TEMP is 0. But the denominator also goes to 0, leading to 0/0 so we need to rule out 0 as well. Well actually the limit of that expression is 0, but the expression as a whole (real and imaginary together) is undefined as TEMP approaches 0, so we need to rule it out anyhow.

Therefore there are no solutions in reals when 25000*x^2-121 is >= 0

[Walter Roberson]

The imaginary case is getting a bit tricky. Could you indicate some typical values for C1 and for the range of x ? And could you say whether this is to be a real-only problem or if it is to involve complex values?