Efficient Solutions for IVP and Root Approximation in Differential Equations

[Robb]

Homework Statement

[/B]
It's been a couple of years since differential equations so I am hoping to find some guidance here. This is for numerical analysis.
Any help would be much appreciated.

Homework Equations

The Attempt at a Solution

 

[phyzguy]

No. When you go from\frac{dv}{dt} = -k v^2
to:\int dv = -\int k v^2 dt
you can't simply integrate the right side as though v is a constant, because v depends on t, and you don't know the functional dependence of v on t. What you need to do is collect the terms in v on one side and the terms in t on the other side so you have: \int \frac{dv}{v^2} = -k \int dtTry going from there.

 

[Robb]

-1/v = -kt
v = 1/kt

or is this more accurate: v(t) = 1/kt ?

 

[phyzguy]

What about the constant of integration and the initial conditions?

 

[Robb]

What about the constant of integration and the initial conditions?

Shoot, I forgot about the constant of integration: v(t) = 1/kt + C
As for the initial conditions I guess I'm not sure. v(0) = v(o). Doesn't this imply division by zero (v(0) = 1/k(0))?

 

[phyzguy]

Shoot, I forgot about the constant of integration: v(t) = 1/kt + C

No! You need to add the constant of integration when you integrate, before you invert both sides.

 

[Ray Vickson]

Shoot, I forgot about the constant of integration: v(t) = 1/kt + C
As for the initial conditions I guess I'm not sure. v(0) = v(o). Doesn't this imply division by zero (v(0) = 1/k(0))?

No: as phyzguy said, the correct statement is $-1/v = -kt + C$.

 

[Robb]

Yep, I got that. So, v(0) = -1/C

 

[phyzguy]

Yep, I got that. So, v(0) = -1/C

Yes, so keep going. Now you should be able to answer the other questions.

 

[Robb]

I now have to approximate the root of K. I'm not sure how to find a starting bracket. I know one way is to turn the equation F(k)=27k - ln abs(30k + 1) into two equations; y = 27k, y = ln abs(30k + 1). I've graphed it but the only intersection I can find is at the point (0,0), which we do not want. Any words of wisdom? Thanks!

 

[phyzguy]

I suggest using Newton's method to find the value of K. It works for finding roots of an equation of the form F(x) = 0. Here you have F(k) = 27k - ln(30k+1) = 0. Start with a guess for k (I tried k = 1) and iterate until it stops changing. What value do you find for k?

 

[Ray Vickson]

I now have to approximate the root of K. I'm not sure how to find a starting bracket. I know one way is to turn the equation F(k)=27k - ln abs(30k + 1) into two equations; y = 27k, y = ln abs(30k + 1). I've graphed it but the only intersection I can find is at the point (0,0), which we do not want. Any words of wisdom? Thanks!

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I refuse to look at your posted images, but am willing to comment on typed material.

In this case you want to find the intersection of the two curves $f(k) = 27 k$ and $g(k) = \ln |30 k + 1|$. For $k \geq 0$ (actually, for $k \geq -1/30$) you can remove the absolute-value signs in $g(k)$. We have $f'(0) = 27$ and $g'(0) = 30$. Since $f(0)=g(0)=0$ and $g$ has higher initial slope, $g(k)$ will exceed $f(k)$ intially. However, since $g(k)$ is a strictly concave function on $(0,\infty)$ (i.e., $g''(k) < 0$ for $k > 0$), and since $g$ does not grow very fast, the two graphs must cross again exactly once at some value of $k > 0$. However, that $k$ will be very small, because already by $k = 1$ we have $f \gg g$. As phyzguy suggested, you can use Newton's method, but another way that looks promising is to expand $g(k)$ as an order-2 Maclauren series of the form $g(k) \approx 30 k - c k^2$ and then solve the resulting quadratic equation.

The graph of $g(k)$ for $k < -1/30$ is a mirror-image of that for $k > -1/30$, so as $k$ decreases below -1/30 the graph rises up again (from $-\infty$) and eventually becomes positive once more. That means that there must be a second root at a value $k < -1/30$.

 

[Robb]

much appreciated.

 

[phyzguy]

Note that the value for k you are searching for must be positive, otherwise the projectile will accelerate without limit.

 

[Ray Vickson]

much appreciated.

For what it is worth: you can solve the equation in terms of the so-called Lambert W- function and so can get an immediate numerical value if you use a computer algebra system such as Maple or Mathematica. You can also get a symbolic and numerical solution using the free on-line program Wolfram Alpha.