Differential equation: I think there is a Gamma Function here

[Saladsamurai]

***Please skip ahead to post # 6 where I have better formulated my question. Thank you! ***

I have an equation

$$AY'' + B\eta^2Y'=0 \qquad(1)$$

where A and B are known constants and Y is a function of η. By using the substitution X = Y' I have reduced the problem to a first order ODE of the form

$$Y' = C_1e^{-\frac{B}{3A}\eta^3 \qquad(2)$$

$$\Rightarrow Y(\eta) = \int_0^\eta C_1e^{-k\eta^3}\,d\eta \qquad(3)$$

where let k = B/3A for compactness.

I am not sure how to integrate this. I know that

$$\Gamma(z) = \int_0^\infty t^{z - 1}e^{-t}\,dt\qquad(4)$$

but I am not really sure how to use this since I don't have a 't' anywhere. Not to mention the bounds of my integral are not infinite.

Any advice on this?

 

[fzero]

Use the change of variables $$t=\eta^3$$ and you can express it in terms of the incomplete Gamma function. Within an additive constant and factors that are easy to determine, it's related to $$\Gamma(1/3,\eta^3)$$. I don't believe that there's a simpler form for this.

 

[Saladsamurai]

Use the change of variables $$t=\eta^3$$ and you can express it in terms of the incomplete Gamma function. Within an additive constant and factors that are easy to determine, it's related to $$\Gamma(1/3,\eta^3)$$. I don't believe that there's a simpler form for this.

Hello again Ok. So if I let $t = \eta^3\rightarrow dt = 3\eta^2\,d\eta\rightarrow d\eta = dt / (3\eta^2)$ Plugging these results into (3) we have

$$Y(t) = C_!\int e^{-kt}\,\frac{dt}{3\eta^2} \qquad(4)$$

Is this really going somewhere? I am pretty bad at these... what do i do with my limits? I know that at eta = 0 -> t = 0 and at eta = eta -> eta = t^(1/3)

I think i am screwing this up

 

[uart]

Hello again Ok. So if I let $t = \eta^3\rightarrow dt = 3\eta^2\,d\eta\rightarrow d\eta = dt / (3\eta^2)$ Plugging these results into (3) we have

$$Y(t) = C_!\int e^{-kt}\,\frac{dt}{3\eta^2} \qquad(4)$$

Is this really going somewhere? I am pretty bad at these... what do i do with my limits? I know that at eta = 0 -> t = 0 and at eta = eta -> eta = t^(1/3)

I think i am screwing this up

If you want to express it as an incomplete gamma then you're best to write it as a definite integral with a dummy variable of integration.

Say you wanted to find the indefinite integral,

$$I(x) = \int \, e^{-x^3} \, dx$$.

Write it instead as,

$$I(x) = \int_{u=0}^{u=x} \, e^{-u^3} \, du + C$$

After substituting $t=u^3$, $dt/du = 3 u^2$, you get :

$$I(x) = \frac{1}{3} \, \int_{t=0}^{t=x^3} \, t^{-2/3} \, e^{-t} \, dt + C$$

$$I(x) = \frac{1}{3} \, \gamma(1/3,x^3) + C$$

Where $\gamma(.,.)$ is the lower incomplete gamma function and "C" is the constant of integration.

 

[Saladsamurai]

Ok folks, this is great info. I am going to plug in the constants A and B that I have as I think it will make this easier. The actual equation that I am looking at right now is:

$$\theta(\eta) = C_1\int e^{(-0.0277\Pr)\eta^3}\,d\eta \qquad(5)$$

where C₁ is some constant (from a previous integration,i.e. a boundary condition) and Pr is a constant (i.e. the Prandtl Number).

I have the boundary conditions on θ:

η = 0: θ = 0
η→∞: θ = 1​

I know that these make the problem easier since now I actually have the infinite integral.

I know that I am supposed to arrive at a final answer of θ'(0) = 0.339*Pr^1/3. Now it is quite possible that (5) is not perfect and that I might not get exactly 0.339. But I would like to at least get as far as getting "something" times Pr^1/3.

So I would like to go ahead and try to get (5) integrated and solve for my C₁. So following fzero's and uart's suggestions we are looking to integrate:

$$\int_0^\infty e^{(-0.0277\Pr)\eta^3}\,d\eta \qquad(6)$$

I guess I should have mentioned that I have boundary conditions as this will probably simplify my problem

If I let $u^3 = 0.0277\Pr\eta^3$ I can now write (6) as

$$\int_0^\infty e^{-u^3}\,du \qquad(7)$$

Now I let

$$t = u^3\rightarrow dt = 3u^2\,du = 3(t^{1/3})^2\,du\rightarrow du = (1/3)*t^{-2/3}\,dt$$

so that (7) becomes:

$$\frac{1}{3}\int_0^\infty t^{-2/3}e^{-t} = \Gamma(1/3)\qquad(8)$$

I feel like I have completely ignored the factor of 0.0277*Pr in the integral. But I am unsure of how to sneak it back in there?

Thanks for the help so far!

EDIT: I should probably just let t = 0.0277*Pr*η³ and it will factor out ... I will try this! I think that I "hid" that factor inside of u and should have taken care of it before evaluating the integral to Γ(1/3)

 

[Saladsamurai]

Ok. I am confusing the heck out of myself here. I am confusing boundary conditions with other things and I need to straighten this out. This is due to my weak background in calculus. Let start from the top. I have a Differential EQ:

$$\frac{d^2\theta}{d\eta^2} + 0.083\Pr\eta^2\frac{d\theta}{d\eta} = 0 \qquad(a)$$

where Pr = a constant.

By letting $X = \frac{d\theta}{d\eta}$ (a) becomes:

$$\frac{dX}{X} = -0.083\Pr\eta^3\,d\eta \qquad(b)$$

$$\Rightarrow \ln X = -0.0277\Pr\eta^3 + C_0$$

$$\Rightarrow X = e^{-0.0277\Pr\eta^3 + C_0}=e^{-0.0277\Pr\eta^3}e^{ C_0}$$

Let

$$e^{C_0} =C_1\qquad(c)$$

and we have

$$X = C_1e^{-0.0277\Pr\eta^3}\qquad(d)$$

But $X = \frac{d\theta}{d\eta}$ so that (d) becomes

$$\int d\theta = C_1\int e^{-0.0277\Pr\eta^3}\,d\eta \qquad(e)$$.This is about where we came in (Fight Club reference). Now as it stands, both integrals in (e) are indefinite and this is where the procedure gets fuzzy for me. I have options:

1) Find the antiderivatives of both sides, which will introduce a 2nd constant of integration and then apply the boundary conditions to solve for C₁ and C₂.

or

2) Impose boundary conditions now while integrating.

The second option is where I get confused. I feel like using option 2 gets me an integral that better fits the definition of the Gamma function, but I am shaky on the procedure.

For boundary conditions, I have:

η = 0: θ = 0
η→∞: θ = 1​

If I write (e) as follows:

$$\int_0^1 d\theta = C_1\int_0^\infty e^{-0.0277\Pr\eta^3}\,d\eta \qquad(f)$$

am I technically using up BOTH of my boundary conditions since I havce asserted in my limits that at η = 0: θ = 0 AND η→∞: θ = 1 ? Because I think this would make it impossible for mr to solve for C₁. But the form of (f) is perfect for getting the Gamma function out of it.

Sorry, it's these little details that kill me . Thank you to anyone who has the patience to go through this with me .

 

[fzero]

To keep from notational confusion, rewrite

$$\int d\theta = C_1\int e^{-0.0277\Pr\eta^3}\,d\eta \qquad(e)$$

in terms of dummy variables

$$\int^\theta_0 d\phi = C_1\int^\eta_0 e^{-0.0277\Pr x^3}\,dx.$$

Doing the integrals, you will find that

$$\theta = C_1 \Gamma(1/3,\eta^3) + C_2$$

You can now apply the boundary conditions: η = 0: θ = 0 fixes $$C_2$$, while η→∞: θ = 1 fixes $$C_1$$.

 

[Saladsamurai]

To keep from notational confusion, rewrite

$$\int d\theta = C_1\int e^{-0.0277\Pr\eta^3}\,d\eta \qquad(e)$$

in terms of dummy variables

$$\int^\theta_0 d\phi = C_1\int^\eta_0 e^{-0.0277\Pr x^3}\,dx.$$

Hi fzero I guess my question is: why should I set the lower limits on each integral to 0 ? Is it because I know that when η = 0, θ = 0 ? (If so, I feel like I am "using up a boundary condition by prescribing these lower limits to be 0).

 

[fzero]

Hi fzero I guess my question is: why should I set the lower limits on each integral to 0 ? Is it because I know that when η = 0, θ = 0 ? (If so, I feel like I am "using up a boundary condition by prescribing these lower limits to be 0).

Yes, it would be more straightforward to just write

$$\int^\theta_{\theta_0} d\phi = C_1\int^\eta_{\eta_0} e^{-0.0277\Pr x^3}\,dx,$$

where $$\theta(\eta_0) = \theta_0$$ is an initial condition. If we did this, we'd find

$$\theta -\theta_0 = C_1 \left( \Gamma(1/3,\eta^3) -\Gamma(1/3,\eta_0^3) \right).$$

So $$C_2$$ is fixed as claimed.

 

[Saladsamurai]

Yes, it would be more straightforward to just write

$$\int^\theta_{\theta_0} d\phi = C_1\int^\eta_{\eta_0} e^{-0.0277\Pr x^3}\,dx,$$

where $$\theta(\eta_0) = \theta_0$$ is an initial condition. If we did this, we'd find

$$\theta -\theta_0 = C_1 \left( \Gamma(1/3,\eta^3) -\Gamma(1/3,\eta_0^3) \right).$$

So $$C_2$$ is fixed as claimed.

Ok. I am not sure that I understand the notation $\Gamma(1/3,\eta^3)$. Why do you have the η³ as an argument? Is that your way of taking care of the 0.0277*Pr part of it?

The reason that I ask, is that I am supposed to simplify this to 0.339*Pr^1/3 and I am just having some trouble getting there.

I need to get from here:

$$\int^\theta_{\theta_0} d\phi = C_1\int^\eta_{\eta_0} e^{-0.0277\Pr x^3}\,dx,$$

to here: 0.339*Pr^1/3,

using η = 0: θ = 0 and η→∞: θ = 1.

If I take my integral:

$$\int^\theta_{\theta_0} d\phi = C_1\int^\eta_{\eta_0} e^{-0.0277\Pr x^3}\,dx,$$

and make the substitution t = 0.0277*Pr*x³ I will eventually arrive at:$$\int^\theta_{\theta_0} d\phi =\frac{1}{3}(0.0277\Pr)^{-1/3}C_1\int_{\eta_o}^\eta t^{-2/3}e^{-t}\,dt \qquad(g)$$

Now the integral on the right side of (g) is almost the Gamma function except that the limits of the integral are not 0 and ∞ as they are in the definition of Gamma. How can I remedy this? Is it enough to assign the limits to be 0 and ∞ and then have it be implied that my solution is valid only on that interval?

 

[fzero]

Ok. I am not sure that I understand the notation $\Gamma(1/3,\eta^3)$. Why do you have the η³ as an argument? Is that your way of taking care of the 0.0277*Pr part of it?

No,

$$\Gamma(s,x) = \int_x^\infty t^{s-1} e^{-t} dt$$

is the incomplete Gamma function. Except perhaps at specific values, it isn't expressible in terms of more elementary functions. Closely related (and more directly applicable to your integral) is the lower incomplete Gamma function

$$\gamma(s,x) = \int_0^x t^{s-1} e^{-t} dt.$$

These are related by the identity

$$\gamma(s,x) + \Gamma(s,x) =\Gamma(s).$$

The reason that I ask, is that I am supposed to simplify this to 0.339*Pr^1/3 and I am just having some trouble getting there.

The solutions to your problem are functions of $$\eta$$, but perhaps that is the value at some particular $$\eta$$.

$$\int^\theta_{\theta_0} d\phi =\frac{1}{3}(0.0277\Pr)^{-1/3}C_1\int_{\eta_o}^\eta t^{-2/3}e^{-t}\,dt \qquad(g)$$

Now the integral on the right side of (g) is almost the Gamma function except that the limits of the integral are not 0 and ∞ as they are in the definition of Gamma. How can I remedy this? Is it enough to assign the limits to be 0 and ∞ and then have it be implied that my solution is valid only on that interval?

The solution, as I said, is a function $$\theta(\eta)$$ that can be expressed in terms of the incomplete Gamma function. It won't be expressible in terms of simpler functions. Typically, you'd resort to numerical approximations for the values of interest. Perhaps there's additional information that would help sort out the value that you're quoting, but we have the general solution that follows from equation (g).

Edit: Also check the powers of $$\eta$$ in the integration limits in (g), since $$t\sim x^3$$.

 

[Saladsamurai]

Hello fzero Thank you for your patience; I appreciate the help. Unfortunately, I am a little slow. I still fail to see some things:

1.) I need to fix my limits to the following:

$$\int^\theta_{\theta_0} d\phi =\frac{1}{3}(0.0277\Pr)^{-1/3}C_1\int_{\eta_o^3}^{\eta^3} t^{-2/3}e^{-t}\,dt \qquad(h)$$

Ok, I think I can deal with that.

2.) You are right, it was a specific value of eta. Moreover, I need to show from all of this that

$$\left ( \frac{d\theta}{d\eta}\right )_{\eta = 0} = 0.339(\Pr)^{1/3} \qquad(i)$$

So I need to be able to solve for my C₁ and any additional constants of integration that come up.

So I need to know how I can move forward from (h) and I am failing miserably to see how to do so. The integral in (h) does not have the same limits as the lower incomplete Gamma function. It runs from arbitrary η_o to η. So how can I tweak (g) so that I can make use of the definition of the lower incomplete Gamma function? And how do I justify those modifications?

 

[fzero]

Hello fzero Thank you for your patience; I appreciate the help. Unfortunately, I am a little slow. I still fail to see some things:

1.) I need to fix my limits to the following:

$$\int^\theta_{\theta_0} d\phi =\frac{1}{3}(0.0277\Pr)^{-1/3}C_1\int_{\eta_o^3}^{\eta^3} t^{-2/3}e^{-t}\,dt \qquad(h)$$

Ok, I think I can deal with that.

2.) You are right, it was a specific value of eta. Moreover, I need to show from all of this that

$$\left ( \frac{d\theta}{d\eta}\right )_{\eta = 0} = 0.339(\Pr)^{1/3} \qquad(i)$$

So I need to be able to solve for my C₁ and any additional constants of integration that come up.

So I need to know how I can move forward from (h) and I am failing miserably to see how to do so. The integral in (h) does not have the same limits as the lower incomplete Gamma function. It runs from arbitrary η_o to η. So how can I tweak (g) so that I can make use of the definition of the lower incomplete Gamma function? And how do I justify those modifications?

Just split the limits of integration up

$$\int_{0}^{\infty} t^{-2/3}e^{-t}\,dt = \int_{0}^{\eta_0^3} t^{-2/3}e^{-t}\,dt+ \int_{\eta_o^3}^{\eta^3} t^{-2/3}e^{-t}\,dt+\int_{\eta^3}^{\infty} t^{-2/3}e^{-t}\,dt .$$

So

$$\Gamma(1/3) = \gamma(1/3,\eta_0^3) + \int_{\eta_o^3}^{\eta^3} t^{-2/3}e^{-t}\,dt + \Gamma(1/3,\eta^3).$$

This gives us the integral in terms of $$\Gamma(1/3,\eta^3)$$ and constants. Solve for $$\theta(\eta)$$ and apply the initial conditions.

 

[Saladsamurai]

Just split the limits of integration up

$$\int_{0}^{\infty} t^{-2/3}e^{-t}\,dt = \int_{0}^{\eta_0^3} t^{-2/3}e^{-t}\,dt+ \int_{\eta_o^3}^{\eta^3} t^{-2/3}e^{-t}\,dt+\int_{\eta^3}^{\infty} t^{-2/3}e^{-t}\,dt .$$

So

$$\Gamma(1/3) = \gamma(1/3,\eta_0^3) + \int_{\eta_o^3}^{\eta^3} t^{-2/3}e^{-t}\,dt + \Gamma(1/3,\eta^3).$$

This gives us the integral in terms of $$\Gamma(1/3,\eta^3)$$ and constants. Solve for $$\theta(\eta)$$ and apply the initial conditions.

So from your equation we have

$$\theta - \theta_0 = \frac{1}{3}(0.0277\Pr)^{-1/3}C_1\int_{\eta_o^3}^{\eta^3} t^{-2/3}e^{-t}\,dt = \frac{1}{3}(0.0277\Pr)^{-1/3}C_1\left [ \Gamma(1/3) -\left( \gamma(1/3,\eta_0^3) + \Gamma(1/3,\eta^3)\right )\right ]$$

$$\Rightarrow \theta(\eta) = \frac{1}{3}(0.0277\Pr)^{-1/3}C_1\left [ \Gamma(1/3) -\left( \gamma(1/3,\eta_0^3) + \Gamma(1/3,\eta^3)\right )\right ] + \theta_0$$

Should there be a second constant of integration somewhere? Or is it somehow included?

 

[fzero]

So from your equation we have

$$\theta - \theta_0 = \frac{1}{3}(0.0277\Pr)^{-1/3}C_1\int_{\eta_o^3}^{\eta^3} t^{-2/3}e^{-t}\,dt = \frac{1}{3}(0.0277\Pr)^{-1/3}C_1\left [ \Gamma(1/3) -\left( \gamma(1/3,\eta_0^3) + \Gamma(1/3,\eta^3)\right )\right ]$$

$$\Rightarrow \theta(\eta) = \frac{1}{3}(0.0277\Pr)^{-1/3}C_1\left [ \Gamma(1/3) -\left( \gamma(1/3,\eta_0^3) + \Gamma(1/3,\eta^3)\right )\right ] + \theta_0$$

Should there be a second constant of integration somewhere? Or is it somehow included?

No, we used definite integrals so there's no integration constant. If we'd performed indefinite integrals, we'd have eliminated the integration constant with the initial condition.

 

[Saladsamurai]

No, we used definite integrals so there's no integration constant. If we'd performed indefinite integrals, we'd have eliminated the integration constant with the initial condition.

Ok. So C1 arose as a result of using an indefinite integral and hence I need to solve for it. We could say that θ₀ is my second constant of integration right?

 

[fzero]

Ok. So C1 arose as a result of using an indefinite integral and hence I need to solve for it.

Yes. Sometimes with a 2nd order DE, instead of being given two values of the function, you're given the value of the function at a point and the value of the derivative somewhere. In that case, we could have used the initial condition on the derivative to eliminate the constant of integration.

We could say that θ₀ is my second constant of integration right?

If you want, but it's really an abuse of terminology. The constant of integration appears in the antiderivative (associated to the indefinite integral). When we perform a definite integral, the integration constant cancels out in the difference between the antiderivatives evaluated at the endpoints of the integral.