How to Solve a Differential Equation Involving Pressure and Velocity?

[fysiikka111]

Homework Statement

$$\frac{dp}{dz}=\mu_{o}\frac{du}{dr}\frac{1}{r}\frac{d}{dr}(r \frac{du}{dr})$$

Homework Equations

The Attempt at a Solution

Multiply by r, and then integrate with respect to r to get:
$$\frac{dp}{dz}\frac{r^{2}}{2}+C_{1}=\mu_{o}ur \frac{du}{dr}$$
Divide by r and integrate again:
$$\frac{dp}{dz}\frac{r^{2}}{4}+C_{1}ln(r)+C_{2}=\mu_{o}u^{2}$$
Are those steps correct? Especially, is the first integration done correctly?
Thanks

 

[praharmitra]

No, it's not correct. To solve this differential eq. change to variable x = ln r.

 

[fysiikka111]

No, it's not correct. To solve this differential eq. change to variable x = ln r.

Thanks for replying. Do you mean if x=ln r, then subsitute r=e^x into the equation?

 

[praharmitra]

yes. Another thing I wanted to ask - Is $$\frac{dp}{dz}$$ a constant? or atleast independent of r?

 

[fysiikka111]

yes. Another thing I wanted to ask - Is $$\frac{dp}{dz}$$ a constant? or atleast independent of r?

Yes, its independent of r. May I ask which part of my first solution was incorrect? Thanks.

 

[praharmitra]

Well, if I understood what you wrote correctly, then you are saying

$$\int r\frac{dp}{dx} dr = \int \mu_0\frac{du}{dr} \frac{d}{dr}\left(r\frac{du}{dr}\right)dr$$

$$\Rightarrow \frac{r^2}{2}\frac{dp}{dx}+C = \int \mu_0\frac{d}{dr}\left(r\frac{du}{dr}\right)du$$

I don't see how you went from the RHS here to the RHS you have written

 

[fysiikka111]

Well, if I understood what you wrote correctly, then you are saying

$$\int r\frac{dp}{dx} dr = \int \mu_0\frac{du}{dr} \frac{d}{dr}\left(r\frac{du}{dr}\right)dr$$

$$\Rightarrow \frac{r^2}{2}\frac{dp}{dx}+C = \int \mu_0\frac{d}{dr}\left(r\frac{du}{dr}\right)du$$

I don't see how you went from the RHS here to the RHS you have written

I see what I did wrong; I canceled both of the dr's when I integrated.
$$\frac{dp}{dz}=\mu_{o} \frac{du}{de^{x}} \frac{1}{e^{x}} \frac{d}{de^{x}}(e^{x} \frac{du}{de^{x}})$$
How would you go about rearranging it now into a function of u? Don't need to show latex, just need a hint. Thanks.

 

[praharmitra]

Use $$de^x = e^x dx$$

 

[fysiikka111]

Use $$de^x = e^x dx$$

Is this correct?
$$\frac{dp}{dz} = \mu_{o} \frac{1}{e^{x}} \frac{du}{dx} \frac{d}{dx} (\frac{du}{dx}) = \frac{\mu_{o}}{de^{x}}du \frac{d}{dx}(\frac{du}{dx})$$

 

[praharmitra]

Is this correct?
$$\frac{dp}{dz} = \mu_{o} \frac{1}{e^{x}} \frac{du}{dx} \frac{d}{dx} (\frac{du}{dx}) = \frac{\mu_{o}}{de^{x}}du \frac{d}{dx}(\frac{du}{dx})$$

No, it should be

$$\frac{dp}{dz} = \mu_0 \frac{1}{e^{3x}} \frac{du}{dx} \frac{d}{dx} (\frac{du}{dx})$$

 

[fysiikka111]

No, it should be

$$\frac{dp}{dz} = \mu_0 \frac{1}{e^{3x}} \frac{du}{dx} \frac{d}{dx} (\frac{du}{dx})$$

I've ended up with:
$$u^{2}\mu_{o} = \frac{dp}{dz} \frac{x^{3}}{6}e^{3x} + C_{1}\frac{x^{2}}{2}e^{3x} + C_{2}xe^{3x} + C_{3}$$
Is that right? Can I say that e^3x=r³? Though I don't believe this is the way its meant to be done as it is unsolvable with three constants. Thanks a lot for your help.

 

[praharmitra]

No you should have only 2 constants. Show me what you have done, I'll tell you where you went wrong

 

[SammyS]

If dp/dz is independent of r, then I suggest multiplying by r² to get:

$$r^2\,\frac{dp}{dz}\,dr=\mu_{o}\frac{du}{dr}\frac{r^2}{r}\frac{d}{dr}(r \frac{du}{dr})dr=\mu_{o}r\,\frac{du}{dr}\,d(r \frac{du}{dr})$$

Integrate both sides to get:

$$\frac{r^3}{3}\frac{dp}{dz}+C=\mu_{0}\,\frac{1}{2}\left(r \frac{du}{dr}\right)^2$$

 

[fysiikka111]

No you should have only 2 constants. Show me what you have done, I'll tell you where you went wrong

$$\begin{multline*}\frac{dp}{dz} = \mu_0 \frac{1}{e^{3x}} \frac{du}{dx} \frac{d}{dx} (\frac{du}{dx})\\<br /> \frac{dp}{dz}x + C_{1} = \mu_0 \frac{1}{e^{3x}} u \frac{d}{dx}(\frac{du}{dx})\\ \frac{dp}{dz} \frac{x^{2}}{2} + C_{1}x + C_{2} = \mu_{o} \frac{1}{e^{3x}}u \frac{du}{dx}\\<br /> u^{2}\mu_{o} = \frac{dp}{dz} \frac{x^{3}}{6}e^{3x} + C_{1}\frac{x^{2}}{2}e^{3x} + C_{2}xe^{3x} + C_{3}\end{multline*}$$
Thanks

 

[praharmitra]

No, you have got several things wrong. Here is how it would go

$$\frac{dp}{dz} = \frac{\mu_0}{e^{3x}}\frac{du}{dx}\frac{d}{dx}\left(\frac{du}{dx}\right)$$
$$\frac{dp}{dz}\int e^{3x}dx = \mu_0\int\frac{du}{dx}d\left(\frac{du}{dx}\right)$$
Now if we call $y = \frac{du}{dx}$. This reads
$$\frac{dp}{dz}\int e^{3x}dx = \mu_0\int ydy \Rightarrow \frac{dp}{dz}\frac{e^{3x}}{3} = \mu_0 \frac{y^2}{2} = \frac{\mu_0}{2}\left(\frac{du}{dx}\right)^2+C_1$$

Thus

$$\frac{du}{dx} = \left[\frac{2}{\mu_0}\left(\frac{dp}{dz}\frac{e^{3x}}{3}-C_1\right)\right]^{1/2}$$

Now you can solve this?

 

[fysiikka111]

No, you have got several things wrong. Here is how it would go

$$\frac{dp}{dz} = \frac{\mu_0}{e^{3x}}\frac{du}{dx}\frac{d}{dx}\left(\frac{du}{dx}\right)$$
$$\frac{dp}{dz}\int e^{3x}dx = \mu_0\int\frac{du}{dx}d\left(\frac{du}{dx}\right)$$
Now if we call $y = \frac{du}{dx}$. This reads
$$\frac{dp}{dz}\int e^{3x}dx = \mu_0\int ydy \Rightarrow \frac{dp}{dz}\frac{e^{3x}}{3} = \mu_0 \frac{y^2}{2} = \frac{\mu_0}{2}\left(\frac{du}{dx}\right)^2+C_1$$

Thus

$$\frac{du}{dx} = \left[\frac{2}{\mu_0}\left(\frac{dp}{dz}\frac{e^{3x}}{3}-C_1\right)\right]^{1/2}$$

Now you can solve this?

Right, thanks.
If $v = w^{3/2}$, where $w = \frac{2}{\mu_0}\left(\frac{dp}{dz}\frac{e^{3x}}{3}-C_1)$, then with chain rule
$$\frac{dv}{dx} = \frac{w^{1/2}e^{3x}}{\mu_{o}}\frac{dp}{dz}$$
Hence
$$\int\frac{dv}{dx}dx = w^{3/2} + C_{2}\rightarrow \int w^{1/2}dx = \frac{\mu_{o}}{e^{3x}\frac{dp}{dz}}(w^{3/2}+C_{2}) = u$$
Would $e^{3x}=r^{3}$?

 

[SammyS]

If dp/dz is independent of r, then I suggest multiplying by r² to get:
...

I probably should have written the following as the result:

$$r^2\,\frac{dp}{dz}=\mu_{o}\frac{du}{dr}\frac{r^2}{r}\frac{d}{dr}\left(r \frac{du}{dr}\right)$$

Integrating with respect to r gives:

$$\int{r^2\,\frac{dp}{dz}}\,dr=\int{\mu_{o}\frac{du}{dr}\frac{r ^2}{r}\frac{d}{dr}\left(r \frac{du}{dr}\right)}dr$$

$$=\int\mu_{o}r\,\frac{du}{dr}\,d\left(r \frac{du}{dr}\right)$$​

$$\frac{r^3}{3}\frac{dp}{dz}+C_1=\mu_{0}\,\frac{1}{2}\left(r \frac{du}{dr}\right)^2$$

This can be solved for du/dr & integrated, to find u.

 

[fysiikka111]

I probably should have written the following as the result:

$$r^2\,\frac{dp}{dz}=\mu_{o}\frac{du}{dr}\frac{r^2}{r}\frac{d}{dr}\left(r \frac{du}{dr}\right)$$

Integrating with respect to r gives:

$$\int{r^2\,\frac{dp}{dz}}\,dr=\int{\mu_{o}\frac{du}{dr}\frac{r ^2}{r}\frac{d}{dr}\left(r \frac{du}{dr}\right)}dr$$

$$=\int\mu_{o}r\,\frac{du}{dr}\,d\left(r \frac{du}{dr}\right)$$​

$$\frac{r^3}{3}\frac{dp}{dz}+C_1=\mu_{0}\,\frac{1}{2}\left(r \frac{du}{dr}\right)^2$$

This can be solved for du/dr & integrated, to find u.

Thanks a lot, that's a good method.