Integrating a partial derivative

[Saladsamurai]

Homework Statement

Well hello!

I am still uncomfortable with partials. In my (fluid mechanics) text we introduce this "stream function" \Psi(x,y) such that u =\partial{\Psi}/\partial{y} and v =-\partial{\Psi}/\partial{x} where u and v are the horizontal and vertical components of the flow velocity.

The author then integrates u =\partial{\Psi}/\partial{y} to obtain \Psi = \int_0^y u\,dy. This does not jive well with me. First of all, I don't think we can just say u =\partial{\Psi}/\partial{y} \rightarrow \Psi = \int u \,dy can we? He just completely ignored the fact that they were partials! Or has he?

Would it be correct to assume that what he means is for us to integrate \int\partial{\Psi} = \int u\,\partial{y} in the same way that we would integrate [/itex]\int d\Psi=\int u \,dy[/itex] under the condition that our constant of integration include the possibility that it is a function of x ?Just trying to reason through this one. Thanks!

 

[lanedance]

exactly so say
<br /> u(x,y) =\partial{\Psi(x,y)}/\partial{y}<br />

then the form of psi will be
<br /> \Psi(x,y)} = \int U(x,y) + g(x) <br />
<br /> U(x,y) = \int u(x,y)<br />
where g is the constant of integration

not that the way psi is written in terms of the definite integral sets it to be zero at y=0

 

[Saladsamurai]

exactly so say
<br /> u(x,y) =\partial{\Psi(x,y)}/\partial{y}<br />

then the form of psi will be
<br /> \Psi(x,y)} = \int U(x,y) + g(x) <br />
<br /> U(x,y) = \int u(x,y)<br />
where g is the constant of integration

not that the way psi is written in terms of the definite integral sets it to be zero at y=0

Right. Ok. I have a feeling he is just "hiding" the g(x) somewhere else, whether it be in an unspoken assumption or elsewhere in the end result.

Out of curiosity: When talking about the integrals of single variable functions and we use Leibnitz notation, we typically 'separate' the equation as dy/dx = f(x)\rightarrow y(x)=\int f(x)\,dx.

Is it proper to do the same with partials: \partial{y}/\partial{x} = f(x)\rightarrow y(x,y)=\int f(x)\,\partial{x}? Where upon integration we let the constant of integration be a function of x.

It is just the notation that I am curious about.

 

[fzero]

The stream function is an example of a potential function, say \Phi(\vec{x}). By potential function, we mean that the components of the gradient \nabla \Phi(\vec{x}) are related to quantities of physical interest, in this case the components of flow velocity. The proper way to compute the potential is via a line integral:

\int_L \nabla \Phi(\vec{x})\cdot d\vec{x} = \Phi(\vec{x}_2) - \Phi(\vec{x}_1),

where L is a smooth path from \vec{x}_1 to \vec{x}_2 and d\vec{x} is the tangent vector at each point of the path.

It might be useful to see if you can apply this definition to the example in the text to see if there's something special about the path chosen there to explain the absence of an integration constant. It's also possible that the potential vanishes at (0,0) and the author is using that implicitly.

Your question about differential equations is also explained by the formula above. We can integrate a function of two variables whenever we can express the equation in the form

M(x,y) dx + N(x,y) dy = \nabla \Phi(x,y) \cdot d\vec{x}.

This is true whenever

\frac{\partial M}{\partial y} = -\frac{\partial N}{\partial x},

which comes from demanding that the curl of a gradient is zero. You can refer to a diff eq text for exact differential equations to see the details.

 

[Saladsamurai]

Hi there fzero! Great information. I will check into this as soon as I trudge through my assignments. I flipped back a few pages and noticed that he imposed the condition that \Psi(x,0) = 0 and looking at your post this looks like it might be what I was missing.

I am familiar (from an engineering mathematics standpoint) with exact diff eqs. So, do we say that we cannot integrate function of two variables when it is not exact?

 

[fzero]

Hi there fzero! Great information. I will check into this as soon as I trudge through my assignments. I flipped back a few pages and noticed that he imposed the condition that \Psi(x,0) = 0 and looking at your post this looks like it might be what I was missing.

I am familiar (from an engineering mathematics standpoint) with exact diff eqs. So, do we say that we cannot integrate function of two variables when it is not exact?

It's sometimes possible that we can use an integrating factor to show that there's a related exact differential. If a differential is completely not exact, then a potential function does not exist, but there can still be some other solution to the differential equation (possibly not expressible analytically). If there is no potential function, then the physics is usually path dependent. For instance if there was no stream function, the integrals of the flow velocity components would not just depend on the endpoints of the line integral, but would depend on the details of the path chosen. Also the line integral over a closed path would not vanish. Presumably most of your course will focus on cases where a potential exists.