Is This Line Integral Differentiable?

[Contingency]

Homework Statement

$\vec { F } \left( x,y \right) =u\left( x,y \right) \hat { i } +v\left( x,y \right) \hat { j }$
$u\left( x,y \right) , v\left( x,y \right)$ are continuous on ℝ²
$\Gamma$ is piecewise smooth.
Is $\psi (x,y){ =\int { \vec { F } \left( x,y \right) \cdot \vec { dr } } }$ differentiable? How can I show this if it is?
ψ is meant to be the line integral over Γ.

Homework Equations

Definition of differentiability; definition of partial derivatives; formula for calculating line integrals under a parametrization.

The Attempt at a Solution

I wanted to show that the partial derivatives of ψ are continuous, which would imply differentiability. I can't deal with ψ as it's defined, only with the formula, and I think differentiability must be independent of some arbitrary parametrization..

 

[STEMucator]

Homework Statement

$\vec { F } \left( x,y \right) =u\left( x,y \right) \hat { i } +v\left( x,y \right) \hat { j }$
$u\left( x,y \right) , v\left( x,y \right)$ are continuous on ℝ²
$\Gamma$ is piecewise smooth.
Is $\psi (x,y){ =\int { \vec { F } \left( x,y \right) \cdot \vec { dr } } }$ differentiable? How can I show this if it is?
ψ is meant to be the line integral over Γ.

Homework Equations

Definition of differentiability; definition of partial derivatives; formula for calculating line integrals under a parametrization.

The Attempt at a Solution

I wanted to show that the partial derivatives of ψ are continuous, which would imply differentiability. I can't deal with ψ as it's defined, only with the formula, and I think differentiability must be independent of some arbitrary parametrization..

So what happens if you parametrize your curve? Remember that it's piecewise smooth, so you might have to break it up a bit. Plug in what you know and remember to choose the interval.

Think about the continuity of u and v.

 

[Contingency]

Could you be a bit more specific?
Parametrizing my curve allows me to use the Line Integral formula, which is an integral of 't' alone (t being a parameter). I still can't see anything as far as proving continuity of partial derivatives goes..

In the original problem, I was to find a formula for ∂ψ/∂x for a given Γ. When I tried to solve it, I did exactly what you said - plugged in all I know etc - but from what I noticed, I only used continuity to guarantee integrability.. The formula I arrived at required ψ be differentiable (used the chain rule).

 

[STEMucator]

Could you be a bit more specific?
Parametrizing my curve allows me to use the Line Integral formula, which is an integral of 't' alone (t being a parameter). I still can't see anything as far as proving continuity of partial derivatives goes..

In the original problem, I was to find a formula for ∂ψ/∂x for a given Γ. When I tried to solve it, I did exactly what you said - plugged in all I know etc - but from what I noticed, I only used continuity to guarantee integrability.. The formula I arrived at required ψ be differentiable (used the chain rule).

Could you show me your work?

 

[Dick]

Could you be a bit more specific?
Parametrizing my curve allows me to use the Line Integral formula, which is an integral of 't' alone (t being a parameter). I still can't see anything as far as proving continuity of partial derivatives goes..

In the original problem, I was to find a formula for ∂ψ/∂x for a given Γ. When I tried to solve it, I did exactly what you said - plugged in all I know etc - but from what I noticed, I only used continuity to guarantee integrability.. The formula I arrived at required ψ be differentiable (used the chain rule).

If you think about the integral dt, you are integrating a piecewise continuous function. Is the result of that necessarily differentiable? Try to think of a counterexample.

 

[Contingency]

If you think about the integral dt, you are integrating a piecewise continuous function. Is the result of that necessarily differentiable? Try to think of a counterexample.

I don't see why my integrand is piecewise continuous.. U, V are everywhere continuous..

My problem is ψ..

 

[Dick]

I don't see why my integrand is piecewise continuous.. U, V are everywhere continuous..

My problem is ψ..

$d \vec r(t)=\vec r'(t) dt$. If you are only given that r(t) is piecewise smooth then r'(t) is only piecewise continuous. Think of the one dimensional example r(t)=|t|. I'd call that piecewise smooth.

 

[Contingency]

I understand that bit now.
Here's my attempt at a solution.
Hopefully you will be able to understand where my question comes from and help me.

 

[Dick]

I understand that bit now.
Here's my attempt at a solution.
Hopefully you will be able to understand where my question comes from and help me.

Yeah, that helps. I was picturing how you were differentiating the psi wrong. But I think the whole 't' thing is leading you astray. H1 and H2 aren't functions of t. t is a dummy variable. When you've got something like $\int_0^1 f(xt,y) d(xt)$ change the variable to x'=xt. Now you've got $\int_0^x f(x',y) d(x')$. What's the x derivative of that?

 

[Contingency]

I don't understand a few things:
First, I don't see d(xt) anywhere, only dt. Where did the integral with d(xt) come from?
Second, can you please explain how is it that I obtain an integral with a variable limit from an integral with a constant limit? Also the bit with x and x', what are they exactly?

As far as what you asked, the derivative of that integral is f(x', t) because of the fundamental theorem.

 

[Dick]

I don't understand a few things:
First, I don't see d(xt) anywhere, only dt. Where did the integral with d(xt) come from?
Second, can you please explain how is it that I obtain an integral with a variable limit from an integral with a constant limit? Also the bit with x and x', what are they exactly?

As far as what you asked, the derivative of that integral is f(x', t) because of the fundamental theorem.

Nope, the derivative can't be f(x',t). x' is a dummy variable of integration. So is t. It's f(x,y). In my example x is the constant. x'=xt. x' is just a change of variables from t. In your formula for H2(t) you have x0*v(x0*t,y) dt. That's the same as v(x0*t,y) d(x0*t) since x0 is constant. You really want to change variables to x'=x0*t.

 

[Contingency]

Okay, gotcha.
So your x is my x₀, and I understand the why the derivative is f(x,y).
I'm still unsure of where the d(xt) came from. I have xU(xt,y)d(t) in my integral. Can you move the x inside the parenthesis because it's a constant? "d" is like a linear operator?

 

[Dick]

Okay, gotcha.
So your x is my x₀, and I understand the why the derivative is f(x,y).
I'm still unsure of where the d(xt) came from. I have xU(xt,y)d(t) in my integral which translates into xd(t). Can you move the x inside the parenthesis because it's a constant? "d" is like a linear operator?

Exactly. You can move x into the d part because it's a constant. d(xt)=xdt. Suggests you do the change of variables x'=xt. So dx'=xdt.

 

[Contingency]

You said that in your example, x is a constant.
Doesn't that mean that $\int _{ 0 }^{ x }{ f\left( x',y \right) dx' }$ is just a number? Why can we suddenly take the derivative with respect to x which is a constant?

I think I have a fundamental problem of understanding here..

 

[Dick]

You said that in your example, x is a constant.
Doesn't that mean that $\int _{ 0 }^{ x }{ f\left( x',y \right) dx' }$ is just a number? Why can we suddenly take the derivative with respect to x which is a constant?

I think I have a fundamental problem of understanding here..

x is a constant with respect to the dt or dx' integration. That's all that I mean by 'constant'.

 

[Contingency]

I think I'm not being clear enough, so I will write down on paper what's bothering me in each step of the problem. Right now I have a mess in my head.
Thanks a lot for bearing with me!

 

[Contingency]

Figured everything out. Thanks a lot for all your help!