# Leibniz rule for differentiating an integral w.r.t a parameter

1. Mar 25, 2014

### richyw

1. The problem statement, all variables and given/known data
I have the function$$u(x,t)=\frac{1}{2c}\int^{x+ct}_{x-ct}g(\xi)d\xi$$where g is continuously differentiable and c is a constant. I need to verify that this is a solution to the wave equation.

2. Relevant equations

My prof gave me the formula$$\frac{d}{dt}\int^{b(t)}_{a(t)}F(y,t)dy=\int^{b(t)}_{a(t)}F_t(y,t)dy+F(b(t),t)b'(t)-F(a(t),t)a'(t)$$

3. The attempt at a solution

I really don't see how I can use this rule when trying to take these derivatives. The $\xi$ is really messing me up. Do I need to use this rule right off the bat on my first derivative (I am trying to take the x derivatives first). It seems like this question is just as easy as using a formula, but I cannot seem to get an answer that works out...

2. Mar 25, 2014

### LCKurtz

That $xi$ is a dummy variable. Would it lessen your confusion if you called it $y$?$$u(x,t)=\frac{1}{2c}\int^{x+ct}_{x-ct}g(y)dy$$Now the integrand looks just like your Leibnitz formula except it's simpler, having no $t$ variable.

3. Mar 25, 2014

### Ray Vickson

Yes, but it does have both x and t in the integration limits. Maybe that is causing the OP's headaches.

4. Mar 25, 2014

### LCKurtz

Could be. He may need an Excedrin before he's done.

5. Mar 25, 2014

### richyw

so the y still didn't do much for me. If I have g(y only), then how all of a sudden I have g(b(t),t)?

6. Mar 25, 2014

### pasmith

More useful is the special case
$$\frac{\partial}{\partial x} \int_{a(x,t)}^{b(x,t)} f(y)\,dy = f(b(x,t))\frac{\partial b}{\partial x} - f(a(x,t))\frac{\partial a}{\partial x}$$

7. Mar 25, 2014

### richyw

ok, so first of all, how can you say the integral in the general formula is zero? because F is a function of y only?

second, working this all out, I get something that looks like the wave equation kinda, but I don't see how I exactly have$u_{tt}=c^2u_{xx}$

Is it okay if I just upload an image instead of typing out my work?

8. Mar 25, 2014

### richyw

Last edited by a moderator: May 6, 2017
9. Mar 25, 2014

### richyw

see on the last two lines, I don't think I have an equality here? also the denominator in the last line is a typo. I know it should be c^2/2c

10. Mar 25, 2014

### LCKurtz

Yes, that's why the integrand with the $F_t$ is zero.

That's why I don't like images. It makes us retype things. In the step where you have$$\frac 1 {2c}(g_t(x+ct)c + g_t(x-ct)c$$you should have$$\frac 1 {2c}g'(x+ct)c - g'(x-ct)(-c)$$The second g should have a negative sign in all your calculations because it came from the lower limit and the last -c is from the chain rule.

Last edited by a moderator: May 6, 2017
11. Mar 25, 2014

### richyw

sorry. I can type stuff out from now on. I'm confused by your primed notation now. what does that mean? even accounting for the negative sign I missed, I do not see how these can be equal, because one I am differentiating wrt x and the other one I am differentiating wrt t.

$$u_{tt}=\frac{c^2}{2c} \left( g_t(x+ct)-g_t(x-ct) \right)$$$$c^2u_{xx}=\frac{c^2}{2c}\left( g_x(x+ct)+g_x(x-ct) \right)$$I don't see $c^2u_{xx}=u_{tt}$ here...

12. Mar 25, 2014

### richyw

wait. am I messing up the chain rule here?

13. Mar 25, 2014

### richyw

ok. I see where I went wrong with the chain rule now, I had to write the chain rule in leibniz notation (with ANOTHER dummy variable) and now I think I have the answer. Thanks a lot for your help.It seems so easy now. Although I guess it's always easy once you have figured it out haha.

14. Mar 25, 2014

### LCKurtz

Once you get the chain rule correct in your $u_t$ and $u_{tt}$ derivatives you will have a + sign where I have the red and your two right sides will be identical.

About the prime, $g$ is a function of one variable. Think of it as $g(u)$ where $u=x\pm ct$ (either way). If you want to differentiate $g$ with respect to $t$ you would use the chain rule$$g_t = g'(u)u_t = g'(x\pm ct)(\pm c)$$
:Although your answers will agree once you fix the chain rule thing, both signs between the two g terms should be negative once you take into account my comment about lower limits (the last sentence in post #10).

Last edited: Mar 25, 2014
15. Mar 25, 2014

### LCKurtz

Please note the editing in post #14. Your final result should have$$\frac{c^2}{2c} \left( g_t(x+ct)-g_t(x-ct) \right)$$on the right side of both.