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

In summary, the function g is continuously differentiable, but the y does not contribute to the wave equation. I need to verify that this is a solution to the wave equation.
  • #1
richyw
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Homework Statement


I have the function[tex]u(x,t)=\frac{1}{2c}\int^{x+ct}_{x-ct}g(\xi)d\xi[/tex]where g is continuously differentiable and c is a constant. I need to verify that this is a solution to the wave equation.

Homework Equations



My prof gave me the formula[tex]\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)[/tex]

The Attempt at a Solution



I really don't see how I can use this rule when trying to take these derivatives. The [itex]\xi[/itex] 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...
 
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  • #2
richyw said:

Homework Statement


I have the function[tex]u(x,t)=\frac{1}{2c}\int^{x+ct}_{x-ct}g(\xi)d\xi[/tex]where g is continuously differentiable and c is a constant. I need to verify that this is a solution to the wave equation.

Homework Equations



My prof gave me the formula[tex]\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)[/tex]

The Attempt at a Solution



I really don't see how I can use this rule when trying to take these derivatives. The [itex]\xi[/itex] 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...

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
LCKurtz said:
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.

Yes, but it does have both x and t in the integration limits. Maybe that is causing the OP's headaches.
 
  • #4
Ray Vickson said:
Yes, but it does have both x and t in the integration limits. Maybe that is causing the OP's headaches.

Could be. He may need an Excedrin before he's done.
 
  • #5
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
richyw said:

Homework Statement


I have the function[tex]u(x,t)=\frac{1}{2c}\int^{x+ct}_{x-ct}g(\xi)d\xi[/tex]where g is continuously differentiable and c is a constant. I need to verify that this is a solution to the wave equation.

Homework Equations



My prof gave me the formula[tex]\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)[/tex]

More useful is the special case
[tex]
\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}[/tex]
 
  • #7
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[itex]u_{tt}=c^2u_{xx}[/itex]

Is it okay if I just upload an image instead of typing out my work?
 
  • #8
http://media.newschoolers.com/uploads/images/17/00/70/22/72/702272.jpeg
 
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  • #9
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
richyw said:
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?

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

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

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

richyw said:
http://media.newschoolers.com/uploads/images/17/00/70/22/72/702272.jpeg

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.
 
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  • #11
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.

[tex]u_{tt}=\frac{c^2}{2c} \left( g_t(x+ct)-g_t(x-ct) \right)[/tex][tex]c^2u_{xx}=\frac{c^2}{2c}\left( g_x(x+ct)+g_x(x-ct) \right)[/tex]I don't see [itex]c^2u_{xx}=u_{tt}[/itex] here...
 
  • #12
wait. am I messing up the chain rule here?
 
  • #13
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
richyw said:
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.

[tex]u_{tt}=\frac{c^2}{2c} \left( g_t(x+ct)\color{red}-g_t(x-ct) \right)[/tex][tex]c^2u_{xx}=\frac{c^2}{2c}\left( g_x(x+ct)+g_x(x-ct) \right)[/tex]I don't see [itex]c^2u_{xx}=u_{tt}[/itex] here...

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$$
g_t = g'(u)u_t = g'(x\pm ct)(\pm c)$$
[Edit]: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).
 
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  • #15
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.
 

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

1. What is the Leibniz rule for differentiating an integral with respect to a parameter?

The Leibniz rule, also known as the differentiation under the integral sign rule, is a method for finding the derivative of an integral with respect to a parameter. It allows us to differentiate an integral without explicitly solving for the integral first.

2. When is the Leibniz rule applicable?

The Leibniz rule is applicable when the integral is of the form ∫ab f(x,t) dt, where f(x,t) is a continuous function of both x and t and both a and b are constants. In other words, the limits of the integral must be constants and not functions of x.

3. How do you apply the Leibniz rule to differentiate an integral?

To apply the Leibniz rule, we first differentiate the integrand f(x,t) with respect to the parameter t. Then, we integrate the resulting expression with respect to t, treating x as a constant. The final result will be the derivative of the integral with respect to x.

4. What are the benefits of using the Leibniz rule?

The Leibniz rule provides a faster and more efficient method for finding the derivative of an integral compared to explicitly solving the integral first. It also allows us to differentiate integrals with complicated expressions and avoid tedious calculations.

5. Are there any limitations or restrictions when using the Leibniz rule?

Yes, the Leibniz rule can only be applied to integrals that meet the criteria mentioned in question 2. Additionally, the function f(x,t) must be continuous and differentiable in both x and t for the rule to be valid.

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