Derivative of Convolution

[lugita15]

Derivative of a Convolution

Homework Statement

How do I find the derivative of a convolution, meaning $$\frac{d}{dt}(f \ast g)(t)$$?

Homework Equations

$$(f \ast g)(t)=\int^{}_{} f(t-\tau)g(\tau)d\tau$$

The Attempt at a Solution

I want to use the fundamental theorem of calculus, but I can't just directly substitute t for $$\tau$$, because that would make$$f(t-\tau)g(\tau)=f(0)g(t)$$, which doesn't make sense. How would I correctly apply the fundamental theorem of calculus in this case?

 

[jbunniii]

If you know basic Fourier transform theory, you can answer the question as follows:

Define

$$h(t) = \frac{d}{dt}(f*g)(t)$$

Then

$$\begin{align*} \hat{h}(\omega) &= \mathcal{F}\{h(t)\} \\ &= \mathcal{F}\{\frac{d}{dt}(f*g)(t)\} \\ &= i\omega \mathcal{F}\{(f*g)(t)\} \\ &= i\omega \hat{f}(\omega)\hat{g}(\omega) \end{align*}$$

You can now associate the $i \omega$ with either $$\hat{f}$$ or $$\hat{g}$$ and invert the Fourier transform to obtain

$$\frac{d}{dt}(f*g)(t) = \left(\left(\frac{d}{dt}f\right) * g\right)(t) = \left(f * \left(\frac{d}{dt}g\right)\right)(t)$$

Now the above is not rigorous, because I don't specify under what conditions the operations are valid. Also, it doesn't answer how to solve the problem without resorting to Fourier transforms. But now you know what answer to work toward (which you could equally well have found by looking in some table, so I don't think I'm giving the game away).

 

[lugita15]

If you know basic Fourier transform theory, you can answer the question as follows:

I'm sorry, but unfortunately I don't know about Fourier transforms, but I do know about Laplace transforms. In fact, it is in the context of solving differential equations using Laplace transforms that I came up with this question. Is there any way that I can use the fundamental theorem of calculus to do this? Intuitively, it seems like this problem is just a simple case of differentiating an integral.

 

[jbunniii]

I'm sorry, but unfortunately I don't know about Fourier transforms, but I do know about Laplace transforms.

You can solve it very similarly by using Laplace transforms. I'll leave the details to you.

Is there any way that I can use the fundamental theorem of calculus

I don't see how that would work, because you are dealing with a definite integral (from $-\infty$ to $\infty$), not an indefinite one.

But notice that the variable of integration is $\tau$, not $t$. Therefore, I claim that you can simply slide the differentiation operator inside the integral as follows:

$$\frac{d}{dt}\int_{-\infty}^{\infty}f(t-\tau) g(\tau) d\tau = \int_{-\infty}^{\infty} \frac{d}{dt} f(t-\tau) g(\tau) d\tau$$

See if you can justify why (and under what conditions) that is valid.

 

[lugita15]

There's no need to use the fundamental theorem of calculus, or Fourier transforms. The integral is over the dummy variable $\tau$ (and the limits of integration do not depend on $t$), so it doesn't matter whether you first integrate w.r.t $\tau$ and then differentiate w.r.t. $t$ or vice versa.

$$\frac{d}{dt}\int_{-\infty}^{\infty}f(t-\tau)g(\tau)d\tau=\int_{-\infty}^{\infty}\frac{d}{dt}\left[f(t-\tau)g(\tau)\right]d\tau\int_{-\infty}^{\infty}f'(t-\tau)g(\tau)d\tau$$

Apparently, we are using different definitions of the covolution. In the definition I am using, the integral goes from 0 to t, not negative infinity to infinity.

 

[jbunniii]

Apparently, we are using different definitions of the covolution. In the definition I am using, the integral goes from 0 to t, not negative infinity to infinity.

That makes the implicit assumption that both $f(t)$ and $g(t)$ are zero for $t < 0$. But that's OK. If they are both zero for negative $t$ then you can equally well set the integration limits to $-\infty$ to $\infty$ without changing the result. Do you see why?

 

[lugita15]

That makes the implicit assumption that both $f(t)$ and $g(t)$ are zero for $t < 0$. But that's OK. If they are both zero for negative $t$ then you can equally well set the integration limits to $-\infty$ to $\infty$ without changing the result. Do you see why?

No, I don't. I see why setting the lower limit to negative infinity rather than zero would make no difference if f and g are zero for t<0. But how can you just arbitrarily change the upper limit from t, a variable, to infinity?

 

[gabbagabbahey]

Apparently, we are using different definitions of the covolution. In the definition I am using, the integral goes from 0 to t, not negative infinity to infinity.

I see...in that case, you might want to make use of the Heaviside theta function:

$$H(x)=\left\{\begin{array}{lr}0, & x<0 \\ 1, & x>0\end{array}\right.$$

by realizing that

$$\int_0^t f(t-\tau)g(\tau)d\tau=\int_0^\infty H(t-\tau)f(t-\tau)g(\tau)d\tau$$

 

[lugita15]

I see...in that case, you might want to make use of the Heaviside theta function:

$$H(x)=\left\{\begin{array}{lr}0, & x<0 \\ 1, & x>0\end{array}\right.$$

by realizing that

$$\int_0^t f(t-\tau)g(t)d\tau=\int_0^\infty H(t-\tau)f(t-\tau)g(\tau)d\tau$$

Why can't the fundamental theorem of calculus just be used directly? If we just leave the limits as 0 and t, then we have an indefinite integral. In other words, the problem is to take the derivative with respect to t of an integral the upper limit of which is t, precisely the case that the fundamental theorem of calculus is supposed to deal with. So why doesn't it work in this case?

 

[jbunniii]

No, I don't. I see why setting the lower limit to negative infinity rather than zero would make no difference if f and g are zero for t<0. But how can you just arbitrarily change the upper limit from t, a variable, to infinity?

Because $f(t) = 0$ for $t < 0$ if and only if $f(t - \tau) = 0$ for $\tau > t$. Therefore I can use any upper endpoint for the integration as long as it is greater than or equal to $t$. If I want one upper endpoint that satisfies this condition no matter what $t$ is, then that endpoint has to be $\infty$.

 

[jbunniii]

Why can't the fundamental theorem of calculus just be used directly? If we just leave the limits as 0 and t, then we have an indefinite integral. In other words, the problem is to take the derivative with respect to t of an integral the upper limit of which is t, precisely the case that the fundamental theorem of calculus is supposed to deal with. So why doesn't it work in this case?

Good question. The answer is that the INTEGRAND also depends on $t$, so the FTC doesn't apply, at least not directly.

 

[lugita15]

Good question. The answer is that the INTEGRAND also depends on $$t$$, so the FTC doesn't apply, at least not directly.

I see what you mean. So in essence we are dealing with an expression of the form $$\frac{\partial}{\partial t}\int^{t}_{0}F(t, \tau) \partial \tau$$. Is there any formula for such a beast?

 

[jbunniii]

I see what you mean. So in essence we are dealing with an expression of the form $$\frac{\partial}{\partial t}\int^{t}_{0}F(t, \tau) \partial \tau$$. Is there any formula for such a beast?

Not in general, as far as I know.

 

[lugita15]

Not in general, as far as I know.

This seems to be a far more interesting topic than the original question I asked. I think I'll start a non-homework thread about it.

 

[gabbagabbahey]

I see what you mean. So in essence we are dealing with an expression of the form $$\frac{\partial}{\partial t}\int^{t}_{0}F(t, \tau) \partial \tau$$. Is there any formula for such a beast?

Sure, you can derive it yourself using the Heaviside theta function...

 

[lugita15]

I started a new thread about it, https://www.physicsforums.com/showthread.php?t=403028".

 

[lugita15]

Sure, you can derive it yourself using the Heaviside theta function...

How would I use the Heaviside theta function in this more general case?

 

[gabbagabbahey]

Same way as before...

$$\int_0^t F(t,\tau)d\tau=\int_0^\infty H(t-\tau)F(t,\tau)d\tau$$

Then take the derivative...

 

[jbunniii]

But H isn't differentiable at 0, so you either need distribution theory to proceed along these lines, or you hand-wave something about a delta function and cross your fingers that the answer turns out valid. I guess it depends on whether you are a mathematician or a physicist.

[Edit]: I'm sure it turns out fine, and can probably be made rigorous by choosing a sequence of differentiable functions $H_n$ that "almost" have the desired property and which converge to $H$, then making an appropriate limiting argument.

 

[lugita15]

$$\int_0^t F(t,\tau)d\tau=\int_0^\infty H(t-\tau)F(t,\tau)d\tau$$

Maybe I'm missing something obvious, but why is this true?

 

[gabbagabbahey]

Maybe I'm missing something obvious, but why is this true?

Because $H(t-\tau)$ is zero for $\tau>t$.

 

[lugita15]

Because $H(t-\tau)$ is zero for $\tau>t$.

OK, that makes sense. So $H(t-\tau)$ is only nonzero between 0 and t.

 

[lugita15]

On the other thread I created, I got the formula $$\frac{\partial}{\partial t}\int^{t}_{0}F(t, \tau) \partial \tau = F(t,t)+\int^{t}_{0}\frac{\partial F}{\partial t} \partial \tau$$. Applying that to the convolution, I get $$\frac{d}{dt}(f \ast g)(t)=f(0)g(t)+\int^{t}_{0}g(\tau) \frac{\partial}{\partial t}(f(t-\tau)) \partial \tau$$. Is that right?

 

[gabbagabbahey]

Looks fine to me...you can rewrite it using the chain rule though:

$$\frac{\partial}{\partial t}f(t-\tau)=f'(t-\tau)\frac{\partial }{\partial t}(t-\tau)=f'(t-\tau)$$

So your integral is just the convolution of $f'$ with $g$, and you are left with

[tex]\frac{d}{dt}(f*g)(t)=f(0)g(t)+(f'*g)(t)[/itex]

 

[lugita15]

Looks fine to me...you can rewrite it using the chain rule though:

$$\frac{\partial}{\partial t}f(t-\tau)=f'(t-\tau)\frac{\partial }{\partial t}(t-\tau)=f'(t-\tau)$$

So your integral is just the convolution of $f'$ with $g$, and you are left with

[tex]\frac{d}{dt}(f*g)(t)=f(0)g(t)+(f'*g)(t)[/itex]

Good, that looks more like what it's supposed to look like. But now I'm worried about the first term. Are we supposed to assume f(0)=0? Because otherwise the first term violates the symmetry of f and g.

 

[gabbagabbahey]

Good, that looks more like what it's supposed to look like. But now I'm worried about the first term. Are we supposed to assume f(0)=0? Because otherwise the first term violates the symmetry of f and g.

Are you sure it violates that symmetry?

$$\begin{aligned}(f*g)(t)=(g*f)(t) &\implies \frac{d}{dt}(f*g)(t)=\frac{d}{dt}(g*f)(t) \\ &\implies f(0)g(t)+(f'*g)(t)= f(t)g(0) +(f*g')(t)\end{aligned}$$