Show that solution is infinitely many times differentiable

In summary, the conversation discusses proving the solution of the heat equation, $u(x,t)=\frac{1}{2 \sqrt{k \pi t}} \int_{-\infty}^{+\infty} e^{-\frac{(x-s)^2}{4kt}} \phi(s) ds$, is infinitely differentiable. The participants consider using Leibniz's integral theorem and the product rule to show that the partial derivatives of $u$ exist. They also discuss the need to address the boundaries of Leibniz's integral theorem and the possibility of swapping limits. There is also mention of checking the partial derivative $\frac{\partial^a \partial^b u(x,t)}{\partial{x}^a \
  • #1
evinda
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Hello! (Wave)

I want to prove that the solution $u$, $u(x,t)=\frac{1}{2 \sqrt{k \pi t}} \int_{-\infty}^{+\infty} e^{-\frac{(x-s)^2}{4kt}} \phi(s) ds, x \in \mathbb{R}, t>0 (\star)$, of the initial value problem for the heat equation, with continuous and bounded initial value $\phi$, is infinitely many times differentiable, for $t>0$.

We suppose that $|\phi(s)| \leq M$.

I found that $\sup_{x \in \mathbb{R}} |u_t(x,t)| \leq C_1 M \frac{1}{t}$ for some constant $C_1$ and $\sup_{x \in \mathbb{R}}|u_x(x,t)| \leq c_2 M \frac{1}{\sqrt{t}}$, for some constant $c_2$.

At the hint it says that we continue showing that all the derivatives of $u$ exist, for a positive $t$, using the following proposition:

Let $\lambda$ be a positive number and $n$ a natural number. Then $\int_{-\infty}^{+\infty} x^{2n} e^{-2 \lambda x^2} dx<+\infty$. How do we use this proposition? (Thinking)
 
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  • #2
Or is there an other way to show that the solution is infinitely many times differentiable? (Thinking)
 
  • #3
Hey evinda! (Wave)

I think we need to check that we can differentiate through the integral sign, and also through the limit to infinity.
Doesn't Leibniz's integral theorem tell us when we can?
I think it would mean that we merely have to be able to differentiate $e^{-\frac{(x-s)^2}{4kt}}$ with respect to both $x$ and $t$ infinitely many times.
And perhaps we need to verify that the integral still converges... perhaps that's where the hint comes in, ensuring the result is bounded. (Thinking)
 
  • #4
I like Serena said:
Hey evinda! (Wave)

I think we need to check that we can differentiate through the integral sign, and also through the limit to infinity.
Doesn't Leibniz's integral theorem tell us when we can?

So, do we apply the following statement?

View attachment 8199
I like Serena said:
I think it would mean that we merely have to be able to differentiate $e^{-\frac{(x-s)^2}{4kt}}$ with respect to both $x$ and $t$ infinitely many times.

Doesn't this only suffice when we differentiate in respect to $x$ ?

Because differentiating in respect to $t$, we also have to take into consideration the term $\frac{1}{2 \sqrt{k \pi t}}$. Or am I wrong ?
 

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  • #5
evinda said:
So, do we apply the following statement?

Yep.

evinda said:
Doesn't this only suffice when we differentiate in respect to $x$ ?
Because differentiating in respect to $t$, we also have to take into consideration the term $\frac{1}{2 \sqrt{k \pi t}}$. Or am I wrong ?

Can't we apply the product rule ad infinitem?
And isn't that term infinitely differentiable? (Wondering)
 
  • #6
I like Serena said:
Yep.

Can't we apply the product rule ad infinitem?
And isn't that term infinitely differentiable? (Wondering)

So, do we just say that from the Leibniz's rule, we have that $\frac{\partial^a(u(x,t))}{\partial{x^a}}=\frac{\partial^a}{\partial{x^a}} \int_{-\infty}^{+\infty} \frac{1}{2 \sqrt{k \pi t}} e^{-\frac{(x-s)^2}{4kt}} \phi(s) ds=\int_{-\infty}^{+\infty} \frac{1}{2\sqrt{k \pi t}} \frac{\partial^a}{\partial{x^a}}\left( e^{-\frac{(x-s)^2}{4kt}}\right) \phi(s) ds $ and this exists for all $a$, since $e^{-\frac{(x-s)^2}{4kt}}$ is infinitely many times differentiable, as an exponential function?

Also, again from the Leibniz's rule, we have that

$$\frac{\partial^a}{\partial{t^a}} u(x,t)=\frac{\partial^a}{\partial{t^a}} \int_{-\infty}^{+\infty} \frac{1}{2 \sqrt{k \pi t}}e^{-\frac{(x-s)^2}{4kt}} \phi(s) ds=\int_{-\infty}^{+\infty} \frac{\partial^a}{\partial{t^a}}\left( \frac{1}{2 \sqrt{k \pi t}} e^{-\frac{(x-s)^2}{4kt}}\right) \phi(s)ds$$

And $\frac{1}{2 \sqrt{k \pi t}} e^{-\frac{(x-s)^2}{4kt}}$ is infinitely many time differentiable as a product of an exponential with a fraction.

Is the above correct? (Wondering)

And do we also have to check the partial derivative $\frac{\partial^a \partial^b u(x,t)}{\partial{x}^a \partial{t}^b}$ ? (Thinking)
 
  • #7
evinda said:
So, do we just say that from the Leibniz's rule, we have that $\frac{\partial^a(u(x,t))}{\partial{x^a}}=\frac{\partial^a}{\partial{x^a}} \int_{-\infty}^{+\infty} \frac{1}{2 \sqrt{k \pi t}} e^{-\frac{(x-s)^2}{4kt}} \phi(s) ds=\int_{-\infty}^{+\infty} \frac{1}{2\sqrt{k \pi t}} \frac{\partial^a}{\partial{x^a}}\left( e^{-\frac{(x-s)^2}{4kt}}\right) \phi(s) ds $ and this exists for all $a$, since $e^{-\frac{(x-s)^2}{4kt}}$ is infinitely many times differentiable, as an exponential function?

Also, again from the Leibniz's rule, we have that

$$\frac{\partial^a}{\partial{t^a}} u(x,t)=\frac{\partial^a}{\partial{t^a}} \int_{-\infty}^{+\infty} \frac{1}{2 \sqrt{k \pi t}}e^{-\frac{(x-s)^2}{4kt}} \phi(s) ds=\int_{-\infty}^{+\infty} \frac{\partial^a}{\partial{t^a}}\left( \frac{1}{2 \sqrt{k \pi t}} e^{-\frac{(x-s)^2}{4kt}}\right) \phi(s)ds$$

And $\frac{1}{2 \sqrt{k \pi t}} e^{-\frac{(x-s)^2}{4kt}}$ is infinitely many time differentiable as a product of an exponential with a fraction.

Is the above correct?

Leibniz's integral theorem states that the boundaries have to be finite. That is, $a(x),b(x)<+\infty$.
So we still need to address that.
I believe we can 'swap' limits if the limits converge though.

That is, I believe the following holds - if the limits that are involved actually exist:
$$\pd {}x \int_{-\infty}^{+\infty} f(x,s)\,ds \\
= \lim_{h\to 0} \frac{\lim_{a\to\infty}\int_{-a}^{+a} f(x+h,s)\,ds - \lim_{a\to\infty}\int_{-a}^{+a} f(x,s)\,ds}{h} \\
= \lim_{a\to\infty} \lim_{h\to 0} \frac{\int_{-a}^{+a} f(x+h,s)\,ds - \int_{-a}^{+a} f(x,s)\,ds}{h} \\
= \lim_{a\to\infty} \pd{}x \int_{-a}^{+a} f(x,s)\,ds \\
= \lim_{a\to\infty} \int_{-a}^{+a} \pd{}x f(x,s)\,ds
$$

evinda said:
And do we also have to check the partial derivative $\frac{\partial^a \partial^b u(x,t)}{\partial{x}^a \partial{t}^b}$ ? (Thinking)

Yep.
 
  • #8
I like Serena said:
Leibniz's integral theorem states that the boundaries have to be finite. That is, $a(x),b(x)<+\infty$.
So we still need to address that.
I believe we can 'swap' limits if the limits converge though.

That is, I believe the following holds - if the limits that are involved actually exist:
$$\pd {}x \int_{-\infty}^{+\infty} f(x,s)\,ds \\
= \lim_{h\to 0} \frac{\lim_{a\to\infty}\int_{-a}^{+a} f(x+h,s)\,ds - \lim_{a\to\infty}\int_{-a}^{+a} f(x,s)\,ds}{h} \\
= \lim_{a\to\infty} \lim_{h\to 0} \frac{\int_{-a}^{+a} f(x+h,s)\,ds - \int_{-a}^{+a} f(x,s)\,ds}{h} \\
= \lim_{a\to\infty} \pd{}x \int_{-a}^{+a} f(x,s)\,ds \\
= \lim_{a\to\infty} \int_{-a}^{+a} \pd{}x f(x,s)\,ds
$$

So in order to do it like that, we have to show that$$\lim_{a \to +\infty} \int_{-a}^a \frac{\partial^{\beta}}{\partial{x^{\beta}}}\left( \frac{1}{2\sqrt{k \pi t}} e^{-\frac{(x-s)^2}{4kt}}\right) \phi(s) ds<+\infty$$

Right?
 
  • #9
evinda said:
So in order to do it like that, we have to show that$$\lim_{a \to +\infty} \int_{-a}^a \frac{\partial^{\beta}}{\partial{x^{\beta}}}\left( \frac{1}{2\sqrt{k \pi t}} e^{-\frac{(x-s)^2}{4kt}}\right) \phi(s) ds<+\infty$$

Right?

Yes. (Thinking)
 
  • #10
I like Serena said:
Yes. (Thinking)

And how could we do this? (Thinking)
 
  • #11
evinda said:
And how could we do this? (Thinking)

Let's try to match the hint with the expression the integral.
With the appropriate substitution we have:
$$\left|\int_{-\infty}^\infty y^{2n}e^{-2\lambda y^2}dy\right| = \left|\int_{-\infty}^\infty (x-s)^{2n}e^{-\frac{(x-s)^2}{4kt}}ds\right| < +\infty
$$
don't we? (Wondering)

And if we take the first derivative with respect to $x$, we get:
$$\frac 1{2\sqrt{k\pi t}}\int_{-\infty}^\infty -\frac{2(x-s)}{4kt}e^{-\frac{(x-s)^2}{4kt}}\phi(s)\,ds
= -\frac 1{4kt\sqrt{k\pi t}}\int_{-\infty}^\infty (x-s)e^{-\frac{(x-s)^2}{4kt}} \phi(s)\,ds
$$
So if we can find an $n$ such that $(x-s)\phi(s) \le (x-s)^{2n}$, we should have what we need don't we?

It is true for $x=s$, and if $x\ne s$ we can make $(x-s)^{2n}$ as big as we want, can't we? (Wondering)
 
  • #12
I like Serena said:
Let's try to match the hint with the expression the integral.
With the appropriate substitution we have:
$$\left|\int_{-\infty}^\infty y^{2n}e^{-2\lambda y^2}dy\right| = \left|\int_{-\infty}^\infty (x-s)^{2n}e^{-\frac{(x-s)^2}{4kt}}ds\right| < +\infty
$$
don't we? (Wondering)

And if we take the first derivative with respect to $x$, we get:
$$\frac 1{2\sqrt{k\pi t}}\int_{-\infty}^\infty -\frac{2(x-s)}{4kt}e^{-\frac{(x-s)^2}{4kt}}\phi(s)\,ds
= -\frac 1{4kt\sqrt{k\pi t}}\int_{-\infty}^\infty (x-s)e^{-\frac{(x-s)^2}{4kt}} \phi(s)\,ds
$$
So if we can find an $n$ such that $(x-s)\phi(s) \le (x-s)^{2n}$, we should have what we need don't we?

It is true for $x=s$, and if $x\ne s$ we can make $(x-s)^{2n}$ as big as we want, can't we? (Wondering)
So according to Leibniz's integral theorem, we have that$$\frac{\partial}{\partial{x}} \left( \frac{1}{2 \sqrt{k \pi t}} \int_{-\infty}^{+\infty} e^{-\frac{(x-s)^2}{4 k t}} \phi(s) ds\right)=\frac{1}{2 \sqrt{k \pi t}} \int_{-\infty}^{+\infty} \frac{\partial}{\partial{x}}\left( e^{-\frac{(x-s)^2}{4kt}}\right) \phi(s) ds$$

if $\int_{-\infty}^{+\infty} \frac{\partial}{\partial{x}}\left( e^{-\frac{(x-s)^2}{4kt}}\right) \phi(s) ds<+\infty$.It holds that $\int_{-\infty}^{+\infty} \frac{\partial}{\partial{x}}\left( e^{-\frac{(x-s)^2}{4kt}}\right) \phi(s) ds=\int_{-\infty}^{+\infty} \frac{-(x-s)}{2kt} e^{-\frac{(x-s)^2}{4kt}} \phi(s) ds=-\frac{1}{2kt} \int_{-\infty}^{+\infty} (x-s) e^{-\frac{(x-s)^2}{4kt}} \phi(s) ds$.And if we find a natural number $n$ such that $(x-s) \phi(s) \leq (x-s)^{2n}$, we will deduce that $\int_{-\infty}^{+\infty} (x-s) e^{-\frac{(x-s)^2}{4kt}} \phi(s) ds$ is finite, and thus also $\int_{-\infty}^{+\infty} \frac{\partial}{\partial{x}}\left( e^{-\frac{(x-s)^2}{4kt}}\right) \phi(s) ds$. And such a $n$ can be found since $\phi$ is bounded.

Have I understood it right?If so, then like that we show that $u_x$ exists, don't we? Then we also do the same for $u_t$, right?

But we want to show that all the derivatives exist. How could we do this? Can we find a general formula for the derivatives of higher degree that we get? (Thinking)
 
  • #13
evinda said:
So according to Leibniz's integral theorem, we have that$$\frac{\partial}{\partial{x}} \left( \frac{1}{2 \sqrt{k \pi t}} \int_{-\infty}^{+\infty} e^{-\frac{(x-s)^2}{4 k t}} \phi(s) ds\right)=\frac{1}{2 \sqrt{k \pi t}} \int_{-\infty}^{+\infty} \frac{\partial}{\partial{x}}\left( e^{-\frac{(x-s)^2}{4kt}}\right) \phi(s) ds$$

if $\int_{-\infty}^{+\infty} \frac{\partial}{\partial{x}}\left( e^{-\frac{(x-s)^2}{4kt}}\right) \phi(s) ds<+\infty$.It holds that $\int_{-\infty}^{+\infty} \frac{\partial}{\partial{x}}\left( e^{-\frac{(x-s)^2}{4kt}}\right) \phi(s) ds=\int_{-\infty}^{+\infty} \frac{-(x-s)}{2kt} e^{-\frac{(x-s)^2}{4kt}} \phi(s) ds=-\frac{1}{2kt} \int_{-\infty}^{+\infty} (x-s) e^{-\frac{(x-s)^2}{4kt}} \phi(s) ds$.And if we find a natural number $n$ such that $(x-s) \phi(s) \leq (x-s)^{2n}$, we will deduce that $\int_{-\infty}^{+\infty} (x-s) e^{-\frac{(x-s)^2}{4kt}} \phi(s) ds$ is finite, and thus also $\int_{-\infty}^{+\infty} \frac{\partial}{\partial{x}}\left( e^{-\frac{(x-s)^2}{4kt}}\right) \phi(s) ds$. And such a $n$ can be found since $\phi$ is bounded.

Have I understood it right?

Yep. (Nod)

evinda said:
If so, then like that we show that $u_x$ exists, don't we? Then we also do the same for $u_t$, right?

But we want to show that all the derivatives exist. How could we do this? Can we find a general formula for the derivatives of higher degree that we get?

Instead of a general formula of the higher degree derivatives, I'd suggest a proof by induction.
Suppose we continue with $u_{xx}$... is there a pattern? (Wondering)
 
  • #14
I like Serena said:
Instead of a general formula of the higher degree derivatives, I'd suggest a proof by induction.
Suppose we continue with $u_{xx}$... is there a pattern? (Wondering)

So is it as follows? (Thinking)

The base case is that $\frac{\partial}{\partial{x}} \left( \frac{1}{2 \sqrt{k \pi t}} \int_{-\infty}^{+\infty} e^{-\frac{(x-s)^2}{4kt}} \phi(s) ds\right)=\frac{1}{2 \sqrt{k \pi t}}\int_{-\infty}^{+\infty} \frac{\partial}{\partial{x}} \left( e^{-\frac{(x-s)^2}{4kt}}\right) \phi(s) ds<+\infty$.Induction hypothesis: We suppose that $\frac{\partial^a}{\partial{x^a}}\left( \frac{1}{2 \sqrt{k \pi t}} \int_{-\infty}^{+\infty} e^{-\frac{(x-s)^2}{4kt}} \phi(s) ds\right)=\frac{1}{2 \sqrt{k \pi t}} \int_{-\infty}^{+\infty} \frac{\partial^a}{\partial{x^a}} \left( e^{-\frac{(x-s)^2}{4kt}}\right) \phi(s) ds<+\infty$.

Induction step:
$$\frac{\partial^{a+1}}{\partial{x^{a+1}}}\left( \frac{1}{2 \sqrt{k \pi t}} \int_{-\infty}^{+\infty} e^{-\frac{(x-s)^2}{4kt}} \phi(s) ds\right)=\frac{\partial}{\partial{x}}\left( \frac{\partial^a}{\partial{x^a}}\left( \frac{1}{2 \sqrt{k \pi t}} \int_{-\infty}^{+\infty} e^{-\frac{(x-s)^2}{4kt}} \phi(s) ds\right)\right)=\frac{\partial}{\partial{x}}\left( \frac{1}{2 \sqrt{k \pi t}} \int_{-\infty}^{+\infty} \frac{\partial^a}{\partial{x^a}} \left( e^{-\frac{(x-s)^2}{4kt}}\right) \phi(s) ds\right) =\frac{1}{2 \sqrt{k \pi t}} \int_{-\infty}^{+\infty}\frac{\partial^{a+1}}{\partial{x^{a+1}}} \left( e^{-\frac{(x-s)^2}{4kt}}\right) \phi(s) ds \ \text{ since}\int_{-\infty}^{+\infty} \frac{\partial^a}{\partial{x^a}} \left( e^{-\frac{(x-s)^2}{4kt}}\right) \phi(s) ds<+\infty.$$

Thus, all the derivatives of $u$ in respect to $x$ exist.
 
  • #15
evinda said:
So is it as follows? (Thinking)

The base case is that $\frac{\partial}{\partial{x}} \left( \frac{1}{2 \sqrt{k \pi t}} \int_{-\infty}^{+\infty} e^{-\frac{(x-s)^2}{4kt}} \phi(s) ds\right)=\frac{1}{2 \sqrt{k \pi t}}\int_{-\infty}^{+\infty} \frac{\partial}{\partial{x}} \left( e^{-\frac{(x-s)^2}{4kt}}\right) \phi(s) ds<+\infty$.Induction hypothesis: We suppose that $\frac{\partial^a}{\partial{x^a}}\left( \frac{1}{2 \sqrt{k \pi t}} \int_{-\infty}^{+\infty} e^{-\frac{(x-s)^2}{4kt}} \phi(s) ds\right)=\frac{1}{2 \sqrt{k \pi t}} \int_{-\infty}^{+\infty} \frac{\partial^a}{\partial{x^a}} \left( e^{-\frac{(x-s)^2}{4kt}}\right) \phi(s) ds<+\infty$.

Induction step:
$$\frac{\partial^{a+1}}{\partial{x^{a+1}}}\left( \frac{1}{2 \sqrt{k \pi t}} \int_{-\infty}^{+\infty} e^{-\frac{(x-s)^2}{4kt}} \phi(s) ds\right)=\frac{\partial}{\partial{x}}\left( \frac{\partial^a}{\partial{x^a}}\left( \frac{1}{2 \sqrt{k \pi t}} \int_{-\infty}^{+\infty} e^{-\frac{(x-s)^2}{4kt}} \phi(s) ds\right)\right)=\frac{\partial}{\partial{x}}\left( \frac{1}{2 \sqrt{k \pi t}} \int_{-\infty}^{+\infty} \frac{\partial^a}{\partial{x^a}} \left( e^{-\frac{(x-s)^2}{4kt}}\right) \phi(s) ds\right) =\frac{1}{2 \sqrt{k \pi t}} \int_{-\infty}^{+\infty}\frac{\partial^{a+1}}{\partial{x^{a+1}}} \left( e^{-\frac{(x-s)^2}{4kt}}\right) \phi(s) ds \ \text{ since}\int_{-\infty}^{+\infty} \frac{\partial^a}{\partial{x^a}} \left( e^{-\frac{(x-s)^2}{4kt}}\right) \phi(s) ds<+\infty.$$

Thus, all the derivatives of $u$ in respect to $x$ exist.

Erm... we have to actually differentiate to tell if we can, shouldn't we?
In the induction step we have an $(n+1)$th derivative now, but we have no clue if the corresponding integral converges, do we? (Worried)

More specifically, we should expand the derivative for the base case as we did in the previous posts and show that it has a pattern such that convergence is guaranteed.
And assuming that pattern in the expansion, we have to show in the induction step that another differentiation results in the same pattern. (Thinking)
 
  • #16
I like Serena said:
Erm... we have to actually differentiate to tell if we can, shouldn't we?
In the induction step we have an $(n+1)$th derivative now, but we have no clue if the corresponding integral converges, do we? (Worried)

More specifically, we should expand the derivative for the base case as we did in the previous posts and show that it has a pattern such that convergence is guaranteed.
And assuming that pattern in the expansion, we have to show in the induction step that another differentiation results in the same pattern. (Thinking)

You mean that we have to show that $\int_{-\infty}^{+\infty} \frac{\partial^a}{\partial{x^a}} \left( e^{-\frac{(x-s)^2}{4kt}}\right) \phi(s) ds<+\infty$ ? (Thinking)
 
  • #17
evinda said:
You mean that we have to show that $\int_{-\infty}^{+\infty} \frac{\partial^a}{\partial{x^a}} \left( e^{-\frac{(x-s)^2}{4kt}}\right) \phi(s) ds<+\infty$ ? (Thinking)

Yes. (Thinking)
 
  • #18
I found that

$$\frac{\partial^2}{\partial{x^2}}\left( e^{-\frac{(x-s)^2}{4kt}}\right)=\frac{e^{-\frac{(x-s)^2}{4kt}}}{2kt}\left( \frac{(x-s)^2}{2kt}-1\right)$$

and

$$\frac{\partial^3}{\partial{x^3}}\left( e^{-\frac{(x-s)^2}{4kt}}\right)=\frac{(x-s)}{kt} \frac{e^{-\frac{(x-s)^2}{4kt}}}{2kt} \left( 1-\frac{1}{2} \left( \frac{(x-s)^2}{2kt}-1\right)\right)$$

So there is no a general formula for $\frac{\partial^a}{\partial{x^a}}\left( e^{-\frac{(x-s)^2}{4kt}}\right)$, is there? (Thinking)
 
  • #19
evinda said:
I found that

$$\frac{\partial^2}{\partial{x^2}}\left( e^{-\frac{(x-s)^2}{4kt}}\right)=\frac{e^{-\frac{(x-s)^2}{4kt}}}{2kt}\left( \frac{(x-s)^2}{2kt}-1\right)$$

and

$$\frac{\partial^3}{\partial{x^3}}\left( e^{-\frac{(x-s)^2}{4kt}}\right)=\frac{(x-s)}{kt} \frac{e^{-\frac{(x-s)^2}{4kt}}}{2kt} \left( 1-\frac{1}{2} \left( \frac{(x-s)^2}{2kt}-1\right)\right)$$

So there is no a general formula for $\frac{\partial^a}{\partial{x^a}}\left( e^{-\frac{(x-s)^2}{4kt}}\right)$, is there? (Thinking)

It looks to me as if it is of the form:
$$ \frac{A}{t^{k_1}}e^{-\frac{(x-s)^2}{4kt}} + \frac{B}{t^{k_2}}(x-s) e^{-\frac{(x-s)^2}{4kt}} + \frac{C}{t^{k_3}}(x-s)^2 e^{-\frac{(x-s)^2}{4kt}} + ...
$$
isn't it? (Wondering)
 
  • #20
Isn't it as follows? (Thinking)

When $a$ is odd then: $$\frac{\partial^a}{\partial{x^a}}\left( e^{-\frac{(x-s)^2}{4kt}}\right)=A(x-s)e^{-\frac{(x-s)^2}{4kt}} + B(x-s)^3 e^{-\frac{(x-s)^2}{4kt}} + C(x-s)^5 e^{-\frac{(x-s)^2}{4kt}} + ...$$

When $a$ is even then: $$\frac{\partial^a}{\partial{x^a}}\left( e^{-\frac{(x-s)^2}{4kt}}\right)==A(x-s)^0e^{-\frac{(x-s)^2}{4kt}} + B(x-s)^2 e^{-\frac{(x-s)^2}{4kt}} + C(x-s)^4 e^{-\frac{(x-s)^2}{4kt}} + ...$$
 
  • #21
evinda said:
Isn't it as follows? (Thinking)

When $a$ is odd then: $$\frac{\partial^a}{\partial{x^a}}\left( e^{-\frac{(x-s)^2}{4kt}}\right)=A(x-s)e^{-\frac{(x-s)^2}{4kt}} + B(x-s)^3 e^{-\frac{(x-s)^2}{4kt}} + C(x-s)^5 e^{-\frac{(x-s)^2}{4kt}} + ...$$

When $a$ is even then: $$\frac{\partial^a}{\partial{x^a}}\left( e^{-\frac{(x-s)^2}{4kt}}\right)==A(x-s)^0e^{-\frac{(x-s)^2}{4kt}} + B(x-s)^2 e^{-\frac{(x-s)^2}{4kt}} + C(x-s)^4 e^{-\frac{(x-s)^2}{4kt}} + ...$$

Ah yes. Perhaps we can make it a bit more general.
How about:
$$ \frac{A}{t^{k_1}}e^{-\frac{(x-s)^2}{4kt}} + \frac{B}{t^{k_2}}(x-s) e^{-\frac{(x-s)^2}{4kt}} + \frac{C}{t^{k_3}}(x-s)^2 e^{-\frac{(x-s)^2}{4kt}} + ...
$$
(Wondering)
 
  • #22
I like Serena said:
Ah yes. Perhaps we can make it a bit more general.
How about:
$$ \frac{A}{t^{k_1}}e^{-\frac{(x-s)^2}{4kt}} + \frac{B}{t^{k_2}}(x-s) e^{-\frac{(x-s)^2}{4kt}} + \frac{C}{t^{k_3}}(x-s)^2 e^{-\frac{(x-s)^2}{4kt}} + ...
$$
(Wondering)

So we have to show that $\int_{-\infty}^{+\infty} \left( \frac{A}{t^{k_1}}e^{-\frac{(x-s)^2}{4kt}} + \frac{B}{t^{k_2}}(x-s) e^{-\frac{(x-s)^2}{4kt}} + \frac{C}{t^{k_3}}(x-s)^2 e^{-\frac{(x-s)^2}{4kt}} + \dots+ \frac{W}{t^{k_a}}(x-s)^a e^{-\frac{(x-s)^2}{4kt}}\right) \phi(s) ds<+\infty $, right?

How could we show this? (Thinking)
 
  • #23
evinda said:
So we have to show that $\int_{-\infty}^{+\infty} \left( \frac{A}{t^{k_1}}e^{-\frac{(x-s)^2}{4kt}} + \frac{B}{t^{k_2}}(x-s) e^{-\frac{(x-s)^2}{4kt}} + \frac{C}{t^{k_3}}(x-s)^2 e^{-\frac{(x-s)^2}{4kt}} + \dots+ \frac{W}{t^{k_a}}(x-s)^a e^{-\frac{(x-s)^2}{4kt}}\right) \phi(s) ds<+\infty $, right?

How could we show this? (Thinking)

Yes.
The integral of the first term corresponds to $u(x,t)$ doesn't it?
Let's make that the base case. That one exists doesn't we? (Wondering)If we can differentiate with respect to $x$ and/or $t$ for a total of $(m-1)$ times, we'll have at most $m$ of those terms won't we? And if we differentiate again, we'll possibly add yet another term of the form $\frac{A_m}{t^{k_m}}(x-s)^{m} e^{-\frac{(x-s)^2}{4kt}}$.
We assume that up to $(m-1)$ the corresponding integral will converge.
Can we use the hint to show that the integral of the term for $m$ will converge as well? (Wondering)
 
  • #24
I like Serena said:
If we can differentiate with respect to $x$ and/or $t$ for a total of $(m-1)$ times, we'll have at most $m$ of those terms won't we? And if we differentiate again, we'll possibly add yet another term of the form $\frac{A_m}{t^{k_m}}(x-s)^{m} e^{-\frac{(x-s)^2}{4kt}}$.
We assume that up to $(m-1)$ the corresponding integral will converge.
Can we use the hint to show that the integral of the term for $m$ will converge as well? (Wondering)

Then we'll have

$$\int_{-\infty}^{+\infty} \frac{\partial^m}{\partial{x^m}}\left( e^{-\frac{(x-s)^2}{4kt}} \right )ds=\int_{-\infty}^{+\infty} \left( \frac{\partial^{m-1}}{\partial{x^{m-1}}}\left( e^{-\frac{(x-s)^2}{4kt}} \right)+\frac{A_m}{t^{k_m}}(x-s)^{m-1} e^{-\frac{(x-s)^2}{4kt}}\right )ds=\int_{-\infty}^{+\infty} \frac{\partial^{m-1}}{\partial{x^{m-1}}}\left( e^{-\frac{(x-s)^2}{4kt}} \right) ds+\int_{-\infty}^{+\infty} \frac{A_m}{t^{k_m}}(x-s)^{m-1} e^{-\frac{(x-s)^2}{4kt}} ds$$

From the induction hypothesis, we have that $\int_{-\infty}^{+\infty} \frac{\partial^{m-1}}{\partial{x^{m-1}}}\left( e^{-\frac{(x-s)^2}{4kt}} \right) ds<+\infty$.From the hint, we have that $\left| \int_{-\infty}^{+\infty} (x-s)^{2m} e^{-\frac{(x-s)^2}{4kt}} ds\right|<+\infty$.

And $(x-s)^{m-1} \leq (x-s)^{2m}$.

Thus $\int_{-\infty}^{+\infty} \frac{A_m}{t^{k_m}}(x-s)^{m-1} e^{-\frac{(x-s)^2}{4kt}} ds<+\infty$.Is it right like that? (Thinking)
 
  • #25
evinda said:
Then we'll have

$$\int_{-\infty}^{+\infty} \frac{\partial^m}{\partial{x^m}}\left( e^{-\frac{(x-s)^2}{4kt}} \right )ds=\int_{-\infty}^{+\infty} \left( \frac{\partial^{m-1}}{\partial{x^{m-1}}}\left( e^{-\frac{(x-s)^2}{4kt}} \right)+\frac{A_m}{t^{k_m}}(x-s)^{m-1} e^{-\frac{(x-s)^2}{4kt}}\right )ds=\int_{-\infty}^{+\infty} \frac{\partial^{m-1}}{\partial{x^{m-1}}}\left( e^{-\frac{(x-s)^2}{4kt}} \right) ds+\int_{-\infty}^{+\infty} \frac{A_m}{t^{k_m}}(x-s)^{m-1} e^{-\frac{(x-s)^2}{4kt}} ds$$

From the induction hypothesis, we have that $\int_{-\infty}^{+\infty} \frac{\partial^{m-1}}{\partial{x^{m-1}}}\left( e^{-\frac{(x-s)^2}{4kt}} \right) ds<+\infty$.From the hint, we have that $\left| \int_{-\infty}^{+\infty} (x-s)^{2m} e^{-\frac{(x-s)^2}{4kt}} ds\right|<+\infty$.

And $(x-s)^{m-1} \leq (x-s)^{2m}$.

Thus $\int_{-\infty}^{+\infty} \frac{A_m}{t^{k_m}}(x-s)^{m-1} e^{-\frac{(x-s)^2}{4kt}} ds<+\infty$.Is it right like that? (Thinking)

I think it's not quite correct yet.
When taking another derivative, we get different terms don't we? (Thinking)
And additionally we get another term of a higher degree.
 
  • #26
I like Serena said:
I think it's not quite correct yet.
When taking another derivative, we get different terms don't we? (Thinking)

You mean the derivative in respect to $t$ ? (Thinking)
 

FAQ: Show that solution is infinitely many times differentiable

1. What does it mean for a solution to be infinitely many times differentiable?

Infinitely many times differentiable means that a function can be differentiated an infinite number of times at any point in its domain. This indicates that the function is very smooth and has no abrupt changes or corners.

2. Why is it important for a solution to be infinitely many times differentiable?

Infinitely many times differentiable solutions are important in many areas of mathematics and physics, particularly in the fields of calculus, differential equations, and dynamical systems. They provide a way to model and analyze complex systems and phenomena with great accuracy.

3. How can we show that a solution is infinitely many times differentiable?

To show that a solution is infinitely many times differentiable, we must demonstrate that the limit of the difference quotient as the step size approaches zero exists. This can be done using various techniques such as using the definition of a derivative, using theorems such as the Mean Value Theorem, or using Taylor series expansions.

4. What are some properties of infinitely many times differentiable solutions?

Infinitely many times differentiable solutions have several important properties, including that they are continuous, have smooth and continuous derivatives of all orders, and have Taylor series expansions that converge to the function at any point in its domain. They also have a unique solution for a given initial condition.

5. Are all solutions infinitely many times differentiable?

No, not all solutions are infinitely many times differentiable. In fact, many real-world phenomena do not have infinitely many times differentiable solutions. This can be due to discontinuities, singularities, or other complex behaviors in the system being modeled.

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