Show that solution is infinitely many times differentiable

[evinda]

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)

 

[evinda]

Or is there an other way to show that the solution is infinitely many times differentiable? (Thinking)

 

[I like Serena]

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)

 

[evinda]

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 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 ?

 

[I like Serena]

So, do we apply the following statement?

Yep.

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)

 

[evinda]

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)

 

[I like Serena]

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:
$$\pd {}x \int_{-\infty}^{+\infty} f(x,s)\,ds \\
= \lim_{h\to 0} \frac{\lim_{a\to\infty}\int_{-a}^{+a} f(x+h,s)\,ds - \lim_{a\to\infty}\int_{-a}^{+a} f(x,s)\,ds}{h} \\
= \lim_{a\to\infty} \lim_{h\to 0} \frac{\int_{-a}^{+a} f(x+h,s)\,ds - \int_{-a}^{+a} f(x,s)\,ds}{h} \\
= \lim_{a\to\infty} \pd{}x \int_{-a}^{+a} f(x,s)\,ds \\
= \lim_{a\to\infty} \int_{-a}^{+a} \pd{}x f(x,s)\,ds
$$

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.

 

[evinda]

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?

 

[I like Serena]

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)

 

[evinda]

Yes. (Thinking)

And how could we do this? (Thinking)

 

[I like Serena]

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)

 

[evinda]

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)

 

[I like Serena]

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)

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)

 

[evinda]

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.

 

[I like Serena]

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)

 

[evinda]

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)

 

[I like Serena]

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)

 

[evinda]

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)

 

[I like Serena]

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)

 

[evinda]

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

 

[I like Serena]

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)

 

[evinda]

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)

 

[I like Serena]

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)

 

[evinda]

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)

 

[I like Serena]

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.

 

[evinda]

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)