Find formula for solution heat equation

[evinda]

Hello! (Wave)

Let $\phi \in C^1(\mathbb{R})$ and periodic.
We consider the problem

$u_t=u_{xx}, x \in \mathbb{R}, \ 0<t<\infty$,

with initial data $\phi$.
I want to compute a formula for a solution $u$ and I want to prove strictly that this formula solves the initial value problem. I also want to show that there is no other solution ($C^1(\mathbb{R})$ and periodic).

I have thought the following.

$u(x,t)=X(x) T(t)$

$u(0,x)=\phi(x)$

$u_t=u_{xx} \Rightarrow X(x) T'(t)=X''(x) T(t) \Rightarrow \frac{T'(t)}{T(t)}=\frac{X''(x)}{X(x)}=-\lambda$

So we have the system

$\left\{\begin{matrix}
X''(x)+\lambda X(x)=0\\
T'(t)+\lambda T(t)=0
\end{matrix}\right.$.

So $T(t)$ is of the form $T(t)=c_1 e^{-\lambda t}$.

The characteristic equation of $X''(x)+\lambda X(x)=0$ is $\mu^2=-\lambda$.

For $\lambda<0$ we get that $X(x)=c_1 e^{\sqrt{-\lambda}x}+c_2 e^{-\sqrt{-\lambda}x}$.

We have that $X(x+T)=c_1 e^{\sqrt{-\lambda}x} e^{\sqrt{-\lambda}T}+c_2 e^{-\sqrt{-\lambda}x}e^{-\sqrt{-\lambda} T}$

This holds only if $ e^{\sqrt{-\lambda}T}=1 \Rightarrow \sqrt{-\lambda}T=0 \Rightarrow T=0$, contradiction.

Am I right so far? (Thinking)

 

[I like Serena]

Hey evinda! (Smile)

Yes, you are right so far for $\lambda<0$.

 

[evinda]

For $\lambda=0$ we have $X(x)=c_1 x+c_2$.

$X(x+T)=c_1(x+T)+c_2= c_1 x+ c_1 T+c_2=c_1 x+ c_2 \Rightarrow c_1 T=0 \Rightarrow c_1=0$.

So for $\lambda=0$ we get that $X(x)T(t)=C$ for some constant $C$.

Do we accept the solution that we get for $\lambda=0$ ?For $\lambda>0$ we have $\mu=\pm \sqrt{\lambda}i$ and thus $X(x)=c_1 \cos{(\sqrt{\lambda}x)}+ c_2 \sin{(\sqrt{\lambda}x)}$.

$X(x)$ is $2 \pi$-periodic, since it contains the trigonometric functions.

So $u(x,t)=C+\sum_{\lambda=1}^{\infty} c_{\lambda}(A_{\lambda} \cos{(\sqrt{\lambda}x)}+B_{\lambda}\sin{(\sqrt{\lambda}x)} ) e^{-\lambda t}$.Right? Or have I done something wrong? (Thinking)

 

[I like Serena]

For $\lambda=0$ we have $X(x)=c_1 x+c_2$.

$X(x+T)=c_1(x+T)+c_2= c_1 x+ c_1 T+c_2=c_1 x+ c_2 \Rightarrow c_1 T=0 \Rightarrow c_1=0$.

So for $\lambda=0$ we get that $X(x)T(t)=C$ for some constant $C$.

Do we accept the solution that we get for $\lambda=0$ ?

It's a valid albeit trivial solution, so we should include it.
Isn't it effectively already included in the case $\lambda >0$ though? (Wondering)

For $\lambda>0$ we have $\mu=\pm \sqrt{\lambda}i$ and thus $X(x)=c_1 \cos{(\sqrt{\lambda}x)}+ c_2 \sin{(\sqrt{\lambda}x)}$.

$X(x)$ is $2 \pi$-periodic, since it contains the trigonometric functions.

So $u(x,t)=C+\sum_{\lambda=1}^{\infty} c_{\lambda}(A_{\lambda} \cos{(\sqrt{\lambda}x)}+B_{\lambda}\sin{(\sqrt{\lambda}x)} ) e^{-\lambda t}$.

Right? Or have I done something wrong?

$\lambda$ is not necessarily a positive integer is it?
Shouldn't we use an integral? (Wondering)

And isn't $c_\lambda$ redundant, since we're already multiplying with arbitrary constants? (Wondering)

 

[evinda]

It's a valid albeit trivial solution, so we should include it.
Isn't it effectively already included in the case $\lambda >0$ though? (Wondering)

How?

$\lambda$ is not necessarily a positive integer is it?
Shouldn't we use an integral? (Wondering)

And isn't $c_\lambda$ redundant, since we're already multiplying with arbitrary constants? (Wondering)

So do we have that $u(x,t)=\int_0^{\infty} [A_{\lambda} \cos{(\sqrt{\lambda}x)}+B_{\lambda} \sin{(\sqrt{\lambda}x)}] e^{-\lambda t} d{\lambda}$ ? (Thinking)

 

[I like Serena]

So do we have that $u(x,t)=\int_0^{\infty} [A_{\lambda} \cos{(\sqrt{\lambda}x)}+B_{\lambda} \sin{(\sqrt{\lambda}x)}] e^{-\lambda t} d{\lambda}$ ?

Oh wait! (Wait)
We need it to be periodic for $t=0$ don't we?
That puts a restriction on it, doesn't it? (Wondering)

 

[evinda]

Oh wait! (Wait)
We need it to be periodic for $t=0$ don't we?
That puts a restriction on it, doesn't it? (Wondering)

What restriction? (Thinking)

 

[I like Serena]

What restriction? (Thinking)

Suppose $\phi$ is periodic with period $\tau$.
Then we must have that $\sqrt{\lambda}(x+\tau)=\sqrt{\lambda}x + 2\pi k$ don't we? (Wondering)

 

[evinda]

Suppose $\phi$ is periodic with period $\tau$.
Then we must have that $\sqrt{\lambda}(x+\tau)=\sqrt{\lambda}x + 2\pi k$ don't we? (Wondering)

Since we have the functions $\cos{x}$, $\sin{x}$, doesn't this imply that the period is $2 \pi$ ? Or am I wrong? (Thinking)

 

[I like Serena]

For $\lambda>0$ we have $\mu=\pm \sqrt{\lambda}i$ and thus $X(x)=c_1 \cos{(\sqrt{\lambda}x)}+ c_2 \sin{(\sqrt{\lambda}x)}$.

$X(x)$ is $2 \pi$-periodic, since it contains the trigonometric functions.

I'm afraid that the period of $X(x)$ is not (necessarily) $2\pi$. (Worried)

Since we have the functions $\cos{x}$, $\sin{x}$, doesn't this imply that the period is $2 \pi$ ? Or am I wrong?

The period of $\cos$ and $\sin$ is indeed $2\pi$.
But $X(x)=c_1 \cos{(\sqrt{\lambda}x)}+ c_2 \sin{(\sqrt{\lambda}x)}$ is then periodic with period $\frac{2\pi}{\sqrt\lambda}$ isn't it?
And if $\phi$ has period $\tau$, then $X(x)$ must have a period that is a multiple of $\tau$ mustn't it? (Wondering)

 

[evinda]

I'm afraid that the period of $X(x)$ is not (necessarily) $2\pi$. (Worried)
The period of $\cos$ and $\sin$ is indeed $2\pi$.
But $X(x)=c_1 \cos{(\sqrt{\lambda}x)}+ c_2 \sin{(\sqrt{\lambda}x)}$ is then periodic with period $\frac{2\pi}{\sqrt\lambda}$ isn't it?
And if $\phi$ has period $\tau$, then $X(x)$ must have a period that is a multiple of $\tau$ mustn't it? (Wondering)

Ok, but how can we include this restriction at the interval? (Thinking)

- - - Updated - - -

The period of $X(x)$ is $\frac{2 \pi}{\sqrt{\lambda}}=\frac{\tau}{k}$, which is a multiple of $\tau$, right? (Thinking)

 

[I like Serena]

Ok, but how can we include this restriction at the interval? (Thinking)

If $\phi$ has period $\tau$, we must have that $X(x+\tau)=X(x)$.
That is:
$$\cos(\sqrt\lambda(x+\tau))=\cos(\sqrt\lambda x) \quad\Rightarrow\quad
\sqrt\lambda(x+\tau)=\sqrt\lambda x + 2\pi k \quad\Rightarrow\quad
\sqrt\lambda=\frac{2\pi k}{\tau}$$
So we should have:
$$u(x,t;\tau) = \sum_{k=1}^\infty \left[A_k \cos\left(\frac{2\pi k}{\tau}x\right) + B_k \sin\left(\frac{2\pi k}{\tau}x\right)\right]e^{-\frac{4\pi^2 k^2}{\tau^2}t}$$
shouldn't we? (Wondering)

 

[evinda]

If $\phi$ has period $\tau$, we must have that $X(x+\tau)=X(x)$.
That is:
$$\cos(\sqrt\lambda(x+\tau))=\cos(\sqrt\lambda x) \quad\Rightarrow\quad
\sqrt\lambda(x+\tau)=\sqrt\lambda x + 2\pi k \quad\Rightarrow\quad
\sqrt\lambda=\frac{2\pi k}{\tau}$$
So we should have:
$$u(x,t;\tau) = \sum_{k=1}^\infty \left[A_k \cos\left(\frac{2\pi k}{\tau}x\right) + B_k \sin\left(\frac{2\pi k}{\tau}x\right)\right]e^{-\frac{4\pi^2 k^2}{\tau^2}t}$$
shouldn't we? (Wondering)

I see... (Nod)
Ans is the trivial solution now included? If so, how do we get it? (Thinking)

 

[I like Serena]

I see...
Ans is the trivial solution now included? If so, how do we get it?

Suppose we lower the sum boundary to $k=0$.
Will we include the trivial solution then? (Wondering)

 

[evinda]

Suppose we lower the sum boundary to $k=0$.
Will we include the trivial solution then? (Wondering)

Yes, we will... (Nod)

And have we proven strictly like that, that the formula that we found solves the initial value problem? Or do we have to substitute it at the given equations? (Thinking)

 

[I like Serena]

Yes, we will...

And have we proven strictly like that, that the formula that we found solves the initial value problem? Or do we have to substitute it at the given equations?

The solution we found solves the given heat equation for a boundary condition with period $\tau$.
We still need to find $A_k$ and $B_k$ for a given $\phi$. Can we? (Wondering)
And we still need to verify that there is no other solution. (Thinking)

 

[evinda]

The solution we found solves the given heat equation for a boundary condition with period $\tau$.
We still need to find $A_k$ and $B_k$ for a given $\phi$. Can we? (Wondering)

We just know that it has to hold that $\sum_{k=0}^{\infty} \left[ A_k \cos{\left( \frac{2k \pi x}{\tau}\right)}+B_k \sin{\left( \frac{2k \pi x}{\tau}\right)}\right]=\phi(x,0)$.

Or is there an additional info? (Thinking) Do we pick a specific $\phi$ ? (Worried)

And we still need to verify that there is no other solution. (Thinking)

Do we show the uniqueness of the solution using the energy method? (Thinking)
Or do we show it somehow else? (Nerd)

 

[I like Serena]

We just know that it has to hold that $\sum_{k=0}^{\infty} \left[ A_k \cos{\left( \frac{2k \pi x}{\tau}\right)}+B_k \sin{\left( \frac{2k \pi x}{\tau}\right)}\right]=\phi(x,0)$.

Or is there an additional info? (Thinking) Do we pick a specific $\phi$ ?

Do we show the uniqueness of the solution using the energy method?
Or do we show it somehow else?

As before, every periodic function can be written uniquely as a Fourier Series.

If $s(x)$ is a function with period $P$, then:
$$s(x)=\frac{a_0}2 + \sum_{n=1}^\infty a_n \cos(2\pi n x/P) + b_n \sin(2\pi n x/P)$$
where:
\begin{cases}a_n = \frac 2P\int_0^P s(x)\cos(2\pi n x/P)\,dx \\
b_n = \frac 2P\int_0^P s(x)\sin(2\pi n x/P)\, dx \end{cases}
(Thinking)

 

[evinda]

As before, every periodic function can be written uniquely as a Fourier Series.

If $s(x)$ is a function with period $P$, then:
$$s(x)=\frac{a_0}2 + \sum_{n=1}^\infty a_n \cos(2\pi n x/P) + b_n \sin(2\pi n x/P)$$
where:
\begin{cases}a_n = \frac 2P\int_0^P s(x)\cos(2\pi n x/P)\,dx \\
b_n = \frac 2P\int_0^P s(x)\sin(2\pi n x/P)\, dx \end{cases}
(Thinking)

So in our case we have that

$$A_k=\frac{2}{\tau} \int_0^T u(x,t;T) \cos{\left( \frac{2 k \pi x}{T}\right)} dx \\ B_k=\frac{2}{T} \int_0^T u(x,t;T) \sin{\left( \frac{2 k \pi x}{T}\right)} dx$$

Right?

 

[I like Serena]

So in our case we have that

$$A_k=\frac{2}{\tau} \int_0^T u(x,t;T) \cos{\left( \frac{2 k \pi x}{T}\right)} dx \\ B_k=\frac{2}{T} \int_0^T u(x,t;T) \sin{\left( \frac{2 k \pi x}{T}\right)} dx$$

Right?

Aren't we trying to find the solution $u(x,t;\tau)$?
Then we cannot use it as input to find it can we? (Worried)

 

[evinda]

Aren't we trying to find the solution $u(x,t;\tau)$?
Then we cannot use it as input to find it can we? (Worried)

Ah right... For $t=0$ we have:

$$\phi(x,0)=\sum_{k=0}^{\infty} \left[ A_k \cos{\left( \frac{2 \pi k x}{T}\right)}+B_k \sin{\left( \frac{2 \pi k x}{T}\right)}\right]$$

where

$$A_k=\frac{2}{T} \int_0^T \phi(x,0) \cos{\left( \frac{2 \pi k x}{T}\right)} dx$$

$$B_k=\frac{2}{T} \int_0^T \phi(x,0) \sin{\left( \frac{2 \pi k x}{T}\right)} dx$$

Right? If so, do we have to write a different formula for $A_0$ ? (Thinking)

 

[I like Serena]

Ah right... For $t=0$ we have:

$$\phi(x,0)=\sum_{k=0}^{\infty} \left[ A_k \cos{\left( \frac{2 \pi k x}{T}\right)}+B_k \sin{\left( \frac{2 \pi k x}{T}\right)}\right]$$

where

$$A_k=\frac{2}{T} \int_0^T \phi(x,0) \cos{\left( \frac{2 \pi k x}{T}\right)} dx$$

$$B_k=\frac{2}{T} \int_0^T \phi(x,0) \sin{\left( \frac{2 \pi k x}{T}\right)} dx$$

Right? If so, do we have to write a different formula for $A_0$ ?

Isn't $ϕ∈C^1(\mathbb R)$? So shouldn't it have only 1 argument? (Wondering)
And yes, we'll have to make an exception for $A_0$ unfortunately.

 

[evinda]

So is it as follows?

$$\phi(x)=\sum_{k=0}^{\infty} \left[ A_k \cos{\left( \frac{2 \pi k x}{T}\right)}+B_k \sin{\left( \frac{2 \pi k x}{T}\right)}\right]$$

where

$$A_k=\frac{2}{T} \int_0^T \phi(x) \cos{\left( \frac{2 \pi k x}{T}\right)} dx, k=1,2, \dots$$

$$A_0=\frac{4}{T} \int_0^T \phi(x) dx$$

$$B_k=\frac{2}{T} \int_0^T \phi(x) \sin{\left( \frac{2 \pi k x}{T}\right)} dx$$

Or am I somewhere wrong? (Thinking)

 

[I like Serena]

So is it as follows?

$$\phi(x)=\sum_{k=0}^{\infty} \left[ A_k \cos{\left( \frac{2 \pi k x}{T}\right)}+B_k \sin{\left( \frac{2 \pi k x}{T}\right)}\right]$$

where

$$A_k=\frac{2}{T} \int_0^T \phi(x) \cos{\left( \frac{2 \pi k x}{T}\right)} dx, k=1,2, \dots$$

$$A_0=\frac{4}{T} \int_0^T \phi(x) dx$$

$$B_k=\frac{2}{T} \int_0^T \phi(x) \sin{\left( \frac{2 \pi k x}{T}\right)} dx$$

Or am I somewhere wrong? (Thinking)

That should be $A_0=\frac{1}{T} \int_0^T \phi(x) dx$.
Or alternatively:
$$\phi(x)=\frac {A_0} 2 + \sum_{k=1}^{\infty} \left[ A_k \cos{\left( \frac{2 \pi k x}{T}\right)}+B_k \sin{\left( \frac{2 \pi k x}{T}\right)}\right]$$
Note that $A_0$ represents the trivial solution that we had included in the solution. (Nerd)

 

[evinda]

That should be $A_0=\frac{1}{T} \int_0^T \phi(x) dx$.
Or alternatively:
$$\phi(x)=\frac {A_0} 2 + \sum_{k=1}^{\infty} \left[ A_k \cos{\left( \frac{2 \pi k x}{T}\right)}+B_k \sin{\left( \frac{2 \pi k x}{T}\right)}\right]$$
Note that $A_0$ represents the trivial solution that we had included in the solution. (Nerd)

Ah yes, right... (Nod)

Does the above procedure suffice for a strict proof that the formula solves the initial value problem? (Thinking) Or do we have to substitute the solution at the given problem?

Also I was sondering if the $u(x,t; \tau)$ that we found is periodic although it contains the exponential function... (Worried)

In order to show the uniqueness, we cannot use the statement with the Fourier series that you mentioned because of the exponential... Or am I wrong?

 

[I like Serena]

Does the above procedure suffice for a strict proof that the formula solves the initial value problem? Or do we have to substitute the solution at the given problem?

We didn't introduce extraneous solutions, such as would happen if we squared something.
And I believe we didn't 'miss' any solutions.
So I think it should suffice for a strict proof.

Generally it's a good idea though to verify if a solution solves the problem statement - if only to make sure we didn't make a mistake.

Also I was sondering if the $u(x,t; \tau)$ that we found is periodic although it contains the exponential function...

Indeed, $u(x,t; \tau)$ is not periodic with respect to $t$.
That wasn't required though, was it?
It seems to me that only the initial value $\phi$ is supposed to be periodic.
As an added bonus $u(x,t; \tau)$ is periodic with respect to $x$ for any given $t$. (Thinking)

In order to show the uniqueness, we cannot use the statement with the Fourier series that you mentioned because of the exponential... Or am I wrong?

How would the exponential make it non-unique? (Wondering)

 

[evinda]

We didn't introduce extraneous solutions, such as would happen if we squared something.
And I believe we didn't 'miss' any solutions.
So I think it should suffice for a strict proof.

Generally it's a good idea though to verify if a solution solves the problem statement - if only to make sure we didn't make a mistake.

(Nod)

Indeed, $u(x,t; \tau)$ is not periodic with respect to $t$.
That wasn't required though, was it?
It seems to me that only the initial value $\phi$ is supposed to be periodic.
As an added bonus $u(x,t; \tau)$ is periodic with respect to $x$ for any given $t$. (Thinking)

No, ii wasn't required... (Shake)

I see... (Nod)

How would the exponential make it non-unique? (Wondering)

So do we justify the uniqueness of the solution as follows? (Thinking)For a given $t$, $u(x,t; \tau)$ is periodic with respect to $x$, because of the given initial data.
So for an arbitrary $t$, $u(x,t; \tau)$ is written uniquelly as a Fourier series.
Thus, there is no other solution $u$ that satisfies our initial value problem.

 

[I like Serena]

No, it wasn't required...

I see...

So do we justify the uniqueness of the solution as follows?

For a given $t$, $u(x,t; \tau)$ is periodic with respect to $x$, because of the given initial data.
So for an arbitrary $t$, $u(x,t; \tau)$ is written uniquelly as a Fourier series.
Thus, there is no other solution $u$ that satisfies our initial value problem.

That seems fine to me. (Nod)

 

[evinda]

That seems fine to me. (Nod)

Great... Thanks a lot! (Smirk)