Find formula for solution heat equation

In summary, the problem considers a function $\phi \in C^1(\mathbb{R})$ that is periodic. The problem is to solve the initial value problem $u_t=u_{xx}$ with initial data $\phi$. The solution is given by $u(x,t)=\sum_{\lambda=1}^{\infty} c_{\lambda}(A_{\lambda} \cos{(\sqrt{\lambda}x)}+B_{\lambda}\sin{(\sqrt{\lambda}x)} ) e^{-\lambda t}$ for $\lambda>0$, $u(x,t)=C$ for $\lambda=0$, and $u(x,t;\tau) = \sum_{k=1}^\in
  • #1
evinda
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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)
 
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  • #2
Hey evinda! (Smile)

Yes, you are right so far for $\lambda<0$.
 
  • #3
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)
 
  • #4
evinda said:
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)

evinda said:
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)
 
  • #5
I like Serena said:
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?
I like Serena said:
$\lambda$ 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)
 
  • #6
evinda said:
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)
 
  • #7
I like Serena said:
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)
 
  • #8
evinda said:
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)
 
  • #9
I like Serena said:
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)
 
  • #10
evinda said:
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)

evinda said:
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)
 
  • #11
I like Serena said:
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)
 
  • #12
evinda said:
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)
 
  • #13
I like Serena said:
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)
 
  • #14
evinda said:
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)
 
  • #15
I like Serena said:
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)
 
  • #16
evinda said:
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)
 
  • #17
I like Serena said:
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)

I like Serena said:
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)
 
  • #18
evinda said:
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)
 
  • #19
I like Serena said:
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?
 
  • #20
evinda said:
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)
 
  • #21
I like Serena said:
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)
 
  • #22
evinda said:
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.
 
  • #23
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)
 
  • #24
evinda said:
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)
 
  • #25
I like Serena said:
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? :confused:
 
  • #26
evinda said:
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. :eek:

evinda said:
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)

evinda said:
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? :confused:

How would the exponential make it non-unique? (Wondering)
 
  • #27
I like Serena said:
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. :eek:

(Nod)

I like Serena said:
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)

I like Serena said:
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.
 
  • #28
evinda said:
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)
 
  • #29
I like Serena said:
That seems fine to me. (Nod)

Great... Thanks a lot! (Smirk)
 

Related to Find formula for solution heat equation

1. What is the heat equation?

The heat equation is a partial differential equation that describes the distribution of heat in a given space over time. It is commonly used in physics and engineering to model heat transfer and diffusion.

2. Why is it important to find the formula for the heat equation?

Knowing the formula for the heat equation allows us to accurately predict how heat will distribute and change over time in a given system. This can be useful in many real-world applications, such as designing heating and cooling systems or studying the behavior of materials under different temperature conditions.

3. How is the formula for the heat equation derived?

The formula for the heat equation is derived from fundamental principles of thermodynamics and Fourier's law of heat conduction. It involves the use of partial derivatives to express the rate of change of temperature with respect to time and position.

4. Are there different versions of the heat equation formula?

Yes, there are different versions of the heat equation formula that can be used depending on the specific situation or boundary conditions. For example, the formula for a one-dimensional system will be different from that of a two-dimensional or three-dimensional system.

5. Can the heat equation formula be solved analytically?

In some cases, yes, the heat equation formula can be solved analytically using mathematical techniques such as separation of variables or Laplace transforms. However, in more complex systems, numerical methods may be necessary to find a solution.

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