# B Why the Fourier series doesn't work to solve any differential equation?

#### jonjacson

I know this may sound as a stupid question but I would like to clarify this.

An arbitrary function f can be expressed in the fourier base of sines and cosines. My question is, Can this method be used to solve any differential equation?

You plug into the unkown function the infinite series and then you get the coefficients from the differential equation itself.

Can anybody tell me why this method may fail?

Thanks!

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#### dRic2

Gold Member
For the little I know, Fourier series can describe only periodic functions. If the solution of the differential equation in non-periodic they won't work

#### RPinPA

Homework Helper
I know this may sound as a stupid question but I would like to clarify this.

An arbitrary function f can be expressed in the fourier base of sines and cosines. My question is, Can this method be used to solve any differential equation?

You plug into the unkown function the infinite series and then you get the coefficients from the differential equation itself.

Can anybody tell me why this method may fail?

Thanks!
It definitely can and does work, under certain conditions (mainly existence of all the relevant series or transforms, see below).

For the little I know, Fourier series can describe only periodic functions.
Periodic functions or functions defined on an interval.

That is, if you want to represent a function between say, -1 and 1, then you can do it with a Fourier series between those endpoints. The resulting Fourier series will be periodic of course, and will replicate the values between -1 and 1 in the regions above and below. If you don't care what the representation does outside of [-1, 1], then that's fine.

However, there is a continuous generalization of the Fourier series called Fourier Transform which results in a value at all real-numbered frequencies, not just multiples of some fundamental. You can Fourier Transform many non-periodic functions. In fact for Fourier Transform you want it to be non-periodic, as the transform involves an integral from -infinity to +infinity and you need a function such that the infinite integral exists.

#### RPinPA

Homework Helper
My question is, Can this method be used to solve any differential equation?
Were you asking if all differential equations can be solved in the Fourier domain? The answer is no. Were you asking if there exist any that can be? The answer is yes, it can be a very powerful and useful technique.

#### dRic2

Gold Member
Periodic functions or functions defined on an interval.
Sorry I forgot to mention. My point is that this works only if you can actually expand the solution in Fourier series, which is not always true.

However, there is a continuous generalization of the Fourier series called Fourier Transform which results in a value at all real-numbered frequencies, not just multiples of some fundamental. You can Fourier Transform many non-periodic functions.
Also this only helps if the solution is "F-transformable" (for example $L^1$).

Were you asking if all differential equations can be solved in the Fourier domain?
In fact this is what I think the OP is asking and, as you said, the answer is no.

#### jonjacson

Sorry I forgot to mention. My point is that this works only if you can actually expand the solution in Fourier series, which is not always true.

Also this only helps if the solution is "F-transformable" (for example $L^1$).

In fact this is what I think the OP is asking and, as you said, the answer is no.
Can you show an example when it is not possible?

For the little I know, Fourier series can describe only periodic functions. If the solution of the differential equation in non-periodic they won't work
BUt if you take -inf to +inf as the period the trick should do the job.

It definitely can and does work, under certain conditions (mainly existence of all the relevant series or transforms, see below).

Periodic functions or functions defined on an interval.

That is, if you want to represent a function between say, -1 and 1, then you can do it with a Fourier series between those endpoints. The resulting Fourier series will be periodic of course, and will replicate the values between -1 and 1 in the regions above and below. If you don't care what the representation does outside of [-1, 1], then that's fine.

However, there is a continuous generalization of the Fourier series called Fourier Transform which results in a value at all real-numbered frequencies, not just multiples of some fundamental. You can Fourier Transform many non-periodic functions. In fact for Fourier Transform you want it to be non-periodic, as the transform involves an integral from -infinity to +infinity and you need a function such that the infinite integral exists.
Ok, so the group of functions whose -inf to +inf integral does not exist in case they are the solution of the differential equation will never be found with this method , is this correct? interesting

Were you asking if all differential equations can be solved in the Fourier domain? The answer is no. Were you asking if there exist any that can be? The answer is yes, it can be a very powerful and useful technique.
What is the fourier domain?

Sorry I forgot to mention. My point is that this works only if you can actually expand the solution in Fourier series, which is not always true.

Also this only helps if the solution is "F-transformable" (for example $L^1$).

In fact this is what I think the OP is asking and, as you said, the answer is no.
Ok, I didn't know it is not always possible to expand a function in terms of fourier series.

#### dRic2

Gold Member
Can you show an example when it is not possible?
I think this may be the easiest

\begin{cases}
\frac {dy}{dx} = 1 \space \space \space x \in ℝ \\
y(0) = 0
\end{cases}

#### jonjacson

I think this may be the easiest

\begin{cases}
\frac {dy}{dx} = 1 \space \space \space x \in ℝ \\
y(0) = 0
\end{cases}
Is that simply a straight line?

#### dRic2

Gold Member
Yes, $y = x$ is a straight line, un unbounded non-periodic function over all real numbers... it can't be expressed using Fourier series

#### RPinPA

Homework Helper
What is the fourier domain?
By that I meant what you get after a Fourier transform. If you start with a function of time, it transforms to a function of frequency. That's the most common, but you can also have spatial Fourier transforms. If you start with a function of x, you end up with a function in "spatial frequency" also called "wave number".

So when dealing with Fourier transforms of time signals or differential equations in time, you'll often hear people refer to "the time domain" and "the frequency domain". For instance, you can transform a time-domain differential equation into a frequency-domain equation which is often simply an algebraic equation, not a differential one. That's because the time derivative in the time domain transforms to a multiplication by $i\omega$ in the frequency domain.

Ok, I didn't know it is not always possible to expand a function in terms of fourier series.
That comment was about Fourier transforms, which require an integral from $-\infty$ to $\infty$. One requirement for such an integral to exist is that the argument goes to 0 in either direction, and there are certainly plenty of functions that don't do that.

#### jonjacson

By that I meant what you get after a Fourier transform. If you start with a function of time, it transforms to a function of frequency. That's the most common, but you can also have spatial Fourier transforms. If you start with a function of x, you end up with a function in "spatial frequency" also called "wave number".

So when dealing with Fourier transforms of time signals or differential equations in time, you'll often hear people refer to "the time domain" and "the frequency domain". For instance, you can transform a time-domain differential equation into a frequency-domain equation which is often simply an algebraic equation, not a differential one. That's because the time derivative in the time domain transforms to a multiplication by $i\omega$ in the frequency domain.

That comment was about Fourier transforms, which require an integral from $-\infty$ to $\infty$. One requirement for such an integral to exist is that the argument goes to 0 in either direction, and there are certainly plenty of functions that don't do that.
Interesting, I didn't know the transformation sends you to a non differential problem.
Now it is crystal clear.
Yes, $y = x$ is a straight line, un unbounded non-periodic function over all real numbers... it can't be expressed using Fourier series

#### AVBs2Systems

Hi

Every linear ordinary differential equation with constant coefficients can be solved using the laplace transform, every linear ODE with constant coefficients is looked at as a system, with stability criteria, a transfer function (the impulse response) and a frequency response if it exists. A smaller subset of these, can be solved using the fourier transform under the condition that the system itself is:
1. BIBO stable, the impulse response must be a function that is fourier transformable. See section below. This relates to the laplace transform and its region of convergence, the laplace transform must have a region of convergence that includes the imaginary axis.
2. Among other conditions...
$$\textbf{The dirac pulse, unit step function and LTI systems}$$
Definition of the dirac pulse $$\delta(t) = \begin{cases} \infty & \text{t  = 0 } \\ 0 & \text{t \ne 0 } \end{cases}$$
Definition of the unit step function:
$$k \epsilon(t) = \begin{cases} k & \text{t \ge 0 } \\ 0 & \text{t \lt 0 } \end{cases}$$
The dirac pulse has a laplace and fourier transform that are both equal, that is, one.
$$\mathcal L \Big[\delta(t) \Big] = 1 = \mathcal F \Big[\delta(t) \Big]$$
The fourier transform of the unit step does not exist, its laplace transform does and is equal to $$\frac{1}{s}$$
An LTI system is one that is having the properties of linearity and time invariance. These are a special class of systems for which the analysis becomes quite simpler. Linear ODES with constant coefficients are linear, and time invariant. Hence they can be completely characterised by the dirac pulse (impulse response). The step response, when the input is a unit step, is also sometimes considered though more in control engineering.
The dirac pulse, inserted into an LTI system, completely characterises the system in both time and frequency domain.
You can see why, because the Fourier and Laplace transform is a constant one, this signal, when inserted into a system subjects that system to all possible frequencies. Hence, it is an essential signal for obtaining the fingerprint of a system. The transfer function of all systems is arrived at, by inserting the dirac pulse into the system, or at least its close real life approximation (see addendum).

$$\textbf{Conditions for a functions fourier transform to exist}$$
The dirichlet criteria must be met: absolute integrabilitiy, bounded variation, and others...
1. $$f(t)$$ must be absolutely integrable: $$\displaystyle \int_{-\infty}^{\infty} \Big | f(t) \Big | \,\,\,\, \text{dt} \in \mathbb{R}$$
2. Bounded variation: $$f(t)$$ must have a finite number of maximma, minima, and discontinuities over any subset of its domain.
3. Others, but these two are the most mentioned (I think they have a higher physical significance).

$$\textbf{Linear ordinary differential equations that are LTI systems: The transfer function} \,\,\,\, H(s) \,\,\,\, and \,\,\,\, H(j \omega)$$
The transfer function is of a linear ordinary differential equation with constant coefficients, derives from the laplace transform of the system. If the region of convergence of this laplace transform includes the imaginary axis, then the fourier transform exists and is equal to the laplace transform evaluated at s = jw.
$$H(s)$$ is called the transfer function of a system. It is the ratio of the laplace transform of output over input, when the input is a dirac pulse.
$$H(j \omega) \tag{The fourier transform of a system}$$ is also referred to as the transfer function but more specifically it is the frequency response of the system. and it also is the ratio of the output over input, when the input is a dirac pulse, or in other words, it is the Laplace transform evaluated at s = jw given the ROC of the Laplace transform includes the imaginary axis.

$$\textbf{Conclusion: Linear O.D.Es with constant coefficients that are BIBO stable can be solved using the fourier transform}$$
At least, you can find the $$H(j \omega)$$.
Systems have a simple relationship in the frequency domain: the output is the input multiplied by the transfer function.
$$Y( j \omega) = H( j \omega ) \cdot X( j \omega)$$
So, if the system is BIBO stable, its fourier transform exists, the frequency response exists, and if the input is Fourier transformable, we can find the output by simply multiplying the input into the frequency response, and taking the inverse fourier transform of $$Y(j \omega)$$
$$\textbf{Solved example: Simple RC circuit with output voltage at the capacitor}$$
The Linear O.D.E with constant coefficients is:
$$\frac{x(t)}{RC} = y'(t) + \frac{ y(t) }{RC}$$
For $$H(s)$$ we assume initial conditions zero, and take the transfer function, we must then make
$$\mathcal L \Bigg( x(t) = \delta(t) \Bigg) = 1 \tag{impulse response}$$
Using the derivative property of the laplace transform we see:
$$H(s) = \dfrac{ \frac{1}{RC} }{s + \frac{1}{RC}} \tag{R.O.C real of s greater than -1/RC}$$
This is a bibo stable system. As, its impulse response in the time domain (taking the inverse laplace transform of H(s) is:
$$h(t) = \frac{1}{RC} \epsilon(t) e^{-\frac{t}{RC} }$$
Hence, the fourier transform at s = jw is:
$$H( j \omega) = \frac{\frac{1}{RC}}{\frac{1}{RC} + j \omega } = \frac{1}{ 1 + j \omega RC } \tag{Frequency response}$$
Hence, the output of this differential equation, or system, is simply:
$$Y(j \omega) = \mathcal F \Big[ x(t) \Big] \cdot H(j \omega)$$
$$\textbf{The fourier transform of differential equations of electrical circuits is the same as the sinusoidal steady state impedence model}$$
Try the voltage divider using impedences, you will get the same transfer function.
$$\textbf{The laplace transform extends the set of systems for which the transfer function exists}$$
The laplace transform extends the set of linear ODEs with constant coefficients for which the transfer function exists, it allows us to solve for unstable systems too. Not only that but in situations where initial conditions are given, the laplace transform takes those initial conditions into account too. In summary, linear ODES with constant coefficients can be solved if the impulse response of the system is fourier transformable, ie it is a bibo stable system. If not, the laplace transform would exist and that can be used to solve the linear ODE.
The laplace transform and other transforms would possibly extend to partial differential equations too, but I am not sure of this. The fourier series, we did not use to solve a differential equation. So to recap, every linear ODE with constant coefficients (a time invariant system) has a Laplace transform, if the system is BIBO stable then it would have a fourier transform too, and the fourier transform can be used to solve the differential equation then.
$$\textbf{Addendum: Real life approximation of the dirac pulse}$$
Because the dirac pulse is not really feasible to generate as a waveform in real life, we take a rectangular approximation:
$$\delta(t) = \displaystyle \lim_{ a \to 0} \Bigg( \frac{1}{a} \text{Rect} \Big({\frac{t}{a} } \Big) \Bigg)$$
This is basically a rectangle compressed in width and elongated in height.

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"Why the Fourier series doesn't work to solve any differential equation?"

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