Please enable JavaScript.
Coggle requires JavaScript to display documents.
Matte 4K (Laplace Transform (ODEs (Shifted Data Proplems (\(y''+y=r(t)\,\…
Matte 4K
Laplace Transform
Basics
S Shifting
\(\mathscr{L}^{-1}(F(s-a))=e^{at}f(t)\)
\(\mathscr{L}(e^{at}f(t))=F(s-a)\)
Existence
\(|f(t)|≤Me^{kt}\)
Piecewise continuous
\(\Rightarrow\) Exists for s > k
Definition
\(F(s)=\mathscr{L}(f)=\int_0^\inf e^{-st}f(t)dt\)
\(f(t)=\mathscr{L}^{-1}(F)\)
Integral
\(\mathscr{L}(\int_0^tf(t)dt)=\frac{F(s)}{s}\)
Derivative
\(\mathscr{L}(f^{(n)})=s^nF(s)-s^{n-1}f(0)-s^{n-2}f'(0)\,-\,...-\,f^{n-1}(0)\)
ODEs
\(\mathscr{L}(ODE)\)
Find expression for Y(s)
\(y(t)=\mathscr{L}^{-1}(Y(s))\)
Partial fraction expansion of Y(s)
Shifted Data Proplems
\(y''+y=r(t)\,\,\,\,\,\,y(t_0)=k_1,\, y'(t_0)=k_2\)
\(t = \tau+t_0\)
\( \lambda(\tau)'' + \lambda(\tau) = r(\tau + t_0)\,\,\,\,\,\,\lambda(0)=k_1,\,\lambda'(0)=k_2\)
\(\mathscr{L}(ODE)\)
Find expression for \( \Lambda(s) \)
Partial fractio expansion of \( \Lambda(s) \)
\( \lambda(\tau)=\mathscr{L}(\Lambda(s))\)
\( y(t) = \lambda(\tau)\,\,\,\,\,\,\tau=t-t_0\)
Variable Coefficients
Useful for linear ODEs of second order with coefficients on the form \(at+b\)
Use differentiation of transform to get a first order ODE for Y, which can be solved analytically e.g. by separation of variables.
Systems of ODEs
Given a system of ODEs for \(y_1\) and \(y_2\)
Find the laplace transform of the equations
Solve algebraically for \(Y_1\) and \(Y_2\)
Inverse laplace to find \(y_1 (t)\) and \(y_2 (t)\)
Unit Step Function
\(u(t-a) \)
\(=0 \) if \( t < a \)
\(=1\) if \(t > a\)
\(\mathscr{L}(u(t-a))=\frac{e^{-as}}{s}\)
Time Shifting
\(f(t-a)u(t-a)\)
\(=0\) if \(t < a\)
\(=f(t-a)\) if \(t>a\)
\(\mathscr{L}(f(t-a)u(t-a))=e^{-as}F(s)\)
Dirac Delta Function
\(\delta(t-a)\)
\(=\infty\) if \(t=a\)
\(=0\) otherwise
\(\int_0^\infty\delta(t-a)\,dt=1\)
\(\int_0^{\infty}g(t)\delta(t-a)\,dt=g(a)\)
\(\mathscr{L}(\delta(t-a))=e^{-as}\)
Convolution
\(h(t)=(f*g)(t)=\int_0^{t}f(\tau)g(t-\tau)\,d\tau\)
\(H(s)=F(s)\cdot G(s)=\mathscr{L}(f)\cdot\mathscr{L}(g)\)
In general \(\,\,f*1≠f\)
Integral Equations
\(y(t)-\int_0^t y(\tau)\cdot g(t-\tau)\,\,d\tau=r(t)\)
\(y-y*g=r\)
\(Y-Y\cdot G=R\)
\(Y=\frac{R}{1-G}\)
\(y(t)=\mathscr{L}(Y)\)
Differentiation and Integration of Transforms
Differentiation
Then \(\mathscr{L}(t\cdot f)=-F'(s)\)
Differentiation of a transform corresponds to the multiplication of the function by \(-t\).
If \(\mathscr{L}(f)=F(s)\)
Integration
If \(\mathscr{L}(f)=F(s)\)
Then \(\mathscr{L}(\frac{f}{t})=\int_s^\infty F(s)\,\,ds\)
Integration of a transform corresponds to the division of the function by \(t\).
Complex Analysis
Complex Functions
\(w=f(z)=u(x,y)+iv(x,y)\)
Analytic Functions
Analyticity
\(f(z)\) is analytic in a domain \(D\) if \(f(z)\) is defined and differentiable for all points of D.
A function that is analytic for all \(z\) is called entire
Cauchy-Riemann Equations
\(f\) is analytic if the Cauchy-Riemann equations are satisfied:
\(u_x=v_y,\,\,\,\,\,\,\,\,u_y=-v_x\)
Harmonic Conjugate Functions
\(u(x,y)\) is harmonic if \( \nabla^2u=0\)
Find \(u_x\) and \(u_y\)
Use C-R equations to find \(v_x\) and \(v_y\)
Integrate \(v_y\) to get \(v=w(x,y)+h(x)\)
Differentiate to get \(v_x\) and compare to C-R \(v_x\) to find \(h'(x)\) and integrate to find \(h(x)\)
\(v(x,y)\) is the harmonic conjugate function to \(u(x,y)\)
Exponential Function
\(e^z=e^x(\cos{y}+i\sin{y})\)
\(|e^{iy}|=1\)
\(e^z≠0\)
Periodicity
\(e^{z+2\pi i}=e^z\)
Trigonometric and Hyperbolic Functions
\(\cos{z}=\frac{1}{2}(e^{iz}+e^{-iz})\)
\(\sin{z}=\frac{1}{2i}(e^{iz}-e^{-iz})\)
\(e^{iz}=\cos{z}+i\sin{z}\)
Equations
\(\cos{z}=a\)
\(e^{2iz}-2ae^{iz}+1=0\)
\(\frac{1}{2}(e^{iz}+e^{-iz})=a\)
\(e^{iz}=e^{-y}e^{ix}=\frac{2a \pm \sqrt{(2a)^2-4}}{2}\)
\(e^{ix}=1 \Rightarrow x=2\pi n\)
\(e^{-y}=\frac{2a \pm \sqrt{(2a)^2-4}}{2}\)
\(\Rightarrow y=-\ln{\frac{2a \pm \sqrt{(2a)^2-4}}{2}}\)
Ans: \(z=±2\pi n-i\ln{\frac{2a \pm \sqrt{(2a)^2-4}}{2}},\,\,\,n=0,1,2,...\)
\(\cosh{z}=\frac{1}{2}(e^z+e^{-z})\)
\(\sinh{z}=\frac{1}{2}(e^z-e^{-z})\)
\(\cosh{iz}=\cos{z}\,\,\,\,\,\,\,\,\cos{iz}=\cosh{z}\)
\(\sinh{iz}=i\sin{z}\,\,\,\,\,\,\,\,\sin{iz}=i\sinh{z}\)
Logarithm
\(w=\ln{z}\) is defined by the relation \(e^{w}=z\)
\(e^w=e^{u+iv}=re^{i\theta}\)
\(e^u=r\,\,\,\,\,\,\,v=\theta\)
\(\ln{z}=\ln{r}+i\theta, ,\,\,\,\,\,\,z=re^{i\theta}\)
Principal value: Ln\(\,z=\ln{|z|}+i\)Arg\(\,z\)
Complex natural logarithm: \(\ln{z}=\) Ln\(\,z±2\pi ni\)
\(\ln{z}\) is analytic, except at 0 and on the negative real axis, and has the derivative
\((\ln{z})'=\frac{1}{z}\)
General Powers
\(z^c=e^{c\ln{z}}=e^{c(\ln{r}+i\theta+2\pi ni)}\)
Principal value: \(z^c=e^{c\, \textrm{Ln} \,z}\)
Complex Line Integral
Indefinite Integration of Analytic Functions
\(f(z)\) analytic in simply connected domain D.
\(\int_{z_0}^{z_1}f(z)\,dz=F(z_1)-F(z_0)\)
\(z_0, z_1 \epsilon D\)
Use of the Path
Let \(C\) be a piecewise smooth path, represented by \(z=z(t)\), where \(a≤t≤b\). Let \(f(z)\) be continuous on \(C\).
\(\int_Cf(z)\,dz=\int_a^bf[z(t)]\dot{z}(t)\,dt\)
\(\oint_C(z-z_0)^m\,dz=2\pi i\) if \(m=-1\)
\(=0\) otherwise
The line integral of non-analytic functions depend on the path \(C\)
\(ML\)-inequality
\(|\int_Cf(z)\,dz|≤ML\)
\(L\) is the length of \(C\), and \(|f(z)|≤M\) everywhere on \(C\)
Cauchy's Integral Theorem
If \(f(z)\) is analytic in a simply connected domain D
\(\oint_Cf(z)\,dz=0\) if \(C\, \epsilon \,D\)
Independence of Path
If \(f(z)\) is analytic in a simply connected domain D, then the integral of \(f(z)\) is independent of the path in D
Principle of Deformation of Path
A closed line integral \(\oint_Cf(z)\) retains its value independent of the path \(C\) as long as \(f(z)\) is analytic in all points in \(C\)
\(\oint_{C_1}\frac{1}{z}\,dz=\oint_{C_2}\frac{1}{z}\,dz=2\pi i,\,\,\,\,\,\,\,\,\,C_1=e^{it},\,\,C_2=2e^{it}\)
Also viable if \(D\) is not simply connected (e.g. singularity) as long as \(f(z)\) is analytic on and inside of \(C\)
Cauchy's Integral Formula
Let \(f(z)\) be analytic in a simply connected domain D. Then
\(f(z_0)=\frac{1}{2\pi i}\oint_C\frac{f(z)}{z-z_0}\,dz\)
\(C\) in \(D\) and \(z_0\) enclosed in \(C\)
Multiply Connected Domains
\(z_0\) inside multiply connected domain \(D\) bounded by e.g. \(C_1\) and \(C_2\) and \(f(z)\) analytic in \(D\)
\(f(z_0)=\frac{1}{2\pi i}\oint_{C_1}\frac{f(z)}{z-z_0}\,dz+\frac{1}{2\pi i}\oint_{C_2}\frac{f(z)}{z-z_0}\,dz\)
Derivatives of Analytic Functions
\(f^{(n)}(z_0)=\frac{n!}{2\pi i}\oint\frac{f(z)}{(z-z_0)^{n+1}}\,dz\)
Cauchy's Inequality
\(|f^{(n)}(z_0)|≤\frac{n!M}{r^n}\)
Liouille's Theorem
If an entire function is bounded in absolute value in the whole complex plane, then this function must be a constant.
Power Series
Tests For Convergence and Divergence
Divergence
If \(\lim_{n\to \infty}z_n≠0\) the series diverges
Comparison Test
If a series \(\sum_{n=1}^{\infty}z_n\) is given and we can find a convergent series \(\sum_{n=1}^{\infty}b_n\) with nonnegative real terms such that \(|z_n|≤b_n|\), then the given series converges absolutely.
The geometric series
\(\sum_{n=0}^{\infty}q^m=\frac{1}{1-q}\) if \(|q|<1\), and diverges if else.
Ratio Test
If \(\lim_{n\to \infty}|\frac{z_{n+1}}{z_n}|=L<1\) the series converges absolutely
If \(L=1\) the series may converge or diverge, and further testing is needed
If \(L>1\) the series converges
Root Test
Similar to the ratio test. Test for \(\lim_{n\to \infty}\sqrt[n]{z_n}=L\)
Rarely useful
The sums of power series are analytic functions
\(\sum_{n=0}^{\infty}a_n(z-z_0)^n\)
Convergence
Every power series converges at the center \(z_0\)
If the series converges for \(z=z_1\), it converges for every \(z\) closer to the center
If it diverges for \(z=z_2\) it diverges for every \(z\) farther away from the center
Radius of Convergence
\(R=\frac{1}{L^*}=\lim_{n\to \infty}\frac{a_n}{a_{n+1}}\)
If \(R=\infty\) the series converges for all \(z\)
If \(R=0\) the series converges only at the center \(z_0\)
The series will then converge int the open disk \(|z-z_0|=R\)
The series may converge on some or all or none of the points on the circle of convergence itself
Functions
If a power series has a radius of convergence \(R>0\) this series represents an analytic function of \(z\)
Since the derived series are power series with the same radius of convergence as the original series, these also represent analytic functions.
Operations on Power Series
Termwise addition of two power with radius of convergence \(R_1\) and \(R_2\) series yields a power series with radius of convergence at least equal to the smallest of \(R_1\) and \(R_2 \)
Termwise multiplication yields the same R as above, and the series will inside the circle with radius \(R\) converge to the sum \(s(z)=f(z)g(z)\)
Derived series (termwise differentiation) will have the same radius of convergence as the originial series
The power series obtained by termwise integration of a given power series will have the same radius of convergence as the original
Taylor and Maclauren Series
\(f(z)=\sum_{n=1}^{\infty}a_n(z-z_0)^n\) where \(a_n=\frac{1}{n!}f^{(n)}(z_0)\)
Every analytic function can be represented by such a power series
A Maclaurin series is a Taylor series with center \(z_0=0\)
To find a Taylor series one can use the definition directly (by finding \(a_n\)) or try to write the function on a form with a known series.
Residue Integration
Laurent Series
\(f(z)=\sum_{n=0}^{\infty}a_n(z-z_0)^n+\sum_{n=1}^{\infty}\frac{b_n}{(z-z_0)^n}\)
A Laurent series of a function \(f(z)\) which is analytic in an annulus with center \(z_0\)
\(a_n=\frac{1}{2\pi i}\oint_C\frac{f(z^*)}{(z^*-z_0)^{n+1}}\,dz^*\)
\(b_n=\frac{1}{2\pi i}\oint_C(z^*-z_0)^{n-1}f(z^*)\,dz^*\)
Where \(C\) lies inside the annulus and encircles the inner circle
More convenient formula:
\(f(z)=\sum_{n=-\infty}^{\infty}a_n(z-z_0)^n\)
\(a_n=\frac{1}{2\pi i}\oint_C\frac{f(z^*)}{(z^*-z_0)^{n+1}}\,dz^*\)
The principle part of a Laurent series is the terms with negative powers (i.e. terms with \(b_n\) coefficients)
Method
Laurent series of \(f(x)\cdot z^n\) when \(f(x)\) has a known Maclaurin series:
Multiply the known Maclaurin series by \(z^n\)
Ex: \(\frac{1}{z^3-z^4}\)
Nonnegative powers: \(=\frac{1}{z^3}\cdot \frac{1}{1-z}=z^{-3}\sum_{n=0}^{\infty}z^n=\sum_{n=0}^{\infty}z^{n-3}\)
Negative powers: \(=\frac{1}{z^4}\cdot \frac{1}{z^{-1}-1}=z^{-4}\frac{-1}{1-z^{-1}}=-z^{-4}\sum_{n=0}^{\infty}z^{-n}=-\sum_{n=0}^{\infty}\frac{1}{z^{n+4}}\)
Poles
If the principal part has finitely many terms \(b_1\) to \(b_m\), then the singularity of \(f(z)\) at \(z=z_0\) is called a pole.
\(m\) is called its order. Poles of the first order are simple poles.
If it has a principal part of infinitely many terms, then \(f(z)\) has an isolated essential singularity at \(z=z_0\)
If \(f(z)\) is analytic and has a pole at \(z=z_0\), then \(f(z) \to \infty\) when \(z \to z_0\)
An analytic function near an isolated essential singularity takes on every value in an arbitrary small neighbourhood of \(z_0\)
Zeros
An analytic function \(f(z)\) has a zero in \(z_0\) if \(f(z_0)=0\)
Zeros are isolated, i.e. no other zeros in the neighbourhood of a zero
Let \(f(z)\) be analytic with a nth order zero at \(z=z_0\). Then \(\frac{h(z)}{f(z)}\) will have nth order poles at \(z=z_0\), provided \(h(z)\) analytic and nonzero at \(z=z_0\)
Analytic at infinity
To check if a function \(f(z)\) is analytic at infinity, let \(g(w)=f(\frac{1}{w})\)
If \(g(w)\) is analytic at \(w=0\), then \(f(z)\) is analytic at infinity, and vice versa if \(g\) is singular.
Method
To find \(b_1\)
Residue at simple pole
(1) \(\mathrm{Res}_{z=z_0}f(z)=\lim_{z\to z_0}(z-z_0)f(z)\)
(2) \(\mathrm{Res}_{z=z_0}f(z)=\frac{p(z_0)}{q'(z_0)}\) assuming that \(f(z)=\frac{p(z)}{q(z)}\) with \(p(z_0)≠0\) and \(q(z)\) has a simple zero at \(z_0\)
Poles of any order at \(z_0\)
The residue of \(f(z)\) at an mth-order pole:
\(\mathrm{Res}_{z=z_0}f(z)=\frac{1}{(m-1)!}\lim_{z\to z_0}[(z-z_0)^mf(z)]^{(m-1)}\)
For \(m=2\): \(\mathrm{Res}_{z=z_0}f(z)=\lim_{z\to z_0}[(z-z_0)^2f(z)]'\)
Solving Integrals
Several Singularities Inside \(C\)
\(\oint_Cf(z)\,dz=2\pi i \sum_{j=1}^k\mathrm{Res}_{z=z_j}f(z)\)
For \(k\) singularities inside \(C\)
\(\int_0^{2\pi}F(\cos{\theta},\sin{\theta})\,d\theta\)
Choose \(z=e^{i\theta}\)
Then \(\cos{\theta}=\frac{1}{2}(z+\frac{1}{z})\) and \(\sin{\theta}=\frac{1}{2i}(z-\frac{1}{z})\)
\(\Rightarrow \int_0^{2\pi}F(\cos{\theta},\sin{\theta})\,d\theta=\oint_Cf(z)\frac{dz}{iz}\)
Solve the new integral by residue integration
\(\int_{-\infty}^{\infty}f(x)\,dx\)
Cauchy's principal value: \(\mathrm{pr.}\,\mathrm{v.}\,\int_{-\infty}^{\infty}f(x)\,dx=\lim_{R\to\infty}\int_{-R}^Rf(x)\,dx\)
\(f(x)=\frac{p(x)}{q(x)}\) is a real rational function with \(q(x)≠0\) for all real \(x\) and \(q(x)\) is of degree at least two units higher than the \(p(x)\) (e.g. \(\frac{1}{x^2+1}\))
Substitute \(x\) with \(z\) and find all poles of \(f(z)\) in the upper half-plane.
Then \((\int_{-\infty}^{\infty}f(x)\,dx=2\pi i\sum\mathrm{Res}\,f(z)\)
If \(\int_0^{\infty}f(x)\,dx\) check if \(f(x)\) is even. Evaluate the integral from -infinity to infinity. The given integral will be half.
Fourier Integrals
Finding the Fourier integral coefficients:
\(\int_{-\infty}^{\infty}f(x)\cos{sx}\,dx+i\int_{-\infty}^{\infty}f(x)\sin{sx}\,dx=\int_{-\infty}^{\infty}f(x)e^{isx}\,dx\)
Let \(\oint_Cf(z)e^{isz}\,dz\)
Then \(\int_{-\infty}^{\infty}f(x)e^{isx}\,dx=2\pi i\sum\mathrm{Res}\,f(z)e^{isz}\)
And \(\int_{-\infty}^{\infty}f(x)\cos{sx}\,dx=2\pi i^2\sum\mathrm{Im\,Res}\,f(z)e^{isz}\)
\(\int_{-\infty}^{\infty}f(x)\sin{sx}\,dx=2\pi \sum\mathrm{Re\,Res}\,f(z)e^{isz}\)
Simple Poles at Real Axis
\(\mathrm{pr.}\,\mathrm{v.}\,\int_{-\infty}^{\infty}f(x)\,dx=2\pi i\sum\mathrm{Res}\,f(z)+\pi i\sum\mathrm{Res}\,f(z)\)
Where the first sum is for all poles in upper half-plane, and the second sum for all simple poles on the \(x\)-axis.
Concept
If \(f(z)\) has a singularity at \(z=z_0\), then \(f(z)\) has a Laurent series given by \(\sum_{n=0}^{\infty}a_n(z-z_0)^n+\frac{b_1}{z-z_0}+\frac{b_2}{(z-z_0)^2}+...\)
\(b_1\) is called the residue of \(f(z)\) at \(z=z_0\) and is given by \(b_1=\frac{1}{2\pi i}\oint_Cf(z)\,dz\)
Given \(\oint_Cf(z)\,dz\). If \(f(z)\) has a singularity inside \(C\) but is otherwise analytic on and inside \(C\), this method is viable.
An integral \(\oint_Cf(z)\,dz\) can then be solved by finding the residue of \(f(z)\) at the singularity inside \(C\)
\(\oint_Cf(z)\,dz=2\pi i\,\mathrm{Res}_{z=z_0}f(z)\)
Conformal Mapping
Conformal mappings preserves angles, except at critical points, and are represented by analytical functions. The critical points are where the derivative of such a function is zero.
Fourier Analysis
Fourier Series
Periodicity
\(f(x+np)=f(x),\,\,\,\,n=1,\,2,\,3,\,...\)
Then f has the period p
f is defined for all real x, except possibly at some points.
\(f(x)=a_0+\sum_{n=1}^{\infty} (a_n \cos{nx}+b_n \sin{nx})\)
Fourier Coefficients
\(a_0=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)\,dx\)
\(a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x) \cos{nx}\,dx\)
\(b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x) \sin{nx}\,dx\)
Convergence and Sum
\(f(x)\) is \(2\pi\) periodic and piecewise continuous in the interval \(-\pi ≤ x ≤\pi\)
Let \(f(x)\) have left- and right-hand derivatives at each point at that interval
Then the Fourier series of \(f(x)\) converges
Its sum is \(f(x)\), except in discontinuous points
There the sum is the average of the left- and right-hand limits of f(x)
Arbitrary Period
Period \(p=2L\)
Series and Coefficients
\(f(x)=a_0+\sum_{n=1}^{\infty} (a_n \cos{\frac{n\pi}{L}x}+b_n \sin{\frac{n\pi}{L}x})\)
\(a_0=\frac{1}{2L}\int_{-L}^{L}f(x)\,dx\)
\(a_n=\frac{1}{L}\int_{-L}^{L}f(x) \cos{\frac{n\pi}{L}x}\,dx\)
\(b_n=\frac{1}{L}\int_{-L}^{L}f(x) \sin{\frac{n\pi}{L}x}\,dx\)
Even and Odd Functions
\(f(x)\) is even (\(f(-x)=f(x)\))
\(f(x)=a_0+\sum_{n=1}^{\infty} a_n \cos{\frac{n\pi}{L}x}\)
\(a_0=\frac{1}{L}\int_{0}^{L}f(x)\,dx\)
\(a_n=\frac{2}{L}\int_{0}^{L}f(x) \cos{\frac{n\pi}{L}x}\,dx\)
\(f(x)\) is odd (\(f(-x)=-f(x)\))
\(f(x)=\sum_{n=1}^{\infty} b_n \sin{\frac{n\pi}{L}x}\)
\(b_n=\frac{2}{L}\int_{0}^{L}f(x) \sin{\frac{n\pi}{L}x}\,dx\)
Sum and Scalar Multiple
Let \(h(x)=f+g\)
Then the Fourier Series of h(x) is
\(h(x)=a_{0h}+\sum_{n=1}^{\infty} (a_{nh} \cos{nx}+b_{nh} \sin{nx})\)
\(a_{0h}=a_{0f}+a_{0g}\)
\(a_{nh}=a_{nf}+a_{ng}\)
\(b_{nh}=b_{nf}+b_{ng}\)
Half-Range Expansions
Given a function defined in the interval \((0, L)\)
The even periodic extension (or Fourier Cosine Series) of \(f\) can be found by calculating \(a_0\) and \(a_n\) for functions of arbitrary periods
The odd periodic extension (or Fourier Sine Series) of \(f\) can be found by calculating \(b_n\) for functions of arbitrary periods
Approximation
Let \(f(x)\) be a function on the interval \(-\pi≤x≤\pi\) that can be represented by a Fourier series.
Then \(f(x)≈A_0+\sum_{n=1}^{N} (A_n \cos{nx}+B_n \sin{nx})\) is an approximation of the given Fourier series.
The square error \(E=\int_{-\pi}^{\pi}(f-F)^2\,dx\) is smallest when the coefficients of the approximation is the Fourier coefficients of \(f\)
The error when the approximation is best is given by \(E^*=\int_{-\pi}^{\pi}f^2\,dx-\pi[2a_0^2+\sum_{n=1}^{N}(a_n^2+b_n^2)]\)
Complex Fourier Series
\(f(x)=\sum_{-\infty}^{\infty} c_n e^{in\pi x/L}\)
\(c_n=\frac{1}{2\pi}\int_{-L}^{L}f(x)\,e^{-in\pi x/L}\,dx\)
Orthogonality of the Trigonometric System
Integrated over an interval of length \(2\pi n\)
\(\int_a^b \cos{nx} \cos{mx}\,dx=0,\,\,\,\,n≠m\)
\(\int_a^b \sin{nx} \sin{mx}\,dx=0,\,\,\,\,n≠m\)
\(\int_a^b \cos{nx} \sin{mx}\,dx=0,\,\,\,\,n,\,m\) integers
If \(n=m\) then the integral is \(\frac{1}{2}(b-a)\)
Fourier Transform
Definition
Inverse Fourier Transform: \(f(w)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\mathscr{f}(x)e^{iwx},dx\)
\(\mathscr{f}=\mathscr{F}(f)\)
\(f=\mathscr{F}^{-1}(\mathscr{f})\)
Fourier Transform: \(\mathscr{f}(w)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{-iwx},dx\)
Existence
Absolutely integrable on the x-axis and piecewise continuous on every finite interval
Transform of the derivative of f
Let \(f(x)\) be continuous on the x-axis and \(f(x)\to 0\) as \(|x|\to \infty\). Let \(f'(x)\) be absolutely integrable on the x-axis. Then
\(\mathscr{F}(f'(x))=iw\mathscr{f}\)
Convolution
\(\mathscr{F}(f*g)=\sqrt{2\pi} \mathscr{F}(f)\cdot \mathscr{F}(g)\)
\((f*g)(x)=\int_{-\infty}^{\infty}\mathscr{f}(w)\mathscr{g}(w)e^{iwx}\,dw\)
Fourier Integral
If \(f(x)\) is piecewise continuous and has a right- and left-hand derivative at every point and the integral \(\int_{-/infty}^{\infty}|f(x)|\,dx\) exists, then \(f(x)\) can be represented by a Fourier integral:
\(f(x)=\int_0^{\infty}[A(w)\cos{wx}+B(w)\sin{wx}]\,dw\)
\(A(w)=\frac{1}{\pi}\int_{-\infty}^{\infty}f(v)\cos{wv}\,dv\)
\(B(w)=\frac{1}{\pi}\int_{-\infty}^{\infty}f(v)\sin{wv}\,dv\)
Odd and Even
If \(f(x)\) is even, \(f(x)=\int_0^{\infty}A(w)\cos{wx}\,dw\)
If \(f(x)\) is odd, \(f(x)=\int_0^{\infty}B(w)\sin{wx}\,dw\)
\(A(w)=\frac{2}{\pi}\int_0^{\infty}f(v)\cos{wv}\,dv\)
\(B(w)=\frac{2}{\pi}\int_0^{\infty}f(v)\sin{wv}\,dv\)
PDEs
One-Dimensional Wave Equation
\(u_{tt}=c^2u_{xx}\)
Boundary Conditions
\(u(0,t)=0\)
\(u(L,t)=0\)
For all \(t\)
Initial Conditions
\(u(x,0)=f(x)\)
\(u_t(x,0)=g(x)\)
Solution
Method of Separating Variables
\(u(x,t)=F(x)\cdot G(t)\)
\(F\ddot{G}=c^2F''G\)
\(\frac{\ddot{G}}{c^2G}=\frac{F''}{F}=k\)
\(F''-kF=0\)
\(\ddot{G}-c^2kG=0\)
Satisfying Boundary Conditions
\(u(0,t)=F(0)G(t)=0,\,\,\,\,u(L,t)=F(L)G(t)=0\)
\(G≠0\Rightarrow F(0)=0,\,\,\,\,\,\,\,F(L)=0\)
Choose \(k=-p^2\)
\(F(x)=A\cos{px}+B\sin{px}\)
\(F(0)=A=0,\,\,\,\,\,\,\,F(L)=B\sin{pL}=0\)
\(\sin{pL}=0\)
\(pL=n\pi\) thus \(p=\frac{n\pi}{L}\)
\(F_n(x)=\sin{\frac{n\pi}{L}x}\,\,\,\,\,\,\,\,\,B=1\)
\(\ddot{G}+\lambda_n^2G=0\) where \(\lambda_n=cp=\frac{cn\pi}{L}\)
\(G_n(t)=B_n\cos{\lambda_nt}+B_n^*\sin{\lambda_nt}\)
\(u_n(x,t)=(B_n\cos{\lambda_nt}+B_n^*\sin{\lambda_nt})\sin{\frac{n\pi}{L}x}\)
These functions are called eigenfunctions, and \(\lambda_n\) are called eigenvalues
Satisfying Initial Conditions
A solution satisfying the initial conditions can be found in the infinite sum of solutions:
\(u(x,t)=\sum_{n=1}^{\infty}u_n(x,t)\)
\(u(x,0)=\sum_{n=1}^{\infty}B_n\sin{\frac{n\pi}{L}x}=f(x)\)
\(B_n=\frac{2}{L}\int_0^{L}f(x)\sin{\frac{n\pi}{L}x}\,dx\)
\(u_t(x,0)=\sum_{n=1}^{\infty}B_n^*\lambda_n\sin{\frac{n\pi}{L}x}=g(x)\)
\(B_n^*\lambda_n=\frac{2}{L}\int_0^{L}g(x)\sin{\frac{n\pi}{L}x}\,dx\)
\(B_n^*=\frac{2}{cn\pi}\int_0^{L}g(x)\sin{\frac{n\pi}{L}x}\,dx\)
d'Alembert's Solution
Transform
\(v=x+ct\,\,\,\,\,\,\,\,w=x-ct\)
\(u=\phi(v)+\psi(w)\)
\(u(x,t)=\phi(x+ct)+\psi(x-ct)\)
Solution
\(u(x,0)=f(x)\,\,\,\,\,\,\,\,u_t(x,0)=g(x)\)
\(u_t(x,t)=c\phi'(x+ct)-c\psi'(x-ct)\)
\(u(x,0)=\phi(x)+\psi(x)=f(x)\)
\(u_t(x,0)=c\phi'(x)-c\psi'(x)=g(x)\)
\(\phi(x)-\psi(x)=k(x_0)+\frac{1}{c}\int_{x_0}^xg(s)\,ds\,\,\,\,\,\,\,\,k(x_0)=\phi(x_0)-\psi(x_0)\)
\(\phi(x)=\frac{1}{2}f(x)+\frac{1}{2c}\int_{x_0}^xg(s)\,ds+\frac{1}{2}k(x_0)\)
\(\psi(x)=\frac{1}{2}f(x)-\frac{1}{2c}\int_{x_0}^xg(s)\,ds-\frac{1}{2}k(x_0)\)
\(u(x,t)=\frac{1}{2}[f(x+ct)+f(x-ct)]+\frac{1}{2c}\int_{x-ct}^{x+ct}g(s)\,ds\)
If \(u_t(x,0)=0\) then \(u(x,t)=\frac{1}{2}[f(x+ct)+f(x-ct)]\)
One-Dimensional Heat Equation
\(u_t=c^2u_{xx}\)
Boundary Conditions
\(u(0,t)=0\)
\(u(L,t)=0\)
Initial Condition
\(u(x,0)=f(x)\)
Product Method
\(F''+p^2F=0\)
\(\dot{G}+c^2p^2G=0\)
Satisfying Boundary Conditions
\(F_n(x)=\sin{\frac{n\pi}{L}x}\) (See 1D Wave Eq)
\(\dot{G}+\lambda_n^2G=0\) where \(\lambda_n=\frac{cn\pi}{L}\)
\(G_n(t)=B_ne^{-\lambda_n^2t}\)
\(u_n(x,t)=B_n\sin{\frac{n\pi}{L}x}e^{-\lambda_n^2t}\)
Satisfying Initial Condition
\(u(x,t)=\sum_{n=1}^{\infty}u_n(x,t)\)
\(u(x,0)=\sum_{=1}^{\infty}B_n\sin{\frac{n\pi}{L}x}=f(x)\)
\(B_n=\frac{2}{L}\int_0^Lf(x)\sin{\frac{n\pi}{L}x}\,dx\)
Insulated Ends
Problem
\(u_x(0,t)=0,\,\,\,\,\,\,\,\,u_x(L,t)=0\)
Given initial conditions
Satisfying Boundary Conditions
\(F'(0)=Bp=0\) and then \(F'(L)=-Ap\sin{pL}\)
\(p=\frac{n\pi}{L},\,\,\,\,\,A=1\) and \(\,\,\,B=0\)
\(F_n(x)=\cos{\frac{n\pi}{L}x}\)
\(G_n(t)=A_ne^{-\lambda_n^2t}\)
\(u_n(x,t)=A_n\cos{\frac{n\pi}{L}x}e^{-\lambda_n^2t}\)
\(F'(x)=-Ap\sin{px}+Bp\cos{px}\)
Satisfying Initial Conditions
\(u(x,t)=\sum_{n=0}^{\infty}u_n(x,t)\)
\(u(x,0)=\sum_{n=0}^{\infty}A_n\cos{\frac{n\pi}{L}x}e^{-\lambda_n^2}=f(x)\)
\(A_0=\frac{1}{L}\int_{0}^{L}f(x)\,dx\)
\(A_n=\frac{2}{L}\int_{0}^{L}f(x) \cos{\frac{n\pi}{L}x}\,dx\)
Infinitely Long Bar
Problem
\(u_t=c^2u_{xx}\)
\(u(x,0)=f(x)=e^{-x^2}\)
Fourier Transform
\(\mathscr{u}_t=\mathscr{F}(u_t)=c^2\mathscr{F}(u_{xx})=c^2i^2w^2\mathscr{F}(u)=-c^2w^2\mathscr{u}\)
\(\mathscr{u}(w,t)=C(w)e^{-c^2w^2t}\)
Initial Condition
\(u(x,0)=f(x)\) thus \(\mathscr{u}(w,0)=\mathscr{f}(w)\)
\(C(w)=\mathscr{u}(w,0)=\mathscr{f}(w)=\mathscr{F}(e^{-x^2})=\frac{1}{\sqrt{2}}e^{-\frac{w^2}{4}}\)
\(\mathscr{u}(w,t)=\mathscr{f}(w)e^{-c^2w^2t}\)
Inverse Transform
\(u(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\mathscr{f}(w)e^{-c^2w^2t}e^{iwx}\,dw\)
Solution by Convolution
\(u(x,t)=(f*g)(x)=\int_{-\infty}^{\infty}\mathscr{f}(w)\mathscr{g}(w)e^{iwx}\,dw\)
Where \(\mathscr{g}(w)=\frac{1}{\sqrt{2\pi}}e^{-c^2w^2t}\)
Find the inverse transform \(g\) of \(\mathscr{g}\)
From the definition of convolution we have: \(u(x,t)=(f*g)(x)=\int_{-\infty}^{\infty}f(p)g(x-p)\,dp\)
Steady Two-Dimensional Heat Equation (Laplace's Equation)
\(u_{xx}+u_{yy}=0\)
Given an area \(R\) bounded by \(x\,\epsilon\,(0,a),\,\,\,y\,\epsilon\,(0,b)\)
Boundary Value Problem
\(u(0,y)=0,\,\,\,\,u(a,y)=0\)
\(u(x,0)=0,\,\,\,\,u(x,b)=f(x)\)
Product Method
Then \(\frac{F''}{F}=-\frac{G''}{G}=-k\)
Choose \(u(x,y)=F(x)\cdot G(y)\)
\(F''+kF=0\)
\(G''-kG=0\)
Eigenfunctions
\(F(0)=F(a)=0\)
\(k=(\frac{n\pi}{a})^2\)
\(G_n(y)=A_ne^{\frac{n\pi}{a}y}+B_ne^{-\frac{n\pi}{a}y}\)
\(F_n(x)=\sin{\frac{n\pi}{a}x}\)
\(u(x,0)=0 \Rightarrow G(0)=0\)
\(B_n=-A_n\)
\(G_n(y)=A_n(e^{\frac{n\pi}{a}y}-e^{-\frac{n\pi}{a}y})=2A_n\sinh{\frac{n\pi}{a}y}\)
\(u_n(x,y)=A_n^*\sinh{\frac{n\pi}{a}y}\sin{\frac{n\pi}{a}x}\)
\(A_n^*=2A_n\)
Complete Solution
\(u(x,y)=\sum_{n=1}^{\infty}u_n(x,y)\)
\(u(x,b)=\sum_{n=1}^{\infty}A_n^*\sinh{\frac{n\pi}{a}b}\sin{\frac{n\pi}{a}x}=f(x)\)
\(A_n^*=\frac{2}{a\sinh{(n\pi b/a)}}\int_0^af(x)\sin{\frac{n\pi}{a}x}\,dx\)