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Advanced Calculus

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Applications of differential equations in real life

Second-order partial differential equations are used to model standing waves on string, the distribution of heat along a solid, or even the quantum behavior of subatomic particles. While they provide useful information about a physical system, partial differential equations are often tricky to solve, requiring the use of advanced techniques beyond those introduced in the IB math HL syllabus.
Wave Equation: 2u(x,t)t2=c22u(x,t)x2\frac{∂^2u(x,t)}{∂t^2}=c^2\frac{∂^2u(x,t)}{∂x^2}
Heat Equation: 2T(x,t)t2=c22T(x,t)x2\frac{∂^2T(x,t)}{∂t^2}=c^2\frac{∂^2T(x,t)}{∂x^2}

Partial Differential Equations (PDEs)

Solution by Separation of Variables

A common technique to solving PDEs is to assume that the multivariate function can be expressed as the product of two single variate functions:
u(x,t)=F(x)G(t)u(x,t)=F(x)G(t)
For example, when solving the Schrodinger equation to find a particle’s wave function Ψ(x,t)Ψ(x,t), which has bivariate input (position and time), the wave function is assumed to be the product of two single variate functions, ψ(x)ψ(x) and φ(t)φ(t), of position and time respectively.

Statistical Interpretation of the Wave Function

abΨ(x,t)2 dx\int_a^b |Ψ(x,t)|^2~dx = probability of finding the particle between aa and b, at time tt.
Figure 1.2: A typical wave function. The shaded area represents the probability of finding the particle between a and b. The particle would be relatively likely to be found near A, and unlikely to be found near B.
iΨ(x,t)t=22m2Ψ(x,t)x2+VΨ(x,t)i\hbar \frac{\partial \Psi(x,t)}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \Psi(x,t)}{\partial x^2} + V\Psi(x,t)
Proceed with solution by setting Ψ(x,t)=ψ(x)φ(t)Ψ(x,t) = ψ(x)φ(t)

Solving Differential Equations Demo

1.
Real life scenario
2.
Mathematical formulation
3.
Find general solution
Ψ(x)=Asinkx+BcoskxΨ(x)=A\sin kx+B\cos kx
4.
Apply boundary conditions
V(x)={0,0xa,, otherwise V(x) = \begin{cases} 0 ,&0\leq x\leq a, \\ \infty,& ~\text{otherwise } \end{cases}
Continuity of Ψ(x)Ψ(x) requires that
Ψ(0)=Ψ(a)=0Ψ(0)=Ψ(a)=0
Ψ(0)=Asin0+Bcos0=B,Ψ(0)=A\sin 0+B\cos 0=B,
So B=0B=0, and hence
Ψ(x)=Asinkx.Ψ(x)=A\sin kx.
5.
Find and interpret specific solution
Ψn(x)=2asin(nπax)Ψ_n(x)=\sqrt{\frac2a}\sin (\frac{n\pi}ax)

Fourier Analysis

A periodic function with period pp, f(x)=f(x+p)f(x) = f(x + p), can be expressed as a discrete sum of trigonometric or exponential functions with specific frequencies. Typically, these frequencies tend to be harmonic, which are integer multiples of the original frequency.
f(x)=a0+n=1[ancos(2πnxL)+bnsin(2πnxL)]f(x)=a_0+\sum_{n=1}^\infty[a_n\cos (\frac{2\pi nx}{L})+b_n\sin (\frac{2\pi nx}L)]
The coefficients are found by calculating the inner product of two functions, the original periodic function f(x)f(x) and the individual terms of the trigonometric or exponential function series. Fourier transformation decomposes the periodic signal f(x)f(x) into its constituent frequencies, each of stands as a dimension of an infinite component vector.
Inner Product <f(x),g(x)>=f(x)g(x) dx<f(x),g(x)>=\int f(x)g(x)~dx
a0=1L0Lf(x) dxa_0=\frac 1L\int _0^Lf(x)~dx
an=2L0Lf(x)cos2πnxL dxa_n=\frac2L \int_0^Lf(x)\cos \frac{2\pi nx}L~dx
bn=2L0Lf(x)sin2πnxL dxb_n=\frac2L\int_0^Lf(x)\sin \frac{2\pi nx}L~dx
In the context of inner products, suppose f(x)f(x) were to be projected onto infinitely many cos2πnxL\cos \frac{2πnx}L axes; ana_n would represent the weighted significance of each cos2πnxL\cos \frac{2\pi nx}Lcomponent comprising f(x)f(x).
an=<f(x),cos2πnxL>a_n=<f(x),\cos\frac{2\pi nx}L>
Colloquially speaking, how much cos2πnxL\cos \frac{2\pi nx}L does f(x)f(x) contain?
Investigation
Identify a periodic signal in real life (ECG diagram of heartbeat, acoustic signal of sound recording, oscilloscope readings of electric circuit, etc.) Extensive use of trigonometric identities, integration techniques, and Euler’s identity (converting trigonometric expressions into exponential expressions) would be involved.
ππsin(mx)sin(nx) dx=πδmn\int_{-\pi}^{\pi}\sin (mx)\sin(nx)~dx=\pi\delta_{mn}
ππcos(mx)cos(nx) dx=πδmn\int_{-\pi}^{\pi} \cos(mx)\cos(nx)~dx=\pi\delta_{mn}
ππsin(mx)cos(nx) dx=0\int_{-\pi}^{\pi}\sin(mx)\cos(nx)~dx=0
ππsin(mx) dx=0\int_{-\pi}^{\pi}\sin(mx)~dx=0
ππcos(mx) dx=0\int_{-\pi}^{\pi}\cos(mx)~dx=0