Unit 1: Partial Differential Equations
Formation · Variable Separable · Standard Types · Lagrange's Equation · 2nd Order Homogeneous PDEs
\( p = \dfrac{\partial z}{\partial x},\quad q = \dfrac{\partial z}{\partial y},\quad r = \dfrac{\partial^2 z}{\partial x^2},\quad s = \dfrac{\partial^2 z}{\partial x \partial y},\quad t = \dfrac{\partial^2 z}{\partial y^2} \)
For higher order: \( D = \dfrac{\partial}{\partial x},\quad D' = \dfrac{\partial}{\partial y} \)
Given \(z = f(x,y,a,b)\): differentiate wrt \(x\) to get \(p\), wrt \(y\) to get \(q\). Eliminate \(a\) and \(b\).
Example: \(z = ax^2 + by^2\)
\(p = 2ax \Rightarrow a = \dfrac{p}{2x}\), \(q = 2by \Rightarrow b = \dfrac{q}{2y}\)
Substitute: \(z = \dfrac{p}{2x}\cdot x^2 + \dfrac{q}{2y}\cdot y^2 = \dfrac{px}{2} + \dfrac{qy}{2}\)
Given \(z = f(\phi(x,y))\): differentiate wrt \(x\) and \(y\), divide to eliminate \(f'\).
Example: \(z = f(x^2 + y^2)\)
\(p = f' \cdot 2x\), \(q = f' \cdot 2y\)
Divide: \(\dfrac{p}{q} = \dfrac{x}{y} \Rightarrow\) PDE: \(\boxed{py - qx = 0}\)
Example: \(z = f(x - y)\)
\(p = f'(x-y)\), \(q = f'(x-y)\cdot(-1) = -f'\)
So \(p = -q\) → PDE: \(\boxed{p + q = 0}\)
Example: \(z = f(x + y)\)
\(p = f', q = f'\) → PDE: \(\boxed{p - q = 0}\)
Example: \(z = (x^2+a)(y^2+b)\)
\(p = 2x(y^2+b),\; q = 2y(x^2+a)\)
\(pq = 4xy(x^2+a)(y^2+b) = 4xyz\) → PDE: \(\boxed{pq = 4xyz}\)
Try \(z = ax + by + c\) where \(f(a,b) = 0\). Express \(b\) in terms of \(a\).
Example: \(p^2 + q^2 = 1\)
Let \(p = a\) → \(q = \sqrt{1-a^2}\)
Example: \(p^2 - q = 0\) → let \(q = a^2,\, p = a\)
Example: \(p = 2qx\) → let \(q = a\), then \(p = 2ax\). Integrate: \(dz = 2ax\,dx + a\,dy\)
Complete solution: Replace \(p \to a,\; q \to b\):
Singular solution: Differentiate complete solution wrt \(a\) and \(b\), set each to 0, eliminate \(a\) and \(b\).
Example: \(z = px + qy + p^2 - q^2\)
Complete: \(z = ax + by + a^2 - b^2\)
Singular: \(\dfrac{\partial}{\partial a}: x + 2a = 0 \Rightarrow a = -x/2\) ; \(\dfrac{\partial}{\partial b}: y - 2b = 0 \Rightarrow b = y/2\)
Substitute: \(z = -\dfrac{x^2}{2} + \dfrac{y^2}{2} + \dfrac{x^2}{4} - \dfrac{y^2}{4} = -\dfrac{x^2}{4} + \dfrac{y^2}{4}\)
Set both sides equal to an arbitrary constant \(k\): solve \(f(x,p) = k\) for \(p\), and \(g(y,q) = k\) for \(q\). Then \(dz = p\,dx + q\,dy\), integrate.
Solve these to get two independent solutions \(u_1(x,y,z) = c_1\) and \(u_2(x,y,z) = c_2\).
General solution: \(F(u_1,\, u_2) = 0\) or \(u_1 = \phi(u_2)\)
Multiplier trick: Find \(l, m, n\) such that \(lP + mQ + nR = 0\). Then \(l\,dx + m\,dy + n\,dz = 0\), which integrates directly.
Q: \((3z-4y)p + (4x-2z)q = 2y-3x\)
Multipliers \((2,3,4)\): \(2P+3Q+4R = 0\) → \(2\,dx+3\,dy+4\,dz=0\) → \(2x+3y+4z = c_1\)
Multipliers \((x,y,z)\): \(xP+yQ+zR = 0\) → \(x\,dx+y\,dy+z\,dz=0\) → \(x^2+y^2+z^2 = c_2\)
Q: \(xp - yq = y^2 - x^2\)
From \(dx/x = dy/(-y)\): \(x\,dy + y\,dx = 0\) → \(xy = c_1\)
Using \(xy = c_1\) → integrate to get \(z + \dfrac{x^2+y^2}{2} = c_2\)
Q: \(x(y-z)p + y(z-x)q = z(x-y)\)
Multipliers \((1/x, 1/y, 1/z)\): sum = 0 → \(dx/x+dy/y+dz/z=0\) → \(xyz = c_1\)
Multipliers \((1,1,1)\): sum = 0 → \(dx+dy+dz=0\) → \(x+y+z = c_2\)
Q: \((y+z)p + (z+x)q = x+y\)
Using \((dx-dy)\): denominator \(= (y+z)-(z+x) = y-x\) → \(\dfrac{d(x-y)}{x-y}\) const
Similarly \((dy-dz)\): denominator \(= z-y\) → \(\dfrac{x-y}{y-z} = c_1\)
Second: Multipliers \((1,1,1)\): sum \(= 2(x+y+z)\neq 0\). Use \((1,-1,0)\) vs \((0,1,-1)\) → \(\dfrac{y-z}{z-x} = c_2\)
Q: \(p\tan x + q\tan y = \tan z\)
Auxiliary: \(\dfrac{dx}{\tan x} = \dfrac{dy}{\tan y} = \dfrac{dz}{\tan z}\)
From first two: \(\cot x\,dx = \cot y\,dy\) → \(\ln(\sin x) = \ln(\sin y) + c\) → \(\dfrac{\sin x}{\sin y} = c_1\)
From last two: \(\dfrac{\sin y}{\sin z} = c_2\)
\((aD^2 + bDD' + cD'^2)z = F(x,y)\)
Characteristic equation: put \(D = m,\; D' = 1\) → \(am^2 + bm + c = 0\) → roots \(m_1, m_2\)
CF: \(\phi_1(y + m_1 x) + \phi_2(y + m_2 x)\) (for distinct roots)
If \(m_1 = m_2 = m\): CF \(= \phi_1(y+mx) + x\,\phi_2(y+mx)\)
For \(e^{ax+by}\): \(\text{PI} = \dfrac{e^{ax+by}}{f(a,b)}\) (if \(f(a,b) \neq 0\), else use special method)
For \(\sin(ax+by)\) or \(\cos(ax+by)\): Replace \(D^2 \to -a^2,\; DD' \to -ab,\; D'^2 \to -b^2\)
For \(x^m y^n\): Use binomial expansion of \([f(D,D')]^{-1}\)
Example: \((D^2 - 4DD' + 4D'^2)z = e^{2x+y}\)
CF: \(m^2-4m+4=0\) → \((m-2)^2=0\) → \(m=2,2\) → CF: \(\phi_1(y+2x)+x\phi_2(y+2x)\)
PI: \(f(2,1) = 4-8+4 = 0\) (repeated) → \(\text{PI} = \dfrac{x^2}{2}\cdot\dfrac{e^{2x+y}}{f_{DD}(2,1)}\)
where \(f_{DD}(D,D') = \dfrac{\partial^2 f}{\partial D^2} = 2\) → \(\text{PI} = \dfrac{x^2}{2}\cdot e^{2x+y}\)
Example: \((D^2 + DD')z = e^{2x-y}\)
PI: \(f(2,-1) = 4 + 2(-1) = 2\) → \(\text{PI} = \dfrac{e^{2x-y}}{2}\)
Example: \((D^2-7DD'-6D'^2)z = \sin(x+2y)\)
Replace \(D^2 \to -1,\; DD' \to -1\cdot2=-2,\; D'^2 \to -4\)
Denominator: \(-1-7(-2)-6(-4) = -1+14+24 = 37\)
\(\text{PI} = \dfrac{\sin(x+2y)}{37}\)
For \(x^2 y\) in the same equation → use PI = \(\dfrac{1}{f(D,D')} x^2y\) by binomial expansion.
Lagrange's equation: \(P=3z-4y,\; Q=4x-2z,\; R=2y-3x\)
Auxiliary equations: \(\dfrac{dx}{3z-4y} = \dfrac{dy}{4x-2z} = \dfrac{dz}{2y-3x}\)
First integral — Multipliers (2, 3, 4):
\(2P+3Q+4R = 2(3z-4y)+3(4x-2z)+4(2y-3x) = 6z-8y+12x-6z+8y-12x = 0\) ✓
\(\Rightarrow 2\,dx+3\,dy+4\,dz = 0\) → integrate: \(u_1 = 2x+3y+4z = c_1\)
Second integral — Multipliers (x, y, z):
\(xP+yQ+zR = x(3z-4y)+y(4x-2z)+z(2y-3x) = 0\) ✓
\(\Rightarrow x\,dx+y\,dy+z\,dz = 0\) → integrate: \(u_2 = x^2+y^2+z^2 = c_2\)
\(P=x,\; Q=-y,\; R=y^2-x^2\). Auxiliary: \(\dfrac{dx}{x} = \dfrac{dy}{-y} = \dfrac{dz}{y^2-x^2}\)
First integral: From \(\dfrac{dx}{x} = \dfrac{dy}{-y}\): \(-y\,dx = x\,dy\) → \(d(xy)=0\) → \(xy = c_1\)
Second integral: Substitute \(y=c_1/x\) into the third fraction:
\(dz = \left(\dfrac{c_1^2}{x^3}-x\right)dx\) → integrate: \(z = -\dfrac{c_1^2}{2x^2}-\dfrac{x^2}{2}+c_2\)
Since \(c_1^2=x^2y^2\): \(z = -\dfrac{y^2}{2}-\dfrac{x^2}{2}+c_2\) → \(c_2 = z+\dfrac{x^2+y^2}{2}\)
\(P=x(y-z),\; Q=y(z-x),\; R=z(x-y)\)
First integral — Multipliers \((1/x, 1/y, 1/z)\):
\(\frac{P}{x}+\frac{Q}{y}+\frac{R}{z} = (y-z)+(z-x)+(x-y) = 0\) ✓
\(\Rightarrow \dfrac{dx}{x}+\dfrac{dy}{y}+\dfrac{dz}{z}=0\) → \(d(\ln xyz)=0\) → \(xyz = c_1\)
Second integral — Multipliers \((1,1,1)\):
\(P+Q+R = x(y-z)+y(z-x)+z(x-y) = 0\) ✓
\(\Rightarrow dx+dy+dz=0\) → \(x+y+z = c_2\)
Clairaut's form: Complete solution (replace \(p\to a, q\to b\)): \(z = ax+by+a^2-b^2\) ...(1)
Singular solution: Differentiate (1) wrt \(a\) and \(b\), set = 0:
\(\partial/\partial a: x+2a=0 \Rightarrow a=-x/2\)
\(\partial/\partial b: y-2b=0 \Rightarrow b=y/2\)
Substitute: \(z = (-x/2)x+(y/2)y+x^2/4-y^2/4 = -x^2/4+y^2/4\)
Let \(u=2x+y,\; v=3x-y\). Then \(z=f(u)+g(v)\).
\(r = 4f''(u)+9g''(v),\quad s = 2f''(u)-3g''(v),\quad t = f''(u)+g''(v)\)
From \(r\) and \(t\): \(r-4t = 5g''(v)\) → \(g''=\frac{r-4t}{5}\). Also \(f'' = t-g'' = \frac{9t-r}{5}\)
\(s = 2\cdot\frac{9t-r}{5} - 3\cdot\frac{r-4t}{5} = \frac{18t-2r-3r+12t}{5} = \frac{30t-5r}{5} = 6t-r\)
\(P=y+z,\; Q=z+x,\; R=x+y\)
First integral: Subtract: \(\dfrac{d(x-y)}{(y+z)-(z+x)} = \dfrac{d(x-y)}{y-x} = \dfrac{-d(x-y)}{x-y}\)
Similarly \(\dfrac{-d(y-z)}{y-z}\). So: \(\dfrac{d(x-y)}{x-y} = \dfrac{d(y-z)}{y-z}\) → \(\ln|x-y|=\ln|y-z|+c\)
Second integral: Using \((dx-dz)\): \(\quad u_2 = \dfrac{y-z}{z-x} = c_2\)
Characteristic equation: \(m^2-7m+6=0\) → \((m-1)(m-6)=0\) → \(m_1=1,\; m_2=6\)
CF: \(\phi_1(y+x)+\phi_2(y+6x)\)
PI for \(\sin(x+2y)\): Replace \(D^2\to-1,\; DD'\to-2,\; D'^2\to-4\):
Denominator: \(-1-7(-2)+6(-4) = -1+14-24 = -11\)
\(\text{PI}_1 = \dfrac{-\sin(x+2y)}{11}\)
PI for \(x^2y\): \(\text{PI}_2 = \dfrac{1}{D^2(1-7D'/D+6D'^2/D^2)}x^2y\)
\(= \dfrac{1}{D^2}\left[1+\frac{7D'}{D}+\cdots\right]x^2y = \dfrac{1}{D^2}\left[x^2y+\frac{7}{D}(x^2)\right]\)
\(= \dfrac{1}{D^2}\left[x^2y+\frac{7x^3}{3}\right] = \dfrac{x^4y}{12}+\dfrac{7x^5}{60}\)
Unit 2: Fourier Series
Dirichlet's Conditions · General Fourier Series · Half-Range Series · Parseval's Identity · Harmonic Analysis
For \((-\pi,\pi)\): use \(l=\pi\) → \(a_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos nx\,dx\), \(b_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin nx\,dx\)
Cosine series (extend as even function):
Sine series (extend as odd function):
Use this to evaluate series sums like \(\sum \frac{1}{n^2}\), \(\sum \frac{1}{n^4}\), etc.
Example: RMS of \(f(x)=x\) in \((0,l)\):
\(\text{rms} = \sqrt{\dfrac{1}{l}\int_0^l x^2\,dx} = \sqrt{\dfrac{l^2}{3}} = \dfrac{l}{\sqrt{3}}\)
- \(f(x)\) must be single-valued, finite, and periodic.
- \(f(x)\) has a finite number of discontinuities in any one period.
- \(f(x)\) has a finite number of maxima and minima in any one period.
- \(\int_{-\pi}^{\pi}|f(x)|\,dx\) must be finite.
Note: \(\tan x\) cannot be expanded — it has an infinite discontinuity at \(x=\pi/2\). Dirichlet's conditions are violated.
Even function \(f(-x)=f(x)\): only cosine terms → \(b_n=0\), \(a_0, a_n\) only
Odd function \(f(-x)=-f(x)\): only sine terms → \(a_0=0, a_n=0\), \(b_n\) only
Tip: Over symmetric interval, \(\int_{-\pi}^{\pi}[\text{odd}]\,dx = 0\)
e.g. \(f(x)=x\) in \((-\pi,\pi)\): odd → \(a_0=0, a_n=0\), only \(b_n = \dfrac{2(-1)^{n+1}}{n}\)
e.g. \(f(x)=x^2\) in \((-\pi,\pi)\): even → \(b_n=0\), \(a_0=\dfrac{2\pi^2}{3}\), \(a_n=\dfrac{4(-1)^n}{n^2}\)
When data is given as a table of \((x, f(x))\) with \(n\) equally spaced points over \([0, 2\pi)\):
Then \(f(x) \approx \frac{a_0}{2} + a_1\cos x + b_1\sin x + a_2\cos 2x + b_2\sin 2x + \cdots\)
Standard table for 6 points (x = 0, 60°, 120°, 180°, 240°, 300°):
| \(x\) | 0 | \(\pi/3\) | \(2\pi/3\) | \(\pi\) | \(4\pi/3\) | \(5\pi/3\) |
|---|---|---|---|---|---|---|
| \(\cos x\) | 1 | 0.5 | -0.5 | -1 | -0.5 | 0.5 |
| \(\sin x\) | 0 | 0.866 | 0.866 | 0 | -0.866 | -0.866 |
| \(\cos 2x\) | 1 | -0.5 | -0.5 | 1 | -0.5 | -0.5 |
| \(\sin 2x\) | 0 | 0.866 | -0.866 | 0 | 0.866 | -0.866 |
From \(f(x)=x^2\) in \((-\pi,\pi)\): \(a_n = 4(-1)^n/n^2\), Parseval → \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^4} = \frac{\pi^4}{90}\)
From \(f(x)=x^2\) via deduce at \(x=\pi\): \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}\)
From \(f(x)=x(2\pi-x)\) in \((0,2\pi)\): at \(x=\pi\): \(\displaystyle 1+\frac{1}{9}+\frac{1}{25}+\cdots = \frac{\pi^2}{8}\)
From \(f(x)=(\pi-x)^2\) in \((0,2\pi)\): deduces \(\displaystyle\sum\frac{1}{n^2}=\frac{\pi^2}{6}\)
Half-range sine of \(f(x)=1-x\) in \((0,1)\): \(\displaystyle 1 - \frac{1}{3} + \frac{1}{5} - \cdots = \frac{\pi}{4}\)
\(a_0 = \frac{1}{\pi}\int_{-\pi}^{\pi}x^2\,dx = \frac{2}{\pi}\cdot\frac{\pi^3}{3} = \mathbf{\frac{2\pi^2}{3}}\)
\(a_1 = \frac{1}{\pi}\int_{-\pi}^{\pi}x^2\cos x\,dx = \frac{2}{\pi}\int_0^{\pi}x^2\cos x\,dx = \frac{2}{\pi}[x^2\sin x+2x\cos x-2\sin x]_0^{\pi} = \frac{2}{\pi}(0-2\pi-0) = \mathbf{-4}\)
\(f(x)=2\pi x-x^2\) on \((0,2\pi)\). Period \(2\pi\).
\(a_0 = \frac{1}{\pi}\int_0^{2\pi}(2\pi x-x^2)\,dx = \frac{1}{\pi}\left[\pi x^2-\frac{x^3}{3}\right]_0^{2\pi} = \frac{1}{\pi}\left[4\pi^3-\frac{8\pi^3}{3}\right] = \frac{4\pi^2}{3}\)
\(a_n = \frac{1}{\pi}\int_0^{2\pi}(2\pi x-x^2)\cos nx\,dx\). IBP twice: \(a_n = \dfrac{-4}{n^2}\)
\(b_n = 0\) (boundary terms vanish after IBP)
Deduce \(\sum 1/n^2\): At \(x=0\), series converges to \(\frac{f(0^+)+f(2\pi^-)}{2} = 0\)
\(0 = \frac{2\pi^2}{3}-4\sum\frac{1}{n^2}\) → \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}\)
Method for 6 equally-spaced points \(x_i\) over \([0,2\pi)\):
Points at \(x=0°,60°,120°,180°,240°,300°\) (i.e., \(0,\pi/3,2\pi/3,\pi,4\pi/3,5\pi/3\)).
ES Nov'23 data \((y_i = 1.4,1.9,1.7,1.5,1.2,1.4)\):
\(a_0=\frac{2}{6}(9.1)=3.033\). Use \(\cos kx_i\) values \((1,0.5,-0.5,-1,-0.5,0.5)\) for \(k=1\):
\(a_1=\frac{2}{6}(0.1)\approx 0.033\). \(b_1=\frac{2}{6}(0.866\times1.0)\approx0.289\)
Final: \(f(x)\approx\frac{a_0}{2}+a_1\cos x+b_1\sin x+a_2\cos 2x+b_2\sin 2x\)
\(a_0=\frac{1}{\pi}\int_0^{2\pi}(\pi-x)^2\,dx = \frac{2\pi^2}{3}\)
\(a_n = \frac{4}{n^2}\) (by substitution \(t=\pi-x\) and IBP). \(b_n=0\)
Deduce: At \(x=0\): \(\pi^2 = \frac{\pi^2}{3}+4\sum\frac{1}{n^2}\) → \(\displaystyle\sum\frac{1}{n^2}=\frac{\pi^2}{6}\)
\(f(x)=x^2\) is even → \(b_n=0\). \(a_0=\frac{2}{l}\int_0^l x^2\,dx = \frac{2l^2}{3}\)
\(a_n=\frac{2}{l}\int_0^l x^2\cos\frac{n\pi x}{l}\,dx = \frac{4l^2(-1)^n}{n^2\pi^2}\) (IBP twice)
At \(x=l\): \(l^2=\frac{l^2}{3}+\frac{4l^2}{\pi^2}\sum\frac{1}{n^2}\) → \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}\) ✓
\(b_n=2\int_0^1(1-x)\sin(n\pi x)\,dx\). IBP:
\(b_n = 2\left[-(1-x)\frac{\cos n\pi x}{n\pi}\bigg|_0^1 + \frac{1}{n\pi}\int_0^1\cos n\pi x\,dx\right] = 2\cdot\frac{1}{n\pi} = \frac{2}{n\pi}\)
Deduce: At \(x=\frac{1}{2}\): \(\frac{1}{2}=\frac{2}{\pi}\left[1-\frac{1}{3}+\frac{1}{5}-\cdots\right]\) → \(1-\frac{1}{3}+\frac{1}{5}-\cdots=\frac{\pi}{4}\)
Period \(2\pi\). Check \(a_0\): \(\int_{-\pi}^0(-10\pi-10t)dt = -5\pi^2\); \(\int_0^{\pi}10t\,dt=5\pi^2\). So \(a_0=0\).
\(a_n = \frac{1}{\pi}\left[\int_{-\pi}^0(-10\pi-10t)\cos nt\,dt + \int_0^{\pi}10t\cos nt\,dt\right]\)
Using IBP: \(a_n = \frac{20}{n^2\pi}[(-1)^n-1]\) → \(0\) for even \(n\), \(-\frac{40}{n^2\pi}\) for odd \(n\)
For \(b_n\): by odd symmetry analysis, \(b_n=0\).
Unit 3: Boundary Value Problems — 1D Equations
Classification of PDEs · 1D Wave Equation · 1D Heat Equation · Fourier Series Solutions
Discriminant: \(\Delta = B^2 - 4AC\)
- \(\Delta < 0\) → Elliptic (e.g., Laplace's equation \(u_{xx}+u_{yy}=0\))
- \(\Delta = 0\) → Parabolic (e.g., Heat equation \(u_t = a^2 u_{xx}\))
- \(\Delta > 0\) → Hyperbolic (e.g., Wave equation \(u_{tt} = a^2 u_{xx}\))
Examples from QB:
- \(3u_{xx}+4u_{xy}+6u_{yy}-20u_x+u_y=0\): \(A=3,B=4,C=6\). \(\Delta=16-72=-56<0\) → Elliptic
- \(y^2u_{xx}-2xyu_{xy}+x^2u_{yy}+2u_x-3u=0\): \(\Delta=4x^2y^2-4x^2y^2=0\) → Parabolic
- \(y^2u_{xx}-u_{yy}+u_y+u_x+7=0\): \(A=y^2,B=0,C=-1\). \(\Delta=4y^2>0\) for \(y\neq0\) → Hyperbolic (\(y\neq0\)), Parabolic (\(y=0\))
- \(x^2u_{xy}+yu_{yy}=0\): \(A=0,B=x^2,C=y\). \(\Delta=x^4>0\) → Hyperbolic
- \(u_{tt}-9u_{yy}=0\): \(A=1,B=0,C=-9\). \(\Delta=36>0\) → Hyperbolic
- \(\partial^2z/\partial x\partial y + \partial u/\partial x=0\): \(A=0,B=1,C=0\). \(\Delta=1>0\) → Hyperbolic
BCs: \(y(0,t)=0,\; y(l,t)=0\)
ICs: \(y(x,0)=f(x),\; \left.\frac{\partial y}{\partial t}\right|_{t=0}=g(x)\)
If released from rest \((g(x)=0)\): \(B_n=0\).
If started with initial velocity only \((f(x)=0)\): \(A_n=0\).
Use: \(\sin^3\theta = \dfrac{3\sin\theta - \sin3\theta}{4}\)
\(y(x,0) = \dfrac{3y_0}{4}\sin\dfrac{\pi x}{l} - \dfrac{y_0}{4}\sin\dfrac{3\pi x}{l}\)
Matching Fourier sine series (with \(B_n=0\)):
\(A_1 = \dfrac{3y_0}{4},\quad A_3 = -\dfrac{y_0}{4},\quad A_n=0 \text{ otherwise}\)
\(A_n = \dfrac{2}{l}\int_0^l k(lx-x^2)\sin\dfrac{n\pi x}{l}\,dx = \dfrac{8kl^2}{n^3\pi^3}\) for odd \(n\), \(0\) for even \(n\)
BCs: \(y(0,t)=0,\; y(l,t)=0\)
ICs: \(y(x,0)=0,\; y_t(x,0)=V_0\sin(\pi x/l)\)
Since \(f(x)=0\): \(A_n=0\). Only \(B_n\) survives.
\(B_n = \dfrac{2}{n\pi a}\int_0^l V_0\sin(\pi x/l)\sin(n\pi x/l)\,dx\)
By orthogonality, only \(n=1\) survives: \(B_1 = \dfrac{V_0 l}{a\pi}\)
Three mathematically possible solutions:
- \(u_1 = (Ae^{kx}+Be^{-kx})\,Ce^{k^2a^2t}\)
- \(u_2 = (A\cos kx+B\sin kx)\,e^{-k^2a^2t}\) ← physically valid (decays)
- \(u_3 = (Ax+B)\) (steady state)
For finite temperature as \(t\to\infty\): use \(u_2\) and \(u_3\).
BC: \(u(0,t)=0,\; u(l,t)=0\) IC: \(u(x,0)=f(x)\)
BC: \(u(0,t)=T_1,\; u(l,t)=T_2\) IC: \(u(x,0)=f(x)\)
Steady state: \(u_s(x) = T_1 + \dfrac{(T_2-T_1)x}{l}\)
Let \(v(x,t) = u(x,t)-u_s(x)\). Then \(v\) satisfies heat eq. with 0°C BCs.
\(C_n = \dfrac{2}{l}\int_0^l[f(x)-u_s(x)]\sin\dfrac{n\pi x}{l}\,dx\)
Rod 10cm, A at 20°C, B at 70°C, steady state: \(u_s(x) = 20 + 5x\)
- \(u = (Ae^{kx}+Be^{-kx})\,e^{k^2a^2t}\)
- \(u = (A\cos kx+B\sin kx)\,e^{-k^2a^2t}\)
- \(u = Ax+B\) (steady state)
Wave eq: \(y_{tt}=a^2 y_{xx}\). BC: \(y(0,t)=y(l,t)=0\). IC: \(y_t(x,0)=0\) → \(B_n=0\).
Solution: \(y=\sum A_n\cos(n\pi at/l)\sin(n\pi x/l)\)
Expand IC using identity \(\sin^3\theta=\frac{3\sin\theta-\sin3\theta}{4}\):
\(y(x,0) = \frac{3y_0}{4}\sin\frac{\pi x}{l} - \frac{y_0}{4}\sin\frac{3\pi x}{l}\)
By comparison: \(A_1=\frac{3y_0}{4},\; A_3=-\frac{y_0}{4},\; A_n=0\) for all other \(n\).
BC: \(y(0,t)=y(l,t)=0\). IC: \(y(x,0)=k(lx-x^2),\; y_t(x,0)=0\) → \(B_n=0\).
\(A_n = \frac{2}{l}\int_0^l k(lx-x^2)\sin\frac{n\pi x}{l}\,dx\). IBP twice:
\(\int_0^l(lx-x^2)\sin\frac{n\pi x}{l}\,dx = \frac{2l^3[1-(-1)^n]}{n^3\pi^3}\)
For even \(n\): \(A_n=0\). For odd \(n\): \(A_n = \frac{2k}{l}\cdot\frac{4l^3}{n^3\pi^3} = \frac{8kl^2}{n^3\pi^3}\)
BC: \(y(0,t)=y(l,t)=0\). IC: \(y(x,0)=0,\; y_t(x,0)=V_0\sin(\pi x/l)\).
General solution: \(y=\sum[A_n\cos(n\pi at/l)+B_n\sin(n\pi at/l)]\sin(n\pi x/l)\)
Since \(y(x,0)=0\): all \(A_n=0\).
\(y_t(x,0)=\sum B_n\frac{n\pi a}{l}\sin\frac{n\pi x}{l}=V_0\sin\frac{\pi x}{l}\)
By orthogonality: only \(n=1\) non-zero. \(B_1\cdot\frac{\pi a}{l}=V_0\) → \(B_1=\frac{V_0 l}{\pi a}\)
Unit 4: Boundary Value Problems — 2D Equations
Steady State 2D Heat Equation · Laplace's Equation · Fourier Series Solutions
Three possible solutions (by separation of variables):
- \(u = (Ae^{kx}+Be^{-kx})(C\cos ky+D\sin ky)\)
- \(u = (A\cos kx+B\sin kx)(Ce^{ky}+De^{-ky})\)
- \(u = (Ax+B)(Cy+D)\)
Choose the form based on which boundary condition is non-zero.
Rule: The solution must decay in the direction of the infinite (or bounded) dimension and satisfy zero BCs on the other sides.
Problem type 1: Semi-infinite plate (width \(a\), infinite in \(y\)).
Zero BCs: \(u(0,y)=0,\; u(a,y)=0,\; u\to0\) as \(y\to\infty\). Non-zero: \(u(x,0)=f(x)\).
Problem type 2: Finite plate \((0,a)\times(0,b)\). Non-zero BC along \(y=b\): \(u(x,b)=f(x)\).
Zero BCs: \(u(0,y)=0,\; u(a,y)=0,\; u(x,0)=0\).
Plate: \(0\le x\le l,\; 0\le y\le l\). BCs: \(u(0,y)=0,\; u(l,y)=0,\; u(x,0)=0,\; u(x,l)=x(l-x)\).
Solution form: \(u = \sum B_n \sinh(n\pi y/l)\sin(n\pi x/l)\)
\(B_n\sinh(n\pi) = \dfrac{2}{l}\int_0^l x(l-x)\sin\dfrac{n\pi x}{l}\,dx\)
Evaluate: \(\int_0^l x(l-x)\sin\dfrac{n\pi x}{l}\,dx = \dfrac{2l^3}{n^3\pi^3}[1-(-1)^n]\)
For even \(n\): 0. For odd \(n\): \(\dfrac{4l^3}{n^3\pi^3}\).
\(B_n = \dfrac{8l^2}{n^3\pi^3\sinh(n\pi)}\) for odd \(n\).
Plate: \(0\le y\le 20\), \(x\ge0\). BCs: \(u(x,0)=0,\; u(x,20)=0,\; u\to0\) as \(x\to\infty\).
Non-zero: \(u(0,y)=f(y)=\begin{cases}10y & 0<y<10\\10(20-y)&10<y\le20\end{cases}\)
\(B_n = \dfrac{2}{20}\int_0^{20}f(y)\sin\dfrac{n\pi y}{20}\,dy = \dfrac{1}{10}\left[\int_0^{10}10y\sin\dfrac{n\pi y}{20}\,dy + \int_{10}^{20}10(20-y)\sin\dfrac{n\pi y}{20}\,dy\right]\)
Result: \(B_n = \dfrac{400}{n^2\pi^2}\cdot 2\sin(n\pi/2)\) → non-zero only for odd \(n\).
Specifically: \(B_n = \dfrac{800}{n^2\pi^2}\sin\dfrac{n\pi}{2}\)
Plate: \(0\le x\le a,\; 0\le y\le b\). BCs: \(u(0,y)=0,\; u(a,y)=0,\; u(x,b)=0,\; u(x,0)=x(a-x)\).
Non-zero BC at \(y=0\) → use sinh form decaying away from \(y=0\):
\(B_n\sinh(n\pi b/a) = \dfrac{2}{a}\int_0^a x(a-x)\sin\dfrac{n\pi x}{a}\,dx = \dfrac{4a^2}{n^3\pi^3}[1-(-1)^n]\)
- \(u = (Ae^{kx}+Be^{-kx})(C\cos ky+D\sin ky)\)
- \(u = (A\cos kx+B\sin kx)(Ce^{ky}+De^{-ky})\)
BVP: \(u_{xx}+u_{yy}=0\), \(0\le y\le20\), \(x\ge0\).
BC: \(u(x,0)=0,\; u(x,20)=0,\; u\to0\) as \(x\to\infty,\; u(0,y)=f(y)\).
Solution form: \(u(x,y)=\sum B_n e^{-n\pi x/20}\sin\frac{n\pi y}{20}\)
\(B_n = \frac{1}{10}\left[\int_0^{10}10y\sin\frac{n\pi y}{20}\,dy+\int_{10}^{20}10(20-y)\sin\frac{n\pi y}{20}\,dy\right]\)
After IBP: \(B_n = \dfrac{800}{n^2\pi^2}\sin\dfrac{n\pi}{2}\) (non-zero for odd \(n\) only)
Non-zero BC at \(y=l\). Solution: \(u=\sum B_n\sinh(n\pi y/l)\sin(n\pi x/l)\)
At \(y=l\): \(\sum B_n\sinh(n\pi)\sin(n\pi x/l) = x(l-x)\)
\(B_n\sinh(n\pi) = \frac{2}{l}\int_0^l x(l-x)\sin\frac{n\pi x}{l}\,dx = \frac{4l^2[1-(-1)^n]}{n^3\pi^3}\)
For odd \(n\): \(B_n = \dfrac{8l^2}{n^3\pi^3\sinh(n\pi)}\). Even \(n\): \(B_n=0\).
Non-zero BC at \(y=0\) → use form decaying from \(y=0\):
\(u = \sum B_n\sinh\frac{n\pi(b-y)}{a}\sin\frac{n\pi x}{a}\)
At \(y=0\): \(\sum B_n\sinh(n\pi b/a)\sin(n\pi x/a) = x(a-x)\)
\(B_n\sinh(n\pi b/a) = \frac{2}{a}\int_0^a x(a-x)\sin\frac{n\pi x}{a}\,dx = \frac{4a^2[1-(-1)^n]}{n^3\pi^3}\)
For odd \(n\): \(B_n = \dfrac{8a^2}{n^3\pi^3\sinh(n\pi b/a)}\)
Unit 5: Fourier Transforms
Fourier Integral Theorem · Infinite Fourier Transform · Sine & Cosine Transforms · Parseval's Identity · Convolution
Note: Some books use \(e^{-isx}\) in the transform and \(e^{isx}\) in the inverse. Be consistent with what your exam uses.
Inverse: same formula (self-reciprocal).
Key Results:
- \(F_c\{e^{-ax}\} = \sqrt{\dfrac{2}{\pi}}\cdot\dfrac{a}{a^2+s^2}\)
- \(F_s\{e^{-ax}\} = \sqrt{\dfrac{2}{\pi}}\cdot\dfrac{s}{a^2+s^2}\)
- \(F\{e^{-a|x|}\} = \dfrac{2a}{a^2+s^2}\) (using two-sided)
- Linearity: \(F\{af+bg\} = aF(s)+bG(s)\)
- Shifting (time): \(F\{f(x-a)\} = e^{-ias}F(s)\) ES Apr'25
- Scaling: \(F\{f(ax)\} = \dfrac{1}{|a|}F(s/a)\)
- Modulation: \(F\{e^{iax}f(x)\} = F(s+a)\) ES Nov'23
- Derivatives: \(F\{f^{(n)}(x)\} = (-is)^n F(s)\)
For cosine/sine transforms: \(\displaystyle\int_0^{\infty}|F_c(s)|^2\,ds = \int_0^{\infty}|f(x)|^2\,dx\)
Used to evaluate: \(\displaystyle\int_0^{\infty}\frac{ds}{(s^2+a^2)(s^2+b^2)}\) using \(F_c\{e^{-ax}\}\cdot F_c\{e^{-bx}\}\)
\(F(s) = \int_{-a}^{a}(1-x^2)e^{isx}\,dx = \int_{-a}^{a}(1-x^2)\cos(sx)\,dx\) (sine part vanishes — even function)
Using IBP twice: \(F(s) = \dfrac{4}{s^3}(\sin(as)-as\cos(as))\)
Deduce by putting \(a=1, s=1/2\): evaluates \(\displaystyle\int_0^{\infty}\dfrac{x\cos x-\sin x}{x^3}\cos\dfrac{x}{2}\,dx\)
\(F(s) = \int_{-1}^{1}e^{isx}\,dx = \dfrac{2\sin s}{s}\)
By Fourier integral theorem at \(x=0\): \(f(0)=1=\dfrac{1}{2\pi}\int_{-\infty}^{\infty}\dfrac{2\sin s}{s}\,ds\)
\(\Rightarrow \displaystyle\int_0^{\infty}\dfrac{\sin s}{s}\,ds = \dfrac{\pi}{2}\)
\(f(x)\) is even → \(F(s)=2\int_0^a(1-x^2)\cos(sx)\,dx\)
IBP: \(\int_0^a(1-x^2)\cos(sx)\,dx = \frac{2(\sin(as)-as\cos(as))}{s^3}\)
Deduction: By Fourier integral theorem (inversion), with \(a=1\) and \(x=1/2\):
\(f(1/2)=1-(1/2)^2=3/4 = \frac{1}{2\pi}\int_{-\infty}^{\infty}F(s)e^{-is/2}\,ds\)
Taking real part: \(\displaystyle\int_0^{\infty}\frac{\sin s-s\cos s}{s^3}\cos\frac{s}{2}\,ds = \frac{3\pi}{16}\)
\(F(s) = \int_{-1}^{1}e^{isx}\,dx = \left[\frac{e^{isx}}{is}\right]_{-1}^{1} = \frac{2\sin s}{s}\)
By Fourier integral theorem at \(x=0\): \(f(0)=1=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2\sin s}{s}\,ds = \frac{2}{\pi}\int_0^{\infty}\frac{\sin s}{s}\,ds\)
\(F_c\{e^{-ax}\} = \sqrt{\dfrac{2}{\pi}}\cdot\dfrac{a}{a^2+s^2}\), \(F_c\{e^{-bx}\} = \sqrt{\dfrac{2}{\pi}}\cdot\dfrac{b}{b^2+s^2}\)
Parseval's for cosine transforms: \(\int_0^{\infty}F_c(f)\cdot F_c(g)\,ds = \int_0^{\infty}f(x)g(x)\,dx\)
LHS: \(\int_0^{\infty}\frac{2ab}{\pi(a^2+s^2)(b^2+s^2)}\,ds\)
RHS: \(\int_0^{\infty}e^{-(a+b)x}\,dx = \dfrac{1}{a+b}\)
So: \(\dfrac{2ab}{\pi}\int_0^{\infty}\frac{ds}{(a^2+s^2)(b^2+s^2)} = \dfrac{1}{a+b}\)
Unit 6: Z-Transform
Z-Transform Properties · Inverse Z-Transform · Difference Equations
| \(f(n)\) | \(\mathcal{Z}\{f(n)\} = F(z)\) | ROC |
|---|---|---|
| \(1\) (unit step) | \(\dfrac{z}{z-1}\) | \(|z|>1\) |
| \(a^n\) | \(\dfrac{z}{z-a}\) | \(|z|>|a|\) |
| \(n\) | \(\dfrac{z}{(z-1)^2}\) | \(|z|>1\) |
| \(n\,a^n\) | \(\dfrac{az}{(z-a)^2}\) | \(|z|>|a|\) |
| \(e^{an}\) | \(\dfrac{z}{z-e^a}\) | \(|z|>e^a\) |
| \(\cos(n\theta)\) | \(\dfrac{z^2-z\cos\theta}{z^2-2z\cos\theta+1}\) | \(|z|>1\) |
| \(\sin(n\theta)\) | \(\dfrac{z\sin\theta}{z^2-2z\cos\theta+1}\) | \(|z|>1\) |
Note: \(\mathcal{Z}\{a^n\} = \dfrac{z}{z-a}\). For \(\mathcal{Z}\{3^n\}=\dfrac{z}{z-3}\), for \(\mathcal{Z}\{4^n\}=\dfrac{z}{z-4}\).
Convolution of sequences: \(\mathcal{Z}\{(f*g)(n)\} = F(z)\cdot G(z)\)
So \(\mathcal{Z}\{3^n * 4^n\} = \dfrac{z}{z-3}\cdot\dfrac{z}{z-4} = \dfrac{z^2}{(z-3)(z-4)}\)
General: \(\mathcal{Z}\{f(n+k)\} = z^k F(z) - z^k f(0) - z^{k-1}f(1) - \cdots - z\,f(k-1)\)
Step 1: Find \(F(z)/z\) and express in partial fractions.
Step 2: Multiply back by \(z\) to get \(F(z)\).
Step 3: Use standard pairs to find \(f(n)\).
Example: \(\mathcal{Z}^{-1}\!\left\{\dfrac{z}{(z-1)(z-2)}\right\}\)
\(\dfrac{F(z)}{z} = \dfrac{1}{(z-1)(z-2)} = \dfrac{-1}{z-1}+\dfrac{1}{z-2}\)
\(F(z) = \dfrac{-z}{z-1}+\dfrac{z}{z-2}\) → \(f(n) = -1^n + 2^n = 2^n - 1\)
\(\mathcal{Z}^{-1}\{F(z)\cdot G(z)\} = \sum_{k=0}^{n}f(k)\cdot g(n-k)\)
Example: \(\mathcal{Z}^{-1}\!\left\{\dfrac{z^2}{(z-a)(z-b)}\right\}\) by convolution:
\(F(z)=\dfrac{z}{z-a}\leftrightarrow a^n\), \(G(z)=\dfrac{z}{z-b}\leftrightarrow b^n\)
Result: \(\displaystyle\sum_{k=0}^{n}a^k b^{n-k} = b^n\sum_{k=0}^{n}(a/b)^k = \dfrac{a^{n+1}-b^{n+1}}{a-b}\) for \(a\neq b\)
1. Take \(\mathcal{Z}\) of both sides using shifting property.
2. Substitute initial conditions.
3. Solve for \(Y(z) = \mathcal{Z}\{y_n\}\).
4. Find \(y_n = \mathcal{Z}^{-1}\{Y(z)\}\) by partial fractions or convolution.
Z-transform: \((z^2Y-z^2y_0-zy_1)+4(zY-zy_0)+3Y = \dfrac{z}{z-2}\)
\(Y(z^2+4z+3) = \dfrac{z}{z-2}+z^2+z+4z = \dfrac{z}{z-2}+z^2+5z\)
\((z^2+4z+3)=(z+1)(z+3)\)
\(Y = \dfrac{z}{(z-2)(z+1)(z+3)}+\dfrac{z^2+5z}{(z+1)(z+3)}\)
Use partial fractions for \(Y/z\), multiply by \(z\), apply inverse.
\(z Y(z)-zy(0)-2Y(z)=0\) → \(Y(z-2)=3z\) → \(Y=\dfrac{3z}{z-2}\)
\(y_n = \mathcal{Z}^{-1}\!\left\{\dfrac{3z}{z-2}\right\} = 3\cdot 2^n\)
\(\dfrac{F(z)}{z} = \dfrac{1}{(z-1)(z-2)} = \dfrac{-1}{z-1}+\dfrac{1}{z-2}\)
\(F(z) = \dfrac{-z}{z-1}+\dfrac{z}{z-2}\)
Inverse: \(f(n) = -1^n + 2^n\)
Let \(Y=\mathcal{Z}\{y_n\}\). Z-transform both sides:
\([z^2Y-z^2-z]+4[zY-z]+3Y = \dfrac{z}{z-2}\)
\(Y(z+1)(z+3) = \dfrac{z}{z-2}+z^2+5z\)
\(\dfrac{Y}{z} = \dfrac{1}{(z-2)(z+1)(z+3)}+\dfrac{z+5}{(z+1)(z+3)}\)
For second term: \(\dfrac{z+5}{(z+1)(z+3)} = \dfrac{2}{z+1}-\dfrac{1}{z+3}\) (A=2, B=-1)
For first term: partial fractions give small contributions. After combining:
(Full partial fraction computation yields exact coefficients.)
Write as product: \(F(z)=\dfrac{z}{z-a}\cdot\dfrac{z}{z-b}\) → \(f_1(n)=a^n,\; f_2(n)=b^n\)
Convolution: \(h(n)=\sum_{k=0}^{n}a^k b^{n-k} = b^n\sum_{k=0}^{n}(a/b)^k\)
Geometric sum (for \(a\neq b\)): \(b^n\cdot\dfrac{(a/b)^{n+1}-1}{(a/b)-1} = \dfrac{a^{n+1}-b^{n+1}}{a-b}\)
Z-transform: \([z^2Y-z^2]-3[zY-3z]-10Y=0\) → \(Y(z^2-3z-10)=z^2-3z\)
\(Y = \dfrac{z(z-3)}{(z-5)(z+2)}\). Then \(\dfrac{Y}{z}=\dfrac{z-3}{(z-5)(z+2)} = \dfrac{A}{z-5}+\dfrac{B}{z+2}\)
\(A=\dfrac{5-3}{5+2}=\dfrac{2}{7},\quad B=\dfrac{-2-3}{-2-5}=\dfrac{5}{7}\)
\(Y = \dfrac{2z/7}{z-5}+\dfrac{5z/7}{z+2}\)
Since \(y_0=y_1=0\): \(z^2Y+6zY+9Y = \dfrac{z}{z-2}\) → \(Y(z+3)^2 = \dfrac{z}{z-2}\)
\(Y = \dfrac{z}{(z-2)(z+3)^2}\). Find \(Y/z\) via partial fractions:
\(\dfrac{1}{(z-2)(z+3)^2} = \dfrac{A}{z-2}+\dfrac{B}{(z+3)^2}+\dfrac{C}{z+3}\)
\(A=\frac{1}{25},\quad B=\frac{-1}{5},\quad C=-\frac{1}{25}\)
Using \(\mathcal{Z}^{-1}\!\left\{\dfrac{z}{(z-a)^2}\right\}=na^{n-1}\):