mth3006 cheat sheet

MTH3006 Methods of Mathematical Physics, Final Exam Cheat Sheet

Created by William Fayers

Good luck and have fun!! :))

0. Reference Tables & Foundational Material

Laplace Transform Table

$f(t)$	$\tilde{f}(s)$		$f(t)$	$\tilde{f}(s)$
$1$	$\frac{1}{s}$		$e^{at}$	$\frac{1}{s-a}$
$t^n$	$\frac{n!}{s^{n+1}}$		$\sin (\omega t)$	$\frac{\omega }{s^2+{\omega }^2}$
$t^p$	$\frac{\Gamma (p+1)}{s^{p+1}}$		$\cos (\omega t)$	$\frac{s}{s^2+{\omega }^2}$
$\sinh (at)$	$\frac{a}{s^2-a^2}$		$\cosh (at)$	$\frac{s}{s^2-a^2}$
$y^{\prime }$	$s\tilde{y}-y(0)$		$y^{\prime \prime }$	$s^2\tilde{y}-sy(0)-y^{\prime }(0)$
Shift theorem: $e^{at}f(t)$	$\tilde{f}(s-a)$		e.g. $\frac{1}{(s+1{)}^2}\to t{e}^{-t}$	(shift $\frac{1}{s^2}\to t$)

Key Proofs & Identities

Potential “show that” / “derive” questions: Γ(p+1)=pΓ(p), double integral lemma, convolution theorems (Laplace & Fourier).

1. Gamma function: $\Gamma (p)={\int }_0^{\infty }{t}^{p-1}{e}^{-t}dt$. Prove $\Gamma (p+1)=p\Gamma (p)$: integration by parts with $u=t^p$, $dv=e^{-t}dt$ gives $[-t^p{e}^{-t}{]}_0^{\infty }+p{\int }_0^{\infty }{t}^{p-1}{e}^{-t}dt=p\Gamma (p)$. Values: $\Gamma (1)=1$, $\Gamma (n+1)=n!$, $\Gamma (\frac{1}{2})=\sqrt{\pi }$.
2. Double integral lemma: ${\int }_a^x{\int }_a^{s}f(z)dzds={\int }_a^x(x-z)f(z)dz$.
   • Method 1 (Region swap): $\underset{\text{inner}}{\underbrace{a\le z\le s}},\underset{\text{outer}}{\underbrace{a\le s\le x}}$ $\Rightarrow$ $\underset{\text{outer}}{\underbrace{a\le z\le x}},\underset{\text{inner}}{\underbrace{z\le s\le x}}$ → ${\int }_a^x(x-z)f(z)dz$.
   • Method 2 (Integration by parts): Let $F(s)={\int }_a^{s}f(z)dz$, so $F^{\prime }(s)=f(s)$, $F(a)=0$.
     • ${\int }_a^{x}F(s)ds=[sF(s){]}_a^x-{\int }_a^{x}sf(s)ds=xF(x)-{\int }_a^{x}sf(s)ds={\int }_a^x(x-s)f(s)ds$.
3. Convolution theorem (Laplace): $\mathcal{L}\{f\ast g\}=\tilde{f}(s)\tilde{g}(s)$.
   • $\mathcal{L}\{f\ast g\}={\int }_0^{\infty }{e}^{-st}\left({\int }_0^{t}f(\tau )g(t-\tau )d\tau \right)dt$. Swap order: ${\int }_0^{\infty }f(\tau )\left({\int }_{\tau }^{\infty }{e}^{-st}g(t-\tau )dt\right)d\tau$.
   • Sub $u=t-\tau$: ${\int }_0^{\infty }f(\tau )e^{-s\tau }\left({\int }_0^{\infty }{e}^{-su}g(u)du\right)d\tau =\tilde{f}(s)\tilde{g}(s)$.
4. Convolution theorem (Fourier): $\widetilde{f\ast g}(\omega )=\sqrt{2\pi }\tilde{f}(\omega )\tilde{g}(\omega )$.
   • Similar: swap order in $\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }{e}^{-i\omega t}\left({\int }_{-\infty }^{\infty }f(\tau )g(t-\tau )d\tau \right)dt$, sub $u=t-\tau$, separate into product.
5. Euler: $e^{i\theta }=\cos \theta +i\sin \theta$, $\sin \theta =\frac{e^{i\theta }-e^{-i\theta }}{2i}$, $\cos \theta =\frac{e^{i\theta }+e^{-i\theta }}{2}$
   • Trig: $(\tan x{)}^{\prime }={\sec }^{2}x$, $\int \tan xdx=-\ln |\cos x|$
6. Hyperbolic functions:
   • Definitions: $\sinh x=\frac{e^x-e^{-x}}{2}$, $\cosh x=\frac{e^x+e^{-x}}{2}$, $\tanh x=\frac{\sinh x}{\cosh x}$, $\text{sech}x=\frac{1}{\cosh x}$
   • Identity: ${\cosh }^{2}x-{\sinh }^{2}x=1$
   • Addition: $\sinh (x\pm y)=\sinh x\cosh y\pm \cosh x\sinh y$, $\cosh (x\pm y)=\cosh x\cosh y\pm \sinh x\sinh y$
   • Derivatives: $(\sinh x{)}^{\prime }=\cosh x$, $(\cosh x{)}^{\prime }=\sinh x$, $(\tanh x{)}^{\prime }={\text{sech}}^{2}x$
   • Integrals: $\int \tanh xdx=\ln \cosh x$
7. Leibniz rule: $\frac{d}{dx}{\int }_{a(x)}^{b(x)}f(x,z)dz=f(x,b)b^{\prime }-f(x,a)a^{\prime }+{\int }_a^b\frac{\partial f}{\partial x}dz$
   • If $f$ doesn’t depend on $x$ (FTC): $\frac{d}{dx}{\int }_0^{x}g(z)dz=g(x)$.
8. U-substitution: $\int f(g(x))g^{\prime }(x)dx=\int f(u)du$ where $u=g(x)$, $du=g^{\prime }(x)dx$.
   • For definite integrals: change limits to $u(a)$ and $u(b)$, or back-substitute at end.
   • E.g. ${\int }_0^{1}x{e}^{x^2}dx$: let $u=x^2$, $du=2xdx$ → $\frac{1}{2}{\int }_0^1{e}^{u}du=\frac{1}{2}(e-1)$.
9. Delta function FT: $\widetilde{\delta (t-a)+\delta (t+a)}(\omega )=\sqrt{\frac{2}{\pi }}\cos (a\omega )$
10. 2nd order constant-coefficient ODEs: For $a{y}^{\prime \prime }+b{y}^{\prime }+cy=0$, solve characteristic equation $a{r}^2+br+c=0$.

Roots	General solution
Real distinct $r_1,r_2$	$A{e}^{r_{1}x}+B{e}^{r_{2}x}$
Real repeated $r$	$(A+Bx)e^{rx}$
Complex $p\pm qi$	$e^{px}(A\cos qx+B\sin qx)$

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Question Topics (ordered by Frequency × marks)

1. Laplace Transforms - Evaluate Integrals (8 Marks, ~2-3× per paper)

1. Method: Match integral with $\tilde{f}(s)={\int }_0^{\infty }f(t)e^{-st}dt$.
   • Identify $f(t)$ and the value of $s$.
   • Use table: $\mathcal{L}\{f(t)\}{|}_{s=a}$ gives the integral value.
2. Example: ${\int }_0^{\infty }{e}^{-at}(t+\sin t)dt$
   • Set $s=a$, use $\mathcal{L}\{t\}=\frac{1}{s^2}$, $\mathcal{L}\{\sin t\}=\frac{1}{s^2+1}$.
   • Result: $\boxed{\frac{1}{a^2}+\frac{1}{a^2+1}}$.

2. Laplace Transforms - Solve ODEs (8-15 marks)

1. Method: Take $\mathcal{L}$ of entire equation.
   • Use $\mathcal{L}\{y^{\prime }\}=s\tilde{y}-y(0)$, $\mathcal{L}\{y^{\prime \prime }\}=s^2\tilde{y}-sy(0)-y^{\prime }(0)$.
   • Solve algebraically for $\tilde{y}(s)$.
   • Apply inverse Laplace (partial fractions or convolution).
2. Example: $x^{\prime }+x=2e^{-t}+e^{-2t}$, $x(0)=0$
   • Transform: $(s+1)\tilde{x}=\frac{2}{s+1}+\frac{1}{s+2}$
   • Solve for $\tilde{x}$, then inverse.

3. Inverse Laplace Transform - Partial Fractions (8-9 marks)

Factor type	Decomposition
Linear $(as+b)$	$\frac{A}{as+b}$
Repeated $(as+b{)}^n$	${\sum }_{k=1}^n\frac{A_k}{(as+b{)}^k}$
Quadratic	$\frac{As+B}{a{s}^2+bs+c}$

1. Method: Decompose → find coefficients (cover-up or equate) → invert each term.
   • Linearity: ${\mathcal{L}}^{-1}\{F+G\}={\mathcal{L}}^{-1}\{F\}+{\mathcal{L}}^{-1}\{G\}$ — handle each fraction separately, don’t combine.
   • ${\mathcal{L}}^{-1}\{\frac{1}{s-a}\}=e^{at}$, ${\mathcal{L}}^{-1}\{\frac{1}{(s-a{)}^2}\}=t{e}^{at}$, ${\mathcal{L}}^{-1}\{\frac{\omega }{s^2+{\omega }^2}\}=\sin \omega t$, etc.

4. Convolution (8-9 marks)

Formula: $(f\ast g)(t)={\int }_0^{t}f(\tau )g(t-\tau )d\tau$

Properties:

• $f\ast g=g\ast f$ (commutative) - choose whichever order makes the integral easier.
• $f\ast (g+h)=f\ast g+f\ast h$ (distributive) - expand sums before convolving.
• $f(t)\ast \delta (t-a)=f(t-a)$ (shifting) - delta functions just shift the argument.

1. For inverse Laplace: When $\tilde{f}(s)=F(s)G(s)$ is a product, use ${\mathcal{L}}^{-1}\{FG\}=f\ast g$.
   • Find $f={\mathcal{L}}^{-1}\{F\}$, $g={\mathcal{L}}^{-1}\{G\}$ → compute convolution integral.
   • E.g. ${\mathcal{L}}^{-1}\{\frac{1}{s(s+1)}\}$: $F=\frac{1}{s}\to 1$, $G=\frac{1}{s+1}\to e^{-t}$ → $(1\ast e^{-t})={\int }_0^t{e}^{-(t-\tau )}d\tau =1-e^{-t}$.
2. Direct convolution: Apply formula directly, use properties to simplify.
   • E.g., $f\ast [\delta (t-a)+\delta (t+a)]=\boxed{f(t-a)+f(t+a)}$ (distributive then shifting).

5. PDEs - Separation of Variables (8 Marks, ~1× per paper)

Separated ODE	General Solution
$\frac{X^{\prime }}{X}=g(x)$	$X=A\exp \left(\int g(x)dx\right)$ — integrate directly
$\frac{X^{\prime \prime }}{X}=0$	$X=Ax+B$
$\frac{X^{\prime \prime }}{X}=\lambda >0$	$X=A{e}^{\sqrt{\lambda }x}+B{e}^{-\sqrt{\lambda }x}$ (or $A\cosh +B\sinh$)
$\frac{X^{\prime \prime }}{X}=-\lambda <0$	$X=A\cos (\sqrt{\lambda }x)+B\sin (\sqrt{\lambda }x)$

(These follow from section 0.10 — characteristic equation method.)

1. Method: Assume $u(x,y)=X(x)Y(y)$ (or $u(x,t)=X(x)T(t)$).
   • Substitute into PDE, divide by $XY$ to separate.
   • Each side equals constant → two ODEs. Solve using table above.
   • Combine: $u=CXY$ (constants from $X$, $Y$ merge into $C$). Apply BC → match functional form to find separation constant, match coefficient to find $C$.
2. Heat/wave (eigenvalue problems): BCs on $X$ determine eigenvalues ${\lambda }_n$, eigenfunctions $X_n$.
   • Start with $X^{\prime \prime }+\lambda X=0$ → $X=A\cos (\sqrt{\lambda }x)+B\sin (\sqrt{\lambda }x)$.
   • Apply 1st BC (e.g., $X(0)=0$ → $A=0$).
   • Apply 2nd BC (e.g., $X(L)=0$ → $\sin (\sqrt{\lambda }L)=0$ → $\sqrt{\lambda }L=n\pi$).
   • Result: ${\lambda }_n={\left(\frac{n\pi }{L}\right)}^2$, $X_n=\sin \frac{n\pi x}{L}$.
   • General solution: $u=\sum C_n{X}_n{T}_n$. Find $C_n$ via orthogonality/Fourier.
3. Example: $u_x=\beta u_y$, $u(0,y)=e^{-y}$
   • Separation: $\frac{X^{\prime }}{X}=\beta \frac{Y^{\prime }}{Y}=\alpha$ → $X=A{e}^{\alpha x}$, $Y=B{e}^{\alpha y/\beta }$ → $u=C{e}^{\alpha x+\alpha y/\beta }$.
   • Apply BC: $u(0,y)=C{e}^{\alpha y/\beta }=e^{-y}$ → $C=1$, $\alpha /\beta =-1$ → $\boxed{u=e^{-\beta x-y}}$.

6. PDEs - Method of Characteristics (8-9 Marks, ~1× per paper)

For $A{u}_x+B{u}_y=C$, write characteristic system $\frac{dx}{A}=\frac{dy}{B}(=\frac{du}{C})$.

1. Homogeneous ($C=0$):
   • Solve $\frac{dx}{A}=\frac{dy}{B}$ → cross-multiply to get $Bdx=Ady$, then integrate both sides.
     • Cross-multiply when $A,B$ are functions of different variables (e.g., $A=A(x)$, $B=B(y)$) to separate.
   • Rearrange to $\alpha (x,y)=c$ (the constant of integration becomes the characteristic constant).
   • General solution: $u=f(\alpha )$ for arbitrary $f$.
   • Find $f$: On BC curve, compute $\alpha$ and $u$ both in terms of one variable (say $x$).
     • Rearrange to get that variable in terms of $\alpha$ → substitute into $u$ → gives $f(\alpha )$.
     • Substitute back: $u=f\left(\alpha (x,y)\right)$.
2. Inhomogeneous ($C\ne 0$):
   • Get $\alpha (x,y)=c_1$ from $\frac{dx}{A}=\frac{dy}{B}$ (same as homogeneous).
   • Also solve $\frac{dx}{A}=\frac{du}{C}$ → integrate: $\int \frac{1}{A}dx=\int \frac{1}{C}du+c_2$ → rearrange to $\beta (x,u)=c_2$.
   • General solution: $c_2=g(c_1)$ → $\beta =g(\alpha )$ for arbitrary $g$.
   • Find $g$: On BC curve, compute $\alpha$ and $\beta$ both in terms of one variable → invert to get $g$.
3. Example: $u_t+e^{-t}{u}_x=0$, $u=e^{-x}$ on $t=0$
   • Chars: $\frac{dt}{1}=\frac{dx}{e^{-t}}$ → $dx=e^{-t}dt$ → $x=-e^{-t}+c$ → $\alpha =x+e^{-t}$.
   • General: $u=f(\alpha )$.
   • BC ($t=0$): $\alpha =x+1$, $u=e^{-x}$. Rearrange: $x=\alpha -1$ → $u=e^{-(\alpha -1)}=e^{1-\alpha }$ → $f(\alpha )=e^{1-\alpha }$.
   • Final: $u=f(x+e^{-t})=\boxed{e^{1-x-e^{-t}}}$.

7. PDEs - Change of Variables (8 Marks, ~1× per paper)

Chain rule formulas for new variables $\xi (x,y)$, $\eta (x,y)$:

• $u_x=u_{\xi }{\xi }_x+u_{\eta }{\eta }_x$, $u_y=u_{\xi }{\xi }_y+u_{\eta }{\eta }_y$
• $u_{xx}=u_{\xi \xi }{\xi }_x^2+2u_{\xi \eta }{\xi }_x{\eta }_x+u_{\eta \eta }{\eta }_x^2(+u_{\xi }{\xi }_{xx}+u_{\eta }{\eta }_{xx}$ if $\xi ,\eta$ nonlinear$)$
• $u_{yy}=u_{\xi \xi }{\xi }_y^2+2u_{\xi \eta }{\xi }_y{\eta }_y+u_{\eta \eta }{\eta }_y^2$
• $u_{xy}=u_{\xi \xi }{\xi }_x{\xi }_y+u_{\xi \eta }({\xi }_x{\eta }_y+{\xi }_y{\eta }_x)+u_{\eta \eta }{\eta }_x{\eta }_y$

1. Method: Compute partials of $\xi ,\eta$ → substitute into PDE → collect coefficients → simplify.
2. Example: Show $\xi =y-x$, $\eta =y+2x$ reduces $u_{xx}-u_{xy}-2u_{yy}=0$ to $u_{\xi \eta }=0$.
   • Compute: ${\xi }_x=-1$, ${\xi }_y=1$, ${\eta }_x=2$, ${\eta }_y=1$.
   • Substitute and simplify.

8. Integral Equations - Separable Kernels (8-9 Marks, ~2× per paper)

Kernel: $K(x,z)$ is the function inside the integral that multiplies $y(z)$, e.g., in $y(x)=f(x)+\lambda \int K(x,z)y(z)dz$.

Type	Limits	$y$ appears
Fredholm	Fixed $a,b$	2nd kind: inside & outside integral
Volterra	Upper limit $x$	1st kind: only inside integral

1. When: $K(x,z)=g(x)h(z)$ (or sum of such products).
2. Method for $y(x)=f(x)+\lambda {\int }_a^{b}g(x)h(z)y(z)dz$:
   • Define $c={\int }_a^{b}h(z)y(z)dz$ (a constant — doesn’t depend on $x$).
   • Rewrite equation: $y(x)=f(x)+\lambda c\cdot g(x)$ — this is your solution in terms of $c$.
   • Substitute $y$ back into definition of $c$: $c={\int }_a^{b}h(z)[f(z)+\lambda c\cdot g(z)]dz$.
   • Expand and solve algebraically for $c$, then substitute back into $y(x)$.
3. Example: $y=\sin x+{\int }_0^1{x}^{2}zy(z)dz$ (here $g(x)=x^2$, $h(z)=z$, $\lambda =1$)
   • Let $c={\int }_0^{1}zy(z)dz$ → $y=\sin x+c{x}^2$.
   • Substitute: $c={\int }_0^{1}z(\sin z+c{z}^2)dz={\int }_0^{1}z\sin zdz+c{\int }_0^1{z}^{3}dz$.
   • Evaluate: $c=(\sin 1-\cos 1)+\frac{c}{4}$ → $\frac{3c}{4}=\sin 1-\cos 1$ → $c=\frac{4(\sin 1-\cos 1)}{3}$.
   • Final: $\boxed{y=\sin x+\frac{4(\sin 1-\cos 1)}{3}{x}^2}$.

9. Integral Equations - Convert to/from ODE (8-9 marks)

1. ODE→Integral: Integrate twice, apply double integral lemma (see section 0.2).
   • Use dummy variables ($s$, $t$, $z$) for integration; original variable ($x$) stays as the limit.
   • E.g., $y^{\prime \prime }+{\omega }^{2}y=f(x)$, $y(0)=0$, $y^{\prime }(0)=v_0$:
   • Rearrange: $y^{\prime \prime }=f(x)-{\omega }^{2}y$.
   • Integrate ($x\to s$): $[y^{\prime }{]}_0^x={\int }_0^{x}f(s)ds-{\omega }^2{\int }_0^{x}y(s)ds$ → $y^{\prime }=v_0+{\int }_0^{x}f(s)ds-{\omega }^2{\int }_0^{x}y(s)ds$.
   • Integrate again ($x\to t$): $y=v_{0}x+{\int }_0^x{\int }_0^{t}f(s)dsdt-{\omega }^2{\int }_0^x{\int }_0^{t}y(s)dsdt$.
   • Apply lemma: $\boxed{y=v_{0}x+{\int }_0^x(x-z)f(z)dz-{\omega }^2{\int }_0^x(x-z)y(z)dz}$ (Volterra).
2. Integral→ODE: Differentiate using Leibniz rule (see section 0.7).
   • E.g., $y=2x+4{\int }_0^x(z-x)y(z)dz$:
   • Leibniz: $y^{\prime }=2+4\left[\underset{=0}{\underbrace{(x-x)y(x)\cdot 1}}+{\int }_0^x\frac{\partial }{\partial x}(z-x)y(z)dz\right]=2-4{\int }_0^{x}y(z)dz$.
   • Differentiate again: $y^{\prime \prime }=-4y(x)$ → $\boxed{y^{\prime \prime }+4y=0}$.
   • ICs: evaluate original equations at $x=0$ → $y(0)=0$, $y^{\prime }(0)=2$.
3. Volterra with convolution kernel $K(x-z)$:
   • Form: $y(x)=f(x)+\lambda {\int }_0^{x}K(x-z)y(z)dz$ — note integral is convolution $K\ast y$.
   • Take Laplace: $\tilde{y}=\tilde{f}+\lambda \tilde{K}\tilde{y}$ → rearrange: $\tilde{y}(1-\lambda \tilde{K})=\tilde{f}$.
   • Solve: $\tilde{y}=\frac{\tilde{f}}{1-\lambda \tilde{K}}$ → inverse Laplace to get $y(x)$.

10. Calculus of Variations - Euler-Lagrange (8-9 Marks, ~1-2× per paper)

E-L equation: For $I[y]={\int }_{x_1}^{x_2}F(x,y,y^{\prime })dx$, stationary/extreme $y$ satisfies $\frac{\partial F}{\partial y}-\frac{d}{dx}\frac{\partial F}{\partial y^{\prime }}=0$.

1. Method: Compute $\frac{\partial F}{\partial y}$, $\frac{\partial F}{\partial y^{\prime }}$, $\frac{d}{dx}(\frac{\partial F}{\partial y^{\prime }})$ → solve ODE → apply BCs.
   • If asked for stationary/extreme value: substitute $y(x)$ back into $I$.
2. Simplified cases (reduce to 1st-order ODE):
   • No $y$ in $F$: $\frac{\partial F}{\partial y}=0$ so $\frac{\partial F}{\partial y^{\prime }}=C$ → solve for $y^{\prime }$ → integrate.
   • No $x$ in $F$: Beltrami identity $F-y^{\prime }\frac{\partial F}{\partial y^{\prime }}=C$ → solve for $y^{\prime }$.
3. Example: $I={\int }_0^1(3x^2{y}^{\prime }+(y^{\prime }{)}^2)dx$, $y(0)=y(1)=0$
   • No $y$ in $F$ → use $\frac{\partial F}{\partial y^{\prime }}=3x^2+2y^{\prime }=C$.
   • Solve: $y^{\prime }=\frac{C-3x^2}{2}$ → integrate → apply BCs → $y=\frac{x(1-x^2)}{2}$, $\boxed{I=-\frac{1}{10}}$.

11. Calculus of Variations - Constrained (Isoperimetric) (8-9 marks)

1. Isoperimetric: Extremize $I=\int Fdx$ subject to $J=\int Gdx=\ell$.
   • Apply E-L to $H=F+\lambda G$.
2. Lagrange multipliers (for functions $f(x,y)$ with constraint $g=c$):
   • Solve $\nabla f=\lambda \nabla g$ and $g=c$.
3. Example: min $x^2+y^2$ subject to $y+x^2=1$
   • Equations: $2x+2\lambda x=0$, $2y+\lambda =0$, $y+x^2=1$.
   • Solution: $\boxed{\min =\frac{3}{4}}$ at $(\pm \frac{1}{\sqrt{2}},\frac{1}{2})$.

12. Fourier Transforms (8-9 Marks, ~1-2× per paper)

Type	Formula	When to use
Full	$\tilde{f}(\omega )=\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }f(t)e^{-i\omega t}dt$	General
Sine	${\tilde{f}}_s(\omega )=\sqrt{\frac{2}{\pi }}{\int }_0^{\infty }f(t)\sin (\omega t)dt$	Odd $f$ on $[0,\infty )$
Cosine	${\tilde{f}}_c(\omega )=\sqrt{\frac{2}{\pi }}{\int }_0^{\infty }f(t)\cos (\omega t)dt$	Even $f$ on $[0,\infty )$

1. Method: Write formula → split integral by ranges → evaluate.
   • Use Euler ($e^{-i\omega t}=\cos \omega t-i\sin \omega t$) when $f$ has trig functions.
2. Special cases (use method above, with these to determine ranges):
   • Absolute value: $|t|=\{\begin{array}{ll}-t& t<0\\ t& t\ge 0\end{array}$ → split at $t=0$.
   • Sifting theorem: ${\int }_{-\infty }^{\infty }\delta (t-a)g(t)dt=g(a)$ → evaluate $e^{-i\omega t}$ at each delta location.

13. Green’s Functions (8-9 Marks, ~1× per paper)

Setup: For ODE $Ly=f(x)$ where $L$ is the differential operator (e.g., $Ly=y^{\prime \prime }-{\alpha }^{2}y$ for $y^{\prime \prime }-{\alpha }^{2}y=f$).

Solution form: $y(x)={\int }_a^{b}G(x,z)f(z)dz$ where $z$ is dummy variable.

• Limits: Use domain from BCs. If $G=0$ for $x<z$, integral reduces to ${\int }_a^x$ (only $z<x$ contributes).

1. Properties of $G$:
   • $G$ satisfies homogeneous ODE ($Ly=0$) in each region $x<z$ and $x>z$.
   • $G$ satisfies homogeneous BCs.
   • $G$ continuous at $x=z$; $\frac{\partial G}{\partial x}$ jumps by $1$ at $x=z$ (for $y^{\prime \prime }$ leading term).
2. Finding $G$ (step-by-step):
   • Step 1: Solve homogeneous ODE $Ly=0$ using characteristic equation (see section 0.10) → get general solution $y=A{y}_1+B{y}_2$.
   • Step 2: Write piecewise (use different constants $a_1,b_1,a_2,b_2$ in each region; $y_1,y_2$ are the independent solutions from Step 1): $G=\{\begin{array}{ll}{a}_1{y}_1+b_1{y}_2& x<z\\ a_2{y}_1+b_2{y}_2& x>z\end{array}$
   • Step 3: Apply BCs to the relevant region (e.g., $x=0$ is in $x<z$ region).
   • Step 4: At $x=z$, apply 2 conditions:
     • Continuity: $a_1{y}_1(z)+b_1{y}_2(z)=a_2{y}_1(z)+b_2{y}_2(z)$ (both pieces equal at $z$).
     • Derivative jump: $(a_2{y}_1^{\prime }(z)+b_2{y}_2^{\prime }(z))-(a_1{y}_1^{\prime }(z)+b_1{y}_2^{\prime }(z))=1$ (for $y^{\prime \prime }+\dots =f$; use $\frac{1}{a(z)}$ if ODE is $a(x)y^{\prime \prime }+\dots =f$).
   • Step 5: Solve for remaining coefficients.
3. If given $G$: Just compute $y=\int G(x,z)f(z)dz$.
4. Example: $y^{\prime \prime }-{\alpha }^{2}y=f$, $y(0)=y^{\prime }(0)=0$
   • Homogeneous: $y=A{e}^{\alpha x}+B{e}^{-\alpha x}$.
   • BCs at $x=0$ ($x<z$ region): $G(0)=0$, $G^{\prime }(0)=0$ → both give $a_1=b_1=0$ → $G=0$ for $x<z$.
   • At $x=z$: continuity gives $a_2{e}^{\alpha z}+b_2{e}^{-\alpha z}=0$; jump gives $\alpha a_2{e}^{\alpha z}-\alpha b_2{e}^{-\alpha z}=1$.
   • Solve: $a_2=\frac{e^{-\alpha z}}{2\alpha }$, $b_2=-\frac{e^{\alpha z}}{2\alpha }$ → $G=\frac{\sinh \alpha (x-z)}{\alpha }$ for $x>z$.
   • Solution: $y={\int }_0^x\frac{\sinh \alpha (x-z)}{\alpha }f(z)dz$.