| item | formula / fact |
|---|---|
| Wronskian (2nd order) | $Wy_1,y_2=\begin{vmatrix}y_1 & y_2\\ y_1' & y_2'\end{vmatrix}=y_1y_2'-y_2y_1'$ |
| Abel’s identity | For $y''+P(x)y'+Q(x)y=0$: $W(x)=W(x_0)\,e^{-\int_{x_0}^x P(s)\,ds}$ ⇒ $W\neq 0 \iff$ linear independence |
Given: $\mathcal{L}\,y(x)=f(x)$
| key point | cheat-sheet line |
|---|---|
| Definition | $\mathcal{L}\,G(x,z)=\delta(x-z), \\ y(x)=\int_a^b G(x,z)\,f(z)\,dz$ |
| Universal conditions on $G$ | Same BCs as $y$ |
| Continuity at $x=z$: $G(z^-,z)=G(z^+,z)$ | |
| Jump in derivative: $\left[ p(x)G'(x,z) \right]_{z^-}^{z^+}=1$ where $\mathcal{L} = \frac{1}{w(x)}\left[ (p(x)y')' + q(x)y \right]$ | |
| Piecewise formula using homogeneous solutions$y_1, y_2$ (with constant Wronskian) | $G(x,z)=\frac{w(z)}{p(z)Wy_1,y_2}\begin{cases} y_1(x)\,y_2(z), & x<z \\ y_1(z)\,y_2(x), & x>z \end{cases}$ |
Given: $y''+P(x)y'+Q(x)y=f(x)$
| step | formula |
|---|---|
| Weight/integrating factor | $w(x)=e^{\int P(x)\,dx}$ |
| Self-adjoint version | $(p(x)\,y')' + q(x)\,y = w(x)\,f(x), \quad \text{with } p(x)=w(x),\; q(x)=w(x)Q(x)$ |
| Green function | Same as in section 1, but now using the new $p(x)$ and two homogeneous solutions that satisfy the BCs. |
Given: $\mathcal{L}\,y(x)=\lambda\,w(x)\,y(x)$
| item | formula |
|---|---|
| Orthogonality | $\int_a^b w(x)\,\phi_m(x)\,\phi_n(x)\,dx = 0 \quad \text{for } m\ne n$ |
| Green function expansion | $G(x,z) = \sum_{n=1}^\infty \frac{\phi_n(x)\,\phi_n(z)}{\lambda_n}, \quad \mathcal{L}\phi_n = \lambda_n w(x)\phi_n(x)$ |
| assuming $\lambda_n \ne 0$ and the $\phi_n$ are $w$-normalised. |
| situation | instant recipe |
|---|---|
| Need $G$ and you know $y_1, y_2$ | Use piecewise Green function formula. Choose $y_1, y_2$ to satisfy one BC each. |
| Only need $y(x)$ once | Variation of parameters: |
| $y(x)=y_1(x)\int \frac{y_2(z)\,f(z)}{W\,p(z)}\,dz - y_2(x)\int \frac{y_1(z)\,f(z)}{W\,p(z)}\,dz$ | |
| Boundary-value + self-adjoint | Use Sturm–Liouville directly. Wronskian is constant, simplifies the algebra. |
| Series solution near regular point | Frobenius method: plug $y = \sum a_k(x-x_0)^{k+r}$ |
| Checking if $\mathcal{L}$ is self-adjoint | Ensure it’s in $(p\,y')' + q\,y$ form and boundary terms vanish: $\left[p(x)\bigl(y_1\,\overline{y_2}' - \overline{y_2}\,y_1'\bigr)\right]_a^b = 0$ |
✅ Tip: Once you compute the homogeneous solutions and identify $$p(x), w(x)$$, you can switch freely between methods 1, 2, and 3.