Differential Equation

• Determine the order of \(y\) by making the equation homogeneous, and substitute \(y = vx^r\).

• Most commonly, introduce \(p\) in the place of \(y'\) as an independent variable, and express the solution in terms of \(p\).
• Clairaut's equation
• D'Alembert's equation

• Introduce \(z\) such that \(y = f(x, z)\).
• Chrystal's equation

• \[ y = xy' + f(y') \] where \(f\) is continuously differentiable.
• It is a particular case of the Lagrange differential equation.

• Differentiate both side: \[ y' = y' + xy'' + f'(y')y'' \implies (x + f'(y'))y'' = 0 \]
• Hence, \[ y'' = 0\lor x + f'(y') = 0. \]

• \[ u = xu_x + yu_y + f(u_x, u_y). \]

• \[ y = f(y')x+g(y') \]

• By the method of parameter introduction.

  • Differentiate both side with respect to \(p = y'\):

    \begin{align*} &p\frac{dx}{dp} = f'(p)x + f(p)\frac{dx}{dp} + g'(p)\\ \implies &[p-f(p)]\frac{dx}{dp} - f'(p)x = g'(p). \end{align*}

• The solution to this linear first order differential equation \(x = \xi(p; C)\) yields the solution:

  \begin{cases} x &= \xi(p; C),\\ y &= f(p)\xi(p; C) + g(p). \\ \end{cases}

• \[ F(y')x+ G(y')y = H(y') \]

• \[ y'^2 + Axy' + By + Cx^2 = 0 \] where \(A, B, C\) are constants.

• Transform by introduction of parameter, \(4By \rightsquigarrow (A^2 - 4C - z^2)x^2\): \[ xz\frac{dz}{dx} = A^2 + AB - 4C \pm Bz - z^2. \]

• D'Alembert Reduction
• The last linearly independent solution can be found using reduction of order.
• For an second order linear homogeneous ordinary differential equation with one of the solutions \(y_1\) known, the other solution can be found by substituting \(y = vy_1\), and solving for \(v\).

• The last linearly independent solution can also be found using Wronskian.
• Wronskian yields the order-reduced inhomogeneous differential equation with respect to the last solution: \[ \sum_{i=1}^n W_{i,j}(x)y_j^{(i-1)} = W(x) \] where \(W_{i,j}\) denotes the \(i,j\)th minor of the Wronskian matrix,
• the particular solution is the last independent solution.

• Given that the coefficients of a monic differential equation are analytic at zero, the solution takes the form of power series: \[ f = \sum_{k=0}^\infty A_kz^k. \]
  • If zero is the regular singular point, Frobenius method is to be applied instead.
• After substitution the coefficients of the reduced power series are required to be zero, yielding the recurrence relation.

• Method of Frobenius
• Method of finding the series solution at the regular singular point of a linear homogeneous differential equation.
• Assume the solution has the form of a Frobenius series: \[ u(z) = z^r\sum_{k=0}^{\infty}A_kz^k \] with \(A_0\neq 0\).

• Algorithm for solving systems of ordinary differential equations.
• It includes the Picard method.
• Parker–Sochacki method - Wikipedia

• Trigonometric Function
  • When the inhomogeneous term is a trigonometric function \(a\cos(\omega t)\), use the trial solution:
    • \[ c_1\cos(\omega t) + c_2\sin(\omega t) \]
  • Or extend the equation to complex domain by replacing \(a\cos(\omega t)\) with \(ae^{i\omega t}\), and use the trial solutions:
    • \[ ce^{i(\omega t - \phi)} \]
      • with \(c\in \mathbb{R}\)
    • \[ ce^{i\omega t} \]
      • with \(c\in \mathbb{C}\)
  • And solve for the undetermined constants.
  • Especially when the differential equation is of the form \(y'' + by' + cy = a\cos(\omega t)\),
    • The extended solution is given by:
    • \[ re^{i(\omega t - \phi)} \]
      • with
      • \[ r = \frac{a}{\sqrt{(c-\omega^2)^2 + (b\omega)^2}} \]
      • \[ \phi = \tan^{-1}\left(\frac{b\omega}{c-\omega^2}\right) \]
    • Or by:
      • \[ re^{i\omega t} \]
      • with
      • \[ c = \frac{a}{c-\omega^2 + ib\omega} \]

• Given an inhomogeneous linear differential equation: \[ \sum_{j=0}^n a_j(x)y^{(j)} = f(x) \] with \(a_n \equiv 1\),

• The derivatives of \(y\) can be arbitrarily constrained with \((n-1)\) constraints:

  \begin{align*} y &= \sum_{i=1}^n u_iy_i\\ y' &= \sum_{i=1}^n u_iy'_i, &\text{Constr. 1: } \sum_{i=1}^n u_i'y_i = 0\\ &\qquad\vdots\\ y^{(k)} &= \sum_{i=1}^n u_iy_i^{(k)}, &\text{Constr. $k$: } \sum_{i=1}^n u_i'y_i^{(k-1)} = 0\\ &\qquad\vdots\\ y^{(n)} &= \sum_{i=1}^n u_i'y_i^{(n-1)} + u_iy_i^{(n)}, &\text{No Constr.}\\ \end{align*}
• The last constraint is found by substituting these into the equation: \[ \sum_{i=1}^n u_i'y_i^{(n-1)} = f(x). \]
• All the constraints can be summerized using Wronskian matrix \(\mathbf{W}\) of homogeneous solutions: \[ \mathbf{W}\begin{bmatrix} u_1' \\ \vdots \\ u_n' \end{bmatrix} = \begin{bmatrix} 0 \\ \vdots \\ f(x) \end{bmatrix}. \]
• Therefore the particular solution can be found with: \[ y = \left({\Large \int} \mathbf{W}^{-1} \begin{bmatrix} 0 \\ \vdots \\ f(x) \end{bmatrix}\mathrm{d}x\right) \cdot \begin{bmatrix} y_1 \\ \vdots \\ y_n \end{bmatrix} \] here \(\mathbf{W}\) is guaranteed to be invertible, since \(y_i\) are linearly independent.
  • For the second order case, the particular solution is given by: \[ y = y_2\int\frac{y_1f}{W}\mathrm{d}x -y_1\int \frac{y_2f}{W}\mathrm{d}x. \]

• The coefficient functions \(p, q, w\) and the derivative \(p'\) are all continuous on \([a, b]\)
• \(p(x) > 0\) and \(w(x) > 0\) for all \(x\in[a,b]\)
• The problem has separated boundary conditions of the form:
  • \[ \alpha_1y(a) + \alpha_2y'(a) = 0 \] with \(\alpha_1, \alpha_2\) are not both 0.
  • \[ \beta_1y(b) + \beta_2y'(b) = 0 \] with \(\beta_1, \beta_2\) are no both 0.

See Time-independent Schrödinger equation

• \[ (1-x^2)y'' - xy' + n^2y = 0 \]
  • Chebyshev Equation, Chebyshev's Equation

• \[ (1-x^2)y''-2xy' + n(n+1)y = 0 \]
  • Legendre polynomials \(P_n(x)\) solve this.
    • Legendre functions of the second kind \(Q_n(x)\) also solve this.
  • The Sturm-Liouville form is: \[ \frac{d}{dx}\left((1-x^2)\frac{d}{dx}\right)y = -\lambda y \] with eigenvalue \(\lambda = n(n+1)\).

• \[ xy'' + (1-x)y' + ny = 0 \]
  • Laguerre polynomials \(L_n(x)\) solve this.
• The Sturm-Liouville form is: \[ \left(xe^{-x}L_n(x)'\right)' = -ne^{-x}L_n(x). \]

• \[ y'' - 2xy' + 2ny = 0 \]

  • Physicist's Hermite polynomials \(H_n(x)\) solve this.
  • The second solution is the confluent hypergeometric functions of the first kind:
    • \[ {}_1F_1(-\tfrac{n}{2};\tfrac{1}{2};x^2) \]
  • The Sturm-Liouville form is: \[ \left(e^{-x^2/2}y'\right)' = -ne^{-x^2/2}y. \]

• \[ x^2y''+xy' + (x^2-\alpha^2)y = 0 \]
  • Solutions are the Bessel function.

• \[ \frac{d^2w}{dz^2} + \left[\frac{1-\alpha - \alpha'}{z-a} + \frac{1-\beta - \beta'}{z-b} + \frac{1-\gamma -\gamma'}{z-c}\right]\frac{dw}{dz} + \left[\frac{\alpha\alpha'(a-b)(a-c)}{z-a} + \frac{\beta\beta'(b-c)(b-a)}{z-b} + \frac{\gamma\gamma'(c-a)(c-b)}{z-c}\right]\frac{w}{(z-a)(z-b)(z-c)} = 0 \]

• It has three regular singular points: \(a,b,c\).

• Cauchy-Euler equation of order \(n\) is of the form: \[ \sum_{k=0}^n a_k x^k y^{(k)}(x) = 0 \]

• The solution is one of the form:
  • \[ \begin{cases} C_1x^{m_1} + C_2x^{m_2}, & \text{distinct roots}\\ (C_1\ln x + C_2)x^{m}, & \text{repeated roots}\\ C_1x^{\alpha}\cos(\beta\ln x) + C_2x^{\alpha}\sin(\beta\ln x), & \text{complex roots}. \end{cases} \]

• Function that facilitate solving a differential equation.

• First order linear inhomogeneous case
• \[ y'' = Ay^r \]
  • \(M = y'\)
  • can be integrated with the factor of \( y' \) multiplied on both sides: \[ \frac{1}{2}(y')^2 = \frac{A}{r+1}y^{r+1} + B. \]
• \[ y'' + 2p(x) y' + (p(x)^2 + p'(x))y = q(x) \]
  • with \(2p(x)\) being \(p_1(x)\).

• If one want to collapse the derivatives into \((M(x)y)^{(n)}\) by multiplying both side by \(M(x)\), the only possible integrating factor is:
  • \[ M(x) = \exp\left(\int \frac{p_1(x)}{n}\,dx\right) \]
  • This restricts coefficients of any lower order derivative terms into a single function.
• The substitution \(y = \mu(x)v(x) = v(x)/M(x)\) removes the \(n-1\) order term.

\begin{align*} y^{(n)} &+ \binom{n}{1}p(x)y^{(n-1)}\\ &+ \binom{n}{2}(p(x)^2 + p'(x))y^{(n-2)} \\ &+\binom{n}{3}(p(x)^3 + 3p(x)p'(x) + p''(x))y^{(n-3)}\\ &+ \binom{n}{4}(p(x)^4 + 6p(x)^2p'(x) + 4p(x)p''(x) + 3p'(x)^2 + p'''(x))y^{(n-4)} \cdots\\ &= \frac{(M(x)y)^{(n)}}{M(x)} = \frac{v^{(n)}}{M(x)} = \mu(x)v^{(n)}\\ \end{align*}

• The set of solutions form a curved hypersurface in the function space.
  • The general solution is not a linear combination of linearly independent solutions.

• \[ Pp + Qq = R \] where \(P, Q, R\) are functions of \(x, y, z\), and \(p = z_x, q = z_y\).

• Assume the solution has the form \(\phi(u, v) = 0\), with \(u, v\) being the functions of \(x, y, z\).
• Observe that taking partial derivative of \(\phi\) with respect to \(x\) and \(y\) yields the differential equation.

\begin{bmatrix} 0 \\ 0 \end{bmatrix} = \begin{bmatrix} u_x + u_z z_x & v_x + v_zz_x \\ u_y + u_zz_y & v_y + v_z z_y \\ \end{bmatrix} \begin{bmatrix} \phi_u \\ \phi_v \end{bmatrix}

• Note that \(z\) is being considered to be the function of \(x\) and \(y\). Since the original differential equation described it as so.
  • The determinant is zero for nonzero derivatives of \(\phi\), Yielding the differential equation: \[ (u_yv_z - u_zv_y)p + (u_zv_x - u_xv_z)q = (u_xv_y - u_yv_x) \iff Pp + Qq = R. \]
    • One can know that \(u\) and \(v\) must satisfy this relation.

• Set \(u = a\) and \(v = b\) as constants, and examine \(x, y, z\) that are on the level sets of \(u\) and \(v\).
  • This is motivated by the implicit function \(\phi(u, v) = 0\), since it is possible.³

  • Take the differential of \(u\) and \(v\):

    \begin{cases} du = u_xdx + u_y dy + u_z dz = 0,\\ dv = v_xdx + v_y dy + v_z dz = 0. \end{cases}
  • From this linear system the ratio relation can be obtained: \[ \frac{dx}{u_yv_z - u_zv_y} = \frac{dy}{u_zv_x - u_xv_z} = \frac{dz}{u_xv_y - u_yv_x}. \]

• Lagrange's System of Ordinary Differential Equations, Auxiliary Equations, Subsidiary Equations, Lagrange-Charpit Equations.
• In other words, \[ \frac{dx}{P} = \frac{dy}{Q} = \frac{dz}{R}. \]

The solution \(\phi(u, v) = 0\) with arbitrary function \(\phi\) is now obtained.

• If the auxiliary equations is not directly integrable, try the mediant of them.
• Reduction of the degree of freedom is possible: \(dx + dy = dt\).

• \[ \sum_{i=1}^na_iu_{x_i} = 0 \]
• The coordinate in the direction of the characteristic curves is \(s = \sum a_ix_i\), since the gradient \(\nabla u\) is orthogonal to the coefficients vector and \(s\) is in the direction of the coefficients.
  • After change of variables, the equation transforms into \[ \left(\sum_{i=1}^na_i^2\right)\hat{u}_s = 0 \]
• The other parameter can be determined, either by inspection, or from the equation of the characteristic curves obtained from the auxiliary equation
  • \[ \frac{dx_i}{a_i} = \frac{dx_j}{a_j} \implies \frac{x_i}{a_i} - \frac{x_j}{a_j} = C \]
  • Since \(C\) can be anything, one can take it (or scalar multiple of it) to be the first independent variable \(t_1\) that is orthogonal to the characteristic curves.
  • Then the next \(t_k\) can be selected to be orthogonal to every previous \(t_n\)

• \[ \sum_{i=1}^n a_iu_{x_i} + q(x_1,\dots, x_n)u = f(x_1, \dots, x_n) \]
• Applying the change of variable \(s = \sum a_ix_i, t_n \text{ orthogonal}\) transforms the equation into an ordinary one in \(s\).
• \[ \left(\sum_{i=1}^na_i^2\right)\hat{u}_s + \hat{q}\hat{u} = \hat{f} \]
• which then can be solved as first order linear inhomogeneous ordinary differential equation.

• The surface of solution \(x_{n+1} = u\) of the differential equation \[ \sum_{i=1}^na_i(x_1,\dots, x_{n}, u) u_{x_i} = a_{n+1}(x_1, \dots, x_n, u) \]

• has the normal vector

  \begin{bmatrix} u_{x_1} \\ \vdots \\ \vphantom{1}u_{x_n} \\ -1 \end{bmatrix}
• since this is the gradient of the equation of the surface \(F(x_1, \dots, x_{n+1}) = u - x_{n+1} = 0\).

• Therefore, the vector field

  \begin{bmatrix} a_1 \\ \vdots \\ \vphantom{1}a_n \\ a_{n+1} \end{bmatrix}

  is always tangent to the surface \(F = 0\),

• and by computing the integral curve \[ \frac{dx_i}{dt} = a_i \] the solution \(u = x_{n+1}(x_1, \dots, x_n)\) is obtained.

• Laplace's Differential Equation

• \[ \nabla^{\cdot 2} F + k^2 F = 0 \]

• Make an ansatz that the solution is of the form: \[ \phi(x_1, x_2,\dots, x_n) = \prod_{i=1}^{n}\phi_i(x_i). \]
• So that the the equation transforms into the the form: \[ F_i(x_i, \phi_i, \phi_i', \dots) = F_j(x_j,\phi_j, \phi_j',\dots) \]
• And then using the independence of the variables, obtain \(n\) ordinary differential equations: \[ F_i(x_i, \phi_i, \phi_i',\dots) = C. \]

• It can be used when the differential equation is linear, and the boundary conditions are separable.
• It might not work for a differential equation with non-constant coefficients. (Chat GPT)

It is applicable if the equation has the form: \[ Tu = Su \] where \( T \) and \( S \) are compact, self-adjoint differential operators in different variables.

• Linear in the dependent variable and its derivatives with the coefficients depending only on the independent variables.
• e.g.
  • \[ a_1(x,y)\frac{\partial^2u}{\partial x^2} + a_2(x,y)\frac{\partial^2u}{\partial x\partial y} + a_3(x,y)\frac{\partial^2u}{\partial y^2} + a_4(x,y)\frac{\partial u}{\partial x} + a_5(x,y)\frac{\partial u}{\partial y} + a_6(x,y)= 0 \]

• Linear only in the highest order derivatives with the coefficients depending only on the independent variables.
• e.g.
  • \[ a_1(x,y)\frac{\partial^2u}{\partial x^2} + a_2(x,y)\frac{\partial^2u}{\partial y^2} + f(x,y,u,u, u_y)= 0 \]

• Linear in the highest order derivatives with the coefficients possibly depending on the dependent variables and its non-highest order derivatives.
• e.g.
  • \[ a_1(x,y,u,u_x, u_y)\frac{\partial^2u}{\partial x^2} + a_2(x,y,u,u_x,u_y)\frac{\partial^2u}{\partial y^2} + f(x,y,u,u_x, u_y)= 0 \]

• One or more of the highest order derivatives is nonlinear.
• e.g. Monge-Ampére equation

\begin{align*} \det(\mathbf{W})' &= \det \begin{bmatrix}y_1&y_2&\cdots&y_n\\ y_1'&y_2'&\cdots&y_n'\\ \vdots&\vdots&&\vdots\\ y_1^{(n-2)}& y_2^{(n-2)} & \cdots & y_n^{(n-2)} \\ y_1^{(n)}&y_2^{(n)}&\cdots&y_n^{(n)}\end{bmatrix} \\[1em] &=\det\left(\begin{bmatrix}1&0&\cdots&0\\ 0&1&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ -p_0(x)&-p_1(x)&\cdots&-p_{n-1}(x)\end{bmatrix}\mathbf{W}\right) \\[1em] &= -p_{n-1}(x)\det\mathbf{W} \end{align*}

• Notice that Wronskian satisfies: \[ \mathbf{W}' = \begin{bmatrix}0&1&0&\cdots&0\\ 0&0&1&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\cdots&1\\ -p_0(x)&-p_1(x)&-p_2(x)&\cdots&-p_{n-1}(x)\end{bmatrix} \mathbf{W} \]
• By the 8, \[ \det(\mathbf{W})' = -p_{n-1}(x)\det{\mathbf{W}} \]

• The solution exists if
  • the boundary is sufficiently smooth
  • the boundary condition \(f\) is Hölder continuous \(C^{1,\alpha}\) with \(\alpha \in (0, 1)\).

• For a second order partial differential equation,
• the maximum is attained at the boundary for any precompact subset of the domain.
• At the maxima, the function \(u\) has to satisfy
  • \(du = 0\)
  • \(\nabla^{\cdot 2}u \le 0\), as a Loewner order

• Applying maximum principle to elliptic PDE, that is, the eigenvalues at each point is positive, the second derivatives must be all zero at the hypothetical maximum.
• If there exists two distinct solution to a linear Dirichlet problem, then the difference is the solution to the homogeneous equation. The difference violates the maximum principle, since it is zero at the boundary and nonzero on the interior.