Differential Equation from Physics ¶ A lot of physical science and engineering is about finding a reduction from a physical problem to differential equations. The newtownian way is to zoom into some local point of the system and derive a set of constraint that holds in the neighborhood. In the case of the vibration of a string, for example, we would find some local segment of string and analyze the tension that acts on it:
T ( sin ( θ 1 ) − sin ( θ 2 ) ) = m ∂ 2 u ∂ t 2 T(\sin(\theta_1)-\sin(\theta_2)) = m\frac{\partial^2 u}{\partial t^2} T ( sin ( θ 1 ) − sin ( θ 2 )) = m ∂ t 2 ∂ 2 u where u u u
∂ 2 u ∂ t 2 = a 2 ∂ 2 u ∂ x 2 \frac{\partial^2 u}{\partial t^2}=a^2 \frac{\partial^2 u}{\partial x^2} ∂ t 2 ∂ 2 u = a 2 ∂ x 2 ∂ 2 u [1]
The Lagrangian way is the opposite, we recognize some global objective the system need to be optimized toward, i.e., minimizing the lagragian L = T − V L=T-V L = T − V T T T kinetic energy and V V V potential energy . Then, by introduce some perturbation, we extract a constraint which also manifest itself as a differential equation. The following sections descirbes this process in detail.
Newtonian mechanics compute the balance of forces locally and obtain differential equation through
F = m d 2 x d t 2 F=m\frac{d^2\mathbf{x}}{dt^2} F = m d t 2 d 2 x Lagrangian mechanics translate the physics into some global optimization problem and reduce to differential equation through the Euler-Lagrange equation
∂ L ∂ y − d d x ∂ L ∂ y ′ = 0 \frac{\partial L}{\partial y}- \frac{d}{dx}\frac{\partial L}{\partial y'}=0 ∂ y ∂ L − d x d ∂ y ′ ∂ L = 0 Functional : A global optimization objective ¶ A functional in calculus of variation is a map from a function to a real number, i.e., J : C ∞ → R J:C^{\infty} \to \mathbb{R} J : C ∞ → R
J [ y ] = ∫ x 1 x 2 f ( x , y , y ′ ) d x J[y] = \int_{x_1}^{x_2} f(x,y,y')dx J [ y ] = ∫ x 1 x 2 f ( x , y , y ′ ) d x The above example suffice to establish the explict connection of (5)
It sometimes also appears to be more complex, for instance,
Depedence on mutliple independent variable
J [ y ] = ∫ Ω f ( x 1 , x 2 , . . . , x n , y , y x 1 , y x 2 , . . . , y x n ) d x J[y] = \int_{\Omega} f(x_1,x_2,...,x_n, y, y_{x_1},y_{x_2},...,y_{x_n}) d{\mathbf{x}} J [ y ] = ∫ Ω f ( x 1 , x 2 , ... , x n , y , y x 1 , y x 2 , ... , y x n ) d x
where Ω ⊂ R n \Omega \subset \mathbb{R}^n Ω ⊂ R n
Dependence on higher derivatives
J [ y ] = ∫ x 1 x 2 f ( x , y , y ( 1 ) , y ( 2 ) , . . . , y ( m ) ) d x J[y] = \int_{x_1}^{x_2} f(x,y,y^{(1)},y^{(2)},...,y^{(m)})dx J [ y ] = ∫ x 1 x 2 f ( x , y , y ( 1 ) , y ( 2 ) , ... , y ( m ) ) d x Multiple Dependent Variable
The three ways of extending the (5)
We now scrutinize how exactly we optimize (10)
Euler-Lagrange Equation ¶ First Integral ¶ Suppose
y ( n ) = f ( x , y , y ( 1 ) , . . . , y ( n − 1 ) ) y^{(n)}=f(x,y,y^{(1)},...,y^{(n-1)}) y ( n ) = f ( x , y , y ( 1 ) , ... , y ( n − 1 ) ) is an ordinary differential equation of order n n n n n n
y 1 = y y 2 = y ( 1 ) ⋮ y n = y ( n − 1 ) \begin{align*}
y_1 &= y \\
y_2 &= y^{(1)} \\
\vdots \\
y_{n} &= y^{(n-1)}
\end{align*} y 1 y 2 ⋮ y n = y = y ( 1 ) = y ( n − 1 ) Denote by:
d y i d x = f i ( x , y 1 , . . . , y n ) \frac{dy_i}{dx}=f_i(x,y_1,...,y_n) d x d y i = f i ( x , y 1 , ... , y n ) Now suppose f f f
Continous about x , y 1 , . . . , y n x,y_1,...,y_n x , y 1 , ... , y n
Continuously differentialbe about y 1 , . . . , y n y_1,...,y_n y 1 , ... , y n
Define the first integral of the n th n^{\text{th}} n th I = I ( x , y 1 , . . . , y n ) I=I(x,y_1,...,y_n) I = I ( x , y 1 , ... , y n )
I ( x , y 1 , . . . , y n ) = Constant I(x,y_1,...,y_n)=\text{Constant} I ( x , y 1 , ... , y n ) = Constant along solution curves Γ = ( y 1 ( x ) , y 2 ( x ) , . . . , y n ( x ) ) \Gamma=(y_1(x),y_2(x),...,y_n(x)) Γ = ( y 1 ( x ) , y 2 ( x ) , ... , y n ( x ))
We are confronted by the higher-order ordinary differential equation:
∂ f ∂ y − d d x ( ∂ f ∂ y ) = 0 \frac{\partial f}{\partial y} -\frac{d}{dx}\left( \frac{\partial f}{\partial y}\right)=0 ∂ y ∂ f − d x d ( ∂ y ∂ f ) = 0 The compute the first integral of this equation is to rewrite the above to obtain an equation of the form:
d d t I ( x , y , y ′ , . . . ) = 0 \frac{d}{dt}I(x,y,y',...)=0 d t d I ( x , y , y ′ , ... ) = 0 The strategy is to compute the difference between (17)
d f d x = ∂ f ∂ x + ∑ j ≥ 0 ∂ f ∂ y ( j ) d y ( j ) d x = ∂ f ∂ x + ∂ f ∂ y d y d x + ∂ f ∂ y ( 1 ) d y ( 1 ) d x + ⋯ \frac{df}{dx}=\frac{\partial f}{\partial x} + \sum_{j\geq 0}\frac{\partial f}{\partial y^{(j)}}\frac{dy^{(j)}}{dx}=\frac{\partial f}{\partial x} +\frac{\partial f}{\partial y}\frac{dy}{dx}+\frac{\partial f}{\partial y^{(1)}}\frac{dy^{(1)}}{dx}+\cdots d x df = ∂ x ∂ f + j ≥ 0 ∑ ∂ y ( j ) ∂ f d x d y ( j ) = ∂ x ∂ f + ∂ y ∂ f d x d y + ∂ y ( 1 ) ∂ f d x d y ( 1 ) + ⋯ If we multiply (17) y ′ y' y ′
d f d x − y ′ { ∂ f ∂ y − d d x ( ∂ f ∂ y ′ ) } = ∂ f ∂ x + y ′ ′ ∂ f ∂ y ′ + y ′ d d x ( ∂ f ∂ y ′ ) − ∑ j ≥ 2 ∂ f ∂ y ( j ) d y ( j ) d x \frac{df}{dx}-y'\left\{\frac{\partial f}{\partial y} - \frac{d}{dx}\left( \frac{\partial f}{\partial y'}\right)\right\} =\frac{\partial f}{\partial x} + \textcolor{cyan}{y''\frac{\partial f}{\partial y'}+ y'\frac{d}{dx}\left( \frac{\partial f}{\partial y'}\right)}-\sum_{j\geq 2}\frac{\partial f}{\partial y^{(j)}}\frac{dy^{(j)}}{dx} d x df − y ′ { ∂ y ∂ f − d x d ( ∂ y ′ ∂ f ) } = ∂ x ∂ f + y ′′ ∂ y ′ ∂ f + y ′ d x d ( ∂ y ′ ∂ f ) − j ≥ 2 ∑ ∂ y ( j ) ∂ f d x d y ( j ) If we assume f f f y , y ′ y,y' y , y ′ f ( y , y ′ ) f(y,y') f ( y , y ′ )
∑ j ≥ 2 ∂ f ∂ y ( j ) d y ( j ) d x = 0 ∂ f ∂ x = 0 \begin{align*}
\sum_{j\geq 2}\frac{\partial f}{\partial y^{(j)}}\frac{dy^{(j)}}{dx} &= 0\\
\frac{\partial f}{\partial x} &=0
\end{align*} j ≥ 2 ∑ ∂ y ( j ) ∂ f d x d y ( j ) ∂ x ∂ f = 0 = 0 Now:
d f d x − y ′ { ∂ f ∂ y − d d x ( ∂ f ∂ y ′ ) } ⏟ 0 = d d x ( y ′ ∂ f ∂ y ′ ) \begin{align*}
\frac{df}{dx}-\underbrace{y'\left\{\frac{\partial f}{\partial y} - \frac{d}{dx}\left( \frac{\partial f}{\partial y'}\right)\right\}}_{0}&= \frac{d}{dx}\left(y'\frac{\partial f}{\partial y'}\right)
\end{align*} d x df − 0 y ′ { ∂ y ∂ f − d x d ( ∂ y ′ ∂ f ) } = d x d ( y ′ ∂ y ′ ∂ f ) Hence:
d d x ( f − y ′ ∂ f ∂ y ′ ) = 0 \frac{d}{dx}\left(f-y'\frac{\partial f}{\partial y'}\right)=0 d x d ( f − y ′ ∂ y ′ ∂ f ) = 0 We have sucesfully obtained the first inetgral:
I ( y , y ) = f − y ′ ∂ f ∂ y ′ I(y,y)=f-y'\frac{\partial f}{\partial y'} I ( y , y ) = f − y ′ ∂ y ′ ∂ f Exercises ¶ Euler-Lagrange Equation for Multiple Independent Variables ¶ Prove that:
Fermat’s Principle [3] ¶ We also define:
Show that light propagates along a stragiht line in homogenous media.
Since n n n
∫ x 1 x 2 1 + y ′ 2 d x \int_{x_1}^{x_2} \sqrt{1+y'^2}dx ∫ x 1 x 2 1 + y ′2 d x Note:
f y = 0 f y ′ = y ′ 1 + y ′ 2 f_y = 0 \quad f_y' = \frac{y'}{\sqrt{1+y'^2}} f y = 0 f y ′ = 1 + y ′2 y ′ The Euler-Lagrange equation is, therefore:
d d x [ y ′ 1 + y ′ 2 ] = 0 y ′ 1 + y ′ 2 = C 1 \begin{align*}
\frac{}{}
\frac{d}{dx}\left[ \frac{y'}{\sqrt{1+y'^2}}\right] &= 0 \\
\frac{y'}{\sqrt{1+y'^2}} &= C_1
\end{align*} d x d [ 1 + y ′2 y ′ ] 1 + y ′2 y ′ = 0 = C 1 Note the L.H.S is a non-constant function of y ′ y' y ′
y ′ = C 2 ⟹ y = C 2 x + C 3 y'= C_2 \Longrightarrow y=C_2x+C_3 y ′ = C 2 ⟹ y = C 2 x + C 3
We consider the first point at ( x 1 , 0 ) (x_1, 0) ( x 1 , 0 ) ( x 2 , y 2 ) (x_2,y_2) ( x 2 , y 2 ) x 1 < 0 < x 2 x_1<0<x_2 x 1 < 0 < x 2
∫ x 1 x 2 1 v ( x ) 1 + y ′ 2 d x \int_{x_1}^{x_2} \frac{1}{v(x)}\sqrt{1+y'^2}dx ∫ x 1 x 2 v ( x ) 1 1 + y ′2 d x where:
v ( x ) = { c / n 1 if x < 0 c / n 2 if x > 0 v(x)=
\begin{cases}
c/n_1 &\text{if } x<0 \\
c/n_2 &\text{if } x>0
\end{cases} v ( x ) = { c / n 1 c / n 2 if x < 0 if x > 0 The Euler-Lagrange equation is:
0 + d d x ( y ′ v ( x ) 1 + y ′ 2 ) = 0 0 + \frac{d}{dx}\left( \frac{y'}{v(x)\sqrt{1+y'^2}} \right)=0 0 + d x d ( v ( x ) 1 + y ′2 y ′ ) = 0 Hence:
y ′ v ( x ) 1 + y ′ 2 = C \frac{y'}{v(x)\sqrt{1+y'^2}} =C v ( x ) 1 + y ′2 y ′ = C For x < 0 x<0 x < 0
1 v ( x ) y ′ 1 + y ′ 2 = n 1 c sin ( θ 1 ) = C 1 \frac{1}{v(x)}\frac{y'}{\sqrt{1+y'^2}} = \frac{n_1}{c}\sin(\theta_1) = C_1 v ( x ) 1 1 + y ′2 y ′ = c n 1 sin ( θ 1 ) = C 1 For x > 0 x>0 x > 0
1 v ( x ) y ′ 1 + y ′ 2 = n 2 c sin ( θ 2 ) = C 2 \frac{1}{v(x)}\frac{y'}{\sqrt{1+y'^2}} = \frac{n_2}{c}\sin(\theta_2) = C_2 v ( x ) 1 1 + y ′2 y ′ = c n 2 sin ( θ 2 ) = C 2 Hence:
n 1 sin ( θ 1 ) = n 2 sin ( θ 2 ) n_1\sin(\theta_1)=n_2\sin(\theta_2) n 1 sin ( θ 1 ) = n 2 sin ( θ 2 ) Question: If we flip the axis we would get n 1 cos θ 1 = n 2 cos θ 2 n_1\cos\theta_1=n_2\cos\theta_2 n 1 cos θ 1 = n 2 cos θ 2
A planar light ray propagates in an layered medium with refractive index n ( x ) n(x) n ( x )
n ( x ) sin ( ψ ( x ) ) = constant n(x)\sin(\psi(x))=\text{constant} n ( x ) sin ( ψ ( x )) = constant by finding the equation for stationary paths for
F 1 = ∫ n ( x ) 1 + y ′ 2 d x F_1=\int n(x)\sqrt{1+y'^2}dx F 1 = ∫ n ( x ) 1 + y ′2 d x (Here the prime denotes differentiation with respect to x.) Repeat this exercise with x , y x,y x , y
F 2 = ∫ n ( y ) 1 + y ′ 2 d x F_2=\int n(y)\sqrt{1+y'^2}dx F 2 = ∫ n ( y ) 1 + y ′2 d x By using suitable definitions of the angle of incidence ψ \psi ψ
Drums and Membranes [3] ¶ The shape of a distorted drumskin is described by the
function h ( x , y ) h(x, y) h ( x , y ) ( x , y ) (x, y) ( x , y )
Consider the area differntial d A = d s x d s y dA=ds_x ds_y d A = d s x d s y d s x = 1 + ∂ x h 2 d x ds_x=\sqrt{1+\partial_x h^2}dx d s x = 1 + ∂ x h 2 d x d s y = 1 + ∂ y h 2 d y ds_y=\sqrt{1+\partial_y h^2}dy d s y = 1 + ∂ y h 2 d y
d A = 1 + ∂ x h 2 + ∂ y h 2 + ( ∂ x h ∂ y h ) 2 d x d y dA=\sqrt{1+\partial_x h^2 + \partial_y h^2 + (\partial_x h\partial_y h)^2}dxdy d A = 1 + ∂ x h 2 + ∂ y h 2 + ( ∂ x h ∂ y h ) 2 d x d y since ∂ x h ∂ y h ∈ O ( d ρ 2 ) \partial_x h\partial_y h\in \mathcal{O}(d\rho^2) ∂ x h ∂ y h ∈ O ( d ρ 2 ) d ρ = d x 2 + d y 2 d\rho = \sqrt{dx^2 + dy^2} d ρ = d x 2 + d y 2
Recall the generalized binomial theorem:
( 1 + x ) α = 1 + α x + α ( α − 1 ) 2 ! x 2 + ⋯ + α ( α − 1 ) . . . ( α − n + 1 ) n ! x n + ⋯ (1+x)^{\alpha}=1+\alpha x + \frac{\alpha(\alpha-1)}{2!}x^2+\cdots + \frac{\alpha(\alpha-1)...(\alpha-n+1)}{n!}x^n + \cdots ( 1 + x ) α = 1 + αx + 2 ! α ( α − 1 ) x 2 + ⋯ + n ! α ( α − 1 ) ... ( α − n + 1 ) x n + ⋯ for ∣ x ∣ < 1 |x|<1 ∣ x ∣ < 1
( 1 + x ) 1 / 2 = 1 + 1 2 x + O ( x 2 ) (1+x)^{1/2}=1+\frac12 x + \mathcal{O}(x^2) ( 1 + x ) 1/2 = 1 + 2 1 x + O ( x 2 ) Assume a small perturbation, in particular, assume ∣ ∇ h ∣ 2 ≪ 1 |\nabla h|^2 \ll 1 ∣∇ h ∣ 2 ≪ 1
1 + ∣ ∇ h ∣ 2 = 1 + 1 2 ∣ ∇ h ∣ 2 + O ( ∣ ∇ h ∣ 4 ) \sqrt{1+|\nabla h|^2} = 1 + \frac12 |\nabla h|^2 + \mathcal{O}(|\nabla h|^4) 1 + ∣∇ h ∣ 2 = 1 + 2 1 ∣∇ h ∣ 2 + O ( ∣∇ h ∣ 4 ) As a result,
A [ h ] = 1 2 ∫ d x d y ∣ ∇ h ∣ 2 + ∫ d x d y ⏟ constant A[h]=\frac12 \int dxdy |\nabla h|^2 + \underbrace{\int dxdy}_{\text{constant}} A [ h ] = 2 1 ∫ d x d y ∣∇ h ∣ 2 + constant ∫ d x d y
We need that the Euler-Lagrange equation for (44) f ( x , y , h , h x , h y ) = h x 2 + h y 2 f(x,y,h, h_x,h_y)=h_x^2 + h_y^2 f ( x , y , h , h x , h y ) = h x 2 + h y 2
∂ f ∂ h + d d x ( ∂ f ∂ h x ) + d d y ( ∂ f ∂ h y ) = 0 0 + 2 ( d h x d x + d h x d x ) = 0 ∇ 2 h = 0 \begin{align*}
\frac{\partial f}{\partial h} + \frac{d}{dx}\left(\frac{\partial f}{\partial h_x}\right) + \frac{d}{dy}\left(\frac{\partial f}{\partial h_y}\right) &= 0 \\
0 + 2\left(\frac{dh_x}{dx} + \frac{dh_x}{dx}\right) &= 0\\
\nabla^2 h &=0
\end{align*} ∂ h ∂ f + d x d ( ∂ h x ∂ f ) + d y d ( ∂ h y ∂ f ) 0 + 2 ( d x d h x + d x d h x ) ∇ 2 h = 0 = 0 = 0
The kinetic term is given by:
T = 1 2 ∫ Ω d x d y ρ 0 h t 2 T=\frac12 \int_{\Omega} dxdy \rho_0h_t^2 T = 2 1 ∫ Ω d x d y ρ 0 h t 2 The potential term is: [4]
V = ∫ Ω d x d y T ∣ ∇ h ∣ 2 V=\int_{\Omega} dxdy T|\nabla h|^2 V = ∫ Ω d x d y T ∣∇ h ∣ 2 Hence the lagrangian is:
L = T − V = ∫ Ω d x d y { 1 2 ρ 0 h t 2 − T ∣ ∇ h ∣ 2 } \begin{align*}
L &= T-V \\
&= \int_{\Omega} dxdy \left\{\frac12 \rho_0h_t^2 - T|\nabla h|^2 \right\}
\end{align*} L = T − V = ∫ Ω d x d y { 2 1 ρ 0 h t 2 − T ∣∇ h ∣ 2 } Computing the various derivatives:
∂ L ∂ h = 0 ∂ L ∂ h t = ρ 0 h t ∂ L ∂ h x = − 2 T h x ∂ L ∂ h y = − 2 T h y \frac{\partial L}{\partial h}=0 \quad \frac{\partial L}{\partial h_t}=\rho_0 h_t \quad \frac{\partial L}{\partial h_x}=-2Th_{x} \quad \frac{\partial L}{\partial h_y}=-2Th_{y} ∂ h ∂ L = 0 ∂ h t ∂ L = ρ 0 h t ∂ h x ∂ L = − 2 T h x ∂ h y ∂ L = − 2 T h y