Partial differential equations
Takehito Yokoyama
Department of Physics, Tokyo Institute of Technology,
2-12-1 Ookayama, Meguro-ku, Tokyo 152-8551, Japan(Dated: January 30, 2014)

PARTIAL DIFFERENTIAL EQUATIONS

Types of partial differential equations are summarized as follows.
∂2u∂2u= c2 2 Wave equation
2∂t
∂x
2∂u
∂u
∂2u
= c2 2 − kWave equation with friction
∂t2
∂x
∂t
∂2u
∂u
∂2u
= c2 2 − k− hu Transmission line equation
2∂t
∂x
∂t4
∂2u2∂ u=−bBeam equation.
∂t2
∂x4
∂u
∂2u
=D 2∂t∂xDiffusion equation∂2u∂u= D 2 − ku Diffusion equation with lateral concentraton loss
∂t
∂x
∂u
∂u
∂2u
=D 2 −v
Diffusion − convection equation
∂t
∂x
∂x
∂u
∂2u
∂u
=D 2 −uBurgers equation.
∂t
∂x
∂x

(1)
(2)
(3)
(4)

(5)
(6)
(7)
(8)

Note

Some notes on the partial differential equations listed above are presented. Some transformations can simplify theequations:
In the transmission line equation, this transformation eliminates the ut termu = e−kt/2 w.

(9)

In the diffusion equation with lateral concentraton loss, this transformation reduces it to the simple diffusion
equationu = e−kt w.

(10)

In the diffusion-convection equation, this transformation reduces it to the simple diffusion equationu = e−v[x−vt/2]/2D w.

(11)

The solution of the Burgers equation can be constructed by the following transformation where ψ obeys the simple
diffusion equation :u = −2D∂ln ψ.
∂x(12)

This transformation is called the Hopf-Cole transformation. The solution of the initial value problem of the Burgers
equation thus reads [u(x, 0) = f (x)] Z ∞Z y∂11(x − y)2u = −2Dln √+exp −f (z)dz dy .
(13)
∂x
4Dt
2D 02 πDt −∞22FIG. 1: Plots of u(x, t). Left: u(x, 0) = e−x . Middle: u(x, 0) = tanh(10x). Right: u(x, 0) = 1 − 2 tanh(10x).

Typical solutions of the Burgers equation are shown in Fig. 1. Interestingly, we see a shock wave solution in theright panel. Intuitively, this is because the velocity of the wave is given by u. Thus, the larger the magnitude of u,
the faster the wave travels. This interpretation is consistent with Fig. 1.
Derivation of the beam equation

Consider a beam along x-axis. We have two assumptions: No cross-sectional deformation:
εyy = εzz = εyz = 0.

(14)

and Bernoulli-Euler assumption (Bernoulli-Euler Beam):
εxy = εxz = 0.

(15)Here, the strain tensor εij is defined asεij =12



∂ui
∂uj+
∂xj
∂xi



.

(16)

ui is the displacement vector. The diagonal elements represent extensional strain while the off-diagonal elements
denote shearing strain.
Let u(x) a displacement along the x-axis, w(x) a deflection along y-axis, and θ(x)(≪ 1) a slope of the beam. Then,
we haveux = u − y sin θ ≃ u − yθ, uy = w + y(1 − cos θ) ≃ w.

(17)

Since εxy = 0, we obtain
dw.
dx

(18)

d2 w
du−y 2.
dx
dx

(19)

θ=
Therefore, we arrive at
εxx =According to the Hooke’s law, we obtain the normal stress of the beam:
d2 w
du−y 2σxx = Eεxx = Edx
dx
where E is the Young’s modulus. The bending moment of the beam M can be calculated asZdud2 w
M=yσxx dA = EJy− EI 2 .
dxdx
A(20)

(21)

3Here, dA is the infinitesimal section area. Jy and I denote the first sectional moment and the sectional moment ofinertia, respectively:ZZ
y 2 dA.
(22)
ydA, I =Jy =AA

If the x-axis goes through a centroid of the section, we have Jy = 0.
Now, consider the equation of motion of the beam:
ρA∂2w∂V=+p2∂t∂x(23)

where ρ is the density, p is the external force, V is the shear force. The equilibrium equation of moments reads∂M= V.
∂x(24)∂4w∂2w=−EI+ p.
∂t2∂x4(25)Thus, we finally obtain the beam equation:
ρA

In the presence of viscous drag, the normal stress of the beam is given by
σxx = Eyκ + ηy∂κ.
∂t(26)
2Here, η is the viscosity coefficient, and κ represents the curvature of the beam, κ = − ∂∂xw2 . The equation of motion ofthe beam then becomes
ρA

∂2w
∂4w
∂5w
= −EI 4 + −ηI+ p.
2∂t
∂x
∂t∂x4

Alternatively, the kinetic and potential energies of the beam, K and U , are respectively given by2
 2 2ZZZZ111∂w∂ w
1dx, U =ρAσxx εxx dAdx =M κdx =EIdx.
K=2
∂t
2
22
∂x2

(27)

(28)

Variation of the action with respect to w also gives the beam equation.
OTHER PARTIAL DIFFERENTIAL EQUATIONS

We know the solutions of ∂u∂t = L[u] for L[u] = ux and uxx . As a next step, consider the case of L[u] = uxxx :∂3u∂u.
=∂t∂x3The solution for u(x, 0) = f (x) can be obtained by Fourier transforming the equation:Z ∞F (x − y, t)f (y)dy,u=−∞
Z ∞
1x
3−1/3.
F (x, t) =exp ik t − ikx dk = (3t)
Ai −
2π −∞
(3t)1/3(29)

(30)
(31)

Here, Ai(x) is the Airy function defined as
Ai(x) =12π 3k
exp i + ikx dk.
3−∞

Z ∞

(32)

Incorporating uux and uxxx terms, we arrive at the KdV equation:∂u∂u ∂ 3 u
= 0.
+u+∂t
∂x ∂x3(33)4FIG. 2: The solutions of the KdV equation for u(x, 0) = 8sech2 (0.5x). Time evolution of the wave from upper left to lowerright.

This equation has a traveling-wave solution given byu(x − vt) = 3vsech2

rv(x − vt + c)2(34)where c is a constant.
The KdV equation can describes solitons. See the solutions of the KdV equation in Fig. 2. Solitary waves propagate
with independent velocities.
∂ψ (dubbed the Miura transformation)
The solution of the following equation can be constructed by u = ψ 2 + ∂x∂u ∂ 3 u∂u=0− 6u+∂t∂x ∂x3(35)∂ψ∂ψ ∂ 3 ψ= 0.
− 6ψ 2+∂t∂x∂x3(36)where ψ satisfies

Finally, let us consider the sine-Gordon equation:2∂2u2∂ u=c− sin u.
∂t2∂x2(37)This equation has a traveling-wave solution of the form u(x − vt). Then, the equation is reduced tou′′ =1c2 − v 2

sin u.

(38)

The solution is then given by−1u(x − vt) = 4 tan
x − vtexp ± √c2 − v 2(39)for c2 > v 2 under the boundary condition cos u → 1 (x → ∞). For c2 = v 2 , the solution is trivial. For c2 < v 2 , theequation is reduced to that of a simple pendulum, which can be written with elliptic functions as we saw in anothernote.