Written: Aug. 2, 2024

In fond memory of our academic father (for DZ) and grandfather (for SBE),

Harry Dym (Jan 26, 1938 - July 18, 2024).

Maple package

• HDequation.txt, a Maple package to investigate, graphically and symbolically, the Harry Dym Equation of Soliton Theory fame

Plots of Selected Solutions of the Harry Dym Equation near the origin

• A plot of the solution with a₀(t)=1, a₁(t)=t, a₂(t)=t/8

• A plot of the solution with a₀(t)=1, a₁(t)=t, a₂(t)=t

• A plot of the solution with a₀(t)=1, a₁(t)=t^2, a₂(t)=t

• A plot of the solution with a₀(t)=1, a₁(t)=t^3, a₂(t)=t^2

• A plot of the solution with a₀(t)=1, a₁(t)=t^4, a₂(t)=t^2

• A plot of the solution with a₀(t)=1, a₁(t)=t, a₂(t)=t^7

• A plot of the solution with a₀(t)=1, a₁(t)=t^3, a₂(t)=t^7

• A plot of the solution with a₀(t)=1, a₁(t)=t^7, a₂(t)=1+t^7

• A plot of the solution with a₀(t)=1, a₁(t)=t, a₂(t)=exp(t)

• A plot of the solution with a₀(t)=1, a₁(t)=t^5, a₂(t)=t^10

• A plot of the solution with a₀(t)=1, a₁(t)=t^6, a₂(t)=t^4

• A plot of the solution with a₀(t)=1, a₁(t)=sin(t)^2, a₂(t)=cos(t)^2

• A plot of the solution with a₀(t)=1, a₁(t)=sech(t)^2, a₂(t)=sech(t)^2

• A plot of the solution with a₀(t)=exp(t), a₁(t)=t , a₂(t)=t

• A plot of the solution with a₀(t)=1, a₁(t)=t^7, a₂(t)=t^7

• A plot of the solution of the KDV equation, u(x,t)= sech(x-4t)^2

Sample Input and Output for HDequation.txt

• If you want to see the first 31 terms in the the variable x, in the "series expansion" of the general (generic, symbolic) solution, u(x,t), of the Harry Dym Equation

  u_t= u u_xxx

  with general initial conditions

  u(0,t)=a₀(t)   u_x(0,t)=a₁(t)   u_xx(0,t)=a₂(t)  
  where, for the sake of clarity we abberviate (d^r/ d t^r)a_i(t) by b_i,r for i=0,1,2
  the input gives the output.

  BTW, this took 74 seconds.

• If you want to see the first 41 terms in the the variable x, in the "series expansion" of the general (generic, symbolic) solution, u(x,t), of the Harry Dym Equation

  u_t= u u_xxx

  with general initial conditions

  u(0,t)=1  u_x(0,t)=a₁(t)   u_xx(0,t)=a₂(t)  
  where, for the sake of clarity we abberviate (d^r/ d t^r)a_i(t) by b_i,r for i=1,2
  the input gives the output.

  BTW, this took 134 seconds.

  ---

• If you want to see the first 31 terms in the the variable x, in the "series expansion" of the general (generic, symbolic) solution, u(x,t), of the KdV Equation

  u_t+u_xxx-6u_x=0

  with general initial conditions

  u(0,t)=a₀(t)   u_x(0,t)=a₁(t)   u_xx(0,t)=a₂(t)  
  where, for the sake of clarity we abberviate (d^r/ d t^r)a_i(t) by b_i,r for i=0,1,2
  the input gives the output.

  BTW, this took 97 seconds.

• If you want to see the first 41 terms in the the variable x, in the "series expansion" of the general (generic, symbolic) solution, u(x,t), of the KdV Equation

  u_t+u_xxx-6u_x=0

  with general initial conditions

  u(0,t)=1   u_x(0,t)=a₁(t)   u_xx(0,t)=a₂(t)  
  where, for the sake of clarity we abberviate (d^r/ d t^r)a_i(t) by b_i,r for i=1,2
  the input gives the output.

  BTW, this took 311 seconds.

Personal Journal of Shalosh B. Ekhad and Zeilberger