restart; read("Jets.s");

SUtKRVRTfk5ld35DQ35mb3J+TWFwbGV+MTN+YXN+b2Z+MDV+T2N0fjIwMTBHNiI=

`Blimit = 25000 ressize = 1000 putsize = 200 maxsize = 100`

coordinates([t,x], [u], 5);

NjZJJXVfNXRHNiJJJ3VfNHRfeEdGJEkodV90dHR4eEdGJEkodV90dHh4eEdGJEkndV90XzR4R0YkSSV1XzV4R0YkSSV1XzR0R0YkSSd1X3R0dHhHRiRJJ3VfdHR4eEdGJEkndV90eHh4R0YkSSV1XzR4R0YkSSZ1X3R0dEdGJEkmdV90dHhHRiRJJnVfdHh4R0YkSSZ1X3h4eEdGJEkldV90dEdGJEkldV90eEdGJEkldV94eEdGJEkkdV90R0YkSSR1X3hHRiQ=

jet(u,t);

SSR1X3RHNiI=

jet(u,x^3*t);

SSd1X3R4eHhHNiI=

jet(u,x^4*t);

SSd1X3RfNHhHNiI=

jet(u,x^7);

LUkkamV0RzYiNiRJInVHRiQqJEkieEdGJCIiKA==

jet(u,x*t)-jet(u,t*x);

IiIh

u_xt-u_tx;

LCZJJXVfeHRHNiIiIiJJJXVfdHhHRiQhIiI=

seq(jet(u,x^i), i=1..5);

NidJJHVfeEc2IkkldV94eEdGJEkmdV94eHhHRiRJJXVfNHhHRiRJJXVfNXhHRiQ=

TD(f,x), TD(f,x^3*t);

NiQtSSNUREc2IjYkSSJmR0YlSSJ4R0YlLUYkNiRGJyomSSJ0R0YlIiIiRigiIiQ=

Total derivatives TD are true di\357\254\200erentiations and accept any algebraic expression as the \357\254\201rst argument. They are linear, and satisfy the Leibniz rule as well as the chain rule: TD(f + g, x); TD(f*g, x); TD(sin(f), x);

LCYtSSNUREc2IjYkSSJmR0YlSSJ4R0YlIiIiLUYkNiRJImdHRiVGKEYp

LCYqJi1JI1RERzYiNiRJImZHRiZJInhHRiYiIiJJImdHRiZGKkYqKiZGKEYqLUYlNiRGK0YpRipGKg==

KiYtSSRjb3NHNiQlKnByb3RlY3RlZEdJKF9zeXNsaWJHNiI2I0kiZkdGKCIiIi1JI1RER0YoNiRGKkkieEdGKEYr

Jets does not automatically evaluate total derivatives of functions, the evaluation of total derivatives can be forced: dependence(f(t,x,u,u_x));

L0kiZkc2IjwmSSJ0R0YkSSJ1R0YkSSJ4R0YkSSR1X3hHRiQ=

TD(f,x);

LUkjVERHNiI2JEkiZkdGJEkieEdGJA==

evalTD(TD(f,x));

LCgqJi1JI3BkRzYiNiRJImZHRiZJInVHRiYiIiJJJHVfeEdGJkYqRiotRiU2JEYoSSJ4R0YmRioqJi1GJTYkRihGK0YqSSV1X3h4R0YmRipGKg==

pd(g,x*u_x^2);

LUkjcGRHNiI2JEkiZ0dGJComSSJ4R0YkIiIiSSR1X3hHRiQiIiM=

pd(TD(f,x),u_xx);

LUkjcGRHNiI2JEkiZkdGJEkkdV94R0Yk

pd(TD(f,x),u_x);

LCYtSSNUREc2IjYkLUkjcGRHRiU2JEkiZkdGJUkkdV94R0YlSSJ4R0YlIiIiLUYoNiRGKkkidUdGJUYt

pd(TD(f,x^2),u_xx);

LCYtSSNUREc2IjYkLUkjcGRHRiU2JEkiZkdGJUkkdV94R0YlSSJ4R0YlIiIjLUYoNiRGKkkidUdGJSIiIg==

Let us verify that a := 2*k*(sech((k)^(1/2)*(x + 4*k*t)))^2;

LCQqJkkia0c2IiIiIi1JJXNlY2hHNiQlKnByb3RlY3RlZEdJKF9zeXNsaWJHRiU2IyomRiQjRiYiIiMsJkkieEdGJUYmKiZGJEYmSSJ0R0YlRiYiIiVGJkYvRi8=

is a solution of the KdV equation KdV := -u_t + u_xxx + 6*u*u_x;

LChJJHVfdEc2IiEiIkkmdV94eHhHRiQiIiIqJkkidUdGJEYnSSR1X3hHRiRGJyIiJw==

subs(u = a, KdV);

LCgtSSRqZXRHNiI2JCwkKiZJImtHRiUiIiItSSVzZWNoRzYkJSpwcm90ZWN0ZWRHSShfc3lzbGliR0YlNiMqJkYpI0YqIiIjLCZJInhHRiVGKiomRilGKkkidEdGJUYqIiIlRipGM0YzRjchIiItRiQ2JEYnKiRGNSIiJEYqKihGKUYqRitGMy1GJDYkRidGNUYqIiM3

dependence(k());

NiQvSSJmRzYiPCZJInRHRiVJInVHRiVJInhHRiVJJHVfeEdGJS9JImtHRiU8Ig==

subs(u = a, KdV);

LCgtSSRqZXRHNiI2JCwkKiZJImtHRiUiIiItSSVzZWNoRzYkJSpwcm90ZWN0ZWRHSShfc3lzbGliR0YlNiMqJkYpI0YqIiIjLCZJInhHRiVGKiomRilGKkkidEdGJUYqIiIlRipGM0YzRjchIiItRiQ2JEYnKiRGNSIiJEYqKihGKUYqRitGMy1GJDYkRidGNUYqIiM3

Problem: The procedure jet requires its \357\254\201rst argument to be a \357\254\201bre variable! Solution: The procedure convert(expression, TD )e\357\254\200ectively replaces u with a, jet(u,x) with TD(a,x), etc: convert(KdV, TD, u = a);

LCwqKkkia0c2IiMiIiYiIiMtSSVzZWNoRzYkJSpwcm90ZWN0ZWRHSShfc3lzbGliR0YlNiMqJkYkIyIiIkYoLCZJInhHRiVGMSomRiRGMUkidEdGJUYxIiIlRjFGMS1JJXNpbmhHRitGLkYxLUklY29zaEdGK0YuISIjIiNPKihGN0YxRjkhIiRGJEYmIiM3KihGNyIiJEYkRiZGOSEiJiEjQyoqRilGMUY3RkFGJEYmRjkhIiVGQyoqRiRGJkYpRkFGN0YxRjlGOyEjWw==

simpl(%);

LCQqKi1JJXNpbmhHNiQlKnByb3RlY3RlZEdJKF9zeXNsaWJHNiI2IyomSSJrR0YpIyIiIiIiIywmSSJ4R0YpRi4qJkYsRi5JInRHRilGLiIiJUYuRi5GLCMiIiZGLywsKiYtSSVzZWNoR0YmRipGLi1JJWNvc2hHRiZGKiIiJCEiJCokRjtGLyEiIiokRiRGL0YvKihGOUYuRiRGL0Y7Ri5GLyomRjlGPUY7Rj1GNEYuRjshIiYhIzc=

simplify(%);

IiIh

E := evalTD(TD(f,x)); F := convert(E, diff, u_x=u1, u_xx=u2); lprint(F); pdsolve(F); The additional arguments indicate a substitution to be applied to jet variables (recall that jet variables are unevaluated function calls while diff requires the indeterminates to be names). G := convert(E, diff); lprint(G); pdsolve(G); lprint(convert(E, diff)); Diff(a,u_x,u_x,u_xx) - pd(a, u_x^2*u_xx);