I would like to change variable in a differential equation within Maxima, e.g.:
(%i1) diff(y(x),x);
d
(%o1) -- (y(x))
dx
(%i2) subst([x=a*z],%);
d
(%o2) ---- (y(a z))
da z
However I do not know how to proceed and get diff(y(a*z),z)/a. The Maxima manual mentions the question, but refers to the at command where this topic is not dealt with.
There are several methods for solve this task in maxima, let's use pdiff. first we solve two odes in maxima
(%i1) eq:'diff(y,x) = -y;
dy
(%o1) -- = - y
dx
(%i2) ode2(eq, y, x);
- x
(%o2) y = %c %e
(%i3) sol: ic1(%, x= 1, y= 8);
1 - x
(%o3) y = 8 %e
maxima has no problem with complex ode like this:
(%i4) eqcv : 'diff(y,x) = (x*y) / (3*x^2 - y^4);
dy x y
(%o4) -- = ---------
dx 2 4
3 x - y
(%i5) ode2(eqcv, y, x);
4 2
y - x
(%o5) - ------- = %c
6
2 y
(%i6) sol: ic1(%, x= 2, y= 1);
4 2
y - x 3
(%o6) - ------- = -
6 2
2 y
but let's use variable changing on this equation using pdiff
y = z ^ a
(%i1) load(pdiff)$
(%i5) eqcv : diff(y(x),x) = (x*y(x)) / (3*x^2 - y(x)^4);
x y(x)
(%o5) y (x) = ------------
(1) 2 4
3 x - y (x)
(%i6) eqcv, y(x) := z(x)^a;
a
a - 1 x z (x)
(%o6) a z(x) z (x) = --------------
(1) 2 4 a
3 x - z(x)
then we can simplify:
(%i7) % / z(x)^(a -1);
x z(x)
(%o7) a z (x) = --------------
(1) 2 4 a
3 x - z(x)
(%i8) % / a;
x z(x)
(%o8) z (x) = ------------------
(1) 2 4 a
a (3 x - z(x) )
and now we can apply a = 1/2
(%i50) %, a = 1/2;
2 x z(x)
(%o50) z (x) = ------------
(1) 2 2
3 x - z (x)
start to solve:
(%i51) eq : %;
2 x z(x)
(%o51) z (x) = ------------
(1) 2 2
3 x - z (x)
(%i52) eq;
2 x z(x)
(%o52) z (x) = ------------
(1) 2 2
3 x - z (x)
(%i56) eq : convert_to_diff(eq);
d 2 x z(x)
(%o56) -- (z(x)) = ------------
dx 2 2
3 x - z (x)
(%i57) depends(z,x);
(%o57) [z(x)]
(%i58) eq : subst(z,z(x),eq);
dz 2 x z
(%o58) -- = ---------
dx 2 2
3 x - z
(%i59) ode2(eq,z,x);
2 2
z - x
(%o59) - ------- = %c
3
z
you can also apply this to partial differential equations example from a mailing list:
I am dealing with the following partial differential equation for the function T(r,t):
diff( T, t ) - T^2 / r^2 * diff( ( r^2 * T^(sigma) * diff(T, r) + q / T ), r ) = 0
I now want to introduce a new variable
eta = r / t^(1/3)
and a new function
Psi(eta) = t^(1/3) * T^(1+sigma)
and I would like Maxima to transform the partial differential equation for T(r,t) stated herebove into a differential equation for Psi(eta).
solution:
(%i1) load(pdiff)$
(%i2) de : diff(f(x,y),x,2) + diff(f(x,y),y,2)$
(%i3) de, f(x,y) := g(x/y);
(%o3) g[(2)](x/y)/y^2+(2*x*g[(1)](x/y))/y^3+(x^2*g[(2)](x/y))/y^4
(%i4) y^2 * %;
(%o4) (g[(2)](x/y)/y^2+(2*x*g[(1)](x/y))/y^3+(x^2*g[(2)](x/y))/y^4)*y^2
(%i5) ratsubst(w,x/y,%);
(%o5) (w^2+1)*g[(2)](w)+2*w*g[(1)](w)
(%i6) nde : convert_to_diff(%);
(%o6) 2*w*('diff(g(w),w,1))+(w^2+1)*('diff(g(w),w,2))
(%i7) depends(g,w)$
(%i8) nde : subst(g,g(w),nde);
(%o8) ('diff(g,w,2))*(w^2+1)+2*('diff(g,w,1))*w
(%i9) ode2(nde,g,w);
(%o9) g=%k1*atan(w)+%k2
so in your case:
(%i11) load(pdiff)$
(%i12) eq : diff(y(x),x);
(%o12) y (x)
(1)
(%i13) eq, y(x) := t(a*x);
(%o13) a t (a x)
(1)