Dear Emmanuel,
The results of your approach might not be pretty. Unfortunatly I think they
seem to be wrong too.
See:
sage: r, t, alpha, delta, r = var('r, t, alpha, delta, r')
sage: k, x, lambda_ = function("k, x, lambda_")
sage: h1(t) = x(t)*e^(-r*t)
sage: D[0](h1)(x(t))
-r*e^(-r*x(t))*x(x(t)) + e^(-r*x(t))*D[0](x)(x(t))
Kindly note the "x(x(t))" in the first term. :-/
Greetings,
Bernd
Bernd Breitschaedel schrieb am Mittwoch, 26. Februar 2025 um 12:43:34 UTC+1:
> Thank you Eric & Emmanuel for your solution proposals.
>
> Regards,
> Bernd
>
> Emmanuel Charpentier schrieb am Dienstag, 25. Februar 2025 um 18:08:15
> UTC+1:
>
>> Complement : Sage’s notation for derivatives is somewhat baroque, and
>> this does not help in the present case.
>>
>> Here, both k and x are functions of t ; writing h as a function of x and
>> k is just an intricate way to write a function of t, unique independent
>> variable.
>>
>> As for notations : what we’d “manually” write in this case would be :
>> [image: \displaystyle{\frac{\partial h}{\partial
>> t}}\,=\,\displaystyle{\frac{\partial h}{\partial x}\,\frac{\partial
>> x}{\partial t}}]
>>
>> and we could deduce
>>
>> [image: \displaystyle{\frac{\partial h}{\partial
>> x}}\,=\,\frac{\frac{\partial h}{\partial t}}{\frac{\partial x}{\partial t}}]
>> .
>>
>> Sage doesn’t have a direct notation for denoting derivative with respect
>> to another function ; instead, it uses the Doperator, which creates
>> derivative operators :
>> sage: t=var("t") sage: h, x=function("h, x") sage: diff(h(x(t)), t)
>> D[0](h)(x(t))*diff(x(t), t) sage: diff(h(x(t)), t)/diff(x(t), t)
>> D[0](h)(x(t))
>>
>> Eric’s notation is an elegant way to work around this problem.
>>
>> The use of standard notations leads to :
>> sage: r, t, alpha, delta, r = var('r, t, alpha, delta, r') sage: k, x,
>> lambda_ = function("k, x, lambda_") sage: h(t) = k(t)*(1-x(t))*e^(-r*t) +
>> lambda_(t)*(k(t)^alpha*x(t)-delta*k(t)) sage: D[0](h)(x(t)) r*(x(x(t)) -
>> 1)*e^(-r*x(t))*k(x(t)) - (x(x(t)) - 1)*e^(-r*x(t))*D[0](k)(x(t)) -
>> e^(-r*x(t))*k(x(t))*D[0](x)(x(t)) + (alpha*k(x(t))^(alpha -
>> 1)*x(x(t))*D[0](k)(x(t)) - delta*D[0](k)(x(t)) +
>> k(x(t))^alpha*D[0](x)(x(t)))*lambda_(x(t)) - (delta*k(x(t)) -
>> k(x(t))^alpha*x(x(t)))*D[0](lambda_)(x(t))
>>
>> [image: \frac{\partial h(t)}{\partial x(t)}\,=\,r
>> {\left(x\left(x\left(t\right)\right) - 1\right)} e^{\left(-r
>> x\left(t\right)\right)} k\left(x\left(t\right)\right) -
>> {\left(x\left(x\left(t\right)\right) - 1\right)} e^{\left(-r
>> x\left(t\right)\right)}
>> \mathrm{D}_{0}\left(k\right)\left(x\left(t\right)\right) - e^{\left(-r
>> x\left(t\right)\right)} k\left(x\left(t\right)\right)
>> \mathrm{D}_{0}\left(x\right)\left(x\left(t\right)\right) + {\left(\alpha
>> k\left(x\left(t\right)\right)^{\alpha - 1} x\left(x\left(t\right)\right)
>> \mathrm{D}_{0}\left(k\right)\left(x\left(t\right)\right) - \delta
>> \mathrm{D}_{0}\left(k\right)\left(x\left(t\right)\right) +
>> k\left(x\left(t\right)\right)^{\alpha}
>> \mathrm{D}_{0}\left(x\right)\left(x\left(t\right)\right)\right)}
>> \lambda\left(x\left(t\right)\right) - {\left(\delta
>> k\left(x\left(t\right)\right) - k\left(x\left(t\right)\right)^{\alpha}
>> x\left(x\left(t\right)\right)\right)}
>> \mathrm{D}_{0}\left(\lambda\right)\left(x\left(t\right)\right).]
>> sage: D[0](h)(k(t)) r*(x(k(t)) - 1)*e^(-r*k(t))*k(k(t)) - (x(k(t)) -
>> 1)*e^(-r*k(t))*D[0](k)(k(t)) - e^(-r*k(t))*k(k(t))*D[0](x)(k(t)) +
>> (alpha*k(k(t))^(alpha - 1)*x(k(t))*D[0](k)(k(t)) - delta*D[0](k)(k(t)) +
>> k(k(t))^alpha*D[0](x)(k(t)))*lambda_(k(t)) - (delta*k(k(t)) -
>> k(k(t))^alpha*x(k(t)))*D[0](lambda_)(k(t))
>>
>> [image: \frac{\partial h(t)}{\partial k(t)}\,=\,r
>> {\left(x\left(k\left(t\right)\right) - 1\right)} e^{\left(-r
>> k\left(t\right)\right)} k\left(k\left(t\right)\right) -
>> {\left(x\left(k\left(t\right)\right) - 1\right)} e^{\left(-r
>> k\left(t\right)\right)}
>> \mathrm{D}_{0}\left(k\right)\left(k\left(t\right)\right) - e^{\left(-r
>> k\left(t\right)\right)} k\left(k\left(t\right)\right)
>> \mathrm{D}_{0}\left(x\right)\left(k\left(t\right)\right) + {\left(\alpha
>> k\left(k\left(t\right)\right)^{\alpha - 1} x\left(k\left(t\right)\right)
>> \mathrm{D}_{0}\left(k\right)\left(k\left(t\right)\right) - \delta
>> \mathrm{D}_{0}\left(k\right)\left(k\left(t\right)\right) +
>> k\left(k\left(t\right)\right)^{\alpha}
>> \mathrm{D}_{0}\left(x\right)\left(k\left(t\right)\right)\right)}
>> \lambda\left(k\left(t\right)\right) - {\left(\delta
>> k\left(k\left(t\right)\right) - k\left(k\left(t\right)\right)^{\alpha}
>> x\left(k\left(t\right)\right)\right)}
>> \mathrm{D}_{0}\left(\lambda\right)\left(k\left(t\right)\right).]
>>
>> Not a pretty sight…
>>
>> Le mardi 25 février 2025 à 11:01:27 UTC+1, egourg...@gmail.com a écrit :
>>
>>> Hi,
>>>
>>> Le lundi 24 février 2025 à 17:39:30 UTC+1, bernd.bre...@gmail.com a
>>> écrit :
>>>
>>> So how should i define H in 'In [2]', so that 'In [4]' shows the
>>> expected result.
>>>
>>>
>>> Basically you should define H as a symbolic expression, not a function,
>>> and make a distinction between functions and symbolic variables.
>>> A solution is
>>>
>>> t = var('t')
>>> r = var('r')
>>> delta = var('delta')
>>> alpha = var('alpha')
>>>
>>> x = function('x')
>>> k = function('k')
>>> lambda_ = function('lambda_')
>>>
>>> X = var('X')
>>> K = var('K')
>>> L = var('L')
>>> H = K*(1 - X)*exp(-r*t) + L*(K^alpha*X - delta*K)
>>>
>>> to_functions = {X: x(t), K: k(t), L: lambda_(t)} # a dictionary to
>>> perform substitutions
>>>
>>> eq1 = (diff(H, X).subs(to_functions) == 0)
>>> eq2 = (diff(H, K).subs(to_functions) == - diff(lambda_(t), t))
>>>
>>> Best wishes,
>>>
>>> Eric. .
>>>
>>
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