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Apologies for the basic question.

I am trying to solve this differential equation for $w(t)$ as a function of $q(t)$: $$ \frac{dw(t)}{dt} = \mu_v \frac{R_{on}}{\delta} \frac{dq(t)}{dt}. $$ Solving with DSolve works fine:

DSolve[D[w[q[t]], t] == (Subscript[μ, v] Subscript[R, on])/ δ q'[t],
 w[q[t]], q[t]]
(* {{w[q[t]] -> C[1] + (q[t] Subscript[R, on] Subscript[μ, v])/δ}} *)

My question is how to impose the boundary condition $q = 0 \implies w = \delta/2$. The naive approach gives me an error:

DSolve[{D[w[q[t]], t] == (Subscript[μ, v] Subscript[R, on])/ δ q'[t],
 w[0] == δ/2}, w[q[t]], q[t]]

Function: parameter specification {q[t]} in Function[{q[t]}, δ/2] should be a symbol or a list of symbols.

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  • $\begingroup$ One problem is mathematical: If $w(t) = \sin t$ and $q(t) = t^2$, then $w(q(t)) = \sin t^2 \ne w(t)$. Thus D[w[q[t]], t] does not represent $dw(t)/dt$. I'm not entirely sure what you wanted, but given your first solution being "fine," perhaps this?: DSolve[{D[w[q[t]], t] == (Subscript[\[Mu], v] Subscript[R, on])/\[Delta] q'[t], w[0] == \[Delta]/2} /. t -> InverseFunction[q][q], w[q], q] $\endgroup$ – Michael E2 Nov 18 '20 at 14:52
  • $\begingroup$ Yes good point. Perhaps I am barking up the wrong tree here, DSolve might just be the wrong tool. I have a relation between the time derivatives of $w$ and $q$ given by the equation in the question, and I want to integrate both sides with respect to time to get a relation between $w$ and $q$. What I meant by "fine" is that it gives me the right relation, but perhaps for the wrong reasons given my derivative of $w$ doesn't make sense... anyway, your solution does give me the answer I was expecting. I'm a little puzzled by the substitution of $t$ for $q^{-1}(q)$ though to be honest. $\endgroup$ – chaffdog Nov 18 '20 at 15:32
  • $\begingroup$ $q=q(t) \Leftrightarrow t = q^{-1}(q)$, abusing notation to treat $q$ as both a variable and a function. $\endgroup$ – Michael E2 Nov 18 '20 at 15:58
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I'm not entirely sure what you want, but given your first solution being "fine," perhaps this?:

DSolve[
 {D[w[q[t]], t] == (Subscript[μ, v] Subscript[R, on])/δ q'[t],
  w[0] == δ/2} /. t -> InverseFunction[q][q],
  w[q], q]

InverseFunction::ifun: Inverse functions are being used. Values may be lost for multivalued inverses.

{{w[q] -> (δ^2 + 2 q Subscript[R, on] Subscript[μ, v])/(2 δ)}}
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