0
$\begingroup$

Following the Symbolic evaluation documentation, I am trying to solve the equation: $y'(x) = a*y(x), \quad y(0)=1$ where I later fix a via FindRoot, requiring that $y(2) = 2$:

ftest[unknown_?NumberQ] := First[y[2] /. NDSolve[{y'[x] == unknown*y[x], y[0] == 1}, y, {x, 0, 4}]] 
FindRoot[ftest[k] - 2, {k, 0, 4}]

This works well, but for some reason, replacing the "k" by "x" inside the Findroot outputs errors? I tried opening a new notebook and/or using Clear, but that did not solve the problem. Why does Mathematica treat x differently?

Improve this question
$\endgroup$
2
$\begingroup$

You're using x within your ftest function, which is getting replaced by FindRoot. To fix this, you need to scope your ftest function. The following works:

ftest[unknown_?NumberQ] := 
 Module[{x}, 
  First[y[2] /. 
    NDSolve[{y'[x] == unknown*y[x], y[0] == 1}, y, {x, 0, 4}]]]
FindRoot[ftest[x] - 2, {x, 0, 4}]

If you read the error message your code outputs, you see something like NDSolve::dsvar: 0. cannot be used as a variable. FindRoot appears to be replacing the x in NDSolve[..., {x, 0, 4}] with 0, resulting in NDSolve[..., {0, 0, 4}], which is where it's trying to use 0 as a variable. Thus, you need to scope your inner x with a Module so that it doesn't get replaced.

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ The OP might note also that ftest[x] translates to the ODE y'[x] == x*y[x], which is quite a bit different than when unknown is a constant. However, in this case, as Carl points out, FindRoot replaces x by a number globally. This is what is meant by this line in the docs: "FindRoot...effectively uses Block to localize variables." $\endgroup$ – Michael E2 Jan 7 '19 at 13:57
  • $\begingroup$ I see. I thought that inside function definition, the variables are automatically made local. Thank you very much for the reply. $\endgroup$ – Patrick.B Jan 7 '19 at 15:32
  • $\begingroup$ @MichaelE2 Thanks a lot for the clarification. I don't have an extremely strong knowledge of the scoping mechanisms myself! $\endgroup$ – Carl Lange Jan 7 '19 at 15:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.