## 8 Reputation

12 years, 259 days
Munich, Germany

## it might not be right...

As you said, the answer will be simplily zero, which is not right. Secondly, the point to give b.c. is to work out the constants automatically. Later, I will add more items in the expression. That could be problematic.

## it might not be right...

As you said, the answer will be simplily zero, which is not right. Secondly, the point to give b.c. is to work out the constants automatically. Later, I will add more items in the expression. That could be problematic.

## Thanks...

Problem seems to have been solved. Thank you.

## Thanks...

Problem seems to have been solved. Thank you.

## Examples...

> restart;
>
> with(PDEtools);
>
> with(Units[Standard]);
> with(ScientificConstants);
>
> #
>
> # Electron velocity distribution:
>
> f[e] := A[e]*exp(-B[e]*(((1/2)*m[e]*v^2)^2+((1/2)*m[e]*v^2*m[e])*`&varpi;`^2*lambda[e]^2)/T[e]^2);
> f[e, 1] := f[e]/A[e];
>
> f[e, 2] := (1/2)*f[e, 1]*m[e]*v^2;
> expr[e, 1] := `assuming`([4*Pi*(int(f[e, 1]*v^2, v = 0 .. infinity))], [positive]);
> expr[e, 2] := `assuming`([4*Pi*(int(f[e, 2]*v^2, v = 0 .. infinity))], [positive]);
> expr[e, lhs] := simplify(expr[e, 1]/expr[e, 2]);
> expr[e, rhs] := 1/T[e];
> eq[e, 1] := expr[e, lhs]/expr[e, rhs] = 1;
>
> # Constants:
> `&varpi;` := 22*10^6*Unit('Hz');
> T[e] := 3*Unit('eV');
> lambda[e] := Unit('mm');
> m[e] := evalf(Constant(m[e], units));
>
> eq[e, 2] := simplify(eq[e, 1]);
> B[e] := fsolve(eq[e, 2], B[e]);
> A[e, res1] := simplify(1/expr[e, 1]);
> A[e, res2] := simplify(T[e]/expr[e, 2]);
>

> restart;
> with(PDEtools);
>
>
> # Electron velocity distribution:
> f[e] := A[e]*exp(-B[e]*(((1/2)*m[e]*v^2)^2+((1/2)*m[e]*v^2*m[e])*`&varpi;`^2*lambda[e]^2)/T[e]^2);
> f[e, 1] := f[e]/A[e];
>
> f[e, 2] := (1/2)*f[e, 1]*m[e]*v^2;
> expr[e, 1] := `assuming`([4*Pi*(int(f[e, 1]*v^2, v = 0 .. infinity))], [positive]);
> expr[e, 2] := `assuming`([4*Pi*(int(f[e, 2]*v^2, v = 0 .. infinity))], [positive]);
> expr[e, lhs] := simplify(expr[e, 1]/expr[e, 2]);
> expr[e, rhs] := 1/T[e];
> eq[e, 1] := expr[e, lhs] = expr[e, rhs];
>
> # Constants:
> `&varpi;` := 22*10^6;
> T[e] := (3*1.6)*10^(-19);
> lambda[e] := 0.1e-2;
> m[e] := 9.1*10^(-31);
>
>
>
> eq[e, 2] := simplify(eq[e, 1]);
> B[e] := fsolve(eq[e, 2], B[e]);
> A[e, res1] := simplify(1/expr[e, 1]);
> A[e, res2] := simplify(T[e]/expr[e, 2]);
>

I am solving two coefficients (A_e and B_e) in one velocity distribution function (3D) with two closures (1: normalization 2: average kinetic energe).  I solved B_e first by dividing 2 closures which eliminate A_e first and then recalculate A_e by substituting B_e to either of two closures. See file question_1 and questions_2 (describing the same question with/without using physics units). In question_2, I do not use units system. However, the A_e obtained from two closures are 3 order different, which makes no sense at all. In question_1 I use units then A_e from the 2nd closures won't return anything. Can anyone explain why it happens and suggest some tricks to avoid it? Thank you in advance.

## @Scott03 Sure. I will upload a maple in...

@Scott03

Sure. I will upload a maple input file in this post to shown what problem I am dealing with. :) Thanks

## I checked. Theoretically expr_1 and expr...

I checked. Theoretically expr_1 and expr_2 should not be zero at all. But thanks for the hint anyway. I will upload the input file so maybe you can have a look.

## I checked. Theoretically expr_1 and expr...

I checked. Theoretically expr_1 and expr_2 should not be zero at all. But thanks for the hint anyway. I will upload the input file so maybe you can have a look.

## @Alejandro Jakubi  Thank you for y...

Thank you for your hint. It is very helpful. I won't stick to this point.

## @Alejandro Jakubi  Thank you for y...

Thank you for your hint. It is very helpful. I won't stick to this point.

## maple gives another result...

actually, B,C,D in my example have already been simplified. In my original work, it contains a long expression. As I use 'dchange', it returns a lim (actually, I tried use 'isolate' to reform the expression but did not really work), which is different form from what you gave in Math7 as well as what I got in Maple if I use a simplified form.

## maple gives another result...

actually, B,C,D in my example have already been simplified. In my original work, it contains a long expression. As I use 'dchange', it returns a lim (actually, I tried use 'isolate' to reform the expression but did not really work), which is different form from what you gave in Math7 as well as what I got in Maple if I use a simplified form.

## here is an example...

To my opinion, those two integrals share the same feature. but why only eq6 can be evaluated? ## But if the form is changed somehow manua...

as in title, I changed the form by put some symbols together and then the maple evaluate, which confuses me a lot. I did not change the kernel of the integral actually.

In fact, I could have done it in a stupid way that first simplify it then resubsitute. But it creats a lot of work

## But if the form is changed somehow manua...

as in title, I changed the form by put some symbols together and then the maple evaluate, which confuses me a lot. I did not change the kernel of the integral actually.

In fact, I could have done it in a stupid way that first simplify it then resubsitute. But it creats a lot of work

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