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My problem is that I am trying to find a Numerical solution to a function of the form f(x,c)== K for x for some given values of c, K, and plot this function over a wide range of values of c.

My function f(x,c) is of the form Integrate[g(x,y,c),{y,a,b}], where a and b are some numerical values. This integration however cannot be evaluated symbolically in x and c,as g is a really complicated function and hence can only be performed numerically.

So if I write my function as f(x_,c_)= NIntegrate[g(x,y,c),{y,a,b}], I get some numerical value of f(x,c) for some numerical values of x and c. (If I write it as Integrate[g(x,y,c),{y,a,b}] I donot get any value for f(x,c)).

But with this form of the function, I cannot get any solution using NSolve, or FindRoot. (FindRoot doesn't converge, and it may not be practical if I plan to plot it for a wide range of values of c).

So at this point, plotting the solutions seems like a far fetched dream. I apologize for not providing details of the functions, as the function g arises from a lot of complicated functions.

Thanks in advance for your help

EDIT:

I decided to post my attempt, and some extent of the functions, based on one of the Solutions by Carl. For my case arh is a known function, and gamma and w are numerical values:

eqn = Inactive[Integrate][
 g1[Hi[mx], a, mx]*mx, {a, 1, arh[\[Gamma], w]}] + 
Inactive[Integrate][
 g2[Hi[mx], a, mx]*mx, {a, arh[\[Gamma], w], \[Infinity]}] ==0.1198/(9.2*10^24);

Distribute /@ D[eqn, mx]

int0[Hi_, mx_?NumberQ] :=NIntegrate[g1[Hi, a, mx], {a, 1, arh[\[Gamma], w]}]
int1[Hi_, mx_?NumberQ] := 
 NIntegrate[
  Derivative[0, 0, 1][g1][Hi, a, mx], {a, 1, arh[\[Gamma], w]}]
int2[Hi_, mx_?NumberQ] := 
 NIntegrate[
  Derivative[1, 0, 0][g1][Hi, a, mx], {a, 1, arh[\[Gamma], w]}]

int02[Hi_, mx_?NumberQ] := 
 NIntegrate[g2[Hi, a, mx], {a, arh[\[Gamma], w], \[Infinity]}]
int12[Hi_, mx_?NumberQ] := 
 NIntegrate[
  Derivative[0, 0, 1][g2][Hi, a, mx], {a, 
   arh[\[Gamma], w], \[Infinity]}]
int22[Hi_, mx_?NumberQ] := 
 NIntegrate[
  Derivative[1, 0, 0][g2][Hi, a, mx], {a, 
   arh[\[Gamma], w], \[Infinity]}]

My Functions are of the form

g1[Hi_, a_, mx_] := 
  a^2/(Hrh[a, \[Gamma], w, Hi]*Trh[\[Gamma], Hi]^3) \[Sigma]vtotal[
    a, \[Gamma], w, Hi, mx, Trh11] nXeq[a, \[Gamma], w, Hi, mx, 
    Trh11]^2;
g2[Hi_, a_, mx_] := 
  a^2/(H[a, \[Gamma], w, Hi]*Trh[\[Gamma], Hi]^3) \[Sigma]vtotal[
    a, \[Gamma], w, Hi, mx, T] nXeq[a, \[Gamma], w, Hi, mx, T]^2;

And I try to Solve:

sol = NDSolveValue[{(int0[Hi[mx], a, mx] + 
        mx (int1[Hi[mx], mx] + 
           Derivative[1][Hi][mx] int2[Hi[mx], mx])) + (int02[Hi[mx], 
         a, mx] + 
        mx (int12[Hi[mx], mx] + 
           Derivative[1][Hi][mx] int22[Hi[mx], mx])) == 0, 
    Hi[10^-7] == 10^-9}, Hi, {mx, .1, 10^-19}];

This returns an error NDSolveValue::ndnum: Encountered non-numerical value for a derivative at mx == 1.`*^-7.

NOTE: Trh, and \sigmavtotal and nXeq are defined previously, and are of the form: \[Sigma]vtotal[a_, \[Gamma]_, w_, Hi_, mx_, T_Symbol] nXeq[a_, \[Gamma]_, w_, Hi_, mx_, T_Symbol] Trh[\[Gamma]_, Hi_]

and the other functions are of the form T[a_, \[Gamma]_, w_, Hi_] and Trh11[a_, \[Gamma]_, w_, Hi_]

I want to plot for mx: 0.1 to 10^-19

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  • 1
    $\begingroup$ Share your attempt otherwise we will be not of much help. $\endgroup$ – zhk Oct 13 '18 at 4:12
  • $\begingroup$ Recommend you provide code that fully runs to where we can replicate your error (I get a different error due to arh not a valid limit of integration) $\endgroup$ – MikeY Dec 13 '18 at 1:06
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Without a concrete example I can only guess at an approach that might work.

g[x_, y_, c_] := 1/Sqrt[x^2 + y^2 + c^2]

a = 0; b = 1;

Functions that use numeric techniques should have their arguments restricted to numeric values.

f[x_?NumericQ, c_?NumericQ] := NIntegrate[g[x, y, c], {y, a, b}]

soln[c_?NumericQ, K_?NumericQ] := x /. FindRoot[f[x, c] == K, {x, 1/2}]

Verifying that soln produces a numeric value

soln[1/2, 1/2]

(* 1.85275 *)

Plotting soln for a range of c and K values

Plot3D[soln[c, K], {c, 0, 1}, {K, 1/10, 1}]

enter image description here

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  • $\begingroup$ I posted more details of my specific question. Thanks in Advance $\endgroup$ – TheInvoker Oct 13 '18 at 19:24
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I would use NDSolveValue for this kind of problem. You are trying to find the dependence of x on c, so let's use x[c] to make this dependence explicit. Then your equation is:

eqn = Inactive[Integrate][g[x[c], y, c], {y, 1, 2}] == k;

where I arbitrarily chose $a=1$ and $b=2$. Let's differentiate this equation to obtain an ODE:

Distribute /@ D[eqn, c] //TeXForm

$\int _1^2x'(c) g^{(1,0,0)}(x(c),y,c)dy+\int _1^2g^{(0,0,1)}(x(c),y,c)dy=0$

Define:

int0[x_, c_?NumberQ] := NIntegrate[g[x, y, c], {y, 1, 2}]
int1[x_, c_?NumberQ] := NIntegrate[Derivative[0,0,1][g][x, y, c], {y, 1, 2}]
int2[x_, c_?NumberQ] := NIntegrate[Derivative[1,0,0][g][x, y, c], {y, 1, 2}]

where the _?NumberQ test prevents premature evaluations of NIntegrate with non-numeric integrands. Finally, let's come up with a sample g:

g[x_, y_, c_] := BesselI[1, x y] BesselJ[1, c y]

Then, the ODE to be solved is:

sol = NDSolveValue[
    {
    int2[x[c], c] x'[c] + int1[x[c], c] == 0,
    x[1] == 1
    },
    x,
    {c, .1, 2}
];

Let's check whether sol satisfies the required equation:

Plot[int0[sol[c], c], {c, .1, 2}, PlotRange -> 1]

enter image description here

And a visualization of the solution:

Plot[sol[c], {c, .1, 2}]

enter image description here

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  • $\begingroup$ This solution looks really promising. The only problem is that I donot know of any initial condition like x[1]==1 . I can guess an order, like x[10^-7]==10^-10. Here is what my eqn looks like:= eqn = Inactive[Integrate][ g1[Hi[mx], a, mx]*mx, {a, 1, arh[[Gamma], w]}] + Inactive[Integrate][ g2[Hi[mx], a, mx]*mx, {a, arh[[Gamma], w], [Infinity]}] == 0.1198/(9.2*10^24); There are two seperate complicated functions g1 and g2 over different known ranges, gamma and w are known, and arh is a known function. I need to find Hi[mx], and I need to plot from mx= 0.1 to 10^-19 $\endgroup$ – TheInvoker Oct 13 '18 at 18:16
  • $\begingroup$ I posted more details by editing the question, thanks in advance $\endgroup$ – TheInvoker Oct 13 '18 at 19:25

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