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Solve a differential equation at specific point?

Posted 7 years ago

Hello, I'm having tremendous troubles getting the solution for a differential equation that should be satisfied when the argument of the function is (u-t), it is important that it is at (u-t) since the equation has other functions of t that should affect the solution if say I use another variable. No matter what I do I get the error that (u-t) can't be used as a variable and I don't know how to get around this. Any help in this area would be highly appreciated it! (Note: I can find a solution for using t as a variable but when replacing by (u-t) the function found with just t as a variable is no longer a solution)

?A[t_]:=?A[t]=?D^2/(t+?D^2/?A);

?B[t_]:=?B[t]=?D^2/(t+?D^2/?B);

FC[t_]:=FC[t]=-2 H[g] (?B[t]/?D^2-(x (?B[t]-?A[t]))/?D^2)
    +((x (x-1))/(2 ?D^2))+2 H[g]^2 ((?B[t]-?A[t])^2/?D^2)+D[H[g],g];

sol=DSolve[FC[t]==0,H[g],g]
POSTED BY: Paula Cocoma
5 Replies
Posted 7 years ago

Sorry, so I mean that the actual differential equation does have only t for the function ga[t] and gb[t] but is when the input of h[.] is u-t that it should be satisfied.

POSTED BY: Paula Cocoma
Posted 7 years ago

Dear Bill,

Thank you for your answer. Unfortunately it would not work because the other functions, ga[t] and gb[t] have still as a parameter time, so replacing back by (t-u) won't solve the equation. I ran the code you very kindly provided and called the solution a new function RC, whatever the solution of my original differential equation should make the testRC=zero, and the solution that mathematica gives for the code you provide doesn't satisfy this :(

RC[t_] := 
  RC[t] = ( (sd^2 + ra t) (sd^2 + 
       rb t) (sd^4 + ra rb t (u - C[1]) + 
        sd^2 (rb (t (1 - x) + x (u - C[1])) + 
           ra (u + x (t - u) + (-1 + x) C[1])))/(2 (ra - 
          rb)^2 sd^6 (t - u + C[1])));
testRC = - 2 RC[u - t] (gb[t]/sd^2 - ( x (gb[t] - ga[t]) )/sd^2) + ( (
    x (x - 1))/(2 sd^2)) + 2 RC[u - t]^2 ((gb[t] - ga[t])^2/sd^2) +  
   D[RC[u - t], t];
    Simplify[testRC]
POSTED BY: Paula Cocoma

Welcome to Wolfram Community! Do not post code as images. Please make sure you know the rules and how to format your code properly, which you can find here: https://wolfr.am/READ-1ST If you do not format code, it may become corrupted and useless to other members. Please EDIT your posts and make sure code blocks start on a new paragraph and look framed and colored like this.

int = Integrate[1/(x^3 - 1), x];
Map[Framed, int, Infinity]

enter image description here

POSTED BY: Moderation Team
Posted 7 years ago

I am assuming I can substitute t->u-tt, solve the equation in tt and then reverse the substitution tt->u-t

ga[t_] := sd^2/(t + sd^2/ra);
gb[t_] := sd^2/(t + sd^2/rb);
f = (-2 h[u-t] (gb[t]/sd^2-x (gb[t]-ga[t])/sd^2)+x (x-1)/(2 sd^2)+2 h[u-t]^2 ((gb[t]-ga[t])^2/sd^2)+D[h[u-t],t]==0) /. t->u-tt;
Simplify[(h[tt] /. DSolve[f, h[tt], tt][[1]]) /. tt->u-t]

Please check this carefully to make certain that I have made no mistakes.

POSTED BY: Bill Simpson
Posted 7 years ago

Just to follow up, if I change the last like to include only H[t] I get the error:

DSolve::litarg: To avoid possible ambiguity, the arguments of the dependent variable in {((-1+x) x)/(2 \[Sigma]D^2)-2 (1/(t+Power[<<2>>] Power[<<2>>])-(x (-Power[<<2>>] Power[<<2>>]+Power[<<2>>] Power[<<2>>]))/\[Sigma]D^2) H[-t+u]+(2 (-(\[Sigma]D^2/Plus[<<2>>])+\[Sigma]D^2/Plus[<<2>>])^2 H[-t+u]^2)/\[Sigma]D^2-(H^\[Prime])[-t+u]==0} should literally match the independent variables.
POSTED BY: Paula Cocoma
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