# Handling recursion limit in transcendental equations

Can anyone tell how do I solve these equations that have a recursion relation?

a = u Exp[-v] + v Exp[u];
v = a Exp[-v]+ u;
u = a Exp[-u]+ v;


I wish to update afirst with an initial value of v=v0,x=x0, after I plug in a to v and u I get recursion in my equations

\$RecursionLimit::reclim2: Recursion depth of 1024 exceeded during evaluation of...


Can someone advice how to solve this problem?

• I would isolate ai.e. a=u*Exp[-v]+v*Exp[u], a=(v-u)*Exp[v], a=-(v-u)*Exp[u]. Comparing the last two equations (v-u)Exp[u]=-(v-u)Exp[v] implying (v-u)=0 since Exp is always positive. This means a=0 is the only possible value of a. Then solving a=0=u*Exp[-u]+u*Exp[u] means u must be zero. So {a,v,u}={0,0,0} is the only solution. It is also obvious if you look at a 3D plot of the surfaces defining a from the three equations. – N.J.Evans Nov 7 '17 at 18:12
• @N.J.Evans Did you know that Exp[I Pi]==-1? – Artes Nov 7 '17 at 18:57
• @Artes, yes. I assumed a,u,v are real. Though admittedly that's not mentioned in OP's post. – N.J.Evans Nov 7 '17 at 19:31

Could be done as below.

Clear[u, v, a];
u[0] = u0;
v[0] = v0;
a[0] = a0;
a[n_] := a[n] = u[n - 1] Exp[-v[n - 1]] + v[n - 1] Exp[u[n - 1]];
v[n_] := v[n] = a[n - 1] Exp[-v[n - 1]] + u[n - 1]
u[n_] := u[n] = a[n - 1] Exp[-u[n - 1]] + v[n - 1];


Example:

In[639]:= u[3]

(* Out[639]=
a0 E^-u0 + v0 + E^(-a0 E^-v0 - u0) (E^-v0 u0 + E^u0 v0) +
E^(-a0 E^-v0 - u0 -
E^(-a0 E^-u0 - v0) (E^-v0 u0 + E^u0 v0)) (E^(
a0 E^-u0 + v0) (a0 E^-v0 + u0) +
E^(-a0 E^-v0 - u0) (a0 E^-u0 + v0)) *)

• RecurrenceTable[] is another possibility: RecurrenceTable[{a[n] == u[n - 1] Exp[-v[n - 1]] + v[n - 1] Exp[u[n - 1]], v[n] == a[n - 1] Exp[-v[n - 1]] + u[n - 1], u[n] == a[n - 1] Exp[-u[n - 1]] + v[n - 1], a[0] == a0, v[0] == v0, u[0] == u0}, {a, v, u}, {n, 3, 3}] – J. M. is in limbo Nov 8 '17 at 0:24