# Why doesn't Assuming or Assumption work?

Posted 9 months ago
1545 Views
|
3 Replies
|
2 Total Likes
|
 In[205]:= eq1 = Assuming[\[Lambda] \[Element] Reals, DSolve[{(\[Phi]^\[Prime]\[Prime])[r1]/\[Phi][r1] == -\[Lambda]^2, k \[Phi][0] == Derivative[1][\[Phi]][0], k \[Phi][l] == Derivative[1][\[Phi]][l]}, \[Phi][r1], r1]] Out[205]= {{\[Phi][r1] -> 0}} In[206]:= eq2 = Assuming[\[Lambda] \[Element] Reals, DSolve[{(\[Phi]^\[Prime]\[Prime])[r1]/\[Phi][r1] == \[Lambda]^2, k \[Phi][0] == Derivative[1][\[Phi]][0], k \[Phi][l] == Derivative[1][\[Phi]][l]}, \[Phi][r1], r1]] Out[206]= {{\[Phi][r1] -> 0}} In[207]:= eq3 = DSolve[{(\[Phi]^\[Prime]\[Prime])[r1]/\[Phi][r1] == -\[Lambda]^2, k \[Phi][0] == Derivative[1][\[Phi]][0], k \[Phi][l] == Derivative[1][\[Phi]][l]}, \[Phi][r1], r1, Assumptions -> \[Lambda] \[Element] Reals] Out[207]= {{\[Phi][r1] -> 0}} In[208]:= eq4 = DSolve[{(\[Phi]^\[Prime]\[Prime])[r1]/\[Phi][r1] == \[Lambda]^2, k \[Phi][0] == Derivative[1][\[Phi]][0], k \[Phi][l] == Derivative[1][\[Phi]][l]}, \[Phi][r1], r1, Assumptions -> \[Lambda] \[Element] Reals] Out[208]= {{\[Phi][r1] -> 0}} In the 4 pieces of codes above, there should appear a solution as c1cos(lambda x) + c2sin(lambda x).
3 Replies
Sort By:
Posted 9 months ago
 I think that the reason is that k is undetermined and Mathematica doesn't know how to deal with k.
Posted 9 months ago
 I get a sensible result this way: DSolve[{\[Phi]''[r1] == -\[Lambda]^2 \[Phi][r1], k \[Phi][0] == \[Phi]'[0], k \[Phi][l] == \[Phi]'[l]}, \[Phi], r1] 
Posted 6 months ago
 Yes, it works, but doesn't include another root, \lambda =-k^2