Mathematica 9 is now available
Student Support Forum
Student Support Forum > General > > "Boundary conditions for PDE"

Post Reply:
Email Address:

Please send email when my message is replied to.

Url (optional):
Message: view original message?
Attachment (optional):
Please answer this:4+1 =

Original Message (ID '273945') By Bill Simpson:
Find the general solution In[1]:= pde=(1/z)D[T[n,z],z]+D[T[n,z],{z,2}]==0; sol=T[n,z]/.DSolve[pde,T[n,z],{n,z}][[1]] Out[2]= Log[z]C[1][n]+C[2][n] Take the general solution, find second derivative, then substitute for z and finally equate to zero In[3]:= Simplify[(D[sol,{z,2}]/.z->5)==0] Out[3]= C[1][n]==0 Take that information and substitute back into the general solution to find the particular solution. In[4]:= sol/.C[1][n]->0 Out[4]= C[2][n] Substitute the particular solution back into your original PDE to verify it. In[5]:= (1/z)D[C[2][n],z]+D[C[2][n],{z,2}]==0 Out[5]= True Please check all this carefully