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Student Support Forum: 'Boundary conditions for PDE' topicStudent Support Forum > General > Archives > "Boundary conditions for PDE"

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Author Comment/Response
Bill Simpson
01/22/13 10:01pm

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

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Subject (listing for 'Boundary conditions for PDE')
Author Date Posted
Boundary conditions for PDE Teresa 01/21/13 9:53pm
Re: Boundary conditions for PDE Bill Simpson 01/22/13 10:01pm
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