Let's assume that the solution will take the form
u=φ(x)G(t)
Substituting this from in the equation, will lead us to
φ(x)G′(t)=φ′′(x)G(t)
G(t)G′(t)=φ(x)φ′′(x)
Since the left hand side depends only of t and the right hand side only of x, the both sides can be equal only if they are equal to some constant.
Let's denote this constant as −λ
Then
G(t)G′(t)=φ(x)φ′′(x)=−λ
Therefore we obtain two ordinary differential equations
G′+λG=0
φ′′+λφ=0
Using boundary condition will lead us to
φ(0)=φ(L)=0
The general solution of the equation is
φ=c1cos(λx)+c2sin(λx)
Applying the first boundary condition gives us
φ(0)=c1=0
Applying the second boundary condition, and using the above result, gives us
φ(L)=c2sin(λL)=0
We will only have a non-trivial solution, if sin(λL)=0 , λ=0.
Hence
λn=(Lπn)2,n∈Z,n=0.
Then for G(t) we will have
G′=−λnG
This is a linear first order differential equation and its solution is
G=Bne−λnt
Hence the solution is
u=n=1∑∞Bne−λntsinλx
The initial condition will lead us to
u(x,0)=n=1∑∞Bnsinλx=x(L−x)
which is the sine Fourier series of x(L−x)
It's coefficients can be obtained using the formula
Bn=L20∫Lx(L−x)sin(Lπn)xdx=−n3π32L2(−2+2cos(nπ)+nπsin(nπ))=−n3π34L2((−1)n−1)
Hence
B2n=0
B2n−1=(2n−1)3π38L2
Therefore we obtain the final form of the function
u=n=1∑∞(2n−1)3π38L2e−λ2n−1tsinλ2n−1x
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