Homogeneous Equation for Mixed Cauchy Problem

We try to solve homogeneous equation for mixed Cauchy problem with null boundary condition of a special form for the parabolic heat equation
$ P(D)u=\frac{\partial u}{\partial t}(x,t)-\frac{\partial^2 u}{\partial x^2}(x,t)=0 $
with initial condition
$ u(x,0)=f(x) $
and boundary condition
$ u(0,t)=0\: $ and $ \frac{\partial u}{\partial x}(1,t)=0 $

Look for a nonzero solution of the form u(x,t)=X(x)T(t).
We get T'(t)X(x)-T(t)X"(x)=0 or
$ \frac{T'(t)}{T(t)}=\frac{X"(x)}{X(x)}=-\lambda^2 $
(*)We get $  T'(t)+\lambda^2T(t)=0 $
(**)$ X"(x)+\lambda^2X(x)=0\: X(0)=X'(1)=0 $
Second equation has a solution of the form
$ X(x)=a\cos(\lambda x)+b\sin(\lambda x) $
$ X(0)=a=0\: X'(1)=b\lambda\cos\lambda =0 $
$ \lambda_n=\frac{(2n+1)\pi}{2} $
So $ X_n(x)=\sin (\frac{(2n+1)\pi}{2}x) $
Equation (*) has solution
$ T_n(t)=A_ne^{-\lambda^2_nt $
or $ T_n(t)=A_ne^{-\frac{(2n+1)^2\pi^2t}{4} $
Look now for a solution of our equation
$ u(x,t)=\sum_{n=1}^{+\infty}T_n(t)X_n(x)=\sum_{n=1}^{+\infty}A_ne^{-\frac{(2n+1)^2\pi^2t}{4}}\sin \frac{(2n+1)\pi}{2}x $
From $ u(x,0)=\sum_{n=1}^{+\infty}A_n\sin \frac{(2n+1)\pi}{2}x=f(x) $ we have
$  A_n=2\int_{0}^{1}f(x)\sin\frac{(2n+1)\pi x}{2}dx $
In a particular case all you have to do is to evaluate $ A_n $ for the given function.

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