I was trying to solve the following heat equation using the Fourier transform
$\frac{du}{dt}=\frac{d^2u}{dx^2}$
With initial boundary condition $\hspace{1cm}u(x,0)=0\hspace{1cm}$ for $\hspace{1cm}0<x<1$
It also has the following boundary conditions:
$u(0,t)=1\hspace{1cm}$ and $\hspace{1cm}u(1,t)=0\hspace{1cm}$ for $\hspace{1cm}t>0$
At first I use the Fourier transform on the heat equation with respect to t and I get:
$\frac{d\hat{u}_{(w,t)}}{dt}=-w^2\hat{u}_{(w,t)}\hspace{0.5cm}$ where $\hat{u}_{(w,t)}\hspace{0.2cm}$ depends on w and t instead of x and t.
Multiplying both sides by $e^{w^2t}\hspace{0.2cm}$ we get
$e^{w^2t}\frac{d\hat{u}_{(w,t)}}{dt}=-w^2e^{w^2t}\hat{u}_{(w,t)}\hspace{0.5cm}\rightarrow\hspace{0.5cm}e^{w^2t}\frac{d\hat{u}_{(w,t)}}{dt}+w^2e^{w^2t}\hat{u}_{(w,t)}=0$
We can see that the previous equation is the derivative with respect to t of:
$\frac{d}{dt}(e^{w^2t}\hat{u}_{(w,t)})=0\hspace{0.5cm}$
By integrating the previous equation we get:
$e^{w^2t}\hat{u}_{(w,t)}=C_{(w)}\hspace{0.5cm}\rightarrow\hspace{0.5cm}\hat{u}_{(w,t)}=C_{(w)}e^{-w^2t}$
Now, from the initial condition we know that:
$u_{(x,0)}=0\hspace{0.5cm}\rightarrow\hspace{0.5cm}\hat{u}_{(w,0)}=0$
So when I evaluate $\hat{u}_{(w,t)}\hspace{0.5cm}$ on t=0 what I get is that $C_{(w)}=0\hspace{0.5cm}\rightarrow\hspace{0.5cm}\hat{u}_{(w,t)}=0$
$\rule{16cm}{0.4pt}$
I couldn't get the solution after this, so thanks to @messenger comment, I decided to use the Laplace transform instead. Here is what I got so far:
$\frac{d^2}{dx^2}\hat{u}_{(x,s)}=s\hat{u}_{(x,s)}-u_{(x,0)}\hspace{0.5cm}\rightarrow\hspace{0.5cm}\frac{d^2}{dx^2}\hat{u}_{(x,s)}=s\hat{u}_{(x,s)}$
Which is an ODE and we know the solution comes in the form of:
$\hat{u}_{(x,s)}=ae^{\sqrt{s}x}+be^{-\sqrt{s}x}=acosh(\sqrt{s}x)+bsinh(\sqrt{s}x)\hspace{0.5cm}$
Taking the Laplace transform on the boundary conditions yields:
$\hat{u}_{(0,s)}=\frac{1}{s}\hspace{0.5cm}$ and $\hat{u}_{(1,s)}=0$
Evaluating on the BC:
$\hat{u}_{(0,s)}=a=\frac{1}{s}\hspace{0.5cm}$ and $\hat{u}_{(1,s)}=\frac{Cosh(\sqrt{s})}{s}+bSinh(\sqrt{s})=0\hspace{0.5cm}\rightarrow\hspace{0.5cm}b=-\frac{Cosh(\sqrt{s})}{sSinh(\sqrt{s})}$
So the function $\hat{u}_{(x,s)}=\frac{Cosh(\sqrt{s}x)}{s}-\frac{Cosh(\sqrt{s})Sinh(\sqrt{s}x)}{sSinh(\sqrt{s})}$
Am I headed in the right direction here?
It's best to have homogeneous conditions at $x=0,1$. So, you can subtract $1-x$ from your $u$ to obtain $v=u-(1-x)$ that satisfies $$ v_t = u_t = u_{xx} = v_{xx} \\ v(x,0)=u(x,0)-(1-x)=x-1 \\ v(0,t)=u(0,t)-1=0 \\ v(1,t)=u(1,t)-0=0 $$ This is a standard separation of variables problem for $v$ because it has homogeneous endpoint conditions on the interval $0 \le x \le 1$. Then, once you have a solution $v$, the desired solution $u$ is $u=v+(1-x)$.