#### 1.2.3 Both ends at zero temperature. Source term depends on $$x$$ only

$$u_{t}=ku_{xx}+Q\left ( x\right ) \tag{1}$$ BC are\begin{align*} u\left ( 0,t\right ) & =0\qquad t>0\\ u\left ( L,t\right ) & =0\qquad t>0 \end{align*}

Initial conditions$u\left ( x,0\right ) =f\left ( x\right ) \qquad 0<x<L$ Solution

Since the boundary conditions are already homogeneous, but there is a source, eigenfunction expansion method is used. Let the solution be$$u\left ( x,t\right ) =\sum _{n}^{\infty }a_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \tag{1A}$$ Where $$\Phi _{n}\left ( x\right )$$ are the eigenfunctions of the corresponding PDE $$u_{t}=ku_{xx}$$ with homogeneous conditions. This was in solved above and found to be$\Phi _{n}\left ( x\right ) =\sin \left ( \sqrt{\lambda _{n}}x\right )$ With the eigenvalues$\lambda _{n}=\left ( \frac{n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots$$\Phi _{n}\left ( x\right ) =\sin \left ( \sqrt{\lambda _{n}}x\right )$ Therefore the original PDE can now be written as$\frac{\partial }{\partial t}\left ( \sum _{n=1}^{\infty }a_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \right ) =k\frac{\partial ^{2}}{\partial x^{2}}\left ( \sum _{n=1}^{\infty }a_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \right ) +Q\left ( x\right )$ Expanding $$Q\left ( x\right )$$ using the same eigenfunctions (hence the name), and the above becomes$\frac{\partial }{\partial t}\left ( \sum _{n=1}^{\infty }a_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \right ) =k\frac{\partial ^{2}}{\partial x^{2}}\left ( \sum _{n=1}^{\infty }a_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \right ) +\sum _{n=1}^{\infty }q_{n}\Phi _{n}\left ( x\right )$ The above simpliﬁes to (term by term diﬀerentiation is justiﬁed here, since $$\Phi _{n}\left ( x\right )$$ satisﬁes the same homogeneous boundary conditions as $$u\left ( x,t\right )$$)$\sum _{n=1}^{\infty }a_{n}^{\prime }\left ( t\right ) \Phi _{n}\left ( x\right ) =k\sum _{n=1}^{\infty }a_{n}\left ( t\right ) \Phi _{n}^{\prime \prime }\left ( x\right ) +\sum _{n=1}^{\infty }q_{n}\Phi _{n}\left ( x\right )$ But $$\Phi _{n}^{\prime \prime }\left ( x\right ) =-\lambda _{n}\Phi _{n}\left ( x\right )$$, since the eigenfunction satisﬁes the original ODE $$X_{n}^{\prime \prime }\left ( x\right ) +\lambda _{n}X\left ( x\right ) =0$$, therefore the above simpliﬁes to$\sum _{n=1}^{\infty }a_{n}^{\prime }\left ( t\right ) \Phi _{n}\left ( x\right ) =-k\sum _{n=1}^{\infty }a_{n}\left ( t\right ) \lambda _{n}\Phi _{n}\left ( x\right ) +\sum _{n=1}^{\infty }q_{n}\Phi _{n}\left ( x\right )$ Or simply \begin{align} a_{n}^{\prime }\left ( t\right ) & =-k\lambda _{n}a_{n}\left ( t\right ) +q_{n}\nonumber \\ a_{n}^{\prime }\left ( t\right ) +k\lambda _{n}a_{n}\left ( t\right ) & =q_{n} \tag{2} \end{align}

This is a ﬁrst order ODE to solve for $$a_{n}\left ( t\right )$$. First $$q_{n}$$ is found by orthogonality. Since $Q\left ( x\right ) =\sum _{n=1}^{\infty }q_{n}\Phi _{n}\left ( x\right )$ Then by orthogonality of eigenfunctions\begin{align*} \int _{0}^{L}Q\left ( x\right ) \Phi _{m}\left ( x\right ) dx & =\int _{0}^{L}\sum _{n=1}^{\infty }q_{n}\Phi _{n}\left ( x\right ) \Phi _{m}\left ( x\right ) dx\\ \int _{0}^{L}Q\left ( x\right ) \Phi _{m}\left ( x\right ) dx & =\int _{0}^{L}q_{n}\Phi _{m}^{2}\left ( x\right ) dx \end{align*}

Since $$\Phi _{m}\left ( x\right ) =\sin \left ( \sqrt{\lambda _{m}}x\right ) =\sin \left ( \frac{m\pi }{L}x\right )$$ in this case (due to the nature of boundary conditions) then $$\int _{0}^{L}\sin ^{2}\left ( \frac{m\pi }{L}x\right ) dx=\frac{L}{2}$$ and the above simpliﬁes to\begin{align*} \int _{0}^{L}Q\left ( x\right ) \Phi _{m}\left ( x\right ) dx & =q_{m}\frac{L}{2}\\ q_{n} & =\frac{2}{L}\int _{0}^{L}Q\left ( x\right ) \Phi _{n}\left ( x\right ) dx \end{align*}

Using the above in (2) gives$a_{n}^{\prime }\left ( t\right ) +k\lambda _{n}a_{n}\left ( t\right ) =q_{n}$ This ODE can now be solved. It solution will be\begin{align*} a_{n}\left ( t\right ) & =a_{n}\left ( 0\right ) e^{-k\lambda _{n}t}+e^{-k\lambda _{n}t}q_{n}\int _{0}^{t}e^{k\lambda _{n}\tau }d\tau \\ & =a_{n}\left ( 0\right ) e^{-k\lambda _{n}t}+\frac{q_{n}}{k\lambda _{n}}e^{-k\lambda _{n}t}\left ( e^{k\lambda _{n}t}-1\right ) \\ & =a_{n}\left ( 0\right ) e^{-k\lambda _{n}t}+\frac{q_{n}}{k\lambda _{n}}-\frac{q_{n}}{k\lambda _{n}}e^{-k\lambda _{n}t} \end{align*}

To ﬁnd $$a_{n}\left ( 0\right )$$, initial conditions are used. At $$t=0$$ the above becomes$f\left ( x\right ) =\sum _{n}^{\infty }a_{n}\left ( 0\right ) \Phi _{n}\left ( x\right )$ By orthogonality\begin{align*} a_{n}\left ( 0\right ) & =\frac{\int _{0}^{L}f\left ( x\right ) \Phi _{n}\left ( x\right ) dx}{\int _{0}^{L}\Phi _{n}^{2}\left ( x\right ) dx}\\ & =\frac{2}{L}\int _{0}^{L}f\left ( x\right ) \Phi _{n}\left ( x\right ) dx \end{align*}

This completes the solution.

Summary of solution

\begin{align*} u\left ( x,t\right ) & =\sum _{n=1}^{\infty }a_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \\ \Phi _{n}\left ( x\right ) & =\sin \left ( \sqrt{\lambda _{n}}x\right ) \\ \lambda _{n} & =\left ( \frac{n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots \\ a_{n}\left ( t\right ) & =a_{n}\left ( 0\right ) e^{-k\lambda _{n}t}+\frac{q_{n}}{k\lambda _{n}}-\frac{q_{n}}{k\lambda _{n}}e^{-k\lambda _{n}t}\\ a_{n}\left ( 0\right ) & =\frac{2}{L}\int _{0}^{L}f\left ( x\right ) \Phi _{n}\left ( x\right ) dx\\ q_{n} & =\frac{2}{L}\int _{0}^{L}Q\left ( x\right ) \Phi _{n}\left ( x\right ) dx \end{align*}

Example

Let $$Q\left ( x\right ) =x,L=1$$ meter, let $$k$$ which is the Thermal diﬀusivity be $$\frac{1}{100}$$ $$\frac{m^{2}}{\sec }$$and let the initial conditions $$f\left ( x\right ) =x\left ( 1-x\right ) .$$

hence the PDE to solve is$$u_{t}=\frac{1}{100}u_{xx}+x \tag{1}$$ BC are\begin{align*} u\left ( 0,t\right ) & =0\qquad t>0\\ u\left ( 1,t\right ) & =0\qquad t>0 \end{align*}

Initial conditions$u\left ( x,0\right ) =x\left ( 1-x\right ) \qquad 0<x<1$ Hence the ﬁnal solution is from above is\begin{align*} u\left ( x,t\right ) & =\sum _{n=1}^{\infty }a_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \\ \Phi _{n}\left ( x\right ) & =\sin \left ( \sqrt{\lambda _{n}}x\right ) \\ \lambda _{n} & =\left ( n\pi \right ) ^{2}\qquad n=1,2,3,\cdots \\ q_{n} & =\frac{2}{L}\int _{0}^{L}x\Phi _{n}\left ( x\right ) dx\\ a_{n}\left ( 0\right ) & =\frac{2}{L}\int _{0}^{L}x\left ( 1-x\right ) \Phi _{n}\left ( x\right ) dx\\ a_{n}\left ( t\right ) & =a_{n}\left ( 0\right ) e^{-k\lambda _{n}t}+\frac{q_{n}}{k\lambda _{n}}-\frac{q_{n}}{k\lambda _{n}}e^{-k\lambda _{n}t} \end{align*}

The solution above was veriﬁed by comparing it to numerical solution. The animation is run for 3 seconds. This animation for $$Q\left ( x\right ) =x$$

This animation below is when $$Q\left ( x\right ) =\sin \left ( 10x\right )$$.

\begin{align*} u\left ( x,t\right ) & =\sum _{n=1}^{\infty }a_{n}\left ( t\right ) \sin \left ( \sqrt{\lambda _{n}}x\right ) \\ \lambda _{n} & =\left ( n\pi \right ) ^{2}\qquad n=1,2,3,\cdots \\ q_{n} & =\frac{2}{L}\int _{0}^{L}\sin \left ( 10x\right ) \sin \left ( \sqrt{\lambda _{n}}x\right ) dx\\ a_{n}\left ( 0\right ) & =\frac{2}{L}\int _{0}^{L}x\left ( 1-x\right ) \sin \left ( \sqrt{\lambda _{n}}x\right ) dx\\ a_{n}\left ( t\right ) & =a_{n}\left ( 0\right ) e^{-k\lambda _{n}t}+\frac{q_{n}}{k\lambda _{n}}-\frac{q_{n}}{k\lambda _{n}}e^{-k\lambda _{n}t} \end{align*}

Code used to the above, including a Manipulate to compare with the numerical solution is below. To use diﬀerence source term, just change the source line and revaluate the code.