#### 3.1.3 Finite domain (bar), right end homogeneous, left end not

##### 3.1.3.1 [207] left end BC depends on time (general case)

problem number 207

Solve the heat equation for $$u(x,t)$$ $u_t = k u_{xx}$ For $$0<x<L$$ and $$t>0$$. The boundary conditions are \begin {align*} u_x(0,t) &= A(t) \\ u(L,t) &= 0 \\ \end {align*}

And initial condition is $$u(x,0)=0$$

Mathematica

$\left \{\left \{u(x,t)\to \underset {K[1]=1}{\overset {\infty }{\sum }}\frac {\sqrt {2} \cos \left (\frac {\pi x (2 K[1]-1)}{2 L}\right ) \left (\frac {4 \sqrt {2} e^{-\frac {k \pi ^2 t (1-2 K[1])^2}{4 L^2}} A(0) L^{3/2}}{\pi ^2 (1-2 K[1])^2}+\int _0^t \frac {4 \sqrt {2} \exp \left (-\frac {k \pi ^2 (1-2 K[1])^2 (t-K[2])}{4 L^2}\right ) L^{3/2} A'(K[2])}{\pi ^2 (1-2 K[1])^2} \, dK[2]\right )}{\sqrt {L}}+A(t) (x-L)\right \}\right \}$

Maple

$u \left (x , t\right ) = \left (-L +x \right ) A (t )+\int _{0}^{t}\left (\Mapleoverset {\infty }{\Mapleunderset {n =0}{\sum }}\frac {8 L \cos \left (\frac {\left (2 n +1\right ) \pi x}{2 L}\right ) \left (\frac {d}{d \tau }A \left (\tau \right )\right ) {\mathrm e}^{-\frac {\left (t -\tau \right ) \pi ^{2} \left (n +\frac {1}{2}\right )^{2} k}{L^{2}}}}{\left (2 n +1\right )^{2} \pi ^{2}}\right )d \tau +8 \left (\Mapleoverset {\infty }{\Mapleunderset {n =0}{\sum }}\frac {A (0) L \cos \left (\frac {\left (2 n +1\right ) \pi x}{2 L}\right ) {\mathrm e}^{-\frac {\pi ^{2} \left (2 n +1\right )^{2} k t}{4 L^{2}}}}{\left (2 n +1\right )^{2} \pi ^{2}}\right )$

Hand solution

This 1D heat PDE has left one end with boundary condition that is time dependent.\begin {align*} u_{t} & =ku_{xx}\qquad 0<x<L,t>0\\ u_{x}\left ( 0,t\right ) & =A\left ( t\right ) \\ u\left ( L,t\right ) & =0\\ u\left ( x,0\right ) & =0 \end {align*}

Solution

Since the boundary condition is not homogeneous, we need to ﬁrst ﬁnd a reference function $$r\left ( x,t\right )$$. Let$r\left ( x,t\right ) =A\left ( t\right ) \left ( x-L\right )$ This function only needs to satisfy the nonhomogeneous boundary conditions given. i.e. $$\frac {\partial r}{\partial x}\left ( 0,t\right ) =A\left ( t\right ) ,r\left ( L,t\right ) =0$$. Now we can write\begin {equation} u\left ( x,t\right ) =v\left ( x,t\right ) +r\left ( x,t\right ) \tag {1} \end {equation} Since $$r\left ( x,t\right )$$ satisﬁes the nonhomogeneous B.C’s, then $$v\left ( x,t\right )$$ satisﬁes the homogeneous boundary conditions. Substituting the above back into the original PDE gives\begin {align} v_{t}+r_{t} & =k\left ( v_{xx}+r_{xx}\right ) \nonumber \\ v_{t}+A^{\prime }\left ( t\right ) \left ( x-L\right ) & =kv_{xx}\nonumber \\ v_{t} & =kv_{xx}+A^{\prime }\left ( t\right ) \left ( L-x\right ) \nonumber \\ & =kv_{xx}+Q\left ( x,t\right ) \tag {2} \end {align}

The PDE $$v_{t}=kv_{xx}+Q\left ( x,t\right )$$ has now homogeneous B.C. $$v_{x}\left ( 0,t\right ) =0,v\left ( L,0\right ) =0$$. Where $$Q\left ( x,t\right ) =A^{\prime }\left ( t\right ) \left ( L-x\right )$$. The method of eigenfunction expansion is now used to solve (2). Let $v\left ( x,t\right ) =\sum _{n=1}^{\infty }a_{n}\left ( t\right ) \Phi _{n}\left ( x\right )$ Substituting this back into (2) gives\begin {align*} \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 ) +Q\left ( x,t\right ) \\ & =k\sum _{n=1}^{\infty }a_{n}\left ( t\right ) \Phi _{n}^{\prime \prime }\left ( x\right ) +\sum _{n=1}^{\infty }q_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \end {align*}

Where $$Q\left ( x,t\right ) =\sum _{n=1}^{\infty }q_{n}\left ( t\right ) \Phi _{n}\left ( x\right )$$. Now, since $$\Phi _{n}^{\prime \prime }\left ( x\right ) =-\lambda _{n}\Phi _{n}\left ( x\right )$$ then the above reduces to\begin {align} \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}\left ( t\right ) \Phi _{n}\left ( x\right ) \nonumber \\ a_{n}^{\prime }\left ( t\right ) +k\lambda _{n}a_{n}\left ( t\right ) & =q_{n}\left ( t\right ) \tag {3} \end {align}

The eigenfunctions $$\Phi _{n}\left ( x\right )$$ come from solving the eigenvalue problem in $$v_{t}=kv_{xx}$$ with homogeneous boundary conditions $$v_{x}\left ( 0,t\right ) =0,v\left ( L,t\right ) =0$$. This was solved before in problem 3.1.1.28 on page 853. The eigenfunctions were found to be $$\Phi _{n}\left ( x\right ) =\cos \left ( \sqrt {\lambda _{n}}x\right ) ,n=1,3,5,\cdots$$ with eigenvalues $$\lambda _{n}=\left ( \frac {n\pi }{2L}\right ) ^{2},n=1,3,5,\cdots$$. Before solving the ODE (3), we need to ﬁrst ﬁnd $$q_{n}\left ( t\right )$$. Orthogonality is now used to ﬁnd $$q_{n}\left ( t\right )$$\begin {align*} Q\left ( x,t\right ) & =\sum _{n=1,3,5,\cdots }^{\infty }q_{n}\left ( t\right ) \cos \left ( \sqrt {\lambda _{n}}x\right ) \\ A^{\prime }\left ( t\right ) \left ( L-x\right ) & =\sum _{n=1,3,5,\cdots }^{\infty }q_{n}\left ( t\right ) \cos \left ( \sqrt {\lambda _{n}}x\right ) \\ \int _{0}^{L}A^{\prime }\left ( t\right ) \left ( L-x\right ) \cos \left ( \sqrt {\lambda _{n}}x\right ) dx & =\frac {L}{2}q_{n}\left ( t\right ) \\ q_{n}\left ( t\right ) & =\frac {2A^{\prime }\left ( t\right ) }{L}\int _{0}^{L}\left ( L-x\right ) \cos \left ( \frac {n\pi }{2L}x\right ) dx\\ & =-\frac {2A^{\prime }\left ( t\right ) }{L}\left ( \frac {4L^{2}\left ( \cos \left ( \frac {n\pi }{2}\right ) -1\right ) }{\pi ^{2}n^{2}}\right ) \qquad n=1,3,5,\cdots \end {align*}

But $$\cos \left ( \frac {n\pi }{2}\right ) -1=-1$$ for $$n=1,3,5,\cdots$$. Hence the above becomes\begin {align*} q_{n}\left ( t\right ) & =\frac {2A^{\prime }\left ( t\right ) }{L}\left ( \frac {4L^{2}}{\pi ^{2}n^{2}}\right ) \qquad n=1,3,5,\cdots \\ & =\frac {8A^{\prime }\left ( t\right ) L}{\pi ^{2}n^{2}} \end {align*}

Hence (3) becomes\begin {align*} a_{n}^{\prime }\left ( t\right ) +k\lambda _{n}a_{n}\left ( t\right ) & =q_{n}\left ( t\right ) \qquad n=1,2,3,\cdots \\ a_{n}^{\prime }\left ( t\right ) +k\left ( \frac {n\pi }{2L}\right ) ^{2}a_{n}\left ( t\right ) & =\frac {8A^{\prime }\left ( t\right ) L}{\pi ^{2}n^{2}}\\ a_{n}^{\prime }\left ( t\right ) +k\lambda _{n}a_{n}\left ( t\right ) & =\frac {8A^{\prime }\left ( t\right ) L}{\pi ^{2}n^{2}} \end {align*}

Integrating factor is $$e^{k\lambda _{n}t}$$. Hence the above becomes$\frac {d}{dt}\left ( a\left ( t\right ) e^{k\lambda _{n}t}\right ) =\frac {8A^{\prime }\left ( t\right ) L}{\pi ^{2}n^{2}}e^{k\lambda _{n}t}$ Integrating gives\begin {align*} a_{n}\left ( t\right ) e^{k\lambda _{n}t} & =\int _{0}^{t}\frac {8A^{\prime }\left ( \tau \right ) L}{\pi ^{2}n^{2}}e^{k\lambda _{n}\tau }d\tau +a_{n}\left ( 0\right ) \\ a_{n}\left ( t\right ) & =a_{n}\left ( 0\right ) e^{-k\lambda _{n}t}+e^{-k\lambda _{n}t}\int _{0}^{t}\frac {8A^{\prime }\left ( \tau \right ) L}{\pi ^{2}n^{2}}e^{k\lambda _{n}\tau }d\tau \end {align*}

Hence\begin {align*} v\left ( x,t\right ) & =\sum _{n=1}^{\infty }a_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \\ & =\sum _{n=1,3,5,\cdots }^{\infty }\left ( a_{n}\left ( 0\right ) e^{-k\lambda _{n}t}+\frac {8L}{\pi ^{2}n^{2}}e^{-k\lambda _{n}t}\int _{0}^{t}A^{\prime }\left ( \tau \right ) e^{k\lambda _{n}\tau }d\tau \right ) \cos \left ( \sqrt {\lambda _{n}}x\right ) \end {align*}

Since $$u\left ( x,t\right ) =v\left ( x,t\right ) +r\left ( x,t\right )$$ then\begin {equation} u\left ( x,t\right ) =A\left ( t\right ) \left ( x-L\right ) +\sum _{n=1,3,5,\cdots }^{\infty }\left ( a_{n}\left ( 0\right ) e^{-k\lambda _{n}t}+\frac {8L}{\pi ^{2}n^{2}}e^{-k\lambda _{n}t}\int _{0}^{t}A^{\prime }\left ( \tau \right ) e^{k\lambda _{n}\tau }d\tau \right ) \cos \left ( \sqrt {\lambda _{n}}x\right ) \tag {4} \end {equation} At $$t=0,u\left ( x,0\right ) =0$$ and the above becomes \begin {align*} 0 & =A\left ( 0\right ) \left ( x-L\right ) +\sum _{n=1,3,5,\cdots }^{\infty }a_{n}\left ( 0\right ) \cos \left ( \sqrt {\lambda _{n}}x\right ) \\ A\left ( 0\right ) \left ( L-x\right ) & =\sum _{n=1,3,5,\cdots }^{\infty }a_{n}\left ( 0\right ) \cos \left ( \sqrt {\lambda _{n}}x\right ) \end {align*}

Applying orthogonality\begin {align*} \int _{0}^{L}\left ( L-x\right ) \cos \left ( \sqrt {\lambda _{n}}x\right ) dx & =a_{n}\left ( 0\right ) \frac {L}{2}\\ a_{n}\left ( 0\right ) & =\frac {2A\left ( 0\right ) }{L}\int _{0}^{L}\left ( L-x\right ) \cos \left ( \sqrt {\lambda _{n}}x\right ) dx\\ & =\frac {2A\left ( 0\right ) }{L}\left ( -\frac {\cos \left ( L\sqrt {\lambda _{n}}\right ) -1}{\lambda _{n}}\right ) \end {align*}

But $$\cos \left ( L\sqrt {\lambda _{n}}\right ) -1=\cos \left ( L\frac {n\pi }{2L}\right ) -1=$$ $$\cos \left ( \frac {n\pi }{2}\right ) -1=-1$$ for $$n=1,3,5,\cdots$$, and the above becomes$a_{n}\left ( 0\right ) =\frac {2A\left ( 0\right ) }{L\lambda _{n}}$ Therefore the solution (4) is $u\left ( x,t\right ) =A\left ( t\right ) \left ( x-L\right ) +\sum _{n=1,3,5,\cdots }^{\infty }e^{-k\lambda _{n}t}\left ( \frac {2A\left ( 0\right ) }{L\lambda _{n}}+\frac {8L}{\pi ^{2}n^{2}}\int _{0}^{t}A^{\prime }\left ( \tau \right ) e^{k\lambda _{n}\tau }\right ) \cos \left ( \sqrt {\lambda _{n}}x\right )$ Where $$\lambda _{n}=\left ( \frac {n\pi }{2L}\right ) ^{2}$$. Hence$u\left ( x,t\right ) =A\left ( t\right ) \left ( x-L\right ) +\sum _{n=1,3,5,\cdots }^{\infty }e^{-k\left ( \frac {n\pi }{2L}\right ) ^{2}t}\left ( \frac {2A\left ( 0\right ) }{L\left ( \frac {n\pi }{2L}\right ) ^{2}}+\frac {8L}{\pi ^{2}n^{2}}\int _{0}^{t}A^{\prime }\left ( \tau \right ) e^{k\left ( \frac {n\pi }{2L}\right ) ^{2}\tau }d\tau \right ) \cos \left ( \frac {n\pi }{2L}x\right )$

________________________________________________________________________________________

##### 3.1.3.2 [208] left end BC depends on time (special case)

problem number 208

Solve the heat equation for $$u(x,t)$$ $u_t = k u_{xx}$ For $$0<x<L$$ and $$t>0$$. The boundary conditions are \begin {align*} u_x(0,t) &= A(t) \\ u(L,t) &= 0 \\ \end {align*}

And initial condition is $$u(x,0)=0$$. Using the following values \begin {align*} L&=5\\ k&=\frac {1}{100}\\ A(t)=e^t \end {align*}

Mathematica

$\left \{\left \{u(x,t)\to \underset {K[1]=1}{\overset {\infty }{\sum }}\frac {40 e^{-\frac {\pi ^2 t (1-2 K[1])^2}{10000}} \cos \left (\frac {1}{10} \pi x (2 K[1]-1)\right ) \left (\pi ^2 (1-2 K[1])^2+10000 e^{\frac {\pi ^2 t (1-2 K[1])^2}{10000}+t}\right )}{\left (\pi ^2 (1-2 K[1])^2+10000\right ) (\pi -2 \pi K[1])^2}+e^t (x-5)\right \}\right \}$

Maple

$u \left (x , t\right ) = \left (\Mapleoverset {\infty }{\Mapleunderset {n =0}{\sum }}\frac {10 \left (2500 \,{\mathrm e}^{t}+\pi ^{2} \left (n +\frac {1}{2}\right )^{2} {\mathrm e}^{-\frac {\pi ^{2} \left (2 n +1\right )^{2} t}{10000}}\right ) \cos \left (\frac {\left (2 n +1\right ) \pi x}{10}\right )}{\left (2500+\left (n +\frac {1}{2}\right )^{2} \pi ^{2}\right ) \pi ^{2} \left (n +\frac {1}{2}\right )^{2}}\right )+\left (x -5\right ) {\mathrm e}^{t}$

Hand solution

This PDE general solution was obtained in problem 3.1.3.1 on page 975 as$u\left ( x,t\right ) =A\left ( t\right ) \left ( x-L\right ) +\sum _{n=1,3,5,\cdots }^{\infty }e^{-k\left ( \frac {n\pi }{2L}\right ) ^{2}t}\left ( \frac {2A\left ( 0\right ) }{L\left ( \frac {n\pi }{2L}\right ) ^{2}}+\frac {8L}{\pi ^{2}n^{2}}\int _{0}^{t}A^{\prime }\left ( \tau \right ) e^{k\left ( \frac {n\pi }{2L}\right ) ^{2}\tau }d\tau \right ) \cos \left ( \frac {n\pi }{2L}x\right )$ When $$A\left ( t\right ) =e^{t}$$, the above becomes$u\left ( x,t\right ) =e^{t}\left ( x-L\right ) +\sum _{n=1,3,5,\cdots }^{\infty }e^{-k\left ( \frac {n\pi }{2L}\right ) ^{2}t}\left ( \frac {2}{L\left ( \frac {n\pi }{2L}\right ) ^{2}}+\frac {8L}{\pi ^{2}n^{2}}\int _{0}^{t}e\left ( \tau \right ) e^{k\left ( \frac {n\pi }{2L}\right ) ^{2}\tau }d\tau \right ) \cos \left ( \frac {n\pi }{2L}x\right )$ But $\int _{0}^{t}e\left ( \tau \right ) e^{k\left ( \frac {n\pi }{2L}\right ) ^{2}\tau }d\tau =\frac {e^{k\left ( \frac {n\pi }{2L}\right ) ^{2}t+t}-1}{k\left ( \frac {n\pi }{2L}\right ) ^{2}+1}$ And the general solution becomes\begin {align*} u\left ( x,t\right ) & =e^{t}\left ( x-L\right ) +\sum _{n=1,3,5,\cdots }^{\infty }e^{-k\left ( \frac {n\pi }{2L}\right ) ^{2}t}\left ( \frac {2}{L\left ( \frac {n\pi }{2L}\right ) ^{2}}+\frac {8L}{\pi ^{2}n^{2}}\frac {e^{k\left ( \frac {n\pi }{2L}\right ) ^{2}t+t}-1}{k\left ( \frac {n\pi }{2L}\right ) ^{2}+1}\right ) \cos \left ( \frac {n\pi }{2L}x\right ) \\ & =e^{t}\left ( x-L\right ) +\sum _{n=1,3,5,\cdots }^{\infty }\left ( \frac {2e^{-k\left ( \frac {n\pi }{2L}\right ) ^{2}t}}{L\left ( \frac {n\pi }{2L}\right ) ^{2}}+\frac {8L}{\pi ^{2}n^{2}}\frac {e^{t}-e^{-k\left ( \frac {n\pi }{2L}\right ) ^{2}t}}{k\left ( \frac {n\pi }{2L}\right ) ^{2}+1}\right ) \cos \left ( \frac {n\pi }{2L}x\right ) \\ & =e^{t}\left ( x-L\right ) +\sum _{n=1,3,5,\cdots }^{\infty }e^{-k\left ( \frac {n\pi }{2L}\right ) ^{2}t}\left ( \frac {2}{L\left ( \frac {n\pi }{2L}\right ) ^{2}}-\frac {8L}{\left ( k\left ( \frac {n\pi }{2L}\right ) ^{2}+1\right ) \pi ^{2}n^{2}}\right ) +\frac {8Le^{t}}{\left ( k\left ( \frac {n\pi }{2L}\right ) ^{2}+1\right ) \pi ^{2}n^{2}}\cos \left ( \frac {n\pi }{2L}x\right ) \\ & =e^{t}\left ( x-L\right ) +\sum _{n=1,3,5,\cdots }^{\infty }\left ( \frac {8L}{n^{2}\pi ^{2}}-\frac {8L}{\left ( k\left ( \frac {n\pi }{2L}\right ) ^{2}+1\right ) \pi ^{2}n^{2}}\right ) e^{-k\left ( \frac {n\pi }{2L}\right ) ^{2}t}+\frac {8Le^{t}}{\left ( k\left ( \frac {n\pi }{2L}\right ) ^{2}+1\right ) \pi ^{2}n^{2}}\cos \left ( \frac {n\pi }{2L}x\right ) \end {align*}

In this problem $$L=5,k=\frac {1}{100}$$, hence the above becomes$u\left ( x,t\right ) =e^{t}\left ( x-5\right ) +\sum _{n=1,3,5,\cdots }^{\infty }\left ( \left ( \frac {40}{n^{2}\pi ^{2}}-\frac {40}{\left ( k\left ( \frac {n\pi }{10}\right ) ^{2}+1\right ) \pi ^{2}n^{2}}\right ) e^{-\frac {n^{2}\pi ^{2}}{10000}t}+\frac {40e^{t}}{\left ( k\left ( \frac {n\pi }{10}\right ) ^{2}+1\right ) \pi ^{2}n^{2}}\right ) \cos \left ( \frac {n\pi }{10}x\right )$ The following is an animation of the solution

Source code used for the above

________________________________________________________________________________________

##### 3.1.3.3 [209] left end BC depends on time (special case)

problem number 209

Solve the heat equation for $$u(x,t)$$ $u_t = k u_{xx}$ For $$0<x<L$$ and $$t>0$$. The boundary conditions are \begin {align*} u_x(0,t) &= A(t) \\ u(L,t) &= 0 \\ \end {align*}

And initial condition is $$u(x,0)=0$$. Using the following values \begin {align*} L&=5\\ k&=\frac {1}{100}\\ A(t)=\sin (t) \end {align*}

Mathematica

$\left \{\left \{u(x,t)\to \underset {K[1]=1}{\overset {\infty }{\sum }}\frac {400000 \cos \left (\frac {1}{10} \pi x (2 K[1]-1)\right ) \left (\pi ^2 \cos (t) (1-2 K[1])^2-e^{-\frac {\pi ^2 t (1-2 K[1])^2}{10000}} \pi ^2 (1-2 K[1])^2+10000 \sin (t)\right )}{\left (\pi ^4 (1-2 K[1])^4+100000000\right ) (\pi -2 \pi K[1])^2}+(x-5) \sin (t)\right \}\right \}$

Maple

$u \left (x , t\right ) = -400000 \left (\Mapleoverset {\infty }{\Mapleunderset {n =0}{\sum }}-\frac {\left (\pi ^{2} \left (n +\frac {1}{2}\right )^{2} \cos (t )-\pi ^{2} \left (n +\frac {1}{2}\right )^{2} {\mathrm e}^{-\frac {\pi ^{2} \left (2 n +1\right )^{2} t}{10000}}+2500 \sin (t )\right ) \cos \left (\frac {\left (2 n +1\right ) \pi x}{10}\right )}{16 \pi ^{2} \left (n +\frac {1}{2}\right )^{2} \left (6250000+\left (n +\frac {1}{2}\right )^{4} \pi ^{4}\right )}\right )+\left (x -5\right ) \sin (t )$

Hand solution

This PDE general solution was obtained in problem 3.1.3.1 on page 975 as$u\left ( x,t\right ) =A\left ( t\right ) \left ( x-L\right ) +\sum _{n=1,3,5,\cdots }^{\infty }e^{-k\lambda _{n}t}\left ( \frac {2A\left ( 0\right ) }{L\lambda _{n}}+\frac {8L}{\pi ^{2}n^{2}}\int _{0}^{t}A^{\prime }\left ( \tau \right ) e^{k\lambda _{n}\tau }d\tau \right ) \cos \left ( \sqrt {\lambda _{n}}x\right )$ Where $$\lambda _{n}=\left ( \frac {n\pi }{2L}\right ) ^{2},n=1,3,5,\cdots$$. When $$A\left ( t\right ) =\sin \left ( t\right )$$, the above becomes$u\left ( x,t\right ) =\sin \left ( t\right ) \left ( x-L\right ) +\sum _{n=1,3,5,\cdots }^{\infty }e^{-k\lambda _{n}t}\left ( \frac {8L}{\pi ^{2}n^{2}}\int _{0}^{t}\cos \left ( \tau \right ) e^{k\lambda _{n}\tau }d\tau \right ) \cos \left ( \sqrt {\lambda _{n}}x\right )$ But $\int _{0}^{t}\cos \left ( \tau \right ) e^{k\lambda _{n}\tau }d\tau =\frac {k\lambda _{n}e^{k\lambda _{n}t}\cos \left ( t\right ) +e^{k\lambda _{n}t}\sin \left ( t\right ) -k\lambda _{n}}{k^{2}\lambda _{n}^{2}+1}$ And the general solution becomes\begin {align*} u\left ( x,t\right ) & =\left ( x-L\right ) \sin \left ( t\right ) +\sum _{n=1,3,5,\cdots }^{\infty }e^{-k\lambda _{n}t}\left ( \frac {8L}{\pi ^{2}n^{2}}\left ( \frac {k\lambda _{n}e^{k\lambda _{n}t}\cos \left ( t\right ) +e^{k\lambda _{n}t}\sin \left ( t\right ) -k\lambda _{n}}{k^{2}\lambda _{n}^{2}+1}\right ) \right ) \cos \left ( \sqrt {\lambda _{n}}x\right ) \\ & =\left ( x-L\right ) \sin \left ( t\right ) +\sum _{n=1,3,5,\cdots }^{\infty }\frac {8L}{\pi ^{2}n^{2}}\left ( \frac {k\lambda _{n}\cos \left ( t\right ) +\sin \left ( t\right ) -k\lambda _{n}e^{-k\lambda _{n}t}}{k^{2}\lambda _{n}^{2}+1}\right ) \cos \left ( \sqrt {\lambda _{n}}x\right ) \end {align*}

In this problem $$L=5,k=\frac {1}{100}$$, hence the above becomes$u\left ( x,t\right ) =\left ( x-5\right ) \sin \left ( t\right ) +\sum _{n=1,3,5,\cdots }^{\infty }\frac {40}{\pi ^{2}n^{2}}\left ( \frac {\frac {1}{100}\left ( \frac {n\pi }{10}\right ) ^{2}\cos \left ( t\right ) +\sin \left ( t\right ) -\frac {1}{100}\left ( \frac {n\pi }{10}\right ) ^{2}e^{-\frac {1}{100}\left ( \frac {n\pi }{10}\right ) ^{2}t}}{\left ( \frac {1}{100}\left ( \frac {n\pi }{10}\right ) ^{2}\right ) ^{2}+1}\right ) \cos \left ( \frac {n\pi }{10}x\right )$ The following is an animation of the solution

Source code used for the above

________________________________________________________________________________________

##### 3.1.3.4 [210] Haberman 8.3.6 (special case)

problem number 210

Problem 8.3.6 from Richard Haberman applied partial diﬀerential equations book, 5th edition

Solve the heat equation for $$u(x,t)$$ $u_t = u_{xx} + \sin (5 x) e^{-2 t}$ For $$0<x<\pi$$ and $$t>0$$. The boundary conditions are \begin {align*} u(0,t) &= 1 \\ u(\pi ,t) &= 0 \\ \end {align*}

Initial condition is $$u(x,0)=0$$

Mathematica

$\left \{\left \{u(x,t)\to \underset {K[1]=1}{\overset {\infty }{\sum }}\left (\sqrt {\frac {2}{\pi }} \int _0^t e^{-K[1]^2 (t-K[2])} \left (\begin {array}{cc} \{ & \begin {array}{cc} e^{-2 K[2]} \sqrt {\frac {\pi }{2}} & K[1]=5 \\ 0 & \text {True} \\\end {array} \\\end {array}\right ) \, dK[2]-\frac {2 e^{-t K[1]^2}}{\pi K[1]}\right ) \sin (x K[1])-\frac {x}{\pi }+1\right \}\right \}$

Maple

$u \left (x , t\right ) = \frac {-23 x -46 \pi \left (\Mapleoverset {\infty }{\Mapleunderset {n =1}{\sum }}\frac {{\mathrm e}^{-n^{2} t} \sin \left (n x \right )}{\pi n}\right )+\pi \left (-{\mathrm e}^{-25 t}+{\mathrm e}^{-2 t}\right ) \sin \left (5 x \right )+23 \pi }{23 \pi }$

Hand solution

This problem has nonhomogeneous B.C. and non-homogenous in the PDE itself (source present). First step is to use reference function to remove the nonhomogeneous B.C. then use the method of eigenfunction expansion on the resulting problem. Let $r\left ( x\right ) =c_{1}x+c_{2}$ At $$x=0,r\left ( x\right ) =1$$, hence $$1=c_{2}$$ and at $$x=\pi ,r\left ( x\right ) =0$$, hence $$0=c_{1}\pi +1$$ or $$c_{1}=-\frac {1}{\pi }$$, therefore$r\left ( x\right ) =1-\frac {x}{\pi }$ Therefore$u\left ( x,t\right ) =v\left ( x,t\right ) +r\left ( x\right )$ Where $$v\left ( x,t\right )$$ solution for the given PDE but with homogeneous B.C., therefore \begin {align} v_{t} & =v_{xx}+e^{-2t}\sin 5x\tag {1}\\ v\left ( 0,t\right ) & =0\nonumber \\ v\left ( \pi ,t\right ) & =0\nonumber \\ v\left ( x,0\right ) & =f\left ( x\right ) \nonumber \\ & =u\left ( x,0\right ) -r\left ( x\right ) \nonumber \\ & =0-\left ( 1-\frac {x}{\pi }\right ) \nonumber \\ & =\frac {x}{\pi }-1\nonumber \end {align}

We now solve (1). This is a PDE with homogeneous B.C. of the form $$v_{t}=v_{xx}+Q\left ( x,t\right )$$. The general solution to above PDE was solved in 3.1.6.4 on page 1135 and the solution is

\begin {equation} v\left ( x,t\right ) =\sum _{n=1}^{\infty }e^{-k\lambda _{n}t}\Phi _{n}\left ( x\right ) \left ( \frac {2}{L}\int _{0}^{L}f\left ( s\right ) \Phi _{n}\left ( s\right ) ds\right ) +\sum _{n=1}^{\infty }e^{-k\lambda _{n}t}\Phi _{n}\left ( x\right ) \left ( \int _{0}^{t}\frac {2}{L}e^{k\lambda _{n}\tau }\left ( \int _{0}^{L}Q\left ( s,\tau \right ) \Phi _{n}\left ( s\right ) ds\right ) d\tau \right ) \tag {2} \end {equation}

Where \begin {align} \Phi _{n}\left ( x\right ) & =\sin \left ( \sqrt {\lambda _{n}}x\right ) \tag {3}\\ \lambda _{n} & =\left ( \frac {n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots \nonumber \end {align}

Replacing $$L=\pi ,f\left ( x\right ) =\frac {x}{\pi }-1,Q\left ( x,t\right ) =e^{-2t}\sin \left ( 5x\right )$$ into (3,2) gives

\begin {align} \Phi _{n}\left ( x\right ) & =\sin \left ( nx\right ) \tag {3A}\\ \lambda _{n} & =n^{2}\qquad n=1,2,3,\cdots \nonumber \end {align}

And

\begin {equation} v\left ( x,t\right ) =\sum _{n=1}^{\infty }e^{-kn^{2}t}\sin \left ( nx\right ) \left ( \frac {2}{\pi }\int _{0}^{\pi }\left ( \frac {s}{\pi }-1\right ) \sin \left ( ns\right ) ds\right ) +\sum _{n=1}^{\infty }e^{-kn^{2}t}\sin \left ( nx\right ) \left ( \int _{0}^{t}\frac {2}{\pi }e^{kn^{2}\tau }e^{-2\tau }\left ( \int _{0}^{\pi }\sin \left ( 5s\right ) \sin \left ( ns\right ) ds\right ) d\tau \right ) \tag {2A} \end {equation}

But $$\int _{0}^{\pi }\left ( \frac {s}{\pi }-1\right ) \sin \left ( ns\right ) ds=\frac {-1}{n}$$ since $$n$$ is integer. And $$\int _{0}^{\pi }\sin 5s\sin \left ( ns\right ) ds=0$$ when $$n\neq 5$$ and for $$n=5$$ it becomes $$\frac {\pi }{2}$$. Using these values in the above gives

\begin {align} v\left ( x,t\right ) & =\sum _{n=1}^{\infty }e^{-kn^{2}t}\sin \left ( nx\right ) \left ( \frac {-2}{\pi n}\right ) +e^{-k\left ( 5\right ) ^{2}t}\sin \left ( 5x\right ) \left ( \int _{0}^{t}\frac {2}{\pi }e^{k\left ( 5\right ) ^{2}\tau }e^{-2\tau }\left ( \frac {\pi }{2}\right ) d\tau \right ) \tag {2C}\\ & =-\frac {2}{\pi }\sum _{n=1}^{\infty }e^{-kn^{2}t}\frac {\sin \left ( nx\right ) }{n}+e^{-25kt}\sin \left ( 5x\right ) \left ( \int _{0}^{t}e^{25k\tau }e^{-2\tau }d\tau \right ) \end {align}

But $$\int _{0}^{t}e^{25k\tau }e^{-2\tau }d\tau =\frac {-1+e^{25kt-2t}}{25k-2}$$ and the above becomes

\begin {align*} v\left ( x,t\right ) & =-\frac {2}{\pi }\sum _{n=1}^{\infty }e^{-kn^{2}t}\frac {\sin \left ( nx\right ) }{n}+e^{-25kt}\sin \left ( 5x\right ) \left ( \frac {-1+e^{25kt-2t}}{25k-2}\right ) \\ & =-\frac {2}{\pi }\sum _{n=1}^{\infty }\frac {1}{n}e^{-kn^{2}t}\sin \left ( nx\right ) +\sin \left ( 5x\right ) \left ( \frac {-e^{-25kt}+e^{-2t}}{25k-2}\right ) \end {align*}

Since $$u\left ( x,t\right ) =v\left ( x,t\right ) +r\left ( x\right )$$ then the ﬁnal solution is

$u\left ( x,t\right ) =\left ( 1-\frac {x}{\pi }\right ) -\left ( \frac {2}{\pi }\sum _{n=1}^{\infty }\frac {1}{n}e^{-kn^{2}t}\sin \left ( nx\right ) \right ) +\sin \left ( 5x\right ) \left ( \frac {-e^{-25kt}+e^{-2t}}{25k-2}\right )$

Animation is below using $$k=1$$, the solution becomes

$u\left ( x,t\right ) =\left ( 1-\frac {x}{\pi }\right ) -\left ( \frac {2}{\pi }\sum _{n=1}^{\infty }\frac {1}{n}e^{-n^{2}t}\sin \left ( nx\right ) \right ) +\sin \left ( 5x\right ) \left ( \frac {e^{-2t}-e^{-25t}}{23}\right )$

Source code used for the above

________________________________________________________________________________________

##### 3.1.3.5 [211] BC depends on time (special case)

problem number 211

added March 8, 2018. Exam problem

Solve the heat equation $u_t= u_{xx}$ For $$0<x<\pi$$ and $$t>0$$. The boundary conditions are \begin {align*} u(0,t) &= t \\ u(\pi ,t) &= 0 \end {align*}

Initial condition is $$u(x,0)=0$$.

Mathematica

$\left \{\left \{u(x,t)\to \underset {n=1}{\overset {\infty }{\sum }}-\frac {\left (2-2 e^{-n^2 t}\right ) \sin (n x)}{n^3 \pi }-\frac {t x}{\pi }+t\right \}\right \}$

Maple

$u \left (x , t\right ) = \frac {12 \pi \left (\Mapleoverset {\infty }{\Mapleunderset {n =1}{\sum }}\frac {{\mathrm e}^{-n^{2} t} \sin \left (n x \right )}{\pi n^{3}}\right )-6 \left (\frac {1}{6} x^{2}+t -\frac {1}{3} \pi x \right ) \left (x -\pi \right )}{6 \pi }$

________________________________________________________________________________________