### 5.2 Exam 2

5.2.1 problem 3
5.2.2 problem 4
5.2.3 problem 5
5.2.4 problem 2 (optional)
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#### 5.2.1 problem 3

5.2.1.1 Part (a)
5.2.1.2 Part(b)
5.2.1.3 Part (c)
5.2.1.4 Part(d)
5.2.1.5 Part(e)
5.2.1.6 Part(f)

##### 5.2.1.1 Part (a)

$$\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}=\nu \frac{\partial ^{2}u}{\partial x^{2}} \tag{1}$$ Let\begin{align} u & =-2\nu \frac{\phi _{x}}{x}\nonumber \\ & =\frac{\partial }{\partial x}\left ( -2\nu \ln \phi \right ) \tag{1A} \end{align}

##### 5.2.1.2 Part(b)

Let $$\psi =-2\nu \ln \phi \tag{2}$$

Hence (1A) becomes

$u=\frac{\partial }{\partial x}\psi$

We now substitute the above back into (1) noting ﬁrst that$\frac{\partial u}{\partial t}=\frac{\partial }{\partial t}\frac{\partial \psi }{\partial x}$ Interchanging the order gives\begin{align*} \frac{\partial u}{\partial t} & =\frac{\partial }{\partial x}\frac{\partial \psi }{\partial t}\\ & =\frac{\partial }{\partial x}\psi _{t} \end{align*}

And\begin{align*} \frac{\partial u}{\partial x} & =\frac{\partial }{\partial x}\frac{\partial \psi }{\partial x}\\ & =\psi _{xx} \end{align*}

And$\frac{\partial ^{2}u}{\partial x^{2}}=\psi _{xxx}$ Hence the original PDE (1) now can be written in term of $$\psi$$ as the new dependent variable as$$\frac{\partial }{\partial x}\psi _{t}+\psi _{x}\left ( \psi _{xx}\right ) =\nu \psi _{xxx} \tag{3}$$ But $\psi _{x}\left ( \psi _{xx}\right ) =\frac{1}{2}\frac{\partial }{\partial x}\left ( \psi _{x}^{2}\right )$ Using the above in (3), then (3) becomes\begin{align*} \frac{\partial }{\partial x}\psi _{t}+\frac{1}{2}\frac{\partial }{\partial x}\left ( \psi _{x}^{2}\right ) & =\nu \psi _{xxx}\\ \frac{\partial }{\partial x}\psi _{t}+\frac{1}{2}\frac{\partial }{\partial x}\left ( \psi _{x}^{2}\right ) -\nu \frac{\partial }{\partial x}\left ( \psi _{xx}\right ) & =0\\ \frac{\partial }{\partial x}\left ( \psi _{t}+\frac{1}{2}\psi _{x}^{2}-\nu \psi _{xx}\right ) & =0 \end{align*}

Therefore $$\psi _{t}+\frac{1}{2}\psi _{x}^{2}-\nu \psi _{xx}=0 \tag{4}$$ But from (2) $$\psi =-2\nu \ln \phi$$, then using this in (4), we now rewrite (4) in terms of $$\phi$$\begin{align*} \frac{\partial }{\partial t}\left ( -2\nu \ln \phi \right ) +\frac{1}{2}\left ( \frac{\partial }{\partial x}\left ( -2\nu \ln \phi \right ) \right ) ^{2}-\nu \frac{\partial ^{2}}{\partial x^{2}}\left ( -2\nu \ln \phi \right ) & =0\\ \left ( -2\nu \frac{\phi _{t}}{\phi }\right ) +\frac{1}{2}\left ( -2\nu \frac{\phi _{x}}{\phi }\right ) ^{2}-\nu \frac{\partial }{\partial x}\left ( -2\nu \frac{\phi _{x}}{\phi }\right ) & =0\\ -2\nu \frac{\phi _{t}}{\phi }+2\nu ^{2}\left ( \frac{\phi _{x}}{\phi }\right ) ^{2}+2\nu ^{2}\frac{\partial }{\partial x}\left ( \frac{\phi _{x}}{\phi }\right ) & =0 \end{align*}

But $$\frac{\partial }{\partial x}\left ( \frac{\phi _{x}}{\phi }\right ) =\frac{\phi _{xx}}{\phi }-\frac{\phi _{x}^{2}}{\phi ^{2}}$$, hence the above becomes\begin{align*} -2\nu \frac{\phi _{t}}{\phi }+2\nu ^{2}\left ( \frac{\phi _{x}}{\phi }\right ) ^{2}+2\nu ^{2}\left ( \frac{\phi _{xx}}{\phi }-\frac{\phi _{x}^{2}}{\phi ^{2}}\right ) & =0\\ -2\nu \frac{\phi _{t}}{\phi }+2\nu ^{2}\left ( \frac{\phi _{x}}{\phi }\right ) ^{2}+2\nu ^{2}\frac{\phi _{xx}}{\phi }-2\nu ^{2}\frac{\phi _{x}^{2}}{\phi ^{2}} & =0\\ -2\nu \frac{\phi _{t}}{\phi }+2\nu ^{2}\frac{\phi _{xx}}{\phi } & =0\\ -\frac{\phi _{t}}{\phi }+\nu \frac{\phi _{xx}}{\phi } & =0 \end{align*}

Since $$\phi \neq 0$$ identically, then the above simpliﬁes to the heat PDE\begin{align} \phi _{t} & =\nu \phi _{xx}\tag{5}\\ \phi \left ( x,0\right ) & =\Phi \left ( x\right ) \nonumber \\ -\infty & <x<\infty \nonumber \end{align}

##### 5.2.1.3 Part (c)

Now we solve (5) for $$\phi \left ( x,t\right )$$ and then convert the solution back to $$u\left ( x,t\right )$$ using the Cole-Hopf transformation. This inﬁnite domain heat PDE has known solution (as $$\phi \left ( \pm \infty ,t\right )$$ is bounded which is $$\phi \left ( x,t\right ) =\int _{-\infty }^{\infty }\Phi \left ( \eta \right ) \frac{1}{\sqrt{4\pi \upsilon t}}\exp \left ( \frac{-\left ( x-\eta \right ) ^{2}}{4\upsilon t}\right ) d\eta \tag{6}$$

##### 5.2.1.4 Part(d)

Now $$\Phi \left ( x\right ) =\phi \left ( x,0\right ) \tag{7}$$ But since $$u\left ( x,t\right ) =\frac{\partial }{\partial x}\left ( -2\nu \ln \phi \right )$$, then integrating\begin{align*} \int _{x_{0}}^{x}u\left ( \alpha ,t\right ) d\alpha & =-2\nu \ln \phi \\ \ln \phi & =\frac{-1}{2\nu }\int _{x_{0}}^{x}u\left ( \alpha ,t\right ) d\alpha \\ \phi \left ( x,t\right ) & =\exp \left ( \frac{-1}{2\nu }\int _{x_{0}}^{x}u\left ( \alpha ,t\right ) d\alpha \right ) \end{align*}

Hence at $$t=0$$ the above becomes\begin{align*} \phi \left ( x,0\right ) & =\exp \left ( \frac{-1}{2\nu }\int _{x_{0}}^{x}u\left ( \alpha ,0\right ) d\alpha \right ) \\ & =\exp \left ( \frac{-1}{2\nu }\int _{x_{0}}^{x}F\left ( \alpha \right ) d\alpha \right ) \end{align*}

Where $$F\left ( x\right ) =u\left ( x,0\right )$$. Hence from the above, comparing it to (6) we see that$$\Phi \left ( x\right ) =\exp \left ( \frac{-1}{2\nu }\int _{x_{0}}^{x}F\left ( \alpha \right ) d\alpha \right ) \tag{8}$$

##### 5.2.1.5 Part(e)

From (6), we found $$\phi \left ( x,t\right ) =\int _{-\infty }^{\infty }\Phi \left ( \eta \right ) \frac{1}{\sqrt{4\pi \upsilon t}}\exp \left ( \frac{-\left ( x-\eta \right ) ^{2}}{4\upsilon t}\right ) d\eta$$. Plugging (8) into this expression gives\begin{align} \phi \left ( x,t\right ) & =\frac{1}{\sqrt{4\pi \upsilon t}}\int _{-\infty }^{\infty }\exp \left ( \frac{-1}{2\nu }\int _{x_{0}}^{\eta }F\left ( \alpha \right ) d\alpha \right ) \exp \left ( \frac{-\left ( x-\eta \right ) ^{2}}{4\upsilon t}\right ) d\eta \nonumber \\ & =\frac{1}{\sqrt{4\pi \upsilon t}}\int _{-\infty }^{\infty }\exp \left ( \frac{-1}{2\nu }\int _{x_{0}}^{\eta }F\left ( \alpha \right ) d\alpha -\frac{\left ( x-\eta \right ) ^{2}}{4\upsilon t}\right ) d\eta \nonumber \\ & =\frac{1}{\sqrt{4\pi \upsilon t}}\int _{-\infty }^{\infty }\exp \left ( \frac{-1}{2\nu }\left [ \int _{x_{0}}^{\eta }F\left ( \alpha \right ) d\alpha +\frac{\left ( x-\eta \right ) ^{2}}{2t}\right ] \right ) d\eta \tag{9} \end{align}

Let $\int _{x_{0}}^{\eta }F\left ( \alpha \right ) d\alpha +\frac{\left ( x-\eta \right ) ^{2}}{2t}=f\left ( \eta ,x,t\right )$ Hence (9) becomes$$\phi \left ( x,t\right ) =\int _{-\infty }^{\infty }\frac{1}{\sqrt{4\pi \upsilon t}}e^{\frac{-f\left ( \eta ,x,t\right ) }{2\nu }}d\eta \tag{10}$$

##### 5.2.1.6 Part(f)

From (10)\begin{align*} \frac{\partial \phi }{\partial x} & =\int _{-\infty }^{\infty }\frac{1}{\sqrt{4\pi \upsilon t}}\frac{\partial }{\partial x}\left ( e^{\frac{-f\left ( \eta ,x,t\right ) }{2\nu }}\right ) d\eta \\ & =\int _{-\infty }^{\infty }\frac{1}{\sqrt{4\pi \upsilon t}}\left ( \frac{\partial }{\partial x}f\left ( \eta ,x,t\right ) e^{\frac{-f\left ( \eta ,x,t\right ) }{2\nu }}\right ) d\eta \\ & =\int _{-\infty }^{\infty }\frac{1}{\sqrt{4\pi \upsilon t}}\left ( \frac{\partial }{\partial x}\left [ \int _{x_{0}}^{\eta }F\left ( \alpha \right ) d\alpha +\frac{\left ( x-\eta \right ) ^{2}}{2t}\right ] e^{\frac{-f\left ( \eta ,x,t\right ) }{2\nu }}\right ) d\eta \end{align*}

Using Leibniz integral rule the above simpliﬁes to$$\frac{\partial \phi }{\partial x}=\int _{-\infty }^{\infty }\frac{1}{\sqrt{4\pi \upsilon t}}\left ( \frac{\left ( x-\eta \right ) }{t}e^{\frac{-f\left ( \eta ,x,t\right ) }{2\nu }}\right ) d\eta \tag{11}$$ But $u=-2\nu \frac{\phi _{x}}{\phi }$ Hence, using (10) and (11) in the above gives$u=-2\nu \frac{\int _{-\infty }^{\infty }\frac{1}{\sqrt{4\pi \upsilon t}}\left ( \frac{\left ( x-\eta \right ) }{t}e^{\frac{-f\left ( \eta ,x,t\right ) }{2\nu }}\right ) d\eta }{\int _{-\infty }^{\infty }\frac{1}{\sqrt{4\pi \upsilon t}}e^{\frac{-f\left ( \eta ,x,t\right ) }{2\nu }}d\eta }$ Hence the solution is$u\left ( x,t\right ) =-2\nu \frac{\int _{-\infty }^{\infty }\frac{\left ( x-\eta \right ) }{t}e^{\frac{-f\left ( \eta ,x,t\right ) }{2\nu }}d\eta }{\int _{-\infty }^{\infty }e^{\frac{-f\left ( \eta ,x,t\right ) }{2\nu }}d\eta }$ Where$f\left ( \eta ,x,t\right ) =\int _{x_{0}}^{\eta }F\left ( \alpha \right ) d\alpha +\frac{\left ( x-\eta \right ) ^{2}}{2t}$

#### 5.2.2 problem 4

5.2.2.1 Part(a)
5.2.2.2 Part(b)

##### 5.2.2.1 Part(a)

\begin{align} \frac{\partial u}{\partial t} & =\upsilon \frac{\partial ^{2}u}{\partial x^{2}}+Q\left ( x,t\right ) \tag{1}\\ 0 & \leq x\leq L\nonumber \\ t & \geq 0\nonumber \end{align}

Initial conditions are$u\left ( x,0\right ) =f\left ( x\right )$ Boundary conditions are\begin{align*} \frac{\partial u\left ( 0,t\right ) }{\partial x} & =0\\ u\left ( L,t\right ) & =A \end{align*}

Multiplying both sides of (1) by $$G\left ( x,t;x_{0},t_{0}\right )$$ and integrating over the domain gives (where in the following $$G$$ is used instead of $$G\left ( x,t;x_{0},t_{0}\right )$$ for simplicity).$$\int _{x=0}^{L}\int _{t=0}^{\infty }Gu_{t}\ dtdx=\int _{x=0}^{L}\int _{t=0}^{\infty }\upsilon u_{xx}G\ dtdx+\int _{x=0}^{L}\int _{t=0}^{\infty }QG\ dtdx \tag{1}$$ For the integral on the LHS, we apply integration by parts once to move the time derivative from $$u$$ to $$G$$$$\int _{x=0}^{L}\int _{t=0}^{\infty }Gu_{t}\ dtdx=\int _{x=0}^{L}\left [ uG\right ] _{t=0}^{\infty }dx-\int _{x=0}^{L}\int _{t=0}^{\infty }G_{t}u\ dtdx \tag{1A}$$ And the ﬁrst integral in the RHS of (1) gives, after doing integration by parts two times on it\begin{align} \int _{x=0}^{L}\int _{t=0}^{\infty }ku_{xx}G\ dtdx & =\int _{t=0}^{\infty }\left [ u_{x}G\right ] _{x=0}^{L}dt-\int _{x=0}^{L}\int _{t=0}^{\infty }\upsilon u_{x}G_{x}\ dtdx\nonumber \\ & =\int _{t=0}^{L}\left [ u_{x}G\right ] _{x=0}^{L}dt-\left ( \int _{t=0}^{\infty }\left [ uG_{x}\right ] _{x=0}^{L}dt-\int _{x=0}^{L}\int _{t=0}^{\infty }\upsilon uG_{xx}\ dtdx\right ) \nonumber \\ & =\int _{t=0}^{\infty }\left ( \left [ u_{x}G\right ] _{x=0}^{L}-\left [ uG_{x}\right ] _{x=0}^{L}\right ) dt+\int _{x=0}^{L}\int _{t=0}^{\infty }\upsilon uG_{xx}\ dtdx\nonumber \\ & =\int _{t=0}^{\infty }\left [ u_{x}G-uG_{x}\right ] _{x=0}^{L}dt+\int _{x=0}^{L}\int _{t=0}^{\infty }\upsilon uG_{xx}\ dtdx\nonumber \\ & =-\int _{t=0}^{\infty }\left [ uG_{x}-u_{x}G\right ] _{x=0}^{L}dt+\int _{x=0}^{L}\int _{t=0}^{\infty }\upsilon uG_{xx}\ dtdx \tag{1B} \end{align}

Substituting (1A) and (1B) back into (1) results in$\int _{x=0}^{L}\left [ uG\right ] _{t=0}^{\infty }dx-\int _{x=0}^{L}\int _{t=0}^{\infty }G_{t}u\ dtdx=\int _{t=0}^{\infty }\left [ u_{x}G-uG_{x}\right ] _{x=0}^{L}dt+\int _{x=0}^{L}\int _{t=0}^{\infty }\upsilon uG_{xx}\ dtdx+\int _{x=0}^{L}\int _{t=0}^{\infty }GQ\ dtdx$ Or$$\int _{x=0}^{L}\int _{t=0}^{\infty }-G_{t}u-\upsilon uG_{xx}\ dtdx=-\int _{x=0}^{L}\left [ uG\right ] _{t=0}^{\infty }dx-\int _{t=0}^{\infty }\left [ uG_{x}-u_{x}G\right ] _{x=0}^{L}dt+\int _{x=0}^{L}\int _{t=0}^{\infty }GQ\ dtdx \tag{2}$$ We now want to choose $$G\left ( x,t;x_{0},t_{0}\right )$$ such that \begin{align} -G_{t}u-\upsilon uG_{xx} & =\delta \left ( x-x_{0}\right ) \delta \left ( t-t_{0}\right ) \nonumber \\ -G_{t}u & =\upsilon uG_{xx}+\delta \left ( x-x_{0}\right ) \delta \left ( t-t_{0}\right ) \tag{3} \end{align}

This way, the LHS of (2) becomes just $$u\left ( x_{0},t_{0}\right )$$. Hence (2) now (after the above choice of $$G$$) reduces to

$$u\left ( x_{0},t_{0}\right ) =-\int _{x=0}^{L}\left [ uG\right ] _{t=0}^{\infty }dx-\int _{t=0}^{t_{0}}\left [ uG_{x}-u_{x}G\right ] _{x=0}^{L}dt+\int _{x=0}^{L}\int _{t=0}^{t_{0}}GQ\ dtdx \tag{4}$$ We now need to ﬁnd the Green function which satisﬁes (3). But (3) is equivalent to solution of problem \begin{align*} -G_{t}u & =\upsilon uG_{xx}\\ G\left ( x,0\right ) & =\delta \left ( x-x_{0}\right ) \delta \left ( t-t_{0}\right ) \\ -\infty & <x<\infty \\ G\left ( x,t;x_{0},t_{0}\right ) & =0\qquad t>t_{0}\\ G\left ( \pm \infty ,t;x_{0},t_{0}\right ) & =0\\ G\left ( x,t_{0};x_{0},t_{0}\right ) & =\delta \left ( x-x_{0}\right ) \end{align*}

The above problem has a known fundamental solution which we found before, but for the forward heat PDE instead of the reverse heat PDE as it is now. The fundamental solution to the forward heat PDE is$G\left ( x,t\right ) =\frac{1}{\sqrt{4\pi \upsilon \left ( t-t_{0}\right ) }}\exp \left ( \frac{-\left ( x-x_{0}\right ) ^{2}}{4\upsilon \left ( t-t_{0}\right ) }\right ) \qquad 0\leq t_{0}\leq t$ Therefore, for the reverse heat PDE the above becomes$$G\left ( x,t\right ) =\frac{1}{\sqrt{4\pi \upsilon \left ( t_{0}-t\right ) }}\exp \left ( \frac{-\left ( x-x_{0}\right ) ^{2}}{4\upsilon \left ( t_{0}-t\right ) }\right ) \qquad 0\leq t\leq t_{0} \tag{5}$$ We now go back to (4) and try to evaluate all terms in the RHS. Starting with the ﬁrst term $$\int _{x=0}^{L}\left [ uG\right ] _{t=0}^{\infty }dx$$. Since $$G\left ( x,\infty ;x_{0},t_{0}\right ) =0$$ then the upper limit is zero. But at lower limit $$t=0$$ we are given that $$u\left ( x,0\right ) =f\left ( x\right )$$, hence this term becomes \begin{align*} \int _{x=0}^{L}\left [ uG\right ] _{t=0}^{\infty }dx & =\int _{x=0}^{L}-u\left ( x,0\right ) G\left ( x,0\right ) dx\\ & =\int _{x=0}^{L}-f\left ( x\right ) G\left ( x,0\right ) dx \end{align*}

Now looking at the second term in RHS of (4), we expand it and ﬁnd$\left [ uG_{x}-u_{x}G\right ] _{x=0}^{L}=\left ( u\left ( L,t\right ) G_{x}\left ( L,t\right ) -u_{x}\left ( L,t\right ) G\left ( L,t\right ) \right ) -\left ( u\left ( 0,t\right ) G_{x}\left ( 0,t\right ) -u_{x}\left ( 0,t\right ) G\left ( 0,t\right ) \right )$ We are told that $$u_{x}\left ( 0,t\right ) =0$$ and $$u\left ( L,t\right ) =A,$$ then the above becomes$$\left [ uG_{x}-u_{x}G\right ] _{x=0}^{L}=AG_{x}\left ( L,t\right ) -u_{x}\left ( L,t\right ) G\left ( L,t\right ) -u\left ( 0,t\right ) G_{x}\left ( 0,t\right ) \tag{5A}$$ There are still two terms above we do not know. We do not know $$u_{x}\left ( L,t\right )$$ and we also do not know $$u\left ( 0,t\right )$$. If we can conﬁgure, using method of images, such that $$G\left ( L,t\right ) =0$$ and $$G_{x}\left ( 0,t\right ) =0$$ then we can get rid of these two terms and end up only with $$\left [ uG_{x}-u_{x}G\right ] _{x=0}^{\infty }=AG_{x}\left ( L,t\right )$$ which we can evaluate once we know what $$G\left ( x,t\right )$$ is.

This means we need to put images on both sides of the boundaries such that to force $$G\left ( L,t\right ) =0$$ and also $$G_{x}\left ( 0,t\right ) =0$$.

We see that this result agrees what we always did, which is, If the prescribed boundary conditions on $$u$$ are such that $$u=A$$, then we want $$G=0$$ there. And if it is $$\frac{\partial u}{\partial x}=A$$, then we want $$\frac{\partial G}{\partial x}=0$$ there. And this is what we conclude here also from the above. In other words, the boundary conditions on Green functions are always the homogeneous version of the boundary conditions given on $$u$$.

To force $$G_{x}\left ( 0,t\right ) =0$$, we need to put same sign images on both sides of $$x=0$$.  So we end up with this

The above makes $$G_{x}\left ( 0,t\right ) =0$$ at $$x=0$$. Now we want to make $$G=0$$ at $$x=L$$. Then we update the above and put a negative image at $$x=2L-x_{0}$$ to the right of $$x=L$$ as follows

But now we see that the image at $$x=-x_{0}$$ has aﬀected condition of $$G=0$$ at $$x=L$$ and will make it not zero as we wanted. So to counter eﬀect this, we have to add another negative image at distance $$x=2L+x_{0}$$ to cancel the eﬀect of the image at $$x=-x_{0}.$$ We end up with this setup

But now we see that the two negative images we added to the right will no longer make $$G_{x}\left ( 0,t\right ) =0$$, so we need to counter eﬀect this by adding two negative images to the left side to keep $$G_{x}\left ( 0,t\right ) =0$$. So we end up with

But now we see that by putting these two images on the left, we no longer have $$G=0$$ at $$x=L$$. So to counter eﬀect this, we have to put copies of these 2 images on the right again but with positive sign, as follows

But now these two images on the right, no longer keep $$G_{x}\left ( 0,t\right ) =0$$, so we have to put same sign images to the left, as follows

And so on. This continues for inﬁnite number of images. Therefore we see from the above, for the positive images, we have the following sum\begin{align*} \sqrt{4\pi \upsilon \left ( t_{0}-t\right ) }G\left ( x,t;x_{0},t_{0}\right ) & =\exp \left ( \frac{-\left ( x-x_{0}\right ) ^{2}}{4\upsilon \left ( t_{0}-t\right ) }\right ) +\exp \left ( \frac{-\left ( x+x_{0}\right ) ^{2}}{4\upsilon \left ( t_{0}-t\right ) }\right ) +\exp \left ( \frac{-\left ( x-\left ( 4L-x_{0}\right ) \right ) ^{2}}{4\upsilon \left ( t_{0}-t\right ) }\right ) \\ & +\exp \left ( \frac{-\left ( x-\left ( -4L+x_{0}\right ) \right ) ^{2}}{4\upsilon \left ( t_{0}-t\right ) }\right ) +\exp \left ( \frac{-\left ( x-\left ( 4L+x_{0}\right ) \right ) ^{2}}{4\upsilon \left ( t_{0}-t\right ) }\right ) \\ & +\exp \left ( \frac{-\left ( x-\left ( -4L-x_{0}\right ) \right ) ^{2}}{4\upsilon \left ( t_{0}-t\right ) }\right ) +\cdots \end{align*}

Or

$$G\left ( x,t;x_{0},t_{0}\right ) =\frac{1}{\sqrt{4\pi \upsilon \left ( t_{0}-t\right ) }}\left ( \sum _{n=-\infty }^{\infty }\exp \left ( \frac{-\left ( x-\left ( 4nL-x_{0}\right ) \right ) ^{2}}{4\upsilon \left ( t_{0}-t\right ) }\right ) +\exp \left ( \frac{-\left ( x-\left ( -4nL+x_{0}\right ) \right ) ^{2}}{4\upsilon \left ( t_{0}-t\right ) }\right ) \right ) \tag{6}$$ The above takes care of the positive images. For negative images, we have this sum of images\begin{align*} \sqrt{4\pi \upsilon \left ( t_{0}-t\right ) }G\left ( x,t;x_{0},t_{0}\right ) & =\exp \left ( \frac{-\left ( x-\left ( 2L-x_{0}\right ) \right ) ^{2}}{4\upsilon \left ( t_{0}-t\right ) }\right ) +\exp \left ( \frac{-\left ( x-\left ( -2L+x_{0}\right ) \right ) ^{2}}{4\upsilon \left ( t_{0}-t\right ) }\right ) \\ & +\exp \left ( \frac{-\left ( x-\left ( 2L+x_{0}\right ) \right ) ^{2}}{4\upsilon \left ( t_{0}-t\right ) }\right ) +\exp \left ( \frac{-\left ( x-\left ( -2L-x_{0}\right ) \right ) ^{2}}{4\upsilon \left ( t_{0}-t\right ) }\right ) \\ & +\exp \left ( \frac{-\left ( x-\left ( 6L-x_{0}\right ) \right ) ^{2}}{4\upsilon \left ( t_{0}-t\right ) }\right ) +\exp \left ( \frac{-\left ( x-\left ( -6L+x_{0}\right ) \right ) ^{2}}{4\upsilon \left ( t_{0}-t\right ) }\right ) \\ & +\exp \left ( \frac{-\left ( x-\left ( 6L+x_{0}\right ) \right ) ^{2}}{4\upsilon \left ( t_{0}-t\right ) }\right ) +\exp \left ( \frac{-\left ( x-\left ( -6L-x_{0}\right ) \right ) ^{2}}{4\upsilon \left ( t_{0}-t\right ) }\right ) \cdots \end{align*}

Or$$\sqrt{4\pi \upsilon \left ( t_{0}-t\right ) }G\left ( x,t;x_{0},t_{0}\right ) =-\left ( \sum _{n=-\infty }^{\infty }\exp \left ( \frac{-\left ( x-\left ( \left ( 4n-2\right ) L+x_{0}\right ) \right ) ^{2}}{4\upsilon \left ( t_{0}-t\right ) }\right ) +\exp \left ( \frac{-\left ( x-\left ( \left ( 4n-2\right ) L-x_{0}\right ) \right ) ^{2}}{4\upsilon \left ( t_{0}-t\right ) }\right ) \right ) \tag{7}$$ Hence the Green function to use is (6)+(7) which gives\begin{align} G\left ( x,t;x_{0},t_{0}\right ) & =\frac{1}{\sqrt{4\pi \upsilon \left ( t_{0}-t\right ) }}\left ( \sum _{n=-\infty }^{\infty }\exp \left ( \frac{-\left ( x-\left ( 4nL-x_{0}\right ) \right ) ^{2}}{4\upsilon \left ( t_{0}-t\right ) }\right ) +\exp \left ( \frac{-\left ( x-\left ( -4nL+x_{0}\right ) \right ) ^{2}}{4\upsilon \left ( t_{0}-t\right ) }\right ) \right ) \nonumber \\ & -\frac{1}{\sqrt{4\pi \upsilon \left ( t_{0}-t\right ) }}\left ( \sum _{n=-\infty }^{\infty }\exp \left ( \frac{-\left ( x-\left ( \left ( 4n-2\right ) L+x_{0}\right ) \right ) ^{2}}{4\upsilon \left ( t_{0}-t\right ) }\right ) +\exp \left ( \frac{-\left ( x-\left ( \left ( 4n-2\right ) L-x_{0}\right ) \right ) ^{2}}{4\upsilon \left ( t_{0}-t\right ) }\right ) \right ) \tag{7A} \end{align}

Using the above Green function, we go back to 5A and ﬁnally are able to simplify it$\left [ uG_{x}-u_{x}G\right ] _{x=0}^{L}=AG_{x}\left ( L,t\right ) -u_{x}\left ( L,t\right ) G\left ( L,t\right ) -u\left ( 0,t\right ) G_{x}\left ( 0,t\right )$

The above becomes now (with the images in place as above)$$\left [ uG_{x}-u_{x}G\right ] _{x=0}^{L}=A\frac{\partial G\left ( L,t;x_{0},t_{0}\right ) }{\partial x} \tag{8}$$ Since now we know what $$G\left ( x,t;x_{0},t_{0}\right )$$, from (7), we can evaluate its derivative w.r.t. $$x.$$ (broken up, so it ﬁts on one page)

\begin{align*} \frac{\partial G\left ( x,t;x_{0},t_{0}\right ) }{\partial x} & =\frac{1}{\sqrt{4\pi \upsilon \left ( t_{0}-t\right ) }}\sum _{n=-\infty }^{\infty }\frac{-\left ( x-\left ( 4nL-x_{0}\right ) \right ) }{2\upsilon \left ( t_{0}-t\right ) }\exp \left ( \frac{-\left ( x-\left ( 4nL-x_{0}\right ) \right ) ^{2}}{4\upsilon \left ( t_{0}-t\right ) }\right ) \\ & +\frac{1}{\sqrt{4\pi \upsilon \left ( t_{0}-t\right ) }}\sum _{n=-\infty }^{\infty }\frac{-\left ( x-\left ( -4nL+x_{0}\right ) \right ) }{2\upsilon \left ( t_{0}-t\right ) }\exp \left ( \frac{-\left ( x-\left ( -4nL+x_{0}\right ) \right ) ^{2}}{4\upsilon \left ( t_{0}-t\right ) }\right ) \\ & -\frac{1}{\sqrt{4\pi \upsilon \left ( t_{0}-t\right ) }}\sum _{n=-\infty }^{\infty }\frac{-\left ( x-\left ( \left ( 4n-2\right ) L+x_{0}\right ) \right ) }{2\upsilon \left ( t_{0}-t\right ) }\exp \left ( \frac{-\left ( x-\left ( \left ( 4n-2\right ) L+x_{0}\right ) \right ) ^{2}}{4\upsilon \left ( t_{0}-t\right ) }\right ) \\ & -\frac{1}{\sqrt{4\pi \upsilon \left ( t_{0}-t\right ) }}\sum _{n=-\infty }^{\infty }\frac{-\left ( x-\left ( \left ( 4n-2\right ) L-x_{0}\right ) \right ) }{2\upsilon \left ( t_{0}-t\right ) }\exp \left ( \frac{-\left ( x-\left ( \left ( 4n-2\right ) L-x_{0}\right ) \right ) ^{2}}{4\upsilon \left ( t_{0}-t\right ) }\right ) \end{align*}

At $$x=L$$, the above derivative becomes\begin{align} \frac{\partial G\left ( L,t;x_{0},t_{0}\right ) }{\partial x} & =\frac{1}{\sqrt{4\pi \upsilon \left ( t_{0}-t\right ) }}\sum _{n=-\infty }^{\infty }\frac{-\left ( L-\left ( 4nL-x_{0}\right ) \right ) }{2\upsilon \left ( t_{0}-t\right ) }\exp \left ( \frac{-\left ( L-\left ( 4nL-x_{0}\right ) \right ) ^{2}}{4\upsilon \left ( t_{0}-t\right ) }\right ) \nonumber \\ & +\frac{1}{\sqrt{4\pi \upsilon \left ( t_{0}-t\right ) }}\sum _{n=-\infty }^{\infty }\frac{-\left ( L-\left ( -4nL+x_{0}\right ) \right ) }{2\upsilon \left ( t_{0}-t\right ) }\exp \left ( \frac{-\left ( L-\left ( -4nL+x_{0}\right ) \right ) ^{2}}{4\upsilon \left ( t_{0}-t\right ) }\right ) \nonumber \\ & -\frac{1}{\sqrt{4\pi \upsilon \left ( t_{0}-t\right ) }}\sum _{n=-\infty }^{\infty }\frac{-\left ( L-\left ( \left ( 4n-2\right ) L+x_{0}\right ) \right ) }{2\upsilon \left ( t_{0}-t\right ) }\exp \left ( \frac{-\left ( L-\left ( \left ( 4n-2\right ) L+x_{0}\right ) \right ) ^{2}}{4\upsilon \left ( t_{0}-t\right ) }\right ) \nonumber \\ & -\frac{1}{\sqrt{4\pi \upsilon \left ( t_{0}-t\right ) }}\sum _{n=-\infty }^{\infty }\frac{-\left ( L-\left ( \left ( 4n-2\right ) L-x_{0}\right ) \right ) }{2\upsilon \left ( t_{0}-t\right ) }\exp \left ( \frac{-\left ( L-\left ( \left ( 4n-2\right ) L-x_{0}\right ) \right ) ^{2}}{4\upsilon \left ( t_{0}-t\right ) }\right ) \tag{9} \end{align}

From (4), we now collect all terms into the solution$$u\left ( x_{0},t_{0}\right ) =-\int _{x=0}^{L}\left [ uG\right ] _{t=0}^{\infty }dx-\int _{t=0}^{t_{0}}\left [ uG_{x}-u_{x}G\right ] _{x=0}^{L}dt+\int _{x=0}^{L}\int _{t=0}^{t_{0}}GQ\ dtdx \tag{4}$$ We found $$\int _{x=0}^{L}\left [ uG\right ] _{t=0}^{\infty }dx=\int _{x=0}^{L}-f\left ( x\right ) G\left ( x,0\right ) dx$$ and now we know what $$G$$ is. Hence we can ﬁnd $$G\left ( x,0;x_{0},t_{0}\right )$$. It is, from (7A)

\begin{align} G\left ( x,0;x_{0},t_{0}\right ) & =\frac{1}{\sqrt{4\pi \upsilon t_{0}}}\left ( \sum _{n=-\infty }^{\infty }\exp \left ( \frac{-\left ( x-\left ( 4nL-x_{0}\right ) \right ) ^{2}}{4\upsilon t_{0}}\right ) +\exp \left ( \frac{-\left ( x-\left ( -4nL+x_{0}\right ) \right ) ^{2}}{4\upsilon t_{0}}\right ) \right ) \nonumber \\ & -\frac{1}{\sqrt{4\pi \upsilon t_{0}}}\left ( \sum _{n=-\infty }^{\infty }\exp \left ( \frac{-\left ( x-\left ( \left ( 4n-2\right ) L+x_{0}\right ) \right ) ^{2}}{4\upsilon t_{0}}\right ) +\exp \left ( \frac{-\left ( x-\left ( \left ( 4n-2\right ) L-x_{0}\right ) \right ) ^{2}}{4\upsilon t_{0}}\right ) \right ) \tag{10} \end{align}

And we now know what $$\left [ uG_{x}-u_{x}G\right ] _{x=0}^{L}$$ is. It is $$A\frac{\partial G\left ( L,t;x_{0},t_{0}\right ) }{\partial x}$$. Hence (4) becomes$u\left ( x_{0},t_{0}\right ) =\int _{x=0}^{L}f\left ( x\right ) G\left ( x,0;x_{0},t_{0}\right ) dx-\int _{t=0}^{t_{0}}A\frac{\partial G\left ( L,t;x_{0},t_{0}\right ) }{\partial x}dt+\int _{x=0}^{L}\int _{t=0}^{t_{0}}G\left ( x,t;x_{0},t_{0}\right ) Q\left ( x,t\right ) \ dtdx$ Changing the roles of $$x_{0},t_{0}$$$$u\left ( x,t\right ) =\int _{x_{0}=0}^{L}f\left ( x_{0}\right ) G\left ( x_{0},t_{0};x,0\right ) dx_{0}-\int _{t=0}^{t}A\frac{\partial G\left ( x_{0},t_{0};L,t\right ) }{\partial x_{0}}dt_{0}+\int _{x_{0}=0}^{L}\int _{t_{0}=0}^{t}G\left ( x_{0},t_{0};x,t\right ) Q\left ( x_{0},t_{0}\right ) \ dt_{0}dx_{0} \tag{11}$$ This completes the solution.

Summary

The solution is $u\left ( x,t\right ) =\int _{x_{0}=0}^{L}f\left ( x_{0}\right ) G\left ( x_{0},t_{0};x,0\right ) dx_{0}-\int _{t=0}^{t}A\frac{\partial G\left ( x_{0},t_{0};L,t\right ) }{\partial x_{0}}dt_{0}+\int _{x_{0}=0}^{L}\int _{t_{0}=0}^{t}G\left ( x_{0},t_{0};x,t\right ) Q\left ( x_{0},t_{0}\right ) \ dt_{0}dx_{0}$ Where $$G\left ( x_{0},t_{0};x,0\right )$$ is given in (10) (after changing roles of parameters):

\begin{align} G\left ( x_{0},t_{0};x,0\right ) & =\frac{1}{\sqrt{4\pi \upsilon t}}\left ( \sum _{n=-\infty }^{\infty }\exp \left ( \frac{-\left ( x_{0}-\left ( 4nL-x\right ) \right ) ^{2}}{4\upsilon t}\right ) +\exp \left ( \frac{-\left ( x_{0}-\left ( -4nL+x\right ) \right ) ^{2}}{4\upsilon t}\right ) \right ) \nonumber \\ & -\frac{1}{\sqrt{4\pi \upsilon t}}\left ( \sum _{n=-\infty }^{\infty }\exp \left ( \frac{-\left ( x_{0}-\left ( \left ( 4n-2\right ) L+x\right ) \right ) ^{2}}{4\upsilon t}\right ) +\exp \left ( \frac{-\left ( x_{0}-\left ( \left ( 4n-2\right ) L-x\right ) \right ) ^{2}}{4\upsilon t}\right ) \right ) \tag{10A} \end{align}

and $$\frac{\partial G\left ( x_{0},t_{0};L,t\right ) }{\partial x_{0}}$$ is given in (9) (after also changing roles of parameters):\begin{align} \frac{\partial G\left ( x_{0},t_{0};L,t\right ) }{\partial x_{0}} & =\frac{1}{\sqrt{4\pi \upsilon \left ( t-t_{0}\right ) }}\sum _{n=-\infty }^{\infty }\frac{-\left ( L-\left ( 4nL-x\right ) \right ) }{2\upsilon \left ( t-t_{0}\right ) }\exp \left ( \frac{-\left ( L-\left ( 4nL-x\right ) \right ) ^{2}}{4\upsilon \left ( t-t_{0}\right ) }\right ) \nonumber \\ & +\frac{1}{\sqrt{4\pi \upsilon \left ( t-t_{0}\right ) }}\sum _{n=-\infty }^{\infty }\frac{-\left ( L-\left ( -4nL+x\right ) \right ) }{2\upsilon \left ( t-t_{0}\right ) }\exp \left ( \frac{-\left ( L-\left ( -4nL+x\right ) \right ) ^{2}}{4\upsilon \left ( t-t_{0}\right ) }\right ) \nonumber \\ & -\frac{1}{\sqrt{4\pi \upsilon \left ( t-t_{0}\right ) }}\sum _{n=-\infty }^{\infty }\frac{-\left ( L-\left ( \left ( 4n-2\right ) L+x\right ) \right ) }{2\upsilon \left ( t-t_{0}\right ) }\exp \left ( \frac{-\left ( L-\left ( \left ( 4n-2\right ) L+x\right ) \right ) ^{2}}{4\upsilon \left ( t-t_{0}\right ) }\right ) \nonumber \\ & -\frac{1}{\sqrt{4\pi \upsilon \left ( t-t_{0}\right ) }}\sum _{n=-\infty }^{\infty }\frac{-\left ( L-\left ( \left ( 4n-2\right ) L-x\right ) \right ) }{2\upsilon \left ( t-t_{0}\right ) }\exp \left ( \frac{-\left ( L-\left ( \left ( 4n-2\right ) L-x\right ) \right ) ^{2}}{4\upsilon \left ( t-t_{0}\right ) }\right ) \tag{9A} \end{align}

and $$G\left ( x_{0},t_{0};x,t\right )$$ is given in (7A), but with roles changed as well to become\begin{align} G\left ( x_{0},t_{0};x,t\right ) & =\frac{1}{\sqrt{4\pi \upsilon \left ( t-t_{0}\right ) }}\left ( \sum _{n=-\infty }^{\infty }\exp \left ( \frac{-\left ( x_{0}-\left ( 4nL-x\right ) \right ) ^{2}}{4\upsilon \left ( t-t_{0}\right ) }\right ) +\exp \left ( \frac{-\left ( x_{0}-\left ( -4nL+x\right ) \right ) ^{2}}{4\upsilon \left ( t-t_{0}\right ) }\right ) \right ) \tag{7AA}\\ & -\frac{1}{\sqrt{4\pi \upsilon \left ( t-t_{0}\right ) }}\left ( \sum _{n=-\infty }^{\infty }\exp \left ( \frac{-\left ( x_{0}-\left ( \left ( 4n-2\right ) L+x\right ) \right ) ^{2}}{4\upsilon \left ( t-t_{0}\right ) }\right ) +\exp \left ( \frac{-\left ( x_{0}-\left ( \left ( 4n-2\right ) L-x\right ) \right ) ^{2}}{4\upsilon \left ( t-t_{0}\right ) }\right ) \right ) \nonumber \end{align}

##### 5.2.2.2 Part(b)

When $$A=0$$, the solution in part (a) becomes$u\left ( x,t\right ) =\int _{x_{0}=0}^{L}f\left ( x_{0}\right ) G\left ( x_{0},t_{0};x,0\right ) dx_{0}+\int _{x_{0}=0}^{L}\int _{t_{0}=0}^{t}G\left ( x_{0},t_{0};x,t\right ) Q\ dt_{0}dx_{0}$ Where $$G\left ( x_{0},t_{0};x,0\right )$$ is given in (10A) in part (a), and $$G\left ( x_{0},t_{0};x,t\right )$$ is given in (7AA) in part (a). Now we ﬁnd the eigenfunction solution for this problem order to compare it with the above green function images solution. Since $$A=0$$ then the PDE now becomes\begin{align} \frac{\partial u}{\partial t} & =\upsilon \frac{\partial ^{2}u}{\partial x^{2}}+Q\left ( x,t\right ) \tag{1}\\ 0 & \leq x\leq L\nonumber \\ t & \geq 0\nonumber \end{align}

Initial conditions$u\left ( x,0\right ) =f\left ( x\right )$ Boundary conditions\begin{align*} \frac{\partial u\left ( 0,t\right ) }{\partial x} & =0\\ u\left ( L,t\right ) & =0 \end{align*}

Since boundary conditions has now become homogenous (thanks for $$A=0$$), we can use separation of variables to ﬁnd the eigenfunctions, and then use eigenfunction expansion. Let the solution be\begin{align} u_{n}\left ( x,t\right ) & =a_{n}\left ( t\right ) \phi _{n}\left ( x\right ) \nonumber \\ u\left ( x,t\right ) & =\sum _{n=1}^{\infty }a_{n}\left ( t\right ) \phi _{n}\left ( x\right ) \tag{1A} \end{align}

Where $$\phi _{n}\left ( t\right )$$ are eigenfunctions for the associated homogeneous PDE $$\frac{\partial u}{\partial t}=\upsilon \frac{\partial ^{2}u}{\partial x^{2}}$$ which can be found from separation of variables. To ﬁnd $$\phi _{n}\left ( x\right )$$, we start by separation of variables. Let $$u\left ( x,t\right ) =X\left ( x\right ) T\left ( t\right )$$ and we plug this solution back to the PDE to obtain\begin{align*} XT^{\prime } & =\upsilon X^{\prime \prime }T\\ \frac{1}{v}\frac{T^{\prime }}{T} & =\frac{X^{\prime \prime }}{X}=-\lambda \end{align*}

Hence the spatial ODE is $$\frac{X^{\prime \prime }}{X}=-\lambda$$ or $$X^{\prime \prime }+\lambda X=0$$ with boundary conditions \begin{align*} X^{\prime }\left ( 0\right ) & =0\\ X\left ( L\right ) & =0 \end{align*}

case $$\lambda =0$$ The solution is $$X=c_{1}+c_{2}x$$. Hence $$X^{\prime }=c_{2}$$. Therefore $$c_{2}=0$$. Hence $$X=c_{1}=0$$. Trivial solution. So $$\lambda =0$$ is not possible.

case $$\lambda >0$$ The solution is $$X=c_{1}\cos \left ( \sqrt{\lambda }x\right ) +c_{2}\sin \left ( \sqrt{\lambda }x\right )$$. and $$X^{\prime }=-\sqrt{\lambda }c_{1}\sin \lambda x+c_{2}\sqrt{\lambda }\cos \lambda x$$. From ﬁrst B.C. at $$x=0$$ we ﬁnd $$0=c_{2}\sqrt{\lambda }$$, hence $$c_{2}=0$$ and the solution becomes $$X=c_{1}\cos \left ( \sqrt{\lambda }x\right )$$. At $$x=L$$, we have $$0=c_{1}\cos \left ( \sqrt{\lambda }L\right )$$ which leads to $$\sqrt{\lambda }L=n\frac{\pi }{2}$$ for $$n=1,3,5,\cdots$$ or \begin{align*} \sqrt{\lambda _{n}} & =\left ( \frac{2n-1}{2}\right ) \frac{\pi }{L}\qquad n=1,2,3,\cdots \\ \lambda _{n} & =\left ( \frac{2n-1}{2}\frac{\pi }{L}\right ) ^{2}\qquad n=1,1,2,3,\cdots \end{align*}

Hence the $$X_{n}\left ( x\right )$$ solution is$X_{n}\left ( x\right ) =c_{n}\cos \left ( \sqrt{\lambda _{n}}x\right ) \qquad n=1,2,3,\cdots$ The time ODE is now solved using the above eigenvalues. (we really do not need to do this part, since $$A_{n}\left ( t\right )$$ will be solved for later, and $$A_{n}\left ( t\right )$$ will contain all the time dependent parts, including those that come from $$Q\left ( x,t\right )$$, but for completion, this is done)\begin{align*} \frac{1}{v}\frac{T^{\prime }}{T} & =-\lambda _{n}\\ T^{\prime }+v\lambda _{n}T & =0\\ \frac{dT}{T} & =-v\lambda _{n}dt\\ \ln \left \vert T\right \vert & =-v\lambda _{n}t+C\\ T & =C_{n}e^{v\lambda _{n}t} \end{align*}

Hence the solution to the homogenous PDE is\begin{align*} u_{n}\left ( x,t\right ) & =X_{n}T_{n}\\ & =c_{n}\cos \left ( \sqrt{\lambda _{n}}x\right ) e^{v\lambda _{n}t} \end{align*}

Where constants of integration are merged into one. Therefore\begin{align} u\left ( x,t\right ) & =\sum _{n=1}^{\infty }X_{n}T_{n}\nonumber \\ & =\sum _{n=1}^{\infty }c_{n}\cos \left ( \sqrt{\lambda _{n}}x\right ) e^{v\lambda _{n}t} \tag{2} \end{align}

From the above we see that $\fbox{\phi _n\left ( x\right ) =\cos \left ( \sqrt{\lambda _n}x\right ) }$ Using this, we now write the solution to $$\frac{\partial u}{\partial t}=\upsilon \frac{\partial ^{2}u}{\partial x^{2}}+Q\left ( x,t\right )$$ using eigenfunction expansion$$u\left ( x,t\right ) =\sum _{n=1}^{\infty }A_{n}\left ( t\right ) \phi _{n}\left ( x\right ) \tag{3}$$ Where now $$A_{n}\left ( t\right )$$ will have all the time dependent terms from $$Q\left ( x,t\right )$$ as well from the time solution from the homogenous PDE $$e^{v\left ( \frac{2n-1}{2}\frac{\pi }{L}\right ) ^{2}t}$$ part. We will solve for $$A_{n}\left ( t\right )$$ now.

In this below, we will expand $$Q\left ( x,t\right )$$ using these eigenfunctions (we can do this, since the eigenfunctions are basis for the whole solution space which the forcing function is in as well). We plug-in (3) back into the PDE, and since boundary conditions are now homogenous, then term by term diﬀerentiation is justiﬁed. The PDE $$\frac{\partial u}{\partial t}=\upsilon \frac{\partial ^{2}u}{\partial x^{2}}+Q\left ( x,t\right )$$ now becomes$$\frac{\partial }{\partial t}\sum _{n=1}^{\infty }A_{n}\left ( t\right ) \phi _{n}\left ( x\right ) =\upsilon \frac{\partial ^{2}}{\partial x^{2}}\sum _{n=1}^{\infty }A_{n}\left ( t\right ) \phi _{n}\left ( x\right ) +\sum _{n=1}^{\infty }q_{n}\left ( t\right ) \phi _{n}\left ( x\right ) \tag{4}$$ Where $$\sum _{n=1}^{\infty }q_{n}\left ( t\right ) \phi _{n}\left ( x\right )$$ is the eigenfunction expansion of $$Q\left ( x,t\right )$$. To ﬁnd $$q_{n}\left ( t\right )$$ we apply orthogonality as follows\begin{align*} Q\left ( x,t\right ) & =\sum _{n=1}^{\infty }q_{n}\left ( t\right ) \phi _{n}\left ( x\right ) \\ \int _{0}^{L}\phi _{m}\left ( x\right ) Q\left ( x,t\right ) dx & =\int _{0}^{L}\left ( \sum _{n=1}^{\infty }q_{n}\left ( t\right ) \phi _{n}\left ( x\right ) \right ) \phi _{m}\left ( x\right ) dx\\ \int _{0}^{L}\phi _{m}\left ( x\right ) Q\left ( x,t\right ) dx & =\sum _{n=1}^{\infty }\int _{0}^{L}q_{n}\left ( t\right ) \phi _{n}\left ( x\right ) \phi _{m}\left ( x\right ) dx\\ & =\int _{0}^{L}q_{m}\left ( t\right ) \phi _{m}^{2}\left ( x\right ) dx\\ & =q_{m}\left ( t\right ) \int _{0}^{L}\cos ^{2}\left ( \sqrt{\lambda _{n}}x\right ) dx \end{align*}

But $\int _{0}^{L}\cos ^{2}\left ( \sqrt{\lambda _{n}}x\right ) dx=\frac{L}{2}$ Hence$$q_{n}\left ( t\right ) =\frac{2}{L}\int _{0}^{L}\phi _{n}\left ( x\right ) Q\left ( x,t\right ) dx \tag{5}$$ Now that we found $$q_{n}\left ( t\right )$$, we go back to (4) and simpliﬁes it more\begin{align*} \sum _{n=1}^{\infty }A_{n}^{\prime }\left ( t\right ) \phi _{n}\left ( x\right ) & =\upsilon \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 ) \\ A_{n}^{\prime }\left ( t\right ) \phi _{n}\left ( x\right ) & =\upsilon A_{n}\left ( t\right ) \phi _{n}^{\prime \prime }\left ( x\right ) +q_{n}\left ( t\right ) \phi _{n}\left ( x\right ) \end{align*}

But since $$\phi _{n}\left ( x\right ) =\cos \left ( \sqrt{\lambda _{n}}x\right )$$ then $\phi _{n}^{\prime }\left ( x\right ) =-\left ( \sqrt{\lambda _{n}}\right ) \sin \left ( \sqrt{\lambda _{n}}x\right )$ And \begin{align*} \phi _{n}^{\prime \prime }\left ( x\right ) & =-\lambda _{n}\cos \left ( \sqrt{\lambda _{n}}x\right ) \\ & =-\lambda _{n}\phi _{n}\left ( x\right ) \end{align*}

Hence the above ODE becomes$A_{n}^{\prime }\phi _{n}=-\upsilon A_{n}\lambda _{n}\phi _{n}+q_{n}\phi _{n}$ Canceling the eigenfunction $$\phi _{n}\left ( x\right )$$ (since not zero) gives$$A_{n}^{\prime }\left ( t\right ) +\upsilon A_{n}\left ( t\right ) \lambda _{n}=q_{n}\left ( t\right ) \tag{6}$$ We now solve this for $$A_{n}\left ( t\right )$$. Integrating factor is \begin{align*} \mu & =\exp \left ( \int \upsilon \lambda _{n}dt\right ) \\ & =e^{\lambda _{n}\upsilon t} \end{align*}

Hence (6) becomes\begin{align} \frac{d}{dt}\left ( \mu A_{n}\left ( t\right ) \right ) & =\mu q_{n}\left ( t\right ) \nonumber \\ e^{\lambda _{n}\upsilon t}A_{n}\left ( t\right ) & =\int _{0}^{t}e^{\lambda _{n}\upsilon s}q_{n}\left ( s\right ) ds+C\nonumber \\ A_{n}\left ( t\right ) & =e^{-\lambda _{n}\upsilon t}\int _{0}^{t}e^{\lambda _{n}\upsilon s}q_{n}\left ( s\right ) ds+Ce^{-\lambda _{n}\upsilon t} \tag{7} \end{align}

Now that we found $$A_{n}\left ( t\right )$$, then the solution (3) becomes$$u\left ( x,t\right ) =\sum _{n=1}^{\infty }\left ( e^{-\left ( \frac{2n-1}{2}\frac{\pi }{L}\right ) ^{2}\upsilon t}\int _{0}^{t}e^{\left ( \frac{2n-1}{2}\frac{\pi }{L}\right ) ^{2}\upsilon s}q_{n}\left ( s\right ) ds+Ce^{-\left ( \frac{2n-1}{2}\frac{\pi }{L}\right ) ^{2}\upsilon t}\right ) \phi _{n}\left ( x\right ) \tag{8}$$ At $$t=0\,$$, we are given that $$u\left ( x,0\right ) =f\left ( x\right )$$, hence the above becomes$f\left ( x\right ) =\sum _{n=1}^{\infty }C\phi _{n}\left ( x\right )$ To ﬁnd $$C$$, we apply orthogonality again, which gives\begin{align*} \int _{0}^{L}f\left ( x\right ) \phi _{m}\left ( x\right ) dx & =\sum _{n=1}^{\infty }\int _{0}^{L}C\phi _{n}\left ( x\right ) \phi _{m}\left ( x\right ) dx\\ & =C\int _{0}^{L}\phi _{m}^{2}\left ( x\right ) dx\\ \int _{0}^{L}f\left ( x\right ) \phi _{m}\left ( x\right ) dx & =\frac{L}{2}C\\ C & =\frac{2}{L}\int _{0}^{L}f\left ( x\right ) \phi _{n}\left ( x\right ) dx \end{align*}

Now that we found $$C$$, then the solution in (8) is complete. It is \begin{align*} u\left ( x,t\right ) & =\sum _{n=1}^{\infty }A_{n}\left ( t\right ) \phi _{n}\left ( x\right ) \\ & =\sum _{n=1}^{\infty }\left ( e^{-\lambda _{n}\upsilon t}\int _{0}^{t}e^{\lambda _{n}\upsilon s}q_{n}\left ( s\right ) ds+\frac{2}{L}e^{-\lambda _{n}\upsilon t}\int _{0}^{L}f\left ( x\right ) \phi _{n}\left ( x\right ) dx\right ) \phi _{n}\left ( x\right ) \end{align*}

Or\begin{align*} u\left ( x,t\right ) & =\sum _{n=1}^{\infty }\left ( \phi _{n}\left ( x\right ) e^{-\lambda _{n}\upsilon t}\int _{0}^{t}e^{\lambda _{n}\upsilon s}q_{n}\left ( s\right ) ds\right ) \\ & +\frac{2}{L}\sum _{n=1}^{\infty }\left ( \phi _{n}\left ( x\right ) e^{-\lambda _{n}\upsilon t}\int _{0}^{L}f\left ( x\right ) \phi _{n}\left ( x\right ) dx\right ) \end{align*}

Summary

The eigenfunction solution is\begin{align*} u\left ( x,t\right ) & =\sum _{n=1}^{\infty }\left ( \phi _{n}\left ( x\right ) e^{-\lambda _{n}\upsilon t}\int _{0}^{t}e^{\lambda _{n}\upsilon s}q_{n}\left ( s\right ) ds\right ) \\ & +\frac{2}{L}\sum _{n=1}^{\infty }\left ( \phi _{n}\left ( x\right ) e^{-\lambda _{n}\upsilon t}\int _{0}^{L}f\left ( x\right ) \phi _{n}\left ( x\right ) dx\right ) \end{align*}

Where\begin{align*} \phi _{n}\left ( x\right ) & =\cos \left ( \sqrt{\lambda _{n}}x\right ) \qquad n=1,2,3,\cdots \\ \lambda _{n} & =\left ( \frac{2n-1}{2}\frac{\pi }{L}\right ) ^{2}\qquad n=1,1,2,3,\cdots \end{align*}

And$q_{n}\left ( t\right ) =\frac{2}{L}\int _{0}^{L}\phi _{n}\left ( x\right ) Q\left ( x,t\right ) dx$ To compare the eigenfunction expansion solution and the Green function solution, we see the following mapping of the two solutions$\overset{\frac{2}{L}\sum _{n=1}^{\infty }\left ( \phi _{n}\left ( x\right ) e^{-\lambda _{n}\upsilon t}\int _{0}^{L}f\left ( x\right ) \phi _{n}\left ( x\right ) dx\right ) }{\overbrace{\int _{0}^{L}f\left ( x_{0}\right ) G\left ( x_{0},t_{0};x,0\right ) dx_{0}}}+\overset{\sum _{n=1}^{\infty }\left ( \phi _{n}\left ( x\right ) e^{-\lambda _{n}\upsilon t}\int _{0}^{t}e^{\lambda _{n}\upsilon s}q_{n}\left ( s\right ) ds\right ) }{\overbrace{\int _{0}^{L}\int _{0}^{t}G\left ( x_{0},t_{0};x,t\right ) Q\left ( x_{0},t_{0}\right ) \ dt_{0}dx_{0}}}$ Where the top expression is the eigenfunction expansion and the bottom expression is the Green function solution using method of images. Where $$G\left ( x_{0},t_{0};x,t\right )$$ in above contains the inﬁnite sums of the images. So the Green function solution contains integrals and inside these integrals are the inﬁnite sums. While the eigenfunction expansions contains two inﬁnite sums, but inside the sums we see the integrals. So summing over the images seems to be equivalent to the operation of summing over eigenfunctions. These two solutions in the limit should of course give the same result (unless I made a mistake somewhere). In this example, I found method of eigenfunction expansion easier, since getting the images in correct locations and sign was tricky to get right.

#### 5.2.3 problem 5

\begin{align} \frac{\partial u\left ( x,t\right ) }{\partial t} & =\frac{i\lambda }{4\pi }\frac{\partial ^{2}u\left ( x,t\right ) }{\partial x^{2}}\tag{1}\\ u\left ( x,0\right ) & =f\left ( x\right ) =\delta \left ( x-a\right ) +\delta \left ( x+a\right ) \nonumber \end{align}

I will Use Fourier transform to solve this, since this is for $$-\infty <x<\infty$$ and the solution $$u\left ( x,t\right )$$ is assumed bounded at $$\pm \infty$$ (or goes to zero there), hence $$u\left ( x,t\right )$$ is square integrable and therefore we can assume it has a Fourier transform.

Let $$U\left ( \xi ,t\right )$$ be the spatial part only Fourier transform of $$u\left ( x,t\right )$$. Using the Fourier transform pairs deﬁned as \begin{align*} U\left ( \xi ,t\right ) & =\mathcal{F}\left ( u\left ( x,t\right ) \right ) =\int _{-\infty }^{\infty }u\left ( x,t\right ) e^{-i2\pi x\xi }dx\\ u\left ( x,t\right ) & =\mathcal{F}^{-1}\left ( U\left ( \xi ,t\right ) \right ) =\int _{-\infty }^{\infty }U\left ( \xi ,t\right ) e^{i2\pi x\xi }d\xi \end{align*}

Therefore, by Fourier transform properties of derivatives\begin{align} \mathcal{F}\left ( \frac{\partial u\left ( x,t\right ) }{\partial x}\right ) & =\left ( 2\pi i\xi \right ) U\left ( \xi ,t\right ) \nonumber \\\mathcal{F}\left ( \frac{\partial ^{2}u\left ( x,t\right ) }{\partial x^{2}}\right ) & =\left ( 2\pi i\xi \right ) ^{2}U\left ( \xi ,t\right ) \tag{2} \end{align}

And$$\mathcal{F}\left ( \frac{\partial u\left ( x,t\right ) }{\partial t}\right ) =\frac{\partial U\left ( \xi ,t\right ) }{\partial t} \tag{3}$$ Where in (3), we just need to take time derivative of $$U\left ( \xi ,t\right )$$ since the transform is applied only to the space part.  Now we take the Fourier transform of the given PDE and using (2,3) relations we obtain the PDE but now in Fourier space.\begin{align} \frac{\partial U\left ( \xi ,t\right ) }{\partial t} & =\left ( \frac{i\lambda }{4\pi }\right ) \left ( 2\pi i\xi \right ) ^{2}U\left ( \xi ,t\right ) \nonumber \\ & =-\left ( \frac{i\lambda }{4\pi }\right ) 4\pi ^{2}\xi ^{2}U\left ( \xi ,t\right ) \nonumber \\ & =\left ( -i\lambda \pi \xi ^{2}\right ) U\left ( \xi ,t\right ) \tag{4} \end{align}

Equation (4) can now be easily solved for $$U\left ( \xi ,t\right )$$ since it is separable.$\frac{\partial U\left ( \xi ,t\right ) }{U\left ( \xi ,t\right ) }=\left ( -i\lambda \pi \xi ^{2}\right ) \partial t$ Integrating\begin{align*} \ln \left \vert U\left ( \xi ,t\right ) \right \vert & =\left ( -i\lambda \pi \xi ^{2}\right ) t+C\\ U\left ( \xi ,t\right ) & =U\left ( \xi ,0\right ) e^{\left ( -i\lambda \pi \xi ^{2}\right ) t} \end{align*}

Where $$U\left ( \xi ,0\right )$$ is the Fourier transform of $$u\left ( x,0\right )$$, the initial conditions, which is $$f\left ( x\right )$$ and is given in the problem.  To go back to spatial domain, we now need to do the inverse Fourier transform. By applying the convolution theorem, we know that multiplication in Fourier domain is convolution in spatial domain, therefore$$\mathcal{F}^{-1}\left ( U\left ( \xi ,t\right ) \right ) =\mathcal{F}^{-1}\left ( U\left ( \xi ,0\right ) \right ) \circledast \mathcal{F}^{-1}\left ( e^{-i\lambda \pi \xi ^{2}t}\right ) \tag{5}$$ But \begin{align*} \mathcal{F}^{-1}\left ( U\left ( \xi ,t\right ) \right ) & =u\left ( x,t\right ) \\\mathcal{F}^{-1}\left ( U\left ( \xi ,0\right ) \right ) & =f\left ( x\right ) \end{align*}

And $$\mathcal{F}^{-1}\left ( e^{-i\lambda \pi \xi ^{2}t}\right ) =\int _{-\infty }^{\infty }e^{-i\lambda \pi \xi ^{2}t}e^{2i\pi x\xi }d\xi \tag{5A}$$ Hence (5) becomes$u\left ( x,t\right ) =f\left ( x\right ) \circledast \mathcal{F}^{-1}\left ( e^{-i\lambda \pi \xi ^{2}t}\right )$ Here, I used Mathematica to help me with the above integral (5A) as I could not ﬁnd it in tables so far2 . Here is the result

Therefore, from Mathematica, we see that$$\mathcal{F}^{-1}\left ( e^{-i\lambda \pi \xi ^{2}t}\right ) =\frac{e^{\frac{i\pi x^{2}}{\lambda t}}}{\sqrt{2\pi }\sqrt{i}\sqrt{\lambda t}} \tag{6}$$ But3 \begin{align*} \sqrt{i} & =\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}\\ & =\cos \left ( \frac{\pi }{4}\right ) +i\sin \left ( \frac{\pi }{4}\right ) \\ & =e^{i\frac{\pi }{4}} \end{align*}

Hence (6) becomes$$\mathcal{F}^{-1}\left ( e^{-i\lambda \pi \xi ^{2}t}\right ) =\frac{1}{e^{i\frac{\pi }{4}}\sqrt{2\pi \lambda t}}e^{i\frac{\pi x^{2}}{\lambda t}} \tag{7}$$ Now we are ready to do the convolution in (5A) since we know everything in the RHS, hence\begin{align} u\left ( x,t\right ) & =f\left ( x\right ) \circledast \frac{1}{e^{i\frac{\pi }{4}}\sqrt{2\pi \lambda t}}e^{i\frac{\pi x^{2}}{\lambda t}}\nonumber \\ & =\left ( \delta \left ( x-a\right ) +\delta \left ( x+a\right ) \right ) \circledast \frac{1}{e^{i\frac{\pi }{4}}\sqrt{2\pi \lambda t}}e^{i\frac{\pi x^{2}}{\lambda t}} \tag{8} \end{align}

Applying convolution integral on (8), which says that\begin{align*} f\left ( x\right ) & =g_{1}\left ( x\right ) \circledast g_{2}\left ( x\right ) \\ & =\int _{-\infty }^{\infty }g_{1}\left ( z\right ) g_{2}\left ( x-z\right ) dz \end{align*}

Therefore (8) becomes\begin{align*} u\left ( x,t\right ) & =\int _{-\infty }^{\infty }\left ( \delta \left ( z-a\right ) +\delta \left ( z+a\right ) \right ) \frac{1}{e^{i\frac{\pi }{4}}\sqrt{2\pi \lambda t}}e^{i\frac{\pi \left ( x-z\right ) ^{2}}{\lambda t}}dz\\ & =\frac{1}{e^{i\frac{\pi }{4}}\sqrt{2\pi \lambda t}}\int _{-\infty }^{\infty }\left ( \delta \left ( z-a\right ) +\delta \left ( z+a\right ) \right ) e^{i\frac{\pi \left ( x-z\right ) ^{2}}{\lambda t}}dz\\ & =\frac{1}{e^{i\frac{\pi }{4}}\sqrt{2\pi \lambda t}}\left ( \int _{-\infty }^{\infty }\delta \left ( z-a\right ) e^{i\frac{\pi \left ( x-z\right ) ^{2}}{\lambda t}}dz+\int _{-\infty }^{\infty }\delta \left ( z+a\right ) e^{i\frac{\pi \left ( x-z\right ) ^{2}}{\lambda t}}dz\right ) \end{align*}

But an integral with delta function inside it, is just the integrand evaluated where the delta argument become zero which is at $$z=a\,\$$and $$z=-a$$ in the above. (This is called the sifting property). Hence the above integrals are now easily found and we obtain the solution$u\left ( x,t\right ) =\frac{1}{e^{i\frac{\pi }{4}}\sqrt{2\pi \lambda t}}\left ( \exp \left ( i\frac{\pi \left ( x-a\right ) ^{2}}{\lambda t}\right ) +\exp \left ( i\frac{\pi \left ( x+a\right ) ^{2}}{\lambda t}\right ) \right )$ The above is the solution we need. But we can simplify it more by using Euler relation.\begin{align*} u\left ( x,t\right ) & =\frac{1}{e^{i\frac{\pi }{4}}\sqrt{2\pi \lambda t}}\left ( \exp \left ( i\frac{\pi \left ( x^{2}+a^{2}-2xa\right ) }{\lambda t}\right ) +\exp \left ( i\frac{\pi \left ( x^{2}+a^{2}+2ax\right ) }{\lambda t}\right ) \right ) \\ & =\frac{1}{e^{i\frac{\pi }{4}}\sqrt{2\pi \lambda t}}\left ( \exp \left ( \frac{i\pi \left ( x^{2}+a^{2}\right ) }{\lambda t}\right ) \exp \left ( \frac{-i2\pi xa}{\lambda t}\right ) +\exp \left ( \frac{i\pi \left ( x^{2}+a^{2}\right ) }{\lambda t}\right ) \exp \left ( \frac{i2\pi ax}{\lambda t}\right ) \right ) \end{align*}

Taking $$\exp \left ( \frac{i\pi \left ( x^{2}+a^{2}\right ) }{\lambda t}\right )$$ as common factor outside results in\begin{align*} u\left ( x,t\right ) & =\frac{\exp \left ( \frac{i\pi \left ( x^{2}+a^{2}\right ) }{\lambda t}\right ) }{e^{i\frac{\pi }{4}}\sqrt{2\pi \lambda t}}\left ( \exp \left ( i\frac{2\pi ax}{\lambda t}\right ) +\exp \left ( -i\frac{2\pi xa}{\lambda t}\right ) \right ) \\ & =\frac{2\exp \left ( \frac{i\pi \left ( x^{2}+a^{2}\right ) }{\lambda t}\right ) }{e^{i\frac{\pi }{4}}\sqrt{2\pi \lambda t}}\left ( \frac{\exp \left ( i\frac{2\pi ax}{\lambda t}\right ) +\exp \left ( -i\frac{2\pi xa}{\lambda t}\right ) }{2}\right ) \\ & =\frac{2\exp \left ( \frac{i\pi \left ( x^{2}+a^{2}\right ) }{\lambda t}\right ) }{e^{i\frac{\pi }{4}}\sqrt{2\pi \lambda t}}\cos \left ( \frac{2\pi ax}{\lambda t}\right ) \\ & =\frac{\sqrt{2}\exp \left ( \frac{i\pi \left ( x^{2}+a^{2}\right ) }{\lambda t}-i\frac{\pi }{4}\right ) }{\sqrt{\pi \lambda t}}\cos \left ( \frac{2\pi ax}{\lambda t}\right ) \\ & =\frac{\sqrt{2}\exp \left ( i\left ( \frac{\pi \left ( x^{2}+a^{2}\right ) }{\lambda t}-\frac{\pi }{4}\right ) \right ) }{\sqrt{\pi \lambda t}}\cos \left ( \frac{2\pi ax}{\lambda t}\right ) \end{align*}

Hence the ﬁnal solution is$$\fbox{u\left ( x,t\right ) =\sqrt{\frac{2}{\pi \lambda t}}\exp \left ( i\frac{\pi \left ( x^2+a^2\right ) }{\lambda t}-\frac{\pi }{4}\right ) \cos \left ( \frac{2\pi a}{\lambda }\frac{x}{t}\right ) } \tag{9}$$

Hence the real part of the solution is$\Re \left ( u\left ( x,t\right ) \right ) =\sqrt{\frac{2}{\pi \lambda t}}\cos \left ( \frac{\pi \left ( x^{2}+a^{2}\right ) }{\lambda t}-\frac{\pi }{4}\right ) \cos \left ( \frac{2\pi a}{\lambda }\frac{x}{t}\right )$ And the imaginary part of the solution is$\Im \left ( u\left ( x,t\right ) \right ) =\sqrt{\frac{2}{\pi \lambda t}}\sin \left ( \frac{\pi \left ( x^{2}+a^{2}\right ) }{\lambda t}-\frac{\pi }{4}\right ) \cos \left ( \frac{2\pi a}{\lambda }\frac{x}{t}\right )$ The $$\frac{\pi }{4}$$ is just a phase shift. Here is a plot of the Real and Imaginary parts of the solution,  using for $$\lambda =600\times 10^{-9}meter,a=1000\times 10^{-9}meter$$ at $$t=1$$ second

We see the rapid oscillations as distance goes away from the origin. This is due to the $$x^{2}$$ term making the radial frequency value increase quickly with $$x$$. We now plot the $$\left \vert u\left ( x,t\right ) \right \vert ^{2}$$.  Looking at solution in (9), and since complex exponential is $$\pm 1$$, then the amplitude is governed by $$\sqrt{\frac{2}{\pi \lambda t}}\cos \left ( \frac{2\pi a}{\lambda }\frac{x}{t}\right )$$ part of the solution. Hence

$\fbox{\left \vert u\left ( x,t\right ) \right \vert ^2=\frac{2}{\pi \lambda t}\cos ^2\left ( \frac{2\pi a}{\lambda }\frac{x}{t}\right ) }$

These plots show the intensity at diﬀerent time values. We see from these plots, that the intensity is well behaved in that it does not have the same rapid oscillations seen in the $$u\left ( x,t\right )$$ solution plots.

Now Comparing argument to cosine in above to standard form in order to ﬁnd the period:$\frac{2\pi a}{\lambda }\frac{x}{t}=2\pi ft$ Where $$f$$ is now in hertz, then when $$x=t$$, we get by comparing terms that\begin{align*} \frac{2\pi a}{\lambda }\frac{1}{t} & =2\pi f\\ \frac{a}{\lambda }\frac{1}{t} & =f \end{align*}

But $$f=\frac{1}{T}$$ where $$T$$ is the period in seconds. Hence $$\frac{a}{\lambda }\frac{1}{t}=\frac{1}{T}$$ or$\fbox{T=\frac{\lambda t}{a}}$ So period on intensity is $$\frac{\lambda t}{a}$$ at $$x=t$$ (why problem statement is saying period is $$\frac{\lambda t}{2a}$$?).

#### 5.2.4 problem 2 (optional)

5.2.4.1 Part (a)
5.2.4.2 Part (b)
5.2.4.3 part (c)

##### 5.2.4.1 Part (a)

\begin{align} \frac{\partial u}{\partial t} & =\nu \frac{\partial ^{2}u}{\partial x^{2}}\tag{1}\\ u\left ( 0,t\right ) & =e^{-i\omega t}\nonumber \end{align}

And $$x\geq 0,u\left ( \infty ,t\right )$$ bounded. Let$u\left ( x,t\right ) =e^{ikx}e^{-i\omega t}$ Hence\begin{align} \frac{\partial u\left ( x,t\right ) }{\partial t} & =-i\omega e^{ikx}e^{-i\omega t}\nonumber \\ & =-i\omega u\left ( x,t\right ) \tag{2} \end{align}

And\begin{align} \frac{\partial u\left ( x,t\right ) }{\partial x} & =ike^{ikx}e^{-i\omega t}\nonumber \\ \frac{\partial ^{2}u\left ( x,t\right ) }{\partial x^{2}} & =-k^{2}e^{ikx}e^{-i\omega t}\nonumber \\ & =-k^{2}u\left ( x,t\right ) \tag{3} \end{align}

Substituting (2,3) into (1) gives$-i\omega u\left ( x,t\right ) =-\nu k^{2}u\left ( x,t\right )$ Since $$u_{\omega }\left ( x,t\right )$$ can not be identically zero (trivial solution), then the above simpliﬁes to$$-i\omega =-\nu k^{2}\nonumber$$ Or$$\fbox{k^2=\frac{i\omega }{\nu }} \tag{4}$$ Writing $\omega =\operatorname{sgn}\left ( \omega \right ) \left \vert \omega \right \vert$ Where $\operatorname{sgn}\left ( \omega \right ) =\left \{ \begin{array} [c]{ccc}+1 & & \omega >0\\ 0 & & \omega =0\\ -1 & & \omega <0 \end{array} \right .$ Then (4) becomes\begin{align*} k^{2} & =\frac{i\ \operatorname{sgn}\left ( \omega \right ) \left \vert \omega \right \vert }{\nu }\\ k & =\pm \frac{\sqrt{i}\ \sqrt{\operatorname{sgn}\left ( \omega \right ) }\sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }} \end{align*}

Since $\sqrt{i}=i^{\frac{1}{2}}=\left ( e^{i\frac{\pi }{2}}\right ) ^{\frac{1}{2}}=e^{i\frac{\pi }{4}}$ Hence $$k$$ can be written as$k=\pm \frac{e^{i\frac{\pi }{4}}\ \sqrt{\operatorname{sgn}\left ( \omega \right ) }\sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}$ case A. Let start with the positive root hence$k=\frac{e^{i\frac{\pi }{4}}\ \sqrt{\operatorname{sgn}\left ( \omega \right ) }\sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}$ Case (A1) $$\omega <0$$ then the above becomes\begin{align*} k & =\frac{ie^{i\frac{\pi }{4}}\ \sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}=\frac{e^{i\frac{\pi }{2}}e^{i\frac{\pi }{4}}\ \sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}=\frac{e^{i\frac{3\pi }{4}}\ \sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}\\ & =\left ( \cos \frac{3}{4}\pi +i\sin \frac{3}{4}\pi \right ) \frac{\ \sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}\\ & =\left ( -\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}\right ) \frac{\ \sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}\\ & =\left ( -\frac{1}{\sqrt{2}}\frac{\ \sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}+i\frac{1}{\sqrt{2}}\frac{\ \sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}\right ) \end{align*}

And the solution becomes\begin{align*} u\left ( x,t\right ) & =\exp \left ( ikx\right ) \exp \left ( -i\omega t\right ) \\ & =\exp \left ( i\left ( -\frac{1}{\sqrt{2}}\frac{\ \sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}+i\frac{1}{\sqrt{2}}\frac{\ \sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}\right ) x\right ) \exp \left ( -i\omega t\right ) \\ & =\exp \left ( \left ( -\frac{1}{\sqrt{2}}\frac{\ \sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}i-\frac{1}{\sqrt{2}}\frac{\ \sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}\right ) x\right ) \exp \left ( -i\omega t\right ) \\ & =\exp \left ( -\frac{1}{\sqrt{2}}\frac{\ \sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}x\right ) \exp \left ( -\frac{1}{\sqrt{2}}\frac{\ \sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}ix\right ) \exp \left ( -i\omega t\right ) \end{align*}

We are told that $$u\left ( \infty ,t\right )$$ is bounded. So for large $$x$$ we want the above to be bounded. The complex exponential present in the above expression cause no issue for large $$x$$ since they are oscillatory trig functions. We then just need to worry about $$\exp \left ( -\frac{1}{\sqrt{2}}\frac{\ \sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}x\right )$$ for large $$x$$. This term will decay for large $$x$$ since $$-\frac{1}{\sqrt{2}}\frac{\ \sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}$$ is negative (assuming $$\nu$$ $$>0$$ always).  Hence positive root hence worked OK when $$\omega <0$$. Now we check if it works OK also when $$\omega >0$$

case A2 When $$\omega >0$$ then $$k$$ now becomes\begin{align*} k & =\frac{e^{i\frac{\pi }{4}}\ \sqrt{\operatorname{sgn}\left ( \omega \right ) }\sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}\\ & =\frac{e^{i\frac{\pi }{4}}\ \sqrt{\omega }}{\sqrt{\nu }}\\ & =\left ( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right ) \frac{\ \sqrt{\omega }}{\sqrt{\nu }}\\ & =\left ( \frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}\right ) \frac{\ \sqrt{\omega }}{\sqrt{\nu }}\\ & =\frac{1}{\sqrt{2}}\frac{\ \sqrt{\omega }}{\sqrt{\nu }}+i\frac{1}{\sqrt{2}}\frac{\ \sqrt{\omega }}{\sqrt{\nu }} \end{align*}

And the solution becomes\begin{align*} u\left ( x,t\right ) & =\exp \left ( ikx\right ) \exp \left ( -i\omega t\right ) \\ & =\exp \left ( i\left ( \frac{1}{\sqrt{2}}\frac{\ \sqrt{\omega }}{\sqrt{\nu }}+i\frac{1}{\sqrt{2}}\frac{\ \sqrt{\omega }}{\sqrt{\nu }}\right ) x\right ) \exp \left ( -i\omega t\right ) \\ & =\exp \left ( \left ( i\frac{1}{\sqrt{2}}\frac{\ \sqrt{\omega }}{\sqrt{\nu }}-\frac{1}{\sqrt{2}}\frac{\ \sqrt{\omega }}{\sqrt{\nu }}\right ) x\right ) \exp \left ( -i\omega t\right ) \\ & =\exp \left ( -\frac{1}{\sqrt{2}}\frac{\ \sqrt{\omega }}{\sqrt{\nu }}x\right ) \exp \left ( i\frac{1}{\sqrt{2}}\frac{\ \sqrt{\omega }}{\sqrt{\nu }}x\right ) \exp \left ( -i\omega t\right ) \end{align*}

We are told that $$u\left ( \infty ,t\right )$$ is bounded. So for large $$x$$ we want the above to be bounded. The complex exponential present in the above expression cause no issue for large $$x$$ since they are oscillatory trig functions. We then just need to worry about $$\exp \left ( -\frac{1}{\sqrt{2}}\frac{\ \sqrt{\omega }}{\sqrt{\nu }}x\right )$$ for large $$x$$. This term will decay for large $$x$$ since $$-\frac{1}{\sqrt{2}}\frac{\ \sqrt{\omega }}{\sqrt{\nu }}$$ is negative (assuming $$\nu$$ $$>0$$ always).  Hence positive root hence worked OK when $$\omega >0$$ as well.

Let check what happens if we use the negative root.

case B. negative root hence$k=-\frac{e^{i\frac{\pi }{4}}\ \sqrt{\operatorname{sgn}\left ( \omega \right ) }\sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}$ Case (A1) $$\omega <0$$ then the above becomes\begin{align*} k & =-\frac{ie^{i\frac{\pi }{4}}\ \sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}=-\frac{e^{i\frac{\pi }{2}}e^{i\frac{\pi }{4}}\ \sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}=-\frac{e^{i\frac{3\pi }{4}}\ \sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}\\ & =-\left ( \cos \frac{3}{4}\pi +i\sin \frac{3}{4}\pi \right ) \frac{\ \sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}\\ & =-\left ( -\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}\right ) \frac{\ \sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}\\ & =-\left ( -\frac{1}{\sqrt{2}}\frac{\ \sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}+i\frac{1}{\sqrt{2}}\frac{\ \sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}\right ) \\ & =\left ( \frac{1}{\sqrt{2}}\frac{\ \sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}-i\frac{1}{\sqrt{2}}\frac{\ \sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}\right ) \end{align*}

And the solution becomes\begin{align*} u\left ( x,t\right ) & =\exp \left ( ikx\right ) \exp \left ( -i\omega t\right ) \\ & =\exp \left ( i\left ( \frac{1}{\sqrt{2}}\frac{\ \sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}-i\frac{1}{\sqrt{2}}\frac{\ \sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}\right ) x\right ) \exp \left ( -i\omega t\right ) \\ & =\exp \left ( \left ( i\frac{1}{\sqrt{2}}\frac{\ \sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}+\frac{1}{\sqrt{2}}\frac{\ \sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}\right ) x\right ) \exp \left ( -i\omega t\right ) \\ & =\exp \left ( +\frac{1}{\sqrt{2}}\frac{\ \sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}x\right ) \exp \left ( \frac{1}{\sqrt{2}}\frac{\ \sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}ix\right ) \exp \left ( -i\omega t\right ) \end{align*}

We are told that $$u\left ( \infty ,t\right )$$ is bounded. So for large $$x$$ we want the above to be bounded. The complex exponential present in the above expression cause no issue for large $$x$$ since they are oscillatory trig functions. We then just need to worry about $$\exp \left ( +\frac{1}{\sqrt{2}}\frac{\ \sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}x\right )$$ for large $$x$$. This term will blow up for large $$x$$ since $$+\frac{1}{\sqrt{2}}\frac{\ \sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}$$ is positive (assuming $$\nu$$ $$>0$$ always). Hence we reject the case of negative sign on $$k$$. And pick\begin{align*} k & =\frac{e^{i\frac{\pi }{4}}\ \sqrt{\operatorname{sgn}\left ( \omega \right ) }\sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}\\ & =\left ( \frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}\right ) \frac{\ \sqrt{\operatorname{sgn}\left ( \omega \right ) }\sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }} \end{align*}

Therefore the solution is\begin{align*} u\left ( x,t;\omega \right ) & =\exp \left ( ikx\right ) \exp \left ( -i\omega t\right ) \\ & =\exp \left ( i\left ( \frac{\left ( \frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}\right ) \ \sqrt{\operatorname{sgn}\left ( \omega \right ) }\sqrt{\left \vert \omega \right \vert }}{\sqrt{\nu }}\right ) x\right ) \exp \left ( -i\omega t\right ) \\ & =\exp \left ( i\left ( \frac{1}{\sqrt{2\nu }}\sqrt{\operatorname{sgn}\left ( \omega \right ) }\sqrt{\left \vert \omega \right \vert }+i\frac{1}{\sqrt{2\nu }}\sqrt{\operatorname{sgn}\left ( \omega \right ) }\sqrt{\left \vert \omega \right \vert }\ \right ) x\right ) \exp \left ( -i\omega t\right ) \\ & =\exp \left ( i\frac{1}{\sqrt{2\nu }}\sqrt{\operatorname{sgn}\left ( \omega \right ) }\sqrt{\left \vert \omega \right \vert }x-\frac{1}{\sqrt{2\nu }}\sqrt{\operatorname{sgn}\left ( \omega \right ) }\sqrt{\left \vert \omega \right \vert }\ x\right ) \exp \left ( -i\omega t\right ) \end{align*}

Hence$\fbox{u\left ( x,t;\omega \right ) =\exp \left ( \frac{-\sqrt{\operatorname{sgn}\left ( \omega \right ) }\sqrt{\left \vert \omega \right \vert }\ }{\sqrt{2\nu }}x\right ) \exp \left ( i\frac{\sqrt{\operatorname{sgn}\left ( \omega \right ) }\sqrt{\left \vert \omega \right \vert }}{\sqrt{2\nu }}x\right ) \exp \left ( -i\omega t\right ) }$ The general solution $$u\left ( x,t\right )$$ is therefore the integral over all $$\omega$$, hence\begin{align*} u\left ( x,t\right ) & =\int _{\omega =-\infty }^{\infty }u\left ( x,t;\omega \right ) \ d\omega \\ & =\int _{-\infty }^{\infty }\exp \left ( -\frac{\sqrt{\operatorname{sgn}\left ( \omega \right ) }\sqrt{\left \vert \omega \right \vert }}{\sqrt{2\nu }}x\right ) \exp \left ( \frac{i\sqrt{\operatorname{sgn}\left ( \omega \right ) }\sqrt{\left \vert \omega \right \vert }}{\sqrt{2\nu }}x\right ) \exp \left ( -i\omega t\right ) \ d\omega \\ & =\int _{-\infty }^{\infty }\exp \left ( -\frac{\sqrt{\operatorname{sgn}\left ( \omega \right ) }\sqrt{\left \vert \omega \right \vert }}{\sqrt{2\nu }}x\right ) \exp \left ( i\left [ \frac{\sqrt{\operatorname{sgn}\left ( \omega \right ) }\sqrt{\left \vert \omega \right \vert }}{\sqrt{2\nu }}x-\omega t\right ] \right ) \ d\omega \\ & =\int _{-\infty }^{\infty }\exp \left ( -\frac{\sqrt{\operatorname{sgn}\left ( \omega \right ) }\sqrt{\left \vert \omega \right \vert }}{\sqrt{2\nu }}x\right ) \exp \left ( ix\left [ \frac{\sqrt{\operatorname{sgn}\left ( \omega \right ) }\sqrt{\left \vert \omega \right \vert }}{\sqrt{2\nu }}-\omega \frac{t}{x}\right ] \right ) \ d\omega \end{align*}

##### 5.2.4.2 Part (b)

From part (a), we found that$$\fbox{u\left ( x,t;\omega \right ) =\exp \left ( -\frac{\sqrt{\operatorname{sgn}\left ( \omega \right ) }\sqrt{\left \vert \omega \right \vert }}{\sqrt{2\nu }}x\right ) \exp \left ( i\frac{\sqrt{\operatorname{sgn}\left ( \omega \right ) }\sqrt{\left \vert \omega \right \vert }}{\sqrt{2\nu }}x\right ) \exp \left ( -i\omega t\right ) } \tag{1}$$ comparing the above to expression given in problem which is$$u\left ( x,t\right ) =\exp \left ( -\sigma x\right ) \ \exp \left ( i\sigma \operatorname{sgn}\left ( \omega \right ) x\right ) \ \exp \left ( -i\omega t\right ) \tag{2}$$ Therefore, by comparing $$\exp \left ( -\sigma x\right )$$ to $$\exp \left ( -\frac{\sqrt{\operatorname{sgn}\left ( \omega \right ) }\sqrt{\left \vert \omega \right \vert }}{\sqrt{2\nu }}x\right )$$ we see that$$\fbox{\sigma =\frac{\sqrt{\operatorname{sgn}\left ( \omega \right ) }\sqrt{\left \vert \omega \right \vert }}{\sqrt{2\nu }}} \tag{3}$$

##### 5.2.4.3 part (c)

Using the solution found in part (a)$u\left ( x,t;\omega \right ) =\exp \left ( -\sigma x\right ) \ \exp \left ( i\sigma \operatorname{sgn}\left ( \omega \right ) x\right ) \ \exp \left ( -i\omega t\right )$ To ﬁnd numerical estimate, assuming $$\omega >0$$ for now \begin{align*} u\left ( x,t;\omega \right ) & =\exp \left ( -\sigma x\right ) \ \exp \left ( i\sigma x\right ) \ \exp \left ( -i\omega t\right ) \\ & =e^{-\sigma x}\left ( \cos \sigma x+i\sin \sigma x\right ) \left ( \cos \omega t-i\sin \omega t\right ) \\ & =e^{-\sigma x}\left ( \cos \sigma x\cos \omega t-i\sin \omega t\cos \sigma x+i\cos \omega t\sin \sigma x+\sin \sigma x\sin \omega t\right ) \\ & =e^{-\sigma x}\left ( \cos \left ( \sigma x\right ) \cos \left ( \omega t\right ) +\sin \left ( \sigma x\right ) \sin \left ( \omega t\right ) +i\left ( \cos \left ( \omega t\right ) \sin \left ( \sigma x\right ) -\sin \left ( \omega t\right ) \cos \left ( \sigma x\right ) \right ) \right ) \\ & =e^{-\sigma x}\left ( \cos \left ( t\omega -x\sigma \right ) -i\sin \left ( t\omega -x\sigma \right ) \right ) \end{align*}

Hence will evaluate \begin{align*} \operatorname{Re}\left ( u\left ( x,t;\omega \right ) \right ) & =e^{-\sigma x}\operatorname{Re}\left ( \cos \left ( t\omega -x\sigma \right ) -i\sin \left ( t\omega -x\sigma \right ) \right ) \\ & =e^{-\sigma x}\cos \left ( t\omega -x\sigma \right ) \end{align*}

I assume here it is asking for numerical estimate. We only need to determine numerical estimate for $$\sigma$$. For $$\omega$$, using the period $$T=24$$ hrs or $$T=86\,400$$ seconds, then $$\omega =\frac{2\pi }{T}$$ is now found. Then we need to determine $$\nu$$, which is thermal diﬀusivity for earth crust. There does not seem to be an agreed on value for this and this value also changed with depth inside the earth crust.  The value I found that seem mentioned more is $$1.2\times 10^{-6}$$ meter$$^{2}$$ per second. Hence \begin{align*} \sigma & =\frac{\sqrt{\frac{2\pi }{\allowbreak 86\,400}}\ }{\sqrt{2\left ( 1.2\times 10^{-6}\right ) }}\\ & =5.505\text{ per meter} \end{align*}

Therefore$\operatorname{Re}\left ( u\left ( x,t;\omega \right ) \right ) =e^{-5.505x}\cos \left ( t\omega -5.505x\right )$ Using $$\omega =\frac{2\pi }{86\,400}=7.29\times 10^{-5}$$ rad/sec. Now we can use the above to estimate ﬂuctuation of heat over $$24$$ hrs period. But we need to ﬁx $$x$$ for each case. Here $$x=0$$ means on the earth surface and $$x$$ say $$10$$, means at depth $$10$$ meters and so on as I understand that $$x$$ is starts at $$0$$ at surface or earth and increases as we go lower into the earth crust. Plotting at the above for $$x=0,0.5,1,1.5,2$$ I see that when $$x>2$$ then maximum value of $$e^{-5.505x}\cos \left ( t\omega -5.505x\right )$$ is almost zero. This seems to indicate a range of heat reach is about little more than $$2$$ meters below the surface of earth.

This is a plot of the ﬂuctuation in temperature at diﬀerent $$x$$ each in separate plot, then later a plot is given that combines them all.

This plot better show the diﬀerence per depth, as it combines all the plots into one.