Partial Differential Equations Tutorial Sheet, #12

Course Notes Chapter

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Tutorials



Problem sheet

Skill Building Questions

Problem 1.

Which of the following are solutions to Laplace’s equation, $\nabla^2 f(\mathbf{x}) = 0$? For those that are not, can you come up with a differential equation involving the Laplacian, $\nabla^2$, that $f(\mathbf{x})$ is a solution of?

(a) $e^{-y} \sin x $

Yes, $\nabla^2 f(\mathbf{x}) = 0$
$\Rightarrow{}\quad$ The Laplacian operator is defined as:
$\Rightarrow{}\nabla^2= \frac{\partial^2 } {\partial x^2}+ \frac{\partial^2} {\partial y^2} + \frac{\partial^2} {\partial z^2}$
$\Rightarrow{}\nabla^2 f(\mathbf{x}) = \frac{\partial^2 }{\partial x^2}{e^{-y} \sin x} + \frac{\partial^2}{\partial y^2}{e^{-y} \sin x}$
$\Rightarrow{} = \frac{\partial }{\partial x}{e^{-y} \cos x} - \frac{\partial}{\partial y}{e^{-y} \sin x}$
$\Rightarrow{} = {e^{-y} \sin x - e^{-y} \sin x} = 0$
NB This is an example for this kind problem in case you get confused by these short answers :)








(b) $\sin\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)$

No, but $\nabla^2 f(\mathbf{x}) + f = 0$








(c) $f(x + i y)$

Yes, $\nabla^2 f(x + i y) = 0$, this becomes a useful result in complex analysis.








(d) $4 \arctan(y/x) + \log(x^2 +y^2)$

Yes, $\nabla^2 f(\mathbf{x}) = 0$








(e) $\frac{x+y}{z}$

No, but $\nabla^2 f(\mathbf{x}) = 2 f / z^2$ would work.








Problem 2.

Show that the function $u\left(x,y\right)=\ln⁡(1+xy^2)$ satisfies the partial differential equation:

\[2 \frac{\partial{} ^2u}{ \partial{} x^2} + y^3 \frac{ \partial{} ^2u }{ \partial{} y \partial{} x } = 0\]
$$ \frac{\partial{}u}{\partial{}x}=\frac{y^2}{1+xy^2}\ ;\ \ \ \frac{ {\partial{}}^2u}{ {\partial{}x}^2}=-\frac{y^2}{ {\left(1+{xy}^2\right)}^2}.y^2=\frac{-y^4}{ {\left(1+{xy}^2\right)}^2} $$
$$ \frac{ {\partial{}}^2u}{\partial{}y\partial{}x}=\frac{\partial{}}{\partial{}y}\left(\frac{\partial{}u}{\partial{}x}\right)=\frac{2y\left(1+xy^2\right)-y^2(2xy)}{ {(1+xy^2)}^2}=\frac{2y+2xy^3-2xy^3}{ { {(1+xy}^2)}^2}=\frac{2y}{ { {(1+xy}^2)}^2} $$
$$ 2\frac{ {\partial{}}^2u}{\partial{}x^2}+y^3\frac{ {\partial{}}^2u}{\partial{}y\partial{}x}=-\frac{2y^4}{ { {(1+xy}^2)}^2}+\frac{ {2y}^4}{ { {(1+xy}^2)}^2}=0 $$








Separation of Variables 0 – 1D Wave Equation

Go through the process of solving the general soultion to the 1D Wave equation, As we did in the lecture notes.

\[\frac{\partial^2 f(x, t)}{\partial t^2} = c^2 \frac{\partial^2 f(x, t)}{\partial x^2}\]
This is solved with more detail in the lecture notes. We'll do a different solution here and solve with complex exponentials. First assume solution, $$ f(x, t) = X(x) T(t) $$ Next plug into the ODE, $$ X(x)T^{\prime\prime}(t) = c^2 X^{\prime\prime}(x)T(t) $$ Divide through by $X(x)T(t)$ $$ \begin{align*} \frac{X(x)T^{\prime\prime}(t)}{X(x)T(t)} &= c^2 \frac{X^{\prime\prime}(x)T(t)}{X(x)T(t)} \\ \frac{T^{\prime\prime}(t)}{T(t)} &= c^2 \frac{X^{\prime\prime}(x)}{X(x)} \end{align*} $$ Assign constants to each group, I've chosen negative second power constants. [See why.](/module-resources/notes/14-partial-differential-equations#better-constants) $$ \begin{align*} \underbrace{\frac{T^{\prime\prime}(t)}{T(t)}}_{-\omega^2} &= c^2 \underbrace{\frac{X^{\prime\prime}(x)}{X(x)}}_{-k^2} \end{align*} $$ Giving, $$ \begin{align*} T^{\prime\prime}(t) &= -\omega^2 T(t) \\ X^{\prime\prime}(x) &= -k^2 X(x) \\ \omega^2 &= c^2 k^2 \end{align*} $$ Let's solve the ODEs, $$ \begin{align*} T(t) &= A e^{i \omega t} + B e^{-i \omega t} \\ X(x) &= C e^{i k x} + D e^{-i k x} \end{align*} $$ Which we can then combine to, $$ f(x, t) = X(x)T(t) = \left[ A e^{i \omega t} + B e^{-i \omega t} \right] \left[ C e^{i k x} + D e^{-i k x} \right] $$ At this point we are done, but we could also insert our dispersion relation, to remove one of the constants. $$ f(x, t) = \left[ A e^{i k c t} + B e^{-i k c t} \right] \left[ C e^{i k x} + D e^{-i k x} \right] $$ Which is the general solution.

Once you have done this, you can return the question later and try to solve it using different trial solutions (Sines and cosines, exponentials, hyperbolic functions) and different choice of constants (powers, sign). See how they each behave and how the choices interact with each other.

Separation of Variables 1 – Choice of constants

In the Better Constants section of the notes, We looked at the 1D wave equation,

\[\frac{\partial^2 f(x, t)}{\partial t^2} = c^2 \frac{\partial^2 f(x, t)}{\partial x^2}\]

and chose ODE constants, $-\omega^2$ and $-k^2$, to solve the PDE.

You were asked to see what happened if instead we used positive squared coefficients, i.e.,

\[\begin{align*} T^{\prime\prime}(t) &= +\gamma^2 T(t) \\ X^{\prime\prime}(x) &= +\kappa^2 X(x) \end{align*}\]

Show, by using a separable solution $f(x, t) = X(x)T(t)$ on the PDE, that we can return the ODEs above, and that their relationship is,

\[\gamma^2 = c^2 \kappa^2\]

Then show by solving the ODEs, that exponentially growing and decaying solutions can be produced. Write the general solution to the PDE in this form.

Let's start with the usual assumtion of a separable solution, $$ f(x, t) = X(x) T(t) $$ Inserting into our PDE, dividing through, we get, $$ \frac{T^{\prime\prime}(t)}{T(t)} = c^2 \frac{X^{\prime\prime}(x)}{X(x)} $$ Now, let's use different positive squared constants, gamma and kappa. $$ \underbrace{\frac{T^{\prime\prime}(t)}{T(t)}}_{\gamma^2} = c^2 \underbrace{\frac{X^{\prime\prime}(x)}{X(x)}}_{\kappa^2} $$ Which we can extract two ODEs and a dispersion relation. $$ \begin{align*} T^{\prime\prime}(t) &= \gamma^2 T(t) \\ X^{\prime\prime}(x) &= \kappa^2 X(x) \\ \gamma^2 &= c^2 \kappa^2 \end{align*} $$ With the dispersion relation being what we were trying to prove. Let's look at the temporal ODE, $$ T^{\prime\prime}(t) = \gamma^2 T(t) $$ This is asking the question, what function when differentiated twice, gives itself times a constant? The hyperbolic trig functions answer this question – try this yourself. But so do growing and decaying exponentials. Let's use $T(t) = e^{\lambda t}$ as our trial solution. The second derivative is, $$ \frac{\partial^2}{\partial t^2} e^{\lambda t} = \lambda^2 e^{\lambda t} $$ And so, $$ \lambda^2 e^{\lambda t} = \gamma^2 e^{\lambda t} $$ hence, $$ \begin{align*} \lambda^2 &= \gamma^2 \\ \lambda &= \pm \gamma \end{align*} $$ So our temporal ODE solution is, $$ T(t) = A e^{\gamma t} + B e^{-\gamma t} $$ The same reasoning applies to the spatial ODE, $$ X(x) = C e^{\kappa x} + D e^{-\kappa x} $$ For a full solution, $$ f(x, t) = X(x)T(t) = \left[ C e^{\kappa x} + D e^{-\kappa x} \right] \left[ A e^{\gamma t} + B e^{-\gamma t} \right] $$ Let's insert the dispersion relation, $$ f(x, t) = \left[ C e^{\kappa x} + D e^{-\kappa x} \right] \left[ A e^{\kappa c t} + B e^{-\kappa c t} \right] $$ These solutions tell us that an initial condition that looks like exponential functions will either grow or decay away at future times. (Depending on the initial time derivative) Do notice, that we indeed return the usual oscillatory solutions for complex or imaginary values of $\kappa$ or $\gamma$.

Separation of Variables 2 – 2D Diffusion Equation

In the notes, we highlighted the 2D Diffusion Equation as an example of using separation of variables.

\[\frac{\partial u(x, y, t)}{\partial t} = \alpha \nabla^2 u(x, y, t)\]

Find the general solution, $u(x, y, t)$, to this PDE.

First, let's write the Laplacian out in full, $$ \frac{\partial}{\partial t} u(x, y, t) = \alpha \frac{\partial^2}{\partial x^2} u(x, y, t) + \alpha \frac{\partial^2}{\partial y^2} u(x, y, t) $$ Next, assume a separated solution of $$ u(x, y, t) = X(x)Y(y)T(t) $$ Then input into the PDE, $$\begin{align*} X(x) Y(y) T^{\prime}(t) &= \alpha X^{\prime\prime}(x) Y(y) T(t) + \alpha X(x) Y^{\prime\prime}(y) T(t) \\ \Rightarrow \; \frac{T^{\prime}(t)}{T(t)} &= \alpha \frac{X^{\prime\prime}(x)}{X(x)} + \alpha \frac{Y^{\prime\prime}(y)}{Y(y)} \end{align*}$$ Each term here can be assigned a constant, For the $T(t)$ term, let's call this $-\gamma$, i.e., $$ T^{\prime}(t) = -\gamma T(t) $$ I've chosen a negative sign in front of the constant here, this is optional, but will eventually mean exponentially decaying solutions for positive $\gamma$ values, rather than unphysical growing ones. For the $X(x)$ and $Y(y)$ constants, let's use, $$\begin{align*} X^{\prime\prime}(x) &= -k_x^2 X(x) \\ Y^{\prime\prime}(y) &= -k_y^2 Y(y) \end{align*}$$ Here I always pick the power of the constant to be as big as the highest order of derivative. In this case, I pick a negative sign too, this will match with the $\gamma$ term and ensure all positive signs in the dispersion relation, which comes out as, $$ \gamma = \alpha \left( k_x^2 + k_y^2 \right) \;. $$ Next we solve the ODEs, $$\begin{align*} &T^{\prime}(t) = -\gamma T(t) \\ \Rightarrow \; &T(t) \propto e^{-\gamma t} \end{align*}$$ And for the spatial parts, we have a choice to solve in terms of sinusoids or complex exponentials, I'll do complex exponentials here. $$\begin{align*} &X^{\prime\prime}(x) = -k_x^2 X(x) \\ \Rightarrow \; &X(x) \propto e^{i k_x x} \\ &Y^{\prime\prime}(y) = -k_y^2 Y(y) \\ \Rightarrow \; &Y(y) \propto e^{i k_y y} \end{align*}$$ The complex exponential solutions are nicer when there's a lot of terms, e.g. in 2d, because the positive and negative solution can be captured with a single exponential term, rather than both a $\sin$ and a $\cos$. The general solution then is the product of all the separated solutions, $$ u(x, y, t) = A e^{i k_x x + i k_y y -\gamma t} $$ with, $$ \gamma = \alpha \left( k_x^2 + k_y^2 \right) $$ As the final answer.









Exam Style Question

Problem 3.

Vibrations on a guitar string can be modelled by the wave equation, $\frac{\partial ^{2}u}{\partial t^{2}}-\frac{\tau\partial ^{2}u}{\rho \partial x^{2}}=0$
where $u(x,t)$ is the displacement of the string from its equilibrium position, $\tau$ is its tension, and $\rho$ is the linear density of the string. This is a clone of the 19-20 progress test.

(a) Use separation of variables to produce linear ODEs for the separated parts. State any relationship between constants you define.

$\Rightarrow{}$ Using $𝑢(x,𝑡)=𝑋(x)𝑇(𝑡)$, the PDE becomes

$X\left ( x \right )T{}''\left ( t \right )-\frac{\tau }{\rho }X{}''\left ( x \right )T\left ( t \right )=0$

$\Rightarrow{}$ And therefore $\frac{T{}''\left ( t \right )}{T\left ( t \right )}-\frac{\tau }{\rho }\frac{X{}''\left ( x \right )}{X\left ( x \right )}=0$

$\Rightarrow{}$Extract ODEs, $\frac{T{}''\left ( t \right )}{T\left ( t \right )}=-\omega ^{2},\frac{X{}''\left ( x \right )}{X\left ( x \right )}=-k^{2}$

$\Rightarrow{}$Then,$\boxed{T{}''\left ( t \right )=-\omega ^{2}T\left ( t \right ),X{}''\left ( x \right )=-k^{2}X\left ( x \right )}$, with $\boxed{\omega ^{2}=\frac{\tau }{\rho }k^{2}}$









(b) Solve the ODEs in terms of sinusoidal solutions.

$\boxed{T(t) = \cos \omega t,\sin \omega t}$ and

$\boxed{X(x) = \cos k x, \sin k x}$

Accept complex exponentials, accept equivalent correct forms, accept sum with coefficients, A cos + B sin etc.









(c) The string has a length L, and is fixed at both ends, such that $𝑢(x = 0, 𝑡) = 𝑢(x = L, 𝑡) = 0$. How does this constrain your solutions? Write a general solution of the PDE, $ 𝑢(x, 𝑡)$, that is subject to these constraints.

Fixed at $x=0$ constrains to $\sin k x$ spatial solutions only.

Fixed at $x=L$ constrains to $k=\frac{n\pi }{L}$.

$\boxed{u\left ( x,t \right )=\left ( A\cos\omega t+B\sin\omega t \right )\sin\frac{n\pi x}{L}}$









(d) Write an expression for the fundamental (lowest) frequency allowed by the string.

Using $\omega ^{2}=\frac{\tau }{\rho }k^{2}$ and $k=\frac{n\pi }{L}$, then $\omega =\sqrt{\frac{\tau }{\rho }}\frac{n\pi }{L}$

Fundamental frequency, set $n=1$.

$\boxed{\omega =\sqrt{\frac{\tau }{\rho }}\frac{\pi }{L}}$









(e) If a guitar has a neck length of 0.65 m, and a string has linear density 5 g/m, what tension does the string need to be to sound an A note at a frequency $\omega = 2\pi × 110$ Hz?

Rearrange for \tau.

$\Rightarrow \tau =\frac{L^{2}\omega ^{2}}{\pi ^{2}}\rho $

Insert given values, $\Rightarrow{}\tau =\frac{\left ( 0.65m \right )^{2}\left ( 2\pi \times 110Hz \right )^{2}}{\pi ^{2}}0.005\frac{kg}{m}$

$\boxed{\tau =102N}$









Extension Questions

Problem 4.

Prove for the 1D diffusion equation, $\frac{ \partial f(x,t)}{ \partial t} = \alpha \frac{ \partial^2 f(x,t)}{\partial x^2}$, that the total area under the curve $f(x,t)$ does not change over time. (You may assume that $\frac{ \partial f(x,t)}{ \partial x} \rightarrow 0$ for $x \rightarrow \pm\infty$).

The area under the curve is calculated by integrating $f(x,t)$ over all x. Integrating both sides of the PDE, $$ \int_{-\infty}^{\infty} \mathrm{d}x \; \frac{ \partial {f(x,t)}}{ \partial t} = \int_{-\infty}^{\infty} \mathrm{d}x \; \alpha \frac{\partial f(x,t)}{\partial x} $$ $\Rightarrow{}$ We can swap the order of differentiation and integration on the LHS and directly integrate the expression on the RHS, $$ \frac{\partial}{\partial t} \int_{-\infty}^{\infty} \mathrm{d}x \; f(x,t) = \alpha \left.\frac{ \partial f(x,t)}{ \partial x}\right| _ {x\rightarrow \infty} - \left.\alpha \frac{ \partial f(x,t)}{ \partial x}\right|_{x\rightarrow -\infty} $$ $\Rightarrow{}$ The x derivatives go to zero at infinity $$ \boxed{ \frac{\partial}{ \partial t} \int_{-\infty}^{\infty} \mathrm{d}x \; f(x,t) = 0 } $$ $\Rightarrow{}$ Therefore the area under the concentration curve does not change in time. The area will be proportional to the numbers of ions or molecules etc. the total number of these will stay the same as they diffuse within the space. The area taken over a region of the space will be proportional to the number of particles in that region.








Problem 5.

Solve the 3D wave equation by separation of variables to show, $u(\mathbf{x}, t) = \exp(i\mathbf{k}\cdot\mathbf{x} - i \omega t)$, is a solution. With $\mathbf{k} = (k_x, k_y, k_z)^T$ What is the resulting relationship between $\mathbf{k}$ and $\omega$?

\[\frac{n^2}{c^2}\frac{\partial^2 u(\mathbf{x}, t)}{\partial t^2} - \nabla^2 u(\mathbf{x}, t) = 0\]
NB: Coordinate vector $\mathbf{x} = (x,y,z)$ and, $u(\mathbf{x}) = (u(x),v(x),w(x))^T$ is a vector field.

$\Rightarrow{}$ Start with a separation of variables solution, let $u(\mathbf{x}, t) = X(x)Y(y)Z(z)T(t)$. then, \begin{align*}\frac{n^2}{c^2}{T''(t)}{X(x)Y(y)Z(z)} -X''(x)Y(y)Z(z)T(t) - Y''(y)X(x)Z(z)T(t)) - Z''(z)X(x)Y(y)T(t) = 0 \end{align*} \begin{align*} \frac{n^2}{c^2}\frac{T''(t)}{T(t)} - \frac{X''(x)}{X(x)} - \frac{Y''(y)}{Y(y)} - \frac{Z''(z)}{Z(z)} = 0 \end{align*} $\Rightarrow{}$ There is some freedom of choice as to how you define the constants here. In order to match to the test solution, $u(\mathbf{x}, t) = \exp(i\mathbf{k}\cdot\mathbf{x} - i \omega t)$, the most straightforward choice is, \begin{align*} \frac{T''(t)}{T(t)} = -\omega^2 \quad \frac{X''(x)}{X(x)} = -k_x^2 \quad \frac{Y''(y)}{Y(y)} = -k_y^2 \quad \frac{Z''(z)}{Z(z)} = -k_z^2 \space . \end{align*} $\Rightarrow{}$ But here we'll solve as if we assumed arbitrary constants instead, \begin{align*} \frac{T''(t)}{T(t)} = a \quad \frac{X''(x)}{X(x)} = b \quad \frac{Y''(y)}{Y(y)} = c \quad \frac{Z''(z)}{Z(z)} = d \space . \end{align*} $\Rightarrow{}$ With $\frac{n^2}{c^2} a - b - c - d = 0$. Solving the ODEs, we get, \begin{align*} T(t) &= A_+ e^{+ \sqrt{a} t} + A_- e^{- \sqrt{a} t} \newline X(x) &= B_+ e^{+ \sqrt{b} x} + B_- e^{- \sqrt{b} t} \end{align*} $\Rightarrow{}$ And equivalently for $Y(y)$ and $Z(z)$. To match with our test solution, $u(\mathbf{x}, t) = \exp(i\mathbf{k}\cdot\mathbf{x} - i \omega t)$, We'll keep the negative exponential from the $T(t)$ function, and the positive exponentials from the spatial functions. This allow us to build the solution, $u(\mathbf{x}, t) \propto \exp(\sqrt{b} x + \sqrt{c} y + \sqrt{d} z - \sqrt{a} t)$, Matching coefficients requires,} $$ \sqrt{a} = i \omega \quad \sqrt{b} = i k_x \quad \sqrt{c} = i k_y \quad \sqrt{d} = i k_z \quad . $$ $\Rightarrow{}$ Which gives, $u(\mathbf{x}, t) \propto \exp(i k_x x + i k_y y + i k_z z - i \omega t)$, or $\exp(i\mathbf{k}\cdot\mathbf{x} - i \omega t)$. and a dispersion relations,} $$ \boxed{ \frac{n^2}{c^2} \omega^2 = k_x^2 + k_y^2 + k_z^2 } $$








Problem 6.

Show that a Gaussian function, $f(x,t) = \frac{1}{\sigma(t)\sqrt{2\pi}} \exp\left( -\frac{x^2}{2\sigma(t)^2} \right)$ , solves the 1D diffusion equation, on the condition that $\sigma(t) = \sqrt{\sigma(0)^2 + 2 \alpha t}$.

This is a show-that question, so you need only show it satisfies the equation, rather than proving it outright (which is done in the course notes). This is an ideal opportunity for the chain rule, rather than directly inserting the form of $\sigma(t)$. First calculate the derivatives of the diffusion equation: \begin{align*} \frac{\partial f(x,t)}{\partial t} &= \frac{1}{\sqrt{2\pi}}\left[ -\frac{1}{\sigma(t)^2} + \frac{x^2}{\sigma(t)^4} \right]\sigma'(t)\exp\left(-\frac{x^2}{2 \sigma(t)^2}\right) \newline \frac{\partial f(x,t)}{\partial x} &= \frac{1}{\sqrt{2\pi}}\left[ -\frac{x}{\sigma(t)^3} \right]\exp\left(-\frac{x^2}{2 \sigma(t)^2}\right) \newline \frac{\partial^2 f(x,t)}{\partial x^2} &= \frac{1}{\sqrt{2\pi}}\left[ -\frac{1}{\sigma(t)^3} +\frac{x^2}{\sigma(t)^5} \right]\exp\left(-\frac{x^2}{2 \sigma(t)^2}\right) \;, \end{align*} $\Rightarrow{}$ and then, \begin{align*} \sigma'(t) &= \frac{\alpha}{\sigma(0)^2 + 2 \alpha t} = \frac{\alpha}{\sigma(t)} \;, \end{align*} $\Rightarrow{}$ Inserting into the diffusion equation, \begin{align*} \frac{f(x,t)}{t} &= \alpha \frac{ \partial^2 f(x,t)}{\partial x^2} \newline \frac{1}{\sqrt{2\pi}}\left[ -\frac{1}{\sigma(t)^2} + \frac{x^2}{\sigma(t)^4} \right]\sigma'(t)\exp\left(-\frac{x^2}{2 \sigma(t)^2}\right) &= \alpha \frac{1}{\sqrt{2\pi}}\left[ -\frac{1}{\sigma(t)^3} +\frac{x^2}{\sigma(t)^5} \right]\exp\left(-\frac{x^2}{2 \sigma(t)^2}\right) \end{align*} $\Rightarrow{}$ Divide common terms, \begin{align*} \left[ -\frac{1}{\sigma(t)^2} + \frac{x^2}{\sigma(t)^4} \right]\sigma'(t) &= \alpha \left[ -\frac{1}{\sigma(t)^3} +\frac{x^2}{\sigma(t)^5} \right] \end{align*} $\Rightarrow{}$ Substitute for $\sigma'(t)$, \begin{align*} \left[ -\frac{1}{\sigma(t)^2} + \frac{x^2}{\sigma(t)^4} \right] \frac{\alpha}{\sigma(t)} &= \alpha \left[ -\frac{1}{\sigma(t)^3} +\frac{x^2}{\sigma(t)^5} \right] \end{align*} $\Rightarrow{}$ LHS is equal RHS, so this is proven.








Problem 7.

For an initial concentration, $f(x,0) = \frac{1}{\sigma_0}\exp\left( -\frac{(x-x_0)^2}{2 \sigma_0^2} \right)$, governed by the diffusion equation, calculate the concentration at $x=0\,\mu\mathrm{m}$ and $t=10\,\mathrm{s}$, for $x_0 = 5\,\mu\mathrm{m}$, $\sigma_0 = 1\,\mu\mathrm{m}$, and $\alpha = 2\,\mu\mathrm{m}^2\mathrm{s}^{-1}$. (Use the solution given in Q6 to help)

$\Rightarrow{}$ First set up a solution to the diffusion equation such that it matches the initial condition. i.e. $$ f(x,t) = \frac{1}{\sqrt{\sigma_0^2 + 2 \alpha t}} \exp\left(-\frac{(x-x0)^2}{2 {\sigma_0^2 + 4 \alpha t}}\right) \;. $$ $\Rightarrow{}$ Then substitute in all constant and variable values, $$ f(x=0,t=10\,\mathrm{s}) = \frac{1}{\sqrt{(1\,\mu\mathrm{m})^2 + 2\times2\,\mu\mathrm{m}^2\mathrm{s}^{-1} \, 10\,\mathrm{s}}} \exp\left(-\frac{(0-5\,\mu\mathrm{m})^2}{2\times {(1\,\mu\mathrm{m})^2 + 4 \times 2\,\mu\mathrm{m}^2\mathrm{s}^{-1} \, 10\,\mathrm{s}}}\right) \; $$ $\Rightarrow{}$ Which reduces to, $$ \boxed{ f(x=0,t=10\,\mathrm{s}) = \frac{1}{\sqrt{41\,\mu\mathrm{m}^2}} \exp\left(-\frac{25\,\mu\mathrm{m}^2}{82\,\mu\mathrm{m}^2}\right) \approx 0.12 \mu\mathrm{m}^{-1} } $$










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Next week, Finite Methods!