Simple Partial Differential Equations Example

I am really enjoying my current partial differential equations class, so I thought I'd share an example problem. Note that this is probably one of the simplest problems in partial DE.

Problem

$$\begin{equation}   \text{(1)    } k^2 \frac{\partial^2 U}{\partial t^2} = \frac{\partial^2 U}{\partial x^2} \end{equation}$$
$$\begin{equation}   \text{(2)    } U \left( 0, t \right) = 0 \end{equation}$$
$$\begin{equation}   \text{(3)    } U \left( L, t \right) = 0 \end{equation}$$
$$\begin{equation}   \text{(4)    } U \left( x, 0 \right) = 0 \end{equation}$$
$$\begin{equation}   \text{(5)    } \frac{\partial U}{\partial t} \left( x, 0 \right) = f \left( x \right) \end{equation}$$

Solution

We are looking for a solution of the form:
$$U = XT$$
Where $X$ is a function of $x$ and $T$ is a function of $t$.

Translating (1) to match the expected solution, we get:
$$k^2 XT^{\prime\prime} = X^{\prime\prime} T$$

Dividing each side by $XT$:
$$k^2 \frac{T^{\prime\prime}}{T} = \frac{X^{\prime\prime}}{X}$$

Since we have a function of only $T$ on the left and only $X$ on the right, we know that these are equal to a constant.
$$k^2 \frac{T^{\prime\prime}}{T} = \frac{X^{\prime\prime}}{X} = constant = \left\{\begin{array}{lr}   0 \\   -\lambda^2 \\   \lambda^2 \end{array}   \right.$$

The notation above says that the constant can either be $0$, some negative number (as forced to be negative by $-\lambda^2$) or some positive number (forced by $\lambda^2$).

Case $constant = 0$:

Finding $X$:
$$\frac{X^{\prime\prime}}{X} = 0 \Rightarrow X^{\prime\prime} = 0$$

Integrating both sides:
$$X^{\prime} = A$$
Where $A$ is an arbitrary constant.

Integrating again:
$$X = A x + B$$
Where $B$ is an arbitrary constant.

Finding $T$:
$$\frac{k^2 T^{\prime\prime}}{T} = 0 \Rightarrow T^{\prime\prime} = 0$$

Integrating both sides:
$$T^{\prime} = C$$
Where $C$ is an arbitrary constant.

Integrating again:
$$T = C t + D$$
Where $D$ is an arbitrary constant.

Since $U = XT$:
$$U = \left(A x + B\right) \left( C t + D \right)$$
Therefore, this satisfies (1).

Looking at (2), we have:
$$U \left( 0, t \right) = 0 \Rightarrow \left(A (0) + B\right) \left( C t + D \right) = 0 \Rightarrow B \left( C t + D \right)= 0$$

This implies that $B = 0$, so:
$$U = A x \left( C t + D \right) = x \left( C t + D \right)$$
satisfies (1) and (2); note that the constant $A$ was absorbed into the other constants (since they are, after all, just arbitrary constants).

Looking at (3), we have:
$$U \left( L, t \right) = 0 \Rightarrow L \left( C t + D \right) = 0$$

To make this true, we can't change the value of $L$ because it is a constraint, so the only option is to make $C = D = 0$.

So,
$$U = 0$$
This, however, is an uninteresting solution for $U$. So, we examine the next possible constant.

Case $constant = -\lambda^2$:

Finding $X$:
$$\frac{X^{\prime\prime}}{X} = -\lambda^2$$

Multiply both sides by $X$:
$$X^{\prime\prime} = -\lambda^2 X$$

Using methods from differential equations (DE), we know that we can solve this by subbing $\alpha^2$ in for $X^{\prime\prime}$ and $1$ in for $X$:
$$\alpha^2 = -\lambda^2$$

Taking the square root of both sides we get:
$$\alpha = \pm \lambda i$$

From DE we know that this fits the form $\alpha = b \pm c i$, where the answer the the DE is:
$$X = e^{bx} \left[ A \cos \left( c x \right) + B \sin \left( c x \right)\right]$$
Where $A$ and $B$ are arbitrary constants.

Therefore:
$$X = e^{0x} \left[ A \cos \left( \lambda x \right) + B \sin \left( \lambda x \right)\right] = \left[ A \cos \left( \lambda x \right) + B \sin \left( \lambda x \right)\right]$$

Finding $T$:

$$k^2 \frac{T^{\prime\prime}}{T} = -\lambda^2$$

Multiply both sides by $\frac{T}{k^2}$:
$$T^{\prime\prime} = -\frac{\lambda^2}{k^2} T$$

Using methods from differential equations (DE), we know that we can solve this by subbing $\beta^2$ in for $T^{\prime\prime}$ and $1$ in for $T$:
$$\beta^2 = -\frac{\lambda^2}{k^2}$$

Taking the square root of both sides we get:
$$\beta = \pm \frac{\lambda}{k} i$$

From DE we know that this also fits the form $\beta = b \pm c i$, where the answer the the DE is:
$$T = e^{bt} \left[ C \cos \left( c t \right) + D \sin \left( c t \right)\right]$$
Where $C$ and $D$ are arbitrary constants (not necessarily the same as $A$ and $B$).

Therefore:
$$T = e^{0t} \left[ C \cos \left( \frac{\lambda}{k} t \right) + D \sin \left( \frac{\lambda}{k} t \right)\right] = \left[ C \cos \left( \frac{\lambda}{k} t \right) + D \sin \left( \frac{\lambda}{k} t \right)\right]$$

Since $U = XT$:
$$U = \left[ A \cos \left( \lambda x \right) + B \sin \left( \lambda x \right)\right] \left[ C \cos \left( \frac{\lambda}{k} t \right) + D \sin \left( \frac{\lambda}{k} t \right)\right]$$
Therefore, this satisfies (1).

Looking at (2), we have:
$$U \left( 0, t \right) = 0 \Rightarrow \left[ A \cos \left( 0 \right) + B \sin \left( 0 \right)\right] \left[ C \cos \left( \frac{\lambda}{k} t \right) + D \sin \left( \frac{\lambda}{k} t \right)\right] = 0$$

$\cos \left( 0 \right) = 1$ and $\sin \left( 0 \right) = 0$, so,
$$A \left[ C \cos \left( \frac{\lambda}{k} t \right) + D \sin \left( \frac{\lambda}{k} t \right)\right] = 0$$

This implies that $A = 0$, so:
$$U = \sin \left( \lambda x \right) \left[ C \cos \left( \frac{\lambda}{k} t \right) + D \sin \left( \frac{\lambda}{k} t \right)\right]$$   
satisfies (1) and (2); note that the constant $B$ was absorbed into the other constants (since they are, after all, just arbitrary constants).

Looking at (3), we have:
$$U \left( L, t \right) = 0 \Rightarrow \sin \left( \lambda L \right) \left[ C \cos \left( \frac{\lambda}{k} t \right) + D \sin \left( \frac{\lambda}{k} t \right)\right] = 0$$

So, we need to pick a value for $\lambda$ that causes the above expression to always be zero. We find that there are an infinite number of them (everywhere $\sin \left( n \pi \right)$ where n is a natural number (better suited to match the solution than an integer). So, because of this, we find that:
$$\lambda = \frac{n \pi}{L},\, \forall n \in \mathbb{N}$$

So, subbing in for $\lambda$, we have:
$$U = \sin \left( \frac{n \pi x}{L} \right) \left[ C \cos \left( \frac{n \pi}{L k} t \right) + D \sin \left( \frac{n \pi}{L k} t \right)\right]$$
Which satisfies (1), (2) and (3).

Looking at (4), we have:
$$U \left( x, 0 \right) = 0 \Rightarrow \sin \left( \frac{n \pi x}{L} \right) \left[ C \cos \left( 0 \right) + D \sin \left( 0 \right)\right] = 0$$

$\cos \left( 0 \right) = 1$ and $\sin \left( 0 \right) = 0$, so,
$$U \left( x, 0 \right) = 0 \Rightarrow \sin \left( \frac{n \pi x}{L} \right) \left( C \right) = 0$$

This implies that $C = 0$, so,
$$U = \sin \left( \frac{n \pi x}{L} \right) \left( D \right) \sin \left( \frac{n \pi}{L k} t \right)$$
Satisfies (1), (2), (3) and (4).

Since constraint (5) is non-zero, I will evaluate it lastly. For now, I claim that:
$$U = \sum_{n = 1}^{\infty} D_n \sin \left( \frac{n \pi x}{L} \right) \sin \left( \frac{n \pi}{L k} t \right)$$
$\forall n \in \mathbb{N}$, also satisfies (1), (2), (3), and (4). This claim is justified because, as show above, constraint (3) is satisfied for all values of $n$. So, for whatever $n$ I pick from the natural numbers ($1,2,3...$, $\mathbb{N}$) the resulting $U$ will also be an answer. So, summing together the infinite answers for $U$ still meets the problem's criteria. $D$ became $D_n$ because $D$ is not necessarily the same thing for any $n$ value.

So, considering constraint (5), we need to first find $\frac{\partial U}{\partial t}$:
$$\frac{\partial U}{\partial t} = \sum_{n = 1}^{\infty} D_n \sin \left( \frac{n \pi x}{L} \right) \left( \frac{n \pi}{L k} \right) \cos \left( \frac{n \pi}{L k} t \right)$$

Looking at the actual constraint, (5):
$$\frac{\partial U}{\partial t} \left( x, 0 \right) = f \left( x \right) \Rightarrow \sum_{n = 1}^{\infty} D_n \sin \left( \frac{n \pi x}{L} \right) \left( \frac{n \pi}{L k} \right) \cos \left( 0 \right) = f \left( x \right)$$

Since, $\cos \left( 0 \right) = 1$:
$$f \left( x \right) = \sum_{n = 1}^{\infty} D_n \sin \left( \frac{n \pi x}{L} \right) \left( \frac{n \pi}{L k} \right)$$

By the Fourier sine series, we know that the coefficients are given by:
$$D_n \left( \frac{n \pi}{L k} \right) = \frac{2}{L} \int_0^L f \left( x \right) \sin \left( \frac{n \pi x}{L} \right) dx$$
We include $\left( \frac{n \pi}{L k} \right)$ because it does not match the Fourier series and needs to be divided out.

Solving for the actual coefficients:
$$D_n = \frac{2 k}{n \pi} \int_0^L f \left( x \right) \sin \left( \frac{n \pi x}{L} \right) dx$$

Where our final $U$ is, as above:
$$U = \sum_{n = 1}^{\infty} D_n \sin \left( \frac{n \pi x}{L} \right) \sin \left( \frac{n \pi}{L k} t \right)$$

Now, we could go on to examine the $constant = \lambda^2$ case. However, it ends up just being $0$ (and is therefore redundant). You can use methods like those above to come to this conclusion for yourself.