# Partial differential equations

#### Hamlings

Hello everyone,

I am currently stuck on this question. I have done a and b, but cannot do c or d or e.

Any help appreciated. Thank you.

The temperature distribution $\Theta(x, t)$ along an insulated metal rod of length $L$ is described by the differential equation $$\frac{\partial^2\Theta}{\partial x^2} = \frac{1}{D} \frac{\partial\Theta}{\partial t} \quad(0 < x < L, \,t > 0),$$ where $D \neq 0$ is a constant. The rod is held at a fixed temperature of 0° C at one end and is insulated at the other end, which gives rise to the boundary conditions $\Theta(0, t) = 0$ and $\Theta_x(L, t) = 0$, for $t > 0$.

The initial temperature distribution in the rod is given by $$\Theta(x, 0) = 0.5 \sin\left(\frac{7\pi x}{2L}\right) \quad(0 ≤ x ≤ L).$$(a) Use the method of separation of variables, with $\Theta(x, t) = X(x) T(t)$, to show that the function $X(x)$ satisfies the differential equation $$X'' − \mu X = 0 \qquad(1)$$ for some constant $\mu$. Write down the corresponding differential equation that $T(t)$ must satisfy.

(b) Write down the boundary conditions that $X(x)$ must satisfy.

Consider the functions $$X_n(x) = \sin(k_nx), \text{ where } k_n = \frac{(2n − 1)\pi}{2L} \text{ and }n = 1,\, 2,\, 3,\, .\, .\, .\, .$$ Show that function $X_n(x)$ satisfies the boundary conditions that you found. Show that $X_n(x)$ satisfies differential equation (1) for some constant $\mu$ (which you should specify).

(c) Solve the differential equation found in part (a) that the function $T(t)$ must satisfy. 

(d) Use your answers to write down a family of product solutions $\Theta_n(x, t) = X_n(x) T_n(t)$ that satisfy the first two boundary conditions. Hence show that the general solution of the partial differential
equation may be expressed as $$\Theta(x,t) = \sum_{n=1}^\infty C_n\exp\left(-\frac{D(2n-1)^2\pi^2t}{4L^2}\right)\sin\left(\frac{(2n-1)\pi x}{2L}\right). $$ (e) Find the particular solution that satisfies the given initial temperature distribution.

#### topsquark

Math Team
What have you been able to do so far? Can you do part a)?

-Dan

#### Hamlings

What have you been able to do so far? Can you do part a)?

-Dan
I've got that T must satisfy
T' - μDT =0

And I've shown that
X''- μX =0

But I'm not sure whether that is correct. This whole thing is quite baffling if I'm honest.

#### topsquark

Math Team
I've got that T must satisfy
T' - μDT =0

And I've shown that
X''- μX =0

But I'm not sure whether that is correct. This whole thing is quite baffling if I'm honest.
This is indeed correct. Good job!

Now, for b).

We know that $$\displaystyle \Theta (0,t ) = X(0) T(t) = 0$$. It makes no sense to say that $$\displaystyle T(t) = 0$$ as that merely gives $$\displaystyle \Theta (x, t) = 0$$, which is trivial. So we let $$\displaystyle X(0) = 0$$.

What does $$\displaystyle \Theta _x( L, t) = 0$$ for $$\displaystyle t > 0$$ tell us?

How about $$\displaystyle \Theta (x, 0) = \frac{1}{2} \sin \left ( \frac{ 7 \pi x }{2L} \right )$$?

-Dan

#### Hamlings

This is indeed correct. Good job!

Now, for b).

We know that $$\displaystyle \Theta (0,t ) = X(0) T(t) = 0$$. It makes no sense to say that $$\displaystyle T(t) = 0$$ as that merely gives $$\displaystyle \Theta (x, t) = 0$$, which is trivial. So we let $$\displaystyle X(0) = 0$$.

What does $$\displaystyle \Theta _x( L, t) = 0$$ for $$\displaystyle t > 0$$ tell us?

How about $$\displaystyle \Theta (x, 0) = \frac{1}{2} \sin \left ( \frac{ 7 \pi x }{2L} \right )$$?

-Dan
I really have no idea when it comes to this bit, I don't truly understand what it means.

#### topsquark

Math Team
One of the main thrusts that Mathematicians employ to solve a problem is to take a new kind of problem and turn it into a problem they already know how to solve. In this case, we are going to attempt to rewrite a partial differential equation into two ordinary differential equations, which we presumably already know how to solve. This particular method has a fair amount of usage in Physics and Engineering. We assume that we can take a solution to the partial differential equation and write it as the product of two functions of a single variable. Sometimes it works, sometimes it doesn't, and sometimes you don't get the full spectrum of possible solutions.

In this case, we are taking $$\displaystyle \Theta (x, t) = X(x) T(t)$$. We are given boundary conditions on $$\displaystyle \Theta$$: $$\displaystyle \Theta (0, t) = 0$$, which means that at some point $$\displaystyle x = 0$$ and any time $$\displaystyle t$$, $$\displaystyle \Theta$$ does not exist. $$\displaystyle \Theta _x (L, t) = 0$$ for $$\displaystyle t > 0$$, which means the first derivative (wrt $$\displaystyle x$$) is zero at some point $$\displaystyle x = L$$. Finally, we know that $$\displaystyle \Theta (x, 0) = \frac{1}{2} \sin \left ( \frac{7 \pi x}{2 L} \right )$$ for $$\displaystyle 0 \leq x \leq L$$, which means at $$\displaystyle t = 0$$, $$\displaystyle \Theta$$ is a sine wave.

We are letting $$\displaystyle \Theta (x, t) = X(x) T(t)$$. In terms of these functions, the initial conditions say
$$\displaystyle \Theta (0, t) = X(0) T(t) = 0 \implies X(0) = 0$$ if we are to have any non-zero solutions to $$\displaystyle \Theta$$ at all.

$$\displaystyle \Theta _x (L, t) = X'(L) T(t) = 0$$ which means that $$\displaystyle X'(L) = 0$$ (again if we want to have non-zero solutions.)

$$\displaystyle \Theta (x, 0) = X(x) T(0) = \frac{1}{2} \sin \left ( \frac{7 \pi x}{2 L} \right )$$ which means that $$\displaystyle T(0)$$ = some constant. (You can usually scale this condition to set $$\displaystyle T(0) = 1$$ for simplicity.)

Note: You could solve the $$\displaystyle X(x)$$ differential equation to find the form for $$\displaystyle X(x)$$, but we actually have been given the solution. You can check that the boundary condition for $$\displaystyle X(x)$$ on the line above satisfies all of the conditions given.

Let us know if/where you are still confused.

-Dan

#### Hamlings

One of the main thrusts that Mathematicians employ to solve a problem is to take a new kind of problem and turn it into a problem they already know how to solve. In this case, we are going to attempt to rewrite a partial differential equation into two ordinary differential equations, which we presumably already know how to solve. This particular method has a fair amount of usage in Physics and Engineering. We assume that we can take a solution to the partial differential equation and write it as the product of two functions of a single variable. Sometimes it works, sometimes it doesn't, and sometimes you don't get the full spectrum of possible solutions.

In this case, we are taking $$\displaystyle \Theta (x, t) = X(x) T(t)$$. We are given boundary conditions on $$\displaystyle \Theta$$: $$\displaystyle \Theta (0, t) = 0$$, which means that at some point $$\displaystyle x = 0$$ and any time $$\displaystyle t$$, $$\displaystyle \Theta$$ does not exist. $$\displaystyle \Theta _x (L, t) = 0$$ for $$\displaystyle t > 0$$, which means the first derivative (wrt $$\displaystyle x$$) is zero at some point $$\displaystyle x = L$$. Finally, we know that $$\displaystyle \Theta (x, 0) = \frac{1}{2} \sin \left ( \frac{7 \pi x}{2 L} \right )$$ for $$\displaystyle 0 \leq x \leq L$$, which means at $$\displaystyle t = 0$$, $$\displaystyle \Theta$$ is a sine wave.

We are letting $$\displaystyle \Theta (x, t) = X(x) T(t)$$. In terms of these functions, the initial conditions say
$$\displaystyle \Theta (0, t) = X(0) T(t) = 0 \implies X(0) = 0$$ if we are to have any non-zero solutions to $$\displaystyle \Theta$$ at all.

$$\displaystyle \Theta _x (L, t) = X'(L) T(t) = 0$$ which means that $$\displaystyle X'(L) = 0$$ (again if we want to have non-zero solutions.)

$$\displaystyle \Theta (x, 0) = X(x) T(0) = \frac{1}{2} \sin \left ( \frac{7 \pi x}{2 L} \right )$$ which means that $$\displaystyle T(0)$$ = some constant. (You can usually scale this condition to set $$\displaystyle T(0) = 1$$ for simplicity.)

Note: You could solve the $$\displaystyle X(x)$$ differential equation to find the form for $$\displaystyle X(x)$$, but we actually have been given the solution. You can check that the boundary condition for $$\displaystyle X(x)$$ on the line above satisfies all of the conditions given.

Let us know if/where you are still confused.

-Dan
For part b

So is the boundary condition that X(x) must satisfy, Θ(x,t) =0 ?

or X(x)=0?

If we let x=0, and plug this into Xn we are given, which gives

sin(0)=0, is this part proved?

and then do we test if the derivative of Xn is 0 at x=L?

If this is correct, I don't have a clue on where to go from here.I really struggle with this topic and find it hard to get my head around it. If you could continue helping it would be greatly appreciated.

Thank you very much for helping, I'm sorry I don't quite fully grasp it yet.