Analytical solution of PDEs

Part of the CFD lecture set.

See also Courant Hilbert I: Section 5

Introduction

Most methods for analytically solving PDEs transform them into systems of ODEs (ordinary differential equations). A fairly comprehensive list of techniques might include:

• Separation of variables - reduces a PDE of $n$ independent variables into $n$ ODEs
• Integral transforms - reduce a PDE of $n$ variables into a PDE of $n-1$ variables (so, useful for 2-variable PDEs)
• Integral equations - changes a PDE into an integral equation, solved using other techniques
• Change of coordinates - changes a PDE into an ODE (or, into an easier PDE) by changing the problem coordinates
• Dependent variable transforms - transform the PDE unknown into a new, easier-to-find unknown
• Perturbation methods - changes a nonlinear problem into a sequence of linear problems that can approximate the nonlinear problem
• Impulse-response technique - decomposes initial and boundary conditions of the problem into simple impulses to find the response to each impulse; the final response is the sum of the simple impulse responses
• Eigenfunction expansion - finds solutions of the PDE in the form of infinite sums of eigenfunctions
• Calculus of variations - reformulates equations into minimization problems, e.g. total energy minimization

Separation of Variables

Restrictions:

• Linear PDEs
• Homogeneous PDEs
• Boundary conditions of the form $\alpha u_{x}(0,t)+\beta u(0,t)=0$ and $\gamma u_{x}(1,t)+\delta u(1,t)=0$, where $\alpha ,\beta ,\gamma ,\delta$ are constants (making the boundary conditions linear and homogeneous).

The principle behind separation of variables is to find solutions of the form:

$u_{n}(x,t)=X_{n}(x)T_{n}(t)$

or,

$u_{n}(x,t)=X_{n}(x)+T_{n}(t)$

In principle, an infinite number of solutions exist which will satisfy the PDE and also satisfy the boundary conditions (hence the subscript $n$).

These solutions are used as building blocks, and the sum of all of these solutions gives the total solution $u(x,t)$.

The idea behind separation of variables (and the reason it only applies to homogeneous problems) is, if one has an equation of the form

${\frac {\partial X}{\partial x}}{\frac {\partial T}{\partial t}}=0$

or,

${\frac {\partial X}{\partial x}}+{\frac {\partial T}{\partial t}}=0$

then each of the terms must be equal to a constant - otherwise the derivative of X would create a function in x that could not be canceled out by the other derivatives.

Given these infinite solutions, and given the boundary conditions, the solution can be found by adding up all the simple solutions $X_{n}(x)T_{n}(t)$ in a way that also matches the initial conditions:

$\displaystyle {\sum _{n=1}^{\infty }}A_{n}X_{n}(x)T_{n}(t)$

or,

$\displaystyle {\sum _{n=1}^{\infty }}A_{n}(X_{n}(x)+T_{n}(t))$

Example 1: First Derivatives

This example uses separation of variables to solve the PDE:

$u_{x}\times u_{y}=0$

This is a straightforward application of separation of variables. A solution of the form $u(x,y)=X(x)\times Y(y)$ will be assumed. Plugging this in:

${\frac {\partial X}{\partial x}}\times {\frac {\partial Y}{\partial y}}=0$

Which means that each derivative must be constant - otherwise (as mentioned above) there would be functions of $x,y,z$ left over:

${\frac {dX}{dx}}=c_{1}$

${\frac {dY}{dy}}=c_{2}$

So, the 1st order PDE has been transformed into a set of 2 ODEs. These are easy to solve, and yield $X=c_{1}x$ and $Y=c_{2}y$, so the solution is:

$u(x,y)=(c_{1}x)(c_{2}y)=c_{3}xy$

Which, when plugged in, satisfies the original PDE.

Example 2: Heat Equation

Separation of variables can be used to solve the heat equation under certain circumstances

What circumstances?

linear homogeneous boundary conditions... linear homogeneous PDE

Consider the heat equation in a normalized domain:

$\theta _{t}=\alpha ^{2}\theta _{xx}\qquad 0<x<1,0<t<\infty$

where $\theta$ stands for temperature.

Given the (linear homogeneous) boundary conditions

$BCs={\begin{cases}\theta (0,t)=0&0<t<\infty \\\theta (1,t)=0&0<t<\infty \end{cases}}$

Now we will use this PDE to describe the temperature profile in a finite rod, with the (relative) temperature at either end fixed at zero, and some initial temperature profile - which is arbitrary - let's call it $\phi (x)$.

Is it homogeneous?

Is this PDE linear? What if $\alpha =\alpha (\theta )$? Can we still use separation of variables?

$\alpha (\theta )$ makes the equation quasi-linear - i.e. NOT LINEAR - so can't use the separation of variables (see Introduction to partial differential equations lecture)

Step 1: Solution to PDE

A solution of the form $X(x)T(t)$ can be assumed, and plugged into the PDE:

$X(x)T'(t)=\alpha ^{2}X''(x)T(t)$

and, dividing each side by $X(x)T(t)$, it becomes

${\frac {T'(t)}{\alpha ^{2}T(t)}}={\frac {X''(x)}{X(x)}}$

which must be equal to a constant, since both expressions are functions of different variables.

${\frac {T'(t)}{\alpha ^{2}T(t)}}={\frac {X''(x)}{X(x)}}=c$

Both sides can be written as ODEs - so the PDE was transformed into two ODEs:

${\begin{aligned}T'-k\alpha ^{2}T&=&0\\X''-kX&=&0\end{aligned}}$

Once we solve these ODEs, the solutions can be multiplied together to give the solution to the PDE.

NOTE: Assuming the initial temperature profile is higher than 0 requires $k$ to be negative - otherwise the temporal behavior of the solution at infinite times would be to increase (not physically possible). It can be replaced with $k=-\lambda ^{2}$, which is guaranteed to be negative.

Thus the ODE system is:

${\begin{aligned}T'+\lambda ^{2}\alpha ^{2}T&=&0\\X''+\lambda ^{2}X&=&0\end{aligned}}$

General solutions can be found for both PDEs, of the form:

${\begin{aligned}T(t)&=&c_{1}\exp {(-\lambda ^{2}\alpha ^{2}t)}\\X(x)&=&c_{2}sin(\lambda x)+c_{3}cos(\lambda x)\end{aligned}}$

So the infinite solutions to our PDE are:

$\theta (x,t)=\exp {(-\lambda ^{2}\alpha ^{2}t)}\left(Asin(\lambda x)+Bcos(\lambda x)\right)$

where $A,B,\lambda$ are arbitrary.

Step 2: Finding solution to PDE and BCs

Let's try these solutions at $t=0$ - our initial condition

Try the function for $A=1,B=1,\lambda =1$ at $x=0$:

${\begin{array}{rcl}\theta (0,0)&=&\exp {(-\lambda ^{2}\alpha ^{2}t)}\left[Asin(\lambda x)+Bcos(\lambda x)\right]\\&=&\exp {(-1^{2}\times \alpha ^{2}\times 0)}\left[(1)sin(1\times 0)+(1)cos(1\times 0)\right]\\&=&1\times (0+1)\\&=&1\end{array}}$

But our boundary condition says $\theta (x=0,t)=0$!

Is the problem with the function?

No, problem is with the values of the constants we chose

We need to find values of $A,B,\lambda$ that will match boundary conditions

Boundary condition 1:

$\theta (x=0,t)=0$

This means:

${\begin{array}{rcl}\theta (x=0,t)&=&\exp {(-\lambda ^{2}\alpha ^{2}t)}\left[Asin(\lambda \times 0)+Bcos(\lambda \times 0)\right]\\\theta (x=0,t)&=&B\exp {(-\lambda ^{2}\alpha ^{2}t)}\\0&=&B\exp {(-\lambda ^{2}\alpha ^{2}t)}\end{array}}$

Which means either $B$ is 0, or $\exp {(-\lambda ^{2}\alpha ^{2}t)}$ is 0.

When is the exponential function 0?

Never - so $B=0$ always, in order to make this true.

Now let's do the same thing for the other boundary condition:

Boundary condition 2:

$\theta (x=1,t)=0$

What do you think we're going to find? Will we learn something about the exponential term? No - it's a function of $t$, and it's never going to be zero anyway.

Will we find something out from the cosine term? No - we know it's gone away permanently.

So we're going to get some information from the sine term.

${\begin{array}{rcl}\theta (x=1,t)&=&\exp {(-\lambda ^{2}\alpha ^{2}t)}\left[Asin(\lambda \times 1)\right]\\0&=&Asin(\lambda )\exp {(-\lambda ^{2}\alpha ^{2}t)}\end{array}}$

Okay! So, $A=0$! All done! That was easy!

Trivial solution - $A=0$ is an example of a trivial solution. It means our initial condition $\phi (x)=0$, so we started out with a uniform temperature of 0.

This leaves one non-zero term:

$sin(\lambda )=0$

which tells us something about $\lambda$ - not about A!

We have to pick functions $\theta =X(x)T(t)$ such that

$\lambda =n\pi$

where $n$ is an integer.

This means we've found an infinite number of functions that satisfy the PDE and the boundary conditions:

$\theta (x,t)=A_{n}\exp {\left(-(n\pi \alpha )^{2}t\right)}sin(n\pi x)$

So the solution to the PDE will be a sum of these simple functions.

Step 3: Solution to PDE, BCs, ICs

The last remaining step is to find a sum of the fundamental solutions, which satisfy the PDE and BCs, that will also satisfy the initial condition. That is, find $A_{n}$ in

$\theta (x,t)=\sum _{n=1}^{\infty }A_{n}\exp {\left(-(n\pi \alpha )^{2}t\right)}sin(n\pi x)$

such that it satisfies the initial condition,

$\theta (x,t=0)=\phi (x)$

This raises a question asked by Fourier: can an arbitrary function $\phi (x)$ be represented as an infinite series of sine functions?

$\theta (x,t=0)=\phi (x)=\sum _{n=1}^{\infty }A_{n}sin(n\pi x)$

(see wikipedia:Fourier series and wikipedia:Fourier transform).

To find values of $A_{n}$, we can use a property called orthogonality.

Orthogonality means that the integral of two functions that are orthogonal is 0.

For sine functions, any sine functions that are not the same are orthogonal, i.e.:

$\int _{0}^{1}sin(m\pi x)sin(n\pi x)dx={\begin{cases}0&m\neq n\\{\frac {1}{2}}&m=n\end{cases}}$

(Convince yourself of this by studying the plots of several sine waves:)

Starting with the expression for the initial condition:

$\phi (x)=A_{1}sin(\pi x)+A_{2}sin(2\pi x)+\dots$

Next, multiplying both sides by $sin(m\pi x)$, with $m$ an arbitrary integer, and integrating from 0 to 1:

${\begin{array}{rcl}\int _{0}^{1}\phi (x)sin(m\pi x)dx&=&A_{m}\int _{0}^{1}sin^{2}(m\pi x)dx\\&=&{\frac {1}{2}}A_{m}\end{array}}$

where all other terms are zero due to orthogonality.

This gives an expression for the coefficients of the Fourier series,

$A_{m}m=2\int _{0}^{1}\phi (x)sin(m\pi x)dx$

Making the final solution:

$\theta (u,x)=\sum _{n=1}^{\infty }A_{n}\exp {(-(n\pi \alpha )^{2}t)}sin(n\pi x)$

with the coefficients given by

$A_{n}=2\int _{0}^{1}\phi (x)sin(n\pi x)dx$

How many people were expecting a different solution? How many are disappointed by this result?

Since we put all this effort into finding this solution, I want to spend some time figuring out what it means.

Let's look at a hypothetical initial temperature profile:

This temperature profile plotted above (for $\alpha =0.01$) is an expression consisting of multiple sine waves, each with a unique frequency. The full expression for the temperature profile is:

${\begin{array}{l}\theta (x,t)=25\exp {(-\pi ^{2}\alpha ^{2}t)}sin(\pi x)\\-4\exp {(-(2)^{2}\pi ^{2}\alpha ^{2}t)}sin(2\pi x)\\+15\exp {(-(3)^{2}\pi ^{2}\alpha ^{2}t)}sin(3\pi x)\\+4\exp {(-(4)^{2}\pi ^{2}\alpha ^{2}t)}sin(4\pi x)\\+8\exp {(-(5)^{2}\pi ^{2}\alpha ^{2}t)}sin(5\pi x)\\+\exp {(-(8)^{2}\pi ^{2}\alpha ^{2}t)}sin(8\pi x)\\+2\exp {(-(20)^{2}\pi ^{2}\alpha ^{2}t)}sin(20\pi x)\\+0.5\exp {(-(21)^{2}\pi ^{2}\alpha ^{2}t)}sin(21\pi x)\\+\exp {(-(25)^{2}\pi ^{2}\alpha ^{2}t)}sin(25\pi x)\\+\exp {(-(28)^{2}\pi ^{2}\alpha ^{2}t)}sin(28\pi x)\\+\exp {(-(35)^{2}\pi ^{2}\alpha ^{2}t)}sin(35\pi x)\\+0.5\exp {(-(100)^{2}\pi ^{2}\alpha ^{2}t)}sin(100\pi x)\end{array}}$

Now, notice that each term has the decaying exponential function (the time function).

For higher frequencies, this exponential function will decay more rapidly. For lower frequencies, it will decay slowly.

This means the high-frequency fluctuations will decay very quickly...

And the low-frequency fluctuations will decay more slowly.

Plotting this solution function for various times yields a time-dependent solution for this function:

Notice how the small-scale (high-frequency) fluctuations are dampened immediately, whereas the moderate length-scale (middle-frequency) fluctuations take longer to be dampened, and finally, the last frequency to be dampened is the $n=1$ frequency (half a sine wave).

This mimics the natural behavior of diffusion of heat - small-scale temperature fluctuations will be dampened first, followed by increasingly larger scale temperature fluctuations.

Combination of Variables

Courant Hilbert II:

Section 3 Part 1 (p.40 of PDF)

References

http://www.exampleproblems.com/wiki/index.php/PDE:Integration_and_Separation_of_Variables

Tannehill

Anderson