Why don't we commonly solve partial differential equations with sums of functions, instead of products?

Question

It's very common to solve partial differential equations via "separable solution", in the following way. Say we have the wave equation,

$$u_t=u_{xx}.$$

We often solve this by assuming a form $u(x,t)=X(x)T(t)$, which gives

$$\frac{T_t}{T}=\frac{X_{xx}}{X}=\lambda,$$

where $\lambda$ is the separation constant, and the final solution looks something like

$$u(x,t)=C\exp (\lambda t -i\sqrt{\lambda} x)$$

However, it seems to me we could just as easily have tried the ansatz $u(x,t)=X(x)+T(t)$. Then our PDE would look like

$$T_t=X_{xx}=\lambda$$

And we would find

$$X(x)=\frac{1}{2}\lambda x^2 + C_2 x + C_3,\qquad T(t)=\lambda t + C_1.$$

Basically, a polynomial solution instead of an exponential one.

Is there any good reason why we often present the first way instead of the second? I guess there are "niceness" properties that the exponentials have, but polynomial solutions are nice in some ways too.

Answer 1

The collection of solutions you find by assuming separability is much richer than the collection of simple polynomial solutions you find by taking $u(x,t) = X(x) + T(t)$. In particular, by taking linear combinations of separable solutions, we can solve any given initial-boundary value problem (IBVP) for the heat equation.

I'll show how this works out in detail. The following material can be found in PDE textbooks such as Boyce and DiPrima.

For example, consider the following IBVP: Find a continuous function $u: [0,1] \times [0,\infty)$ such that $u$ satisfies the heat equation $u_t = u_{xx}$ on $\Omega = (0,1) \times (0,\infty)$ as well as the boundary conditions $u(0,t) = u(1,t) = 0$ (for all $t > 0$) and the initial condition $u(x,0) = f(x)$. Here $f:[0,1] \to \mathbb R$ is any given continuous, piecewise differentiable function that satisfies $f(0) = f(1) = 0$. (Other assumptions on $f$ are also possible, but the restrictions on $f$ are always very mild.)

First we find separable solutions to the heat equation which satisfy the given boundary conditions. If a separable function $u(x,t) = X(x) T(t)$ satisfies the heat equation on $\Omega$, then there must exist a constant $\lambda$ such that $$ T'(t)/T(t) = X''(x)/X(x) = \lambda $$ for all $(x,t) \in \Omega$. If $\lambda$ were positive, our boundary conditions could not be satisfied. If $\lambda = 0$ then $u(x,t) = 0$, a trivial solution. So we assume that $\lambda$ is negative. It follows that $T(t) = T_0 e^{\lambda t}$ and $X(x) = a \cos(\sqrt{-\lambda} x) + b \sin(\sqrt{-\lambda} x)$ for some constants $T_0, a$, and $b$. Now plugging in the boundary conditions $u(0,t) = u(1,t) = 0$, we find that $a = 0$ and $b \sin(\sqrt{-\lambda}) = 0$. Assuming that $b \neq 0$, it follows that $\sqrt{-\lambda} = n \pi$ for some positive integer $n$. So the separable solutions we have found are $$ u_n(x,t) = c e^{- n^2 \pi^2 t} \sin(n \pi x) \quad \text{for } n = 1, 2, \ldots. $$

Now we solve the IBVP given above by seeking a solution of the form $$ u(x,t) = \sum_{n = 1}^\infty c_n e^{-n^2 \pi^2 t} \sin(n \pi x). $$ This $u$ automatically satisfies $u_t = u_{xx}$ as well as $u(0,t) = u(1,t) = 0$ for all $t > 0$. The only issue is to make $u$ satisfy the initial condition $$ u(x,0) = f(x). $$ This initial condition can only be satisfied if there exist constants $c_n$ (for $n = 1, 2, \ldots$) such that $$ f(x) = \sum_{n=1}^\infty c_n \sin(n \pi x). $$ Now here is the amazing part: it turns out that any function $f$ satisfying the above conditions can be represented like this! That is a fact from Fourier analysis. Fourier made this claim, optimistically, but no one believed him at first. If I understand the history correctly, this is the thought process that led to the discovery of Fourier series --- a momentous moment in the history of math. Why math tends to work out so nicely is a deep mystery, a glimpse of perfection in the universe.

So by taking linear combinations of the separable functions we are able to solve our IBVP. This wouldn't have worked if we had tried to find a solution $u$ that is a linear combination of the simple polynomial solutions mentioned in the question. In particular, the step where we have to represent $f$ would have failed.

Answer 2

Partial differential equations (PDEs) differ from ordinary differential equations (ODEs) that involve functions of only one variable. However, this difference makes PDEs appreciably more difficult to solve. In fact, the vast majority of PDE cannot be solved analytically and those classes of special PDEs that can be solved analytically invariably involve converting the PDE into one or more ODEs and then solving independently. One of these approaches is the method of separation of variables or, Fourier's method or, product rule or, the method of eigenfunction expansion. It is one of the most widely used techniques to solve partial differential equations and is based on the assumption that the solution of the equation is separable, that is, the final solution can be represented as a product of several functions, each of which is only dependent upon a single independent variable. Daniel Bernoulli invented this technique in the $1700$s. We do not know the actual reason behind the ansatz i.e., why we have to choose the form of the method of separation of variables as $~u(x,y) = X(x)Y(y)~$. But it works good. If this assumption is incorrect, then clear violations of mathematical principles will be obvious from the analysis.

The method of separation of variables is used when the partial differential equation and the boundary conditions are linear and homogeneous. The advantage of the product method is that it transforms a partial differential equation, which we do not know how to solve, into two ordinary differential equations.

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