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In an ordinary differential equation, we only need real-valued initial conditions.

Now consider the heat equation $\delta_t u=\delta_{xx} u$.

In partial differential equations we instead need function-valued initial conditions. E.g. $u(x,0)=sin(x)$. This I understand.

However, we apparently also need boundary conditions, such as $Lim_{x\to - \infty}u(x,t)=0$ and the same for positive infinity.

Why do we need this? Why doesn't the initial condition already constrain the function $u(x,t)$ enough to give a specific function?

(I primarily have the heat equation in mind here)

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  • $\begingroup$ That's the heat equation. Thinking about the physical object we're modeling can help to understand the need for both initial and boundary values. $\endgroup$ – littleO Apr 5 '17 at 6:55
  • $\begingroup$ Sorry, of course. I mean the heat equation $\endgroup$ – user56834 Apr 5 '17 at 7:06
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I wouldn't differ initial conditions from other forms of boundary conditions. Also, ordinary differential equations are nothing but partial differential equations with one-dimensional domain.

As you stated yourself, the boundary conditions are usually formulated so that one is able to prove existence and uniqueness of solutions. Theherefore suitable boundary conditions depend on the context. These examples may show that this is not always the same:

  1. $u \colon [0,1] \to \mathbb{R}$, $u' = 1$. Fixing the value of $u(0)$ (or $u(1)$) determines the solution.

  2. $u \colon [0,1] \to \mathbb{R}$, $u'' = 0$ (Laplace equation). Fixing $u(0)$ is not enough, but fixing both $u(0)$ and $u(1)$ (i.e. $u$ on the whole boundary of the domain) determines the solution (which is just a linear function). In the context of ODEs one would usually fix $u(0)$ and $u'(0)$, which is another suitable boundary condition.

  3. $u \colon B(0,1) \to \mathbb{R}$, $\frac{\partial^2 u}{\partial x_1^2} + \ldots + \frac{\partial^2 u}{\partial x_n^2} = 0$ (Laplace equation). Just as above, fixing values of $u$ on the whole boundary $\partial B(0,1)$ determines the solution.

  4. $u \colon \mathbb{R} \times \mathbb{R}^+ \to \mathbb{R}$, $\frac{\partial u}{\partial x_1} + \frac{\partial u}{\partial x_2} = 0$ (exemplary transport equation). This time fixing $u(x_1,0)$ for all $x_1 \in \mathbb{R}$ (i.e. values on the boundary) determines the solution (which by the way is $u(x_1,x_2) = u(x_1-x_2,0)$).

Maybe you had some particular PDE in mind?

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  • $\begingroup$ yes I had the heat equation in mind, sorry for the confusion. $\endgroup$ – user56834 Apr 6 '17 at 6:33

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