基于变分法求解给定二阶常微分方程的变分表达式推导问询

Hey there! Let's break down exactly how that variational expression your textbook gave comes together, step by step. We'll start from the original ODE and build up to the $\delta J$ formula using integration by parts (the key tool here) and basic variational calculus rules.

Step 1: Start with the ODE and Variational Core Idea #

Your original differential equation is:
$$\frac{d^2u}{dx2} - u = -x \quad (0<x<1)$$
with boundary conditions $u(0)=0$, $u(1)=0$.

In variational calculus, we want to link this ODE to the extremum of a functional $J[u]$ (a function that takes another function as input). The core rule here is: if $J[u]$ has an extremum at the solution $u(x)$, then its variation $\delta J$ (a measure of how $J$ changes when we tweak $u$ slightly by $\delta u$) must be zero.

First, let's rearrange the ODE to set it equal to zero:
$$-\frac{d^2u}{dx2} + u - x = 0$$
If we multiply both sides by the variation $\delta u$ (a small, arbitrary function that satisfies $\delta u(0)=\delta u(1)=0$, since $u$ is fixed at the boundaries) and integrate over $[0,1]$, we get:
$$\int_0^1 \left(-\frac{d^2u}{dx2} + u - x\right)\delta u dx = 0$$
This is the condition we need to connect to $\delta J$.

Step 2: Use Integration by Parts to Handle the Second Derivative #

The tricky term here is the second derivative $-\frac{d^2u}{dx2}$. We need to "move" the derivative off of $u$ and onto $\delta u$ using integration by parts, which follows the rule:
$$\int_a^b v , dw = \left[vw\right]_a^b - \int_a^b w , dv$$

Let's apply this to the term $-\int_0^1 \frac{d^2u}{dx2}\delta u dx$:

• Let $v = \delta u$ (so $dv = \frac{d(\delta u)}{dx}dx = \delta\left(\frac{du}{dx}\right)dx$—important: variation and differentiation commute, so $\delta\left(\frac{du}{dx}\right) = \frac{d(\delta u)}{dx}$)
• Let $dw = \frac{d^2u}{dx2}dx$ (so $w = \frac{du}{dx}$)

Plugging into the integration by parts formula:
$$-\int_0^1 \frac{d^2u}{dx2}\delta u dx = -\left[\frac{du}{dx}\delta u\right]_0^1 + \int_0^1 \frac{du}{dx} \cdot \delta\left(\frac{du}{dx}\right)dx$$

Step 3: Reconstruct the Functional $J[u]$ #

Now substitute this result back into our original integral equation:
$$\int_0^1 \left(-\frac{d^2u}{dx2} + u - x\right)\delta u dx = -\left[\frac{du}{dx}\delta u\right]_0^1 + \int_0^1 \frac{du}{dx}\cdot\delta\left(\frac{du}{dx}\right)dx + \int_0^1 (u - x)\delta u dx$$

Notice that the last two integrals are exactly the variation of a functional $J[u]$! Let's verify:

• $\delta\left(\frac{1}{2}\left(\frac{du}{dx}\right)^2\right) = \frac{du}{dx}\delta\left(\frac{du}{dx}\right)$ (power rule for variations)
• $\delta\left(\frac{1}{2}u^2\right) = u\delta u$
• $\delta(xu) = x\delta u$

Combining these, the sum of the two integrals becomes:
$$\int_0^1 \left[\delta\left(\frac{1}{2}\left(\frac{du}{dx}\right)^2\right) + \delta\left(\frac{1}{2}u^2\right) - \delta(xu)\right]dx$$

Since variation and integration commute, we can pull the $\delta$ outside the integral:
$$\delta\int_0^1 \left(\frac{1}{2}\left(\frac{du}{dx}\right)^2 + \frac{1}{2}u^2 - xu\right)dx = \delta J$$

Here, the functional is:
$$J[u] = \int_0^1 \left(\frac{1}{2}\left(\frac{du}{dx}\right)^2 + \frac{1}{2}u^2 - xu\right)dx$$

Step 4: Rearrange to Get the Textbook Expression #

Now just rearrange the equation from Step 3 to solve for $\delta J$:
$$\delta J = \int_0^1 \left(-\frac{d^2u}{dx2} + u - x\right)\delta u dx + \left[\frac{du}{dx}\delta u\right]_0^1$$

And that's exactly the expression your textbook provided! A quick note on the boundary term: since our boundary conditions fix $u(0)=u(1)=0$, the variation $\delta u$ must be zero at $x=0$ and $x=1$ (we can't change the function at fixed boundaries). So $\left[\frac{du}{dx}\delta u\right]_0^1 = \frac{du}{dx}(1)\delta u(1) - \frac{du}{dx}(0)\delta u(0) = 0$, which means the variational condition $\delta J=0$ reduces exactly to the original ODE holding everywhere in $(0,1)$.

内容的提问来源于stack exchange，提问作者Loukit Khemka