For \begin{align} p_0 (x) \frac{d^2 u}{dx^2} + p_1(x) \frac{du}{dx} + p_2(x) u = f(x) \end{align}

The adjoint equation is \begin{align} v(x)[p_0 (x) \frac{d^2 u}{dx^2} + p_1(x) \frac{du}{dx} + p_2(x) u ] = \frac{d}{dx} \left( A(x) \frac{du}{dx} + B(x) u \right) \end{align} For some $A(x)$ and $B(x)$. I want to find the differential equation for $v(x)$.

The equation for $v(x)$ is called adjoint differential equation, how this and original equation is related with?

$i.e$, If we know the solution for $v(x)$ then we also know $u(x)$?


Very short answer: introducing the so-called inner product \begin{equation} \langle f,g \rangle = \int_a^b f(x) g(x) \text{d}x, \end{equation} and introducing the 'operator $L$ which acts on a function $u$ as \begin{equation} L u = p_0(x) u''(x) + p_1(x) u'(x) + p_2(x) u(x), \end{equation} we can look at the inner product $\langle L u,v\rangle$ for some function $v(x)$. Now, we can ask ourselves if we can find the so-called adjoint operator $L^*$, for which the following holds: \begin{equation} \langle L u,v \rangle = \langle u, L^* v \rangle. \end{equation} If you write down what this means in terms of the actual form of $L$ and the actual form of the inner product, which is an integral, you see that you can try to apply integration by parts (possibly multiple times) to convert a term in the integral such as $u'' v$ into something like $u v''$.

This is all properly and clearly defined in terms of function spaces, in particular Hilbert spaces. Fore more information, I would advise you search for these terms, in particular in the context of Sturm-Liouville problems.


\begin{align} v(x) \left[ p_0 (x) u'' + p_1 (x) u' + p_2(x) u\right] = \frac{d}{dx} ( A(x) u' + Bu) \end{align} Find the differential equation for $v(x)$.

equating left and right hand side \begin{align} ( vp_0 - A ) u'' + ( v p_1 - (vp_0)' - B) u' + (vp_2 - B')u=0 \end{align} Thus \begin{align} vp_0 = A, \quad vp_1 = A'+B, \quad vp_2 = B' \end{align} eliminating $A$, and $B$ we have \begin{align} &(vp_0)'' - (vp_1)' + vp_2=0 \\ & p_0 v'' + (2p_0'-p_1)v' + (p_0'' - p_1' + p_2)v=0 \end{align} The differential equation for $v$ is called the adjoint differential equations. This $v(x)$ is the integration factor for original equation $p_0 u'' + p_1 u' + p_2 u =f$, giving \begin{align} v p_0 u' - (vp_0)' u + vp_1 u = \int v f dx + C \end{align}


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