Sobolev Space definition with example

 Sobolev spaces are essential in functional analysis and partial differential equations.

 The space $H^1(-1,1)$ is a specific Sobolev space defined on the interval $(-1,1)$ on the real line.

Definition of $H^1(-1,1)$

The Sobolev space $H^1(-1,1)$ includes functions $u$ that satisfy both the function itself and its first derivative being in $L^2(-1,1)$. Specifically, the space is defined as follows:

•    A function $u$ belongs to $L^2(-1,1)$ if:

  $${\int }_{-1}^1|u(x){|}^{2}dx<\mathrm{\infty }$$

•   The first weak derivative $u^{\prime }$ must also be in $L^2(-1,1)$, which means:

  $${\int }_{-1}^1|u^{\prime }(x){|}^{2}dx<\mathrm{\infty }$$


Weak derivatives are used to accommodate functions that might not be differentiable in the classical sense everywhere on the interval, including functions that are continuous and differentiable almost everywhere but may have points of derivative discontinuity.

Norm in $H^1(-1,1)$}

The norm in the Sobolev space $H^1(-1,1)$ is defined to incorporate both the function and its derivative, given by:
$$\Vert u{\Vert }_{H^1}={({\int }_{-1}^1|u(x){|}^2+|u^{\prime }(x){|}^{2}dx)}^{1/2}$$

Importance and Applications

Sobolev spaces such as $H^1(-1,1)$ are crucial in the study of partial differential equations (PDEs), as they frequently arise as the solution spaces for various PDEs. Understanding Sobolev spaces is foundational for advanced studies in fields such as mathematical physics, engineering, and applied mathematics.

Example Function for $H^1(-1,1)$

Working with a concrete example clarifies above definitions.

Consider the function $u(x)=x^2$ as an example for a function in the Sobolev space $H^1(-1,1)$. We will verify that both $u$ and its first derivative $u^{\prime }$ belong to $L^2(-1,1)$.

\Verification of $u\in L^2(-1,1)$
First, we check if the function $u(x)=x^2$ is square integrable over $(-1,1)$:

$${\int }_{-1}^1|u(x){|}^{2}dx={\int }_{-1}^1{x}^{4}dx$$

Calculate the integral:

$${\int }_{-1}^1{x}^{4}dx={\left[\frac{x^5}{5}\right]}_{-1}^1=\frac{1}{5}-(-\frac{1}{5})=\frac{2}{5}$$

This result confirms that $u(x)=x^2$ is in $L^2(-1,1)$ as the integral is finite ($<\mathrm{\infty }$).

Verification of $u^{\prime }\in L^2(-1,1)$
Next, compute the derivative of $u(x)$ and check its square integrability:

$$u^{\prime }(x)=2x$$

Check if $u^{\prime }(x)=2x$ is square integrable:

$${\int }_{-1}^1|u^{\prime }(x){|}^{2}dx={\int }_{-1}^1(2x{)}^{2}dx=4{\int }_{-1}^1{x}^{2}dx$$

Computing the integral:

$${\int }_{-1}^1{x}^{2}dx={\left[\frac{x^3}{3}\right]}_{-1}^1=\frac{1}{3}-(-\frac{1}{3})=\frac{2}{3}$$

Thus:

$$4{\int }_{-1}^1{x}^{2}dx=4\cdot \frac{2}{3}=\frac{8}{3}$$

Since this integral is also finite, $u^{\prime }(x)=2x$ is in $L^2(-1,1)$.

Thus, the function $u(x)=x^2$ and its derivative $u^{\prime }(x)=2x$ are both in $L^2(-1,1)$, confirming that $u(x)=x^2$ is a valid example of a function in the Sobolev space $H^1(-1,1)$.