Q.14.11-1MP - Mathematical Methods in Physical Sciences

Question

Question: Verify that the given function is harmonic, and find a function   of which it is the real part. Hint: Use Problem 2.64. For Problem 2, see Chapter 2, Section 17, Problem 19.

$\mathbf{ln}\sqrt{{\left(1+x\right)}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}$

Step-by-Step Solution

Answer

Verified that the given function is harmonic and the function is:

$\mathrm{f}\left(\mathrm{z}\right)=\ln \left(1+\mathrm{z}\right)+\mathrm{C}$

Use the given information for the calculation

Given function is,  .

$\ln \sqrt{{(1 + x)}^2 +y^2}$ 

Let,  $f\left(z\right) = u + iv$.

 

Here $u=\ln \sqrt{{\left(1+x\right)}^2+y^2}$.

 

Now it has to verify given function is harmonic.

 

If the given function is satisfied $\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}=0$ then the function is harmonic.

 

Now,

 $$\begin{gathered} \frac{\partial u}{\partial x}=\frac{1}{2}\frac{1}{{\left(1+x\right)}^2+y^2}\times 2\left(1+x\right) \\ =\frac{\left(1+x\right)}{{\left(1+x\right)}^2+y^2} \\ \frac{{\partial }^{2}u}{\partial x^2}=\frac{{\left(1+x\right)}^2-y-2\left(1+x\right).\left(1+x\right)}{{\left(1+x\right)}^2+y^2} \\ =\frac{{\left(1+x\right)}^2-y^2}{{\left[{\left(1+x\right)}^2+y^2\right]}^2} \end{gathered}$$……. (1)

 

Also,

 $$\begin{gathered} \frac{\partial u}{\partial x}=\frac{1}{2}\frac{1}{{\left(1+x\right)}^2+y^2}\times 2y \\ =\frac{y}{{\left(1+x\right)}^2+y^2} \\ \frac{{\partial }^{2}u}{\partial x^2}=\frac{{\left(1+x\right)}^2-y^2}{{\left[{\left(1+x\right)}^2+y^2\right]}^2} \end{gathered}$$……. (2)

Add equations (1) and (2)

Adding equations (1) and (2), it has  $\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial }=0$

 

It can be shown that, if   is a harmonic function which is defined at   $z_0=x_0+i{y}_0$, then an analytic function of which u is the real part is given as follows:

 $f\left(z\right)=2u\left(\begin{array}{cc}\frac{z+\overline{z_0}}{2},& \frac{z-\overline{z_0}}{2i}\end{array}\right)+cons\tan t$

 

Let $z_0=0$  then

 

$$\begin{gathered} u\left(0,0\right)=\ln \sqrt{{\left(1+0\right)}^2+0^2} \\ =\ln 1 \\ =0 \end{gathered}$$

 

Therefore, obtain:

 

$$\begin{gathered} f\left(z\right)=2\ln \sqrt{{\left(1+\frac{z}{2}\right)}^2-\frac{z^2}{4}}+cons\tan t \\ f\left(z\right)=\ln \left(1+z\right)+C \end{gathered}$$