Helmholtz Equation

Helmholtz equation
$$\nabla^2\psi+k^2\psi=0\ \ \ \ \ \mbox{(1)}$$
is extremely important in physics. Solving many physically important partial differential equations such as heat equation, wave equation (Klein-Gordon equation), Maxwell’s equations, and Schrödinger equation, etc. often require solving Helmholtz equation (1).

In this notes, we discuss how to solve Helmholtz equation using separation of variables in rectangular, cylindrical, and spherical coordinate systems. The solutions we discuss here will be used when you solve boundary value problems associated with Helmholtz equation.

Helmholtz Equation in Rectangular Coordinates

Assume that $\psi(x,y,z)=X(x)Y(y)Z(z)$. Then the equation (1) becomes
$$YZ\frac{d^2X}{dx^2}+XZ\frac{d^2Y}{dy^2}+XY\frac{d^2Z}{dz^2}+k^2XYZ=0.\ \ \ \ \ \mbox{(2)}$$
Dividing (2) by $XYZ$, we obtain
$$\frac{1}{X}\frac{d^2X}{dx^2}+\frac{1}{Y}\frac{d^2Y}{dy^2}+\frac{1}{Z}\frac{d^2Z}{dz^2}+k^2=0.\ \ \ \ \mbox{(3)}$$
Let us write (3) as
$$\frac{1}{X}\frac{d^2X}{dx^2}=-\frac{1}{Y}\frac{d^2Y}{dy^2}-\frac{1}{Z}\frac{d^2Z}{dz^2}-k^2.\ \ \ \ \ \mbox{(4)}$$
Now we have a paradox. The LHS of (4) depends only on the $x$-variable while the RHS of (4) depends on $y$ and $z$-variables. One way to to avoid this paradox is to assume that the LHS and the RHS of (4) is a constant, say $-l^2$. If you are wondering why we choose a negative constant, the reason comes from physics. For a physical reason, we need an oscillating solution which can be obtained by choosing a negative separation constant. Often boundary conditions for Helmholtz equation lead to a trivial solution for a positive separation constant. Continuing a similar process, we separate Helmholtz equation into three ordinary differential equations:
\begin{align*}
\frac{1}{X}\frac{d^2 X}{dx^2}&=-l^2,\\
\frac{1}{Y}\frac{d^2Y}{dy^2}&=-m^2,\\
\frac{1}{Z}\frac{d^2Z}{dz^2}&=-n^2,
\end{align*}
where $k^2=l^2+m^2+n^2$.

Each mode is given by
$$\psi_{lmn}(x,y,z)=X_l(x)Y_m(y)Z_n(z)$$ and the most general solution is given by the linear combination of the modes
$$\psi(x,y,z)=\sum_{i,m,n}a_{lmn}\psi_{lmn}(x,y,z).$$

Helmholtz Equation in Cylindrical Coordinates

In cylindrical coordinate system $(\rho,\varphi,z)$, Helmholtz equation (1) is written as
$$\frac{1}{\rho}\frac{\partial}{\partial\rho}\left(\rho\frac{\partial\psi}{\partial\rho}\right)+\frac{1}{\rho^2}\frac{\partial^2\psi}{\partial\varphi^2}+\frac{\partial^2\psi}{\partial z^2}+k^2\psi=0.\ \ \ \ \ \mbox{(5)}$$

We assume that $\psi(\rho,\varphi,z)=P(\rho)\Phi(\varphi)Z(z)$. Then (5) can be written as
$$\frac{\Phi Z}{\rho}\frac{\partial}{\partial\rho}\left(\rho\frac{\partial\psi}{\partial\rho}\right)+\frac{PZ}{\rho^2}\frac{\partial^2\psi}{\partial\varphi^2}+P\Phi\frac{\partial^2\psi}{\partial z^2}+k^2=0.\ \ \ \ \ \mbox{(6)}$$
As we have done in rectangular coordinate system, by introducing the separation constants we can separate (6) into three ordinary differential equations
\begin{align*}
\frac{d^2Z}{dz^2}=l^2z,\\
\frac{d^2\Phi}{d\phi^2}=-m^2\Phi,\\
\rho\frac{d}{d\rho}\left(\rho\frac{dP}{d\rho}\right)+(n^2\rho^2-m^2)P=0,\ \ \ \ \ \mbox{(7)}
\end{align*}
where $n^2=k^2+l^2$. The last equation (7) is Bessel’s differential equation.

The general solution of Helmholtz equation in cylindrical coordinates is given by
$$\psi(\rho,\varphi,z)=\sum_{m,n}a_{mn}P_{mn}(\rho)\Phi_m(\varphi)Z_n(z).$$

Helmholtz Equation in Spherical Coordinates

In spherical coordinates $(r,\theta,\varphi)$, Helmholtz equation (1) is written as
$$\frac{1}{r^2\sin\theta}\left[\sin\theta\frac{\partial}{\partial r}\left(r^2\frac{\partial\psi}{\partial r}\right)+\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial\psi}{\partial\theta}\right)+\frac{1}{\sin\theta}\frac{\partial^2\psi}{\partial\varphi^2}\right]=-k^2\psi.\ \ \ \ \ \mbox{(8)}$$
Assume that $\psi(r,\theta,\varphi)=R(r)\Theta(\theta)\phi(\varphi)$. Then (8) can be written as
$$\frac{1}{Rr^2}\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)+\frac{1}{\Theta r^2\sin\theta}\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta}{d\theta}\right)+\frac{1}{\Phi r^2\sin^2\theta}\frac{d^2\Phi^2}{d\varphi^2}=-k^2.\ \ \ \ \ \mbox{(9)}$$
By introducing separation constants, (9) is separated into three ordinary differential equations
\begin{align*}
\frac{1}{\Phi}\frac{d^2\Phi}{d\varphi^2}=-m^2,\\
\frac{1}{\sin\theta}\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta}{d\theta}\right)+\left(Q-\frac{m^2}{\sin^2\theta}\right)\Theta=0,\ \ \ \ \ \mbox{(10)}\\
\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)+\left(k^2-\frac{Q}{r^2}\right)R=0.\ \ \ \ \ \mbox{(11)}
\end{align*}
The second equation (10) is the associated Legendre equation with $Q=l(l+1)$. The third equation (11) is spherical Bessel equation with $k^2>0$.

The general solution of Helmholtz equation (8) is then given by
$$\psi(r,\theta,\varphi)=\sum_{Q,m}R_Q(r)\Theta_{Qm}(\theta)\Phi_m(\varphi).$$

The restriction that $k^2$ be a constant is unnecessary. For instance the separation process will still be possible for $k^2=f(r)$. If $k^2=f(r)$, (11) is the associated Laguerre equation. The associated Laguerre equation is appeared in the hydrogen atom problem in quantum mechanics.

Connection Forms

Let $E_1, E_2, E_3$ be an arbitrary frame field on $\mathbb{E}^3$. At each $v\in T_p\mathbb{E}^3$, $\nabla_v E_i\in T_p\mathbb{E}^3$, $i=1,2,3$. So, there exists uniquely 1-forms $\omega_{ij}:T_p\mathbb{E}^3\longrightarrow\mathbb{R}$, $i,j=1,2,3$ such that
\begin{align*}
\nabla_vE_1&=\omega_{11}(v)E_1(p)+\omega_{12}(v)E_2(p)+\omega_{13}(v)E_3(p),\\
\nabla_vE_2&=\omega_{21}(v)E_1(p)+\omega_{22}(v)E_2(p)+\omega_{23}(v)E_3(p),\\
\nabla_vE_3&=\omega_{31}(v)E_1(p)+\omega_{32}(v)E_2(p)+\omega_{33}(v)E_3(p)
\end{align*}
for each $v\in T_p\mathbb{E}^3$. These equations are called the connection equations of the frame field $E_1$, $E_2$, $E_3$. One can clearly see that $\omega_{ij}$ is determined by
$$\omega_{ij}(v)=\nabla_v E_i\cdot E_j(p).$$ The 1-forms $\omega_{ij}$ are called the connection forms of the frame field $E_1,E_2,E_3$. Often the matrix $\omega=(\omega_{ij})$ is called the connection 1-form of the frame field $E_1,E_2,E_3$. The linearity of $\omega_{ij}$ is due to the linearity of the covariant derivative $\nabla E_i$.

Proposition. The matrix $\omega$ is a skew symmetric matrix, i.e. $\omega+{}^t\omega=0$.

Proof. Since $E_i\cdot E_j=0$, the directional derivative $v[E_i\cdot E_j]=0$. On the other hand, by Leibniz rule,
\begin{align*}
v[E_i\cdot E_j]&=\nabla_vE_i\cdot E_j(p)+E_i(p)\cdot \nabla_vE_j\\
&=\omega_{ij}(v)+\omega_{ji}(v).
\end{align*}
Hence,
\begin{equation}\label{eq:skewsymm}\omega_{ij}+\omega_{ji}=0.\end{equation}

If $i=j$ in \eqref{eq:skewsymm}, we get $\omega_{ii}=0$. So, the connection 1-form $\omega$ is written as
$$\omega=\begin{pmatrix}
0 & \omega_{12} & \omega_{13}\\
-\omega_{12} & 0 &\omega_{23}\\
-\omega_{13} & -\omega_{23} & 0
\end{pmatrix}.$$

Remark. The set of all $3\times 3$ skew symmetric matrices is denoted by $\mathfrak{o}(3)$. It is the Lie algebra of the orthogonal group $\mathrm{O}(3)$. The orthogonal group $\mathrm{O}(3)$ is the set of all $3\times 3$ orthogonal matrices and it is a Lie group. Recall that a square matrix $A$ is orthogonal if and only if $A\cdot{}^tA=I$, i.e. $A^{-1}={}^tA$.

The connection equations of the frame field $E_1$, $E_2$, $E_3$
\begin{equation}\label{eq:connecteqns}\nabla_VE_i=\sum_i\omega_{ij}(V)E_j,\ i=1,2,3\end{equation}
where $V$ is a vector field on $\mathbb{E}^3$ become
$$\begin{array}{ccccccc}
\nabla_VE_1&=&&&\omega_{12}(V)E_2&+&\omega_{13}(V)E_3,\\
\nabla_VE_2&=&-\omega_{12}(V)E_1& & &+&\omega_{23}(V)E_3,\\
\nabla_VE_3&=&-\omega_{13}(V)E_1&-&\omega_{23}(V)E_2.
\end{array}
$$
The connections equations are in fact a generalization of the Frenet-Serret formulas.

Let $Y$ be a vector field defined on a region containing a curve $\alpha(t)$. Then $Y_\alpha(t):=Y(\alpha(t))$ defined a vector field on the curve $\alpha(t)$. Then one can easily see that
$$\nabla_{\dot\alpha(t)}Y=\frac{d}{dt}Y_\alpha(t).$$
Let $\alpha(t)$ be a curve with unit speed. Let $E_1=T$, $E_2=N$, $E_3=B$. Then
\begin{align*}
\omega_{12}&=\nabla_{\dot\alpha_(t)}E_1\cdot E_2=\dot T\cdot N=(\kappa N)\cdot N=\kappa,\\
\omega_{13}&=\nabla_{\dot\alpha_(t)}E_1\cdot E_3=\dot T\cdot B=0,\\
\omega_{23}&=\nabla_{\dot\alpha_(t)}E_2\cdot E_3=\dot N\cdot B=(-\kappa T+\tau B)=\tau.
\end{align*}
The connection equations \eqref{eq:connecteqns} are then nothing but the Frenet-Serret formulas
$$\begin{array}{ccccccc}
\dot T&=&&&\kappa N&&\\
\dot N&=&-\kappa T& & &+&\tau B\\
\dot B&=&&-&\tau N.
\end{array}
$$

The frame $E_1,E_2,E_3$ can be written in terms of the natural frame $U_1,U_2,U_3$ as
\begin{align*}
E_1&=a_{11}U_1+a_{12}U_2+a_{13}U_3,\\
E_2&=a_{21}U_1+a_{22}U_2+a_{23}U_3,\\
E_3&=a_{31}U_1+a_{32}U_2+a_{33}U_3.
\end{align*}
Each real-valued function $a_{ij}:\mathbb{E}^3\longrightarrow\mathbb{R}$ is uniquely determined by $a_{ij}=E_i\cdot U_j$. The matrix $A=(a_{ij})$ is called the attitude matrix (also called rotation matrix or orientation matrix) of the frame field $E_1,E_2,E_3$. One can clearly see that the attitude matrix $A$ is an orthogonal matrix. In the above remark, I mentioned that the set of all $3\times $ skew symmetric matrices is the Lie algebra $\mathfrak{o}(3)$. The Lie algebra $\mathfrak{g}$ of a Lie group $G$ is defined to be the tangent space $T_e G$ to $G$ at the identity element $e$. (A Lie group is a differentiable manifold, so it make sense to talk about tangent spaces to $G$.)

Let us define a curve $\gamma: \mathbb{R}\longrightarrow\mathrm{O}(3)$ by
$$\gamma(t)=A(t)\cdot{}^tA(0).$$
Then $\gamma(0)=I$.
Hence $\dot{\gamma}(0)=\frac{dA(t)}{dt}|_{t=0}\cdot{}^tA(0)$ is a tangent vector to $\mathrm{O}(3)$ at the identity matrix $I$. That is, $\dot{\gamma}(0)\in\mathfrak{o}(3)$. Hence one can easily expect that the following theorem holds.

Theorem. If $A=(a_{ij})$ is the attitude matrix and $\omega=(\omega_{ij})$ the connection 1-form of a frame field $E_1, E_2, E_3$, then
$$\omega=dA\cdot{}^tA$$
or equivalently
$$\omega_{ij}=\sum_k da_{ik} \cdot a_{jk}\ \mbox{for}\ i,j=1,2,3.$$

Proof. For each $v\in T_p\mathbb{E}^3$,
$$\omega_{ij}(v)=\nabla_vE_i\cdot E_j(p).$$
In terms of the natural field $U_i$, $i=1,2,3$,
$$E_i=\sum_ka_{ik}U_k,\ i=1,2,3.$$
So,
\begin{align*}
\nabla_vE_i&=\sum_k v[a_{ik}]U_k(p)\\
&=\sum_k da_{ik} U_k(p).
\end{align*}
Hence,
$$\omega_{ij}=\sum_k da_{ik}a_{jk},$$
i.e.
$$\omega=dA\cdot{}^tA.$$

Remark. In general, if $G$ is a Lie group then its Lie algebra $\mathfrak{g}$ is given by the set of differential $1$-forms
$$\mathfrak{g}=\{g^{-1}dg:\ g\in G\}=\{(dg^{-1})g:\ g\in G\}.$$

Example. Let us compute the connection forms of the cylindrical frame field. The attitude matrix is
$$A=\begin{pmatrix}
\cos\theta & \sin\theta & 0\\
-\sin\theta & \cos\theta & 0\\
0 & 0 & 1
\end{pmatrix}.$$ Thus
$$dA=\begin{pmatrix}
-\sin\theta d\theta & \cos\theta d\theta & 0\\
-\cos\theta d\theta & -\sin\theta d\theta & 0\\
0 & 0 & 0
\end{pmatrix}.$$
Hence,
\begin{align*}
\omega&=dA\cdot{}^tA\\
&=\begin{pmatrix}
-\sin\theta d\theta & \cos\theta d\theta & 0\\
-\cos\theta d\theta & -\sin\theta d\theta & 0\\
0 & 0 & 0
\end{pmatrix}\begin{pmatrix}
\cos\theta & -\sin\theta & 0\\
\sin\theta & \cos\theta & 0\\
0 & 0 & 1\end{pmatrix}\\
&=\begin{pmatrix}
0 & d\theta & 0\\
-d\theta & 0 & 0\\
0 & 0 & 0
\end{pmatrix}.
\end{align*}
The connection equations of the cylindrical frame field are then
\begin{align*}
\nabla_VE_1&=d\theta(V)E_2=V[\theta]E_2,\\
\nabla_VE_2&=-d\theta(V)E_1=-V[\theta]E_1,\\
\nabla_VE_3&=0
\end{align*}
for all vector fields $V$. As expected the vector field $E_3$ is parallel.

Frame Fields

In Euclidean 3-space $\mathbb{E}^3$, we have naturally defined frame $U_1(p)$, $U_2(p)$, $U_3(p)$ for each $p\in\mathbb{E}^3$, where $U_1=(1,0,0)$, $U_2(0,1,0)$, $U_3=(0,0,1)$. The frame $U_1$, $U_2$, $U_3$ (as vector fields) is called the natural frame. As a generalization of the natural frame, we can define

Definition. Vector fields $E_1$, $E_2$, $E_3$ on $\mathbb{E}^3$ constitute a frame field on $\mathbb{E}^3$ provided
$$E_i\cdot E_j=\delta_{ij},\ i,j=1,2,3$$
where $\delta_{ij}$ is the Kronecker’s delta.

There are two important examples of frame fields: the cylindrical frame field and the spherical frame field.

Example. [The Cylindrical Frame Field]

Let $(r,\theta,z)$ be the usual cylindrical coordinates on $\mathbb{E}^3$.

Fig. 1 The Cylindrical Frame

We find a unit vector field in the direction in which each coordinate increases. For $r$, this is
$$E_1=\cos\theta U_1+\sin\theta U_2.$$
For $\theta$, we find
$$E_2=-\sin\theta U_1+\cos\theta U_2.$$ Finally for $z$, it is clearly
$$E_3=U_3.$$

Example. [The spherical Frame Field]

Let $(\rho,\theta,\varphi)$ be the usual spherical coordinates.

Fig. 2 The Spherical Frame

One can find the spherical frame $F_1$, $F_2$, $F_3$ using the cylindrical frame $E_1$, $E_2$, $E_3$. Clearly
$$F_2=E_2=-\sin\theta U_1+\cos\theta U_2.$$

Fig 3. The Spherical Frame

As one can see in the Figure 3, $F_1$ and $F_3$ are obtained as
\begin{align*}
F_1&=\cos\varphi E_1+\sin\varphi E_3\\
&=\cos\varphi(\cos\theta U_1+\sin\theta U_2)+\sin\varphi U_3,\\
F_3&=-\sin\varphi E_1+\cos\varphi E_3\\
&=-\sin\varphi(\cos\theta U_1+\sin\theta U_2)+\cos\varphi U_3.
\end{align*}
Hence,
\begin{align*}
F_1&=\cos\varphi\cos\theta U_1+\cos\varphi\sin\theta U_2+\sin\varphi U_3,\\
F_2&=-\sin\theta U_1+\cos\theta U_3,\\
F_3&=-\sin\varphi\cos\theta U_1-\sin\varphi\sin\theta U_2+\cos\varphi U_3.
\end{align*}

Solving Linear Inequalities

Linear Inequalities:

Solving linear inequalities is as easy as solving linear equations. You only need to know the following principles.

For a given inequality,

  1. Adding the same number to each side of the inequality does not change the symbol $<$.
  2. Subtracting the same number from each side of the inequality does not change the symbol $<$.
  3. Multiplying or dividing the inequality by the same positive number does not change the symbol $<$.
  4. Multiplying or dividing the inequality by the same negative number reverses the symbol $<$ from, say, $<$ to $>$.

The principle 4 can be easily understood with a simple example, say everybody would agree that $1<2$. If one multiplies the inequality by $-1$, it is true that $-1>-2$, i.e. the symbol $<$ reverses.

Example. Solve the inequality
$$3x-5<6-2x.$$

Solution. Adding 5 to each side of the inequality results
$$3x<11-2x.\ \ \ \ \ \mbox{(1)}$$
Adding $2x$ to each side of (1) results
$$5x<11.\ \ \ \ \ \mbox{(2)}$$
Dividing the inequality (2) by the same positive number 5 results
$$x<\frac{11}{5}.$$
There are various ways to write the solution. Normally $x<\frac{11}{5}$ would suffice but it can be written more formally as the solution set
$$\left\{x|\ x<\frac{11}{5}\right\}.$$
In interval notation, the solution can be written as
$$\left(-\infty,\frac{11}{5}\right).$$
Graphically it can be represented as


Example. Solve the inequality
$$13-7x\geq 10x-4.$$

Solution. Subtracting 13 from each side of the inequality results
$$-7x\geq 10x-17.\ \ \ \ \ \mbox{(3)}$$
Subtracting $10x$ from each side of (3) results
$$-17x\geq -17.\ \ \ \ \ \mbox{(4)}$$
Dividing each side of (4) by the same negative number $-17$ results
$$x\leq 1.$$

Note that the symbol $\geq$ has been reversed to $\leq$. The solution set can be written as $\{x|\ x\leq 1\}$ or $(-\infty,1]$ in interval notation. Graphically it can be represented as


Compound Inequalities:

A compound inequality is two inequalities joined by a conjunction AND or OR.

Example. Solve $-3<2x+5\leq 7$.

Solution. Notice that the compund inequality is formed by
$$-3<2x+5\ \mbox{and}\ 2x+5\leq 7.$$
Subtract 5 from each side of the given inequality.
$$-8<2x\leq 2.\ \ \ \ \ \mbox{(5)}$$
Devide (5) by 2.
$$-4<x\leq 1.$$
Hence the solution set is $\{x|\ -4<x\leq 1\}$ or $(-4,1]$ in interval notation. The graph of the solution set is given by


Example. Solve $2x-5\leq -7$ or $2x-5>1$.

Solution. Add 5 to eqch side of the two inequalities:
$$2x\leq -2\ \mbox{or}\ 2x>6.\ \ \ \ \ \mbox{(6)}$$
Divide each side of the two inequalities (6) by 2:
$$x\leq -1\ \mbox{or}\ x>3.$$
The solution set is $\{x|\ x\leq -1\ \mbox{or}\ x>3\}$ or $(-\infty,-1]\cup(3,\infty)$ in interval notation. The graph of the solution set is given by

More Equation Solving

Rational Equations:

The Recipe of Solving Rational Equations

  1. First find the LCD (Least Common Denominator), i.e. the Least Common Multiple of all denominators.
  2. Multiply your rational equation by the LCD.
  3. Solve the resulting equation (usually a linear equation or a quadratic equation).
  4. Plug solutions you obtained from STEP 3 into the original rational equation for $x$ to see if they all satisfy.

Example. Solve
$$\frac{1}{3x+6}-\frac{1}{x^2-4}=\frac{3}{x-2}.$$

Solution. STEP 1. $3x+6=3(x+2)$ and $x^2-4=(x+2)(x-2)$. Thus the LCD is $3(x+2)(x-2)$.

STEP 2. Multiply the equation the by LCD.
$$\frac{1}{3x+6}\cdot 3(x+2)(x-2)-\frac{1}{x^2-4}\cdot 3(x+2)(x-2)=\frac{3}{x-2}\cdot 3(x+2)(x-2)$$
which results the linear equation.
$$(x-2)-3=3\cdot 3(x+2).$$
The solution of this linear equation is $x=-\frac{23}{8}$.

Radical Equations:

The Recipe of Solving Radical Equations

  1. First isolate one radical term in one side.
  2. Square both sides of the equation.
  3. If all radical are gone, solve the resulting equation (usually linear or quadratic). If not (in case the radical equation had two radical terms), repeat the steps 1 and 2.
  4. Plug solutions you obtained from STEP 3 into the original radical equation for $x$ to see if they all satisfy.

Example. Solve $5+\sqrt{x+7}=x$.

Solution. STEP 1. Isolate the radical term $\sqrt{x+7}$ in the LHS.
$$\sqrt{x+7}=x-5.$$

STEP 2. Square both sides of the resulting equation.
$$x+7=(x-5)^2.$$
This is simplified to the quadratic equation
$$x^2-11x+18=0$$ which is factored to
$$(x-9)(x-2)=0.$$
Hence we obtain the two solutions $x=2,9$. However, not all these solutions may satisfy the original radical equation.

STEP 3. If $x=9$, then $$\mbox{LHS}=5+\sqrt{9+7}=9=\mbox{RHS}.$$ However, if $x=2$ then
$$\mbox{LHS}=5+\sqrt{x+7}=8\ne 2=\mbox{RHS}.$$ Therefore $x=9$ is the only solution to the radical equation.

Example. Solve $\sqrt{x-3}+\sqrt{x+5}=4$.

Solution. STEP 1. Isolate on radical term in one side, say isolate $\sqrt{x-3}$ in the LHS.
$$\sqrt{x-3}=4-\sqrt{x+5}.$$

STEP 2. Square both side of the resulting equation.
\begin{align*}
x-3&=(4-\sqrt{x+5})^2\\
&=16-8\sqrt{x+5}+(x+5)\\
&=x+21-8\sqrt{x+5}.
\end{align*}

Repeat STEP 1. Isolate the radical term $\sqrt{x+5}$ in the RHS.
$$3=\sqrt{x+5}.$$

Repeat STEP 2. Square both sides of the resulting equation.
$$9=x+5.$$
Hence we obtain the solution $x=4$.

One can readily check that $x=4$ satisfies the original radical equation.