Fourier Transform Solution for PDE

We are given a partial differential equation (PDE) and asked to find how the Fourier modes ϕ_k(y) depend on y when using a specific Fourier transform. We need to find the exponents α and β in the solution form y^α and y^β.

The given PDE is:

\[{\rm{ - }}\frac{{{\partial ^2}ϕ \left( {x,y} \right)}}{{\partial {y^2}}}\, - \,\frac{1}{{{y^2}}}\frac{{{\partial ^2}ϕ \left( {x,y} \right)}}{{\partial {x^2}}} + \frac{{{m^2}}}{{{y^2}}}ϕ \left( {x,y} \right) = 0\]

The given Fourier transform relation is:

\[ϕ(x, y) = \int\limits_{-\infty}^{\infty} {{{\rm{e}}^{{\rm{ikx}}}}} {ϕ _{\rm{k}}}\left( {\rm{y}} \right){\rm{dk}}\]

We need to substitute the Fourier transform into the PDE. First, let's find the required partial derivatives of ϕ(x, y) with respect to x and y.

Partial Derivatives using Fourier Transform

Differentiating ϕ(x, y) with respect to x:

\[\frac{{\partial ϕ}}{{\partial x}} = \frac{\partial }{{\partial x}}\int\limits_{-\infty}^{\infty} {{{\rm{e}}^{{\rm{ikx}}}}} {ϕ _{\rm{k}}}\left( {\rm{y}} \right){\rm{dk}} = \int\limits_{-\infty}^{\infty} {\frac{\partial }{{\partial x}}{{\rm{e}}^{{\rm{ikx}}}}} {ϕ _{\rm{k}}}\left( {\rm{y}} \right){\rm{dk}} = \int\limits_{-\infty}^{\infty} {(ik){{\rm{e}}^{{\rm{ikx}}}}} {ϕ _{\rm{k}}}\left( {\rm{y}} \right){\rm{dk}}\]

Differentiating again with respect to x:

\[\frac{{{\partial ^2}ϕ}}{{\partial {x^2}}} = \frac{\partial }{{\partial x}}\int\limits_{-\infty}^{\infty} {(ik){{\rm{e}}^{{\rm{ikx}}}}} {ϕ _{\rm{k}}}\left( {\rm{y}} \right){\rm{dk}} = \int\limits_{-\infty}^{\infty} {(ik)^2{{\rm{e}}^{{\rm{ikx}}}}} {ϕ _{\rm{k}}}\left( {\rm{y}} \right){\rm{dk}} = \int\limits_{-\infty}^{\infty} {(-k^2){{\rm{e}}^{{\rm{ikx}}}}} {ϕ _{\rm{k}}}\left( {\rm{y}} \right){\rm{dk}}\]

Differentiating ϕ(x, y) with respect to y:

\[\frac{{\partial ϕ}}{{\partial y}} = \frac{\partial }{{\partial y}}\int\limits_{-\infty}^{\infty} {{{\rm{e}}^{{\rm{ikx}}}}} {ϕ _{\rm{k}}}\left( {\rm{y}} \right){\rm{dk}} = \int\limits_{-\infty}^{\infty} {{{\rm{e}}^{{\rm{ikx}}}}} \frac{\partial }{{\partial y}}{ϕ _{\rm{k}}}\left( {\rm{y}} \right){\rm{dk}}\]

Since ϕ_k depends only on y, the partial derivative becomes an ordinary derivative:

\[\frac{{\partial ϕ}}{{\partial y}} = \int\limits_{-\infty}^{\infty} {{{\rm{e}}^{{\rm{ikx}}}}} \frac{{dϕ _{\rm{k}}}}{{dy}}\left( {\rm{y}} \right){\rm{dk}}\]

Differentiating again with respect to y:

\[\frac{{{\partial ^2}ϕ}}{{\partial {y^2}}} = \frac{\partial }{{\partial y}}\int\limits_{-\infty}^{\infty} {{{\rm{e}}^{{\rm{ikx}}}}} \frac{{dϕ _{\rm{k}}}}{{dy}}\left( {\rm{y}} \right){\rm{dk}} = \int\limits_{-\infty}^{\infty} {{{\rm{e}}^{{\rm{ikx}}}}} \frac{{{d^2}ϕ _{\rm{k}}}}{{d{y^2}}}\left( {\rm{y}} \right){\rm{dk}}\]

Substitute into the PDE

Now, substitute these derivatives back into the original PDE:

\[{\rm{ - }}\int\limits_{-\infty}^{\infty} {{{\rm{e}}^{{\rm{ikx}}}}} \frac{{{d^2}ϕ _{\rm{k}}}}{{d{y^2}}}{\rm{dk}} \, - \,\frac{1}{{{y^2}}}\int\limits_{-\infty}^{\infty} (-k^2) {{\rm{e}}^{{\rm{ikx}}}} {ϕ _{\rm{k}}}\left( {\rm{y}} \right){\rm{dk}} + \frac{{{m^2}}}{{{y^2}}}\int\limits_{-\infty}^{\infty} {{\rm{e}}^{{\rm{ikx}}}} {ϕ _{\rm{k}}}\left( {\rm{y}} \right){\rm{dk}} = 0\]

We can combine the terms under a single integral:

\[\int\limits_{-\infty}^{\infty} {{\rm{e}}^{{\rm{ikx}}}}\left[ - \frac{{{d^2}ϕ _{\rm{k}}}}{{d{y^2}}}\left( {\rm{y}} \right) + \frac{k^2}{{{y^2}}}{ϕ _{\rm{k}}}\left( {\rm{y}} \right) + \frac{{{m^2}}}{{{y^2}}}{ϕ _{\rm{k}}}\left( {\rm{y}} \right) \right]{\rm{dk}} = 0\]

Ordinary Differential Equation for ϕ_k(y)

For this integral to be zero for all values of x, the expression inside the square brackets must be zero for each value of k (this follows from the properties of the Fourier transform). This gives us an ordinary differential equation (ODE) for ϕ_k(y):

\[ - \frac{{{d^2}ϕ _{\rm{k}}}}{{d{y^2}}} + \frac{k^2}{{{y^2}}}{ϕ _{\rm{k}}} + \frac{{{m^2}}}{{{y^2}}}{ϕ _{\rm{k}}} = 0\]

Rearranging the terms and multiplying by \(-y^2\):

\[y^2 \frac{{{d^2}ϕ _{\rm{k}}}}{{d{y^2}}} - (k^2 + m^2){ϕ _{\rm{k}}} = 0\] \[y^2 \frac{{{d^2}ϕ _{\rm{k}}}}{{d{y^2}}} + 0 \cdot y \frac{{dϕ _{\rm{k}}}}{{dy}} - (k^2 + m^2){ϕ _{\rm{k}}} = 0\]

This is a homogeneous linear second-order ODE with variable coefficients, specifically a Cauchy-Euler equation.

Solving the ODE for ϕ_k(y)

For a Cauchy-Euler equation of the form \(ay^2y'' + byy' + cy = 0\), we assume a solution of the form \(y(y) = y^\lambda\).

Substituting ϕ_k(y) = y^λ into the ODE:

• \(\frac{{dϕ _{\rm{k}}}}{{dy}} = \lambda y^{\lambda-1}\)
• \(\frac{{{d^2}ϕ _{\rm{k}}}}{{d{y^2}}} = \lambda (\lambda-1) y^{\lambda-2}\)

Substitute these into the ODE \(y^2 \frac{{{d^2}ϕ _{\rm{k}}}}{{d{y^2}}} - (k^2 + m^2){ϕ _{\rm{k}}} = 0\):

\[y^2 \left( \lambda (\lambda-1) y^{\lambda-2} \right) - (k^2 + m^2) y^{\lambda} = 0\] \[\lambda (\lambda-1) y^{\lambda} - (k^2 + m^2) y^{\lambda} = 0\]

Since \(y > 0\) in the given domain, \(y^{\lambda} \neq 0\). We can divide by \(y^{\lambda}\) to get the characteristic equation:

\[\lambda (\lambda-1) - (k^2 + m^2) = 0\] \[\lambda^2 - \lambda - (k^2 + m^2) = 0\]

Finding the Values of α and β

This is a quadratic equation for λ. We can solve it using the quadratic formula, \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=1\), \(b=-1\), and \(c=-(k^2 + m^2)\).

\[\lambda = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-(k^2 + m^2))}}{2(1)}\] \[\lambda = \frac{1 \pm \sqrt{1 + 4(k^2 + m^2)}}{2}\] \[\lambda = \frac{1}{2} \pm \frac{1}{2}\sqrt{1 + 4(k^2 + m^2)}\]

The two possible values for λ are:

• \(\alpha = \frac{1}{2} + \frac{1}{2}\sqrt{1 + 4(k^2 + m^2)}\)
• \(\beta = \frac{1}{2} - \frac{1}{2}\sqrt{1 + 4(k^2 + m^2)}\)

These are the exponents describing the dependence of ϕ_k(y) on y.

Comparing these values with the given options, we find that Option 3 matches our result.

The values of α and β are:

\[\frac{1}{2}\, + \,\frac{1}{2}\sqrt {1\, + \,4\left( {{k^2}\, + \,{m^2}} \right)} \quad \text{and} \quad \frac{1}{2}{\mkern 1mu} - {\mkern 1mu} \frac{1}{2}\sqrt {1{\mkern 1mu} + {\mkern 1mu} 4\left( {{k^2}{\mkern 1mu} + {\mkern 1mu} {m^2}} \right)}\]