The Laplacian of a Scalar Evaluate the Laplacian of the following scalar: Φ = ce^ky cos(kx − ωt). (E5.5.1)

Question

The Laplacian of a Scalar

Evaluate the Laplacian of the following scalar:

[latex]\phi=c e^{k y} \cos (k x-\omega t) .[/latex] (E5.5.1)

Accepted Answer

The necessary derivatives for forming the Laplacian, [latex] abla^{2} \phi[/latex], are:

[latex]\begin{aligned}\frac{\partial \phi}{\partial x} &=-c k e^{k y} \sin (k x-\omega t), & \frac{\partial^{2} \phi}{\partial x^{2}} &=-c k^{2} e^{k y} \cos (k x-\omega t), \ \frac{\partial \phi}{\partial y} &=c k e^{k y} \cos (k x-\omega t), & \frac{\partial^{2} \phi}{\partial y^{2}} &=c k^{2} e^{k y} \cos (k x-\omega t). (E5.5.2)\end{aligned}[/latex]

The Laplacian of [latex]\phi[/latex] is therefore:

[latex]\frac{\partial^{2} \phi}{\partial x^{2}}+\frac{\partial^{2} \phi}{\partial y^{2}}=-c k^{2} e^{k y} \cos (k x-\omega t)+c k^{2} e^{k y} \cos (k x-\omega t)=0.[/latex] (E5.5.3)

The scalar [latex]\phi[/latex] is closely related to the potential function for an irrotational wave traveling on the surface of deep water, to be discussed further in Section 7.11. Note that although time [latex]t[/latex] is one of the three independent variables, it is not involved in the formation of the Laplacian.

Question 5.5: The Laplacian of a Scalar Evaluate the Laplacian of the foll...

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