Partial Differential Equations of Physics Questions and Answers

Q: Two students solved the Poisson's equation \(\nabla^{2}\varphi = -\frac{\rho_{0}}{\varepsilon_{0}}\) (the symbols have usual meanings). They obtained two different answers \(\varphi_{1}(x) = -\frac{1}{2}\frac{\rho_{0}x^{2}}{\varepsilon_{0}}\) and \(\varphi_{2}(y) = -\frac{1}{2}\frac{\rho_{0}y^{2}}{\varepsilon_{0}}\). What is the reason ?

On the key, the reason is that the boundary conditions are different in the two cases.Both functions satisfy Poisson's equation: each has \(\nabla^{2}\varphi = -\rho_{0}/\varepsilon_{0}\). Their difference\[ \begin{aligned} &\varphi_{1} - \varphi_{2} \\ &= -\frac{\rho_{0}}{2\varepsilon_{0}}(x^{2} - y^{2}) \end{aligned} \]is a solution of Laplace's equation. By the uniqueness theorem, Poisson's equation has exactly one solution once the boundary conditions are fixed, so two different solutions can only arise if the two students imposed different boundary conditions.Neither answer is wrong, so 'one of them was wrong' fails. Isotropy of space does not make two different functions equal. Adding a Laplace solution describes how the answers differ, but what allows that difference is the change in boundary conditions, which is the reason the key accepts.

Q: A general solution of the partial differential equation\[ \begin{aligned} &\frac{\partial^{2}f(x,y)}{\partial x^{2}} \\ &\quad - \frac{\partial^{2}f(x,y)}{\partial y^{2}} = 0 \end{aligned} \]can be expressed in terms of arbitrary functions u and v as :

The general solution is \(f(x,y) = u(x-y) + v(x+y)\).The equation is the one-dimensional wave equation with unit speed. In the characteristic variables \(\xi = x - y\) and \(\eta = x + y\) it becomes \(\partial^{2}f/\partial\xi\,\partial\eta = 0\), whose solution is a sum of an arbitrary function of each variable. This is d'Alembert's solution.Check directly: for \(u(x-y)\), \(f_{xx} = u''\) and \(f_{yy} = (-1)^{2}u'' = u''\), so \(f_{xx} - f_{yy} = 0\).The form \(u(x+iy) + v(x-iy)\) solves Laplace's equation \(f_{xx} + f_{yy} = 0\) instead. For \(u(2x \pm y)\) or \(u(x \pm 2y)\), \(f_{xx}\) and \(f_{yy}\) differ by a factor of 4, so they solve wave equations with speed 2 or 1/2, not this one.

Two students solved the Poisson's equation \(\nabla^{2}\varphi = -\frac{\rho_{0}}{\varepsilon_{0}}\) (the symbols have usual meanings). They obtained two different answers \(\varphi_{1}(x) = -\frac{1}{2}\frac{\rho_{0}x^{2}}{\varepsilon_{0}}\) and \(\varphi_{2}(y) = -\frac{1}{2}\frac{\rho_{0}y^{2}}{\varepsilon_{0}}\). What is the reason ?

A.Space is isotropic, hence x and y are equivalent
B.We can add solution of Laplace's equation to both
C.The boundary conditions are different in the two cases
D.One of them was wrong

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

On the key, the reason is that the boundary conditions are different in the two cases.

Both functions satisfy Poisson's equation: each has \(\nabla^{2}\varphi = -\rho_{0}/\varepsilon_{0}\). Their difference

\[ \begin{aligned} &\varphi_{1} - \varphi_{2} \\ &= -\frac{\rho_{0}}{2\varepsilon_{0}}(x^{2} - y^{2}) \end{aligned} \]

is a solution of Laplace's equation. By the uniqueness theorem, Poisson's equation has exactly one solution once the boundary conditions are fixed, so two different solutions can only arise if the two students imposed different boundary conditions.

Neither answer is wrong, so 'one of them was wrong' fails. Isotropy of space does not make two different functions equal. Adding a Laplace solution describes how the answers differ, but what allows that difference is the change in boundary conditions, which is the reason the key accepts.

Discuss in Forum

A general solution of the partial differential equation

\[ \begin{aligned} &\frac{\partial^{2}f(x,y)}{\partial x^{2}} \\ &\quad - \frac{\partial^{2}f(x,y)}{\partial y^{2}} = 0 \end{aligned} \]

can be expressed in terms of arbitrary functions u and v as :

A.\(f(x,y) = u(x-y) + v(x+y)\)
B.\(f(x,y) = u(x+iy) + v(x-iy)\)
C.\(f(x,y) = u(2x-y) + v(2x+y)\)
D.\(f(x,y) = u(x+2y) + v(x-2y)\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The general solution is \(f(x,y) = u(x-y) + v(x+y)\).

The equation is the one-dimensional wave equation with unit speed. In the characteristic variables \(\xi = x - y\) and \(\eta = x + y\) it becomes \(\partial^{2}f/\partial\xi\,\partial\eta = 0\), whose solution is a sum of an arbitrary function of each variable. This is d'Alembert's solution.

Check directly: for \(u(x-y)\), \(f_{xx} = u''\) and \(f_{yy} = (-1)^{2}u'' = u''\), so \(f_{xx} - f_{yy} = 0\).

The form \(u(x+iy) + v(x-iy)\) solves Laplace's equation \(f_{xx} + f_{yy} = 0\) instead. For \(u(2x \pm y)\) or \(u(x \pm 2y)\), \(f_{xx}\) and \(f_{yy}\) differ by a factor of 4, so they solve wave equations with speed 2 or 1/2, not this one.

Discuss in Forum

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Partial Differential Equations of Physics Questions and Answers

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