First Order Partial Differential Equations Questions and Answers

Answer: B

The equation is \(\dfrac{\partial^{2} z}{\partial y^{2}} + \dfrac{\partial^{2} z}{\partial x^{2}}\) = 0.

Differentiate \(z = a e^{bx}\sin by\):

\[ \begin{aligned} z_{x} &= b z, \quad z_{xx} = b^{2} z \\ z_{y} &= ab\, e^{bx}\cos by \\ z_{yy} &= -ab^{2} e^{bx}\sin by = -b^{2} z \end{aligned} \]

Adding, \(z_{xx} + z_{yy} = 0\): the constants \(a\) and \(b\) are eliminated, and \(z\) satisfies Laplace's equation.

Option A would need \(z_{x} = b^{2}z\), but \(z_{x} = bz\). Options C and D bring in a factor \(x\) that the differentiation never produces.

Answer: C

The solution is \(z = \phi_{1}(x + y) + x\phi_{2}(x + y)\).

With \(D = \partial/\partial x\) and \(D' = \partial/\partial y\), the equation is \((D^{2} - 2DD' + D'^{2})z = (D - D')^{2}z = 0\). For \((D - mD')^{2}z = 0\) the general solution is \(\phi_{1}(y + mx) + x\phi_{2}(y + mx)\); here \(m = 1\), so the argument is \(x + y\) in both terms.

Check: if \(z = \phi(x + y)\), then \(z_{xx} = z_{xy} = z_{yy} = \phi''\) and \(\phi'' - 2\phi'' + \phi'' = 0\).

Option A has \(x - y\), which belongs to \(D + D'\), a different factor. Option B mixes the two arguments, and option D's \(x^{2}\) would arise only from a cubed factor.

Answer: B

The intersections are circles.

The Pfaffian equation \(x\,dx + y\,dy + z\,dz = 0\) integrates to \(x^{2} + y^{2} + z^{2} = c_{1}\): spheres centred at the origin.

The linear equation \(yp - xq = 0\) has Lagrange subsidiary equations \(dx/y = dy/(-x) = dz/0\), giving \(z = \text{const}\) and \(x^{2} + y^{2} = \text{const}\). Its integral surfaces are \(z = \phi(x^{2} + y^{2})\), surfaces of revolution about the \(z\)-axis.

A sphere centred on the axis meets a surface of revolution about that axis in curves that are themselves symmetric about the axis and lie on the sphere: horizontal circles \(z = \text{const}\), \(x^{2} + y^{2} = \text{const}\). The characteristics of \(yp - xq = 0\) are exactly such circles. So the locus is not a parabola, straight line or helix (a helix is not symmetric under rotation about its axis).

Answer: D

All three, (I), (II) and (III), are true.

For a first-order PDE and a given initial curve, the Cauchy problem can turn out three ways, depending on how the curve sits relative to the characteristics:

• If the curve is nowhere characteristic (the transversality determinant is nonzero), there is a unique integral surface through it (II).
• If the curve is a characteristic curve, infinitely many integral surfaces pass through it (III).
• If the projection is characteristic but the data along it are not compatible, no integral surface passes through it (I).

For example, for \(z_{x} = 0\) the curve \(x = t,\ y = 0,\ z = t\) admits no solution, the curve \(x = 0,\ y = t,\ z = t\) gives exactly \(z = y\), and the curve \(x = t,\ y = 0,\ z = 0\) lies on every surface \(z = \phi(y)\) with \(\phi(0) = 0\). Options listing only two cases leave one out.

Answer: A

There is a unique integral surface through \(x^{2} = z,\ y = 0\).

The general solution of \(yp - xq = 0\) is \(z = \phi(x^{2} + y^{2})\), because the characteristics satisfy \(dx/y = dy/(-x)\), giving \(x^{2} + y^{2} = \text{const}\), and \(z\) is constant along them.

On the curve, \(y = 0\) and \(z = x^{2}\), so \(\phi(x^{2}) = x^{2}\), that is \(\phi(t) = t\) for every \(t \ge 0\). This fixes \(\phi\) on all the values \(x^{2} + y^{2}\) can take, so

\[ z = x^{2} + y^{2} \]

is the only solution, a paraboloid. The curve meets each characteristic circle exactly where it crosses the \(xz\)-plane, which is why the data determine \(\phi\) once and for all. So 'finitely many', 'infinitely many' and 'none' are wrong.

Answer: B

The equations are compatible on D if \(\dfrac{\partial(f, g)}{\partial(p, q)} \neq 0\) and \(dz = p\,dx + q\,dy\) is integrable.

Compatibility means the two equations have a common one-parameter family of solutions. The Jacobian condition \(\partial(f, g)/\partial(p, q) \neq 0\) is what lets us solve \(f = 0, g = 0\) for \(p\) and \(q\) as functions \(p = \phi(x, y, z)\), \(q = \psi(x, y, z)\) by the implicit function theorem. The common solutions are then the integrals of \(dz = \phi\,dx + \psi\,dy\), which exist when this Pfaffian equation is integrable; the integrability condition reduces to \([f, g] = 0\) in Jacobi bracket form.

The Jacobians with respect to \((x, y)\), \((p, x)\) or \((q, y)\) do not let us solve for \(p\) and \(q\), so they are not the right condition.

Answer: C

(I) and (II) are both true.

For the quasi-linear equation \(Pp + Qq = R\), the characteristic curves are the integral curves of the direction field \((P, Q, R)\).

(I) A surface swept out by a one-parameter family of characteristics has \((P, Q, R)\) tangent to it at every point, which is exactly the condition \(Pp + Qq = R\). So it is an integral surface.

(II) Conversely, through every point of an integral surface the characteristic curve through that point lies on the surface (a standard theorem in Sneddon's treatment), so every integral surface is a union of characteristics.

These two facts together are the basis of the method of characteristics, so the options denying either one are wrong.

Answer: C

Two first-order equations are compatible if they have common solution.

That is the definition used in the theory of first-order PDEs (Sneddon): \(f(x, y, z, p, q) = 0\) and \(g(x, y, z, p, q) = 0\) are compatible when they have solutions in common, in practice a one-parameter family; it is not required that every solution of one also solve the other. This is the idea behind Charpit's method, which looks for a second equation compatible with the given one.

The first two options demand that one solution set contain the other, which is far stronger and is not the definition. The last option is the opposite of compatibility.

Answer: B

The normals are orthogonal.

The surfaces given by \(P\,dx + Q\,dy + R\,dz = 0\) are those whose normal at each point is parallel to \((P, Q, R)\): every displacement \((dx, dy, dz)\) along such a surface is perpendicular to that vector.

The integral surfaces of Lagrange's equation \(Pp + Qq = R\) have normal \((p, q, -1)\), and the equation says \((P, Q, R) \cdot (p, q, -1) = 0\). So \((P, Q, R)\) is tangent to them and their normals are perpendicular to \((P, Q, R)\).

One family has normals along \((P, Q, R)\) and the other has normals perpendicular to it, so the normals of the two families are orthogonal, and the two families of surfaces cut each other at right angles. 'Collinear' and 'parallel' describe the same family, not the two.

Answer: C

The relation that is not a general solution is x² + y² + z² = f(y/x).

Lagrange's auxiliary equations are \(\frac{dx}{xz} = \frac{dy}{-yz} = \frac{dz}{y^{2} - x^{2}}\). The first pair gives \(\frac{dx}{x} + \frac{dy}{y} = 0\), so \(xy = c_{1}\). Using multipliers x, y, z, the numerator \(x \cdot xz - y \cdot yz + z(y^{2} - x^{2}) = 0\), so \(x^{2} + y^{2} + z^{2} = c_{2}\).

The general solution is any relation between \(c_{1} = xy\) and \(c_{2} = x^{2} + y^{2} + z^{2}\). Option A is one directly. Since \((x + y)^{2} + z^{2} = c_{2} + 2c_{1}\), options B and D are also relations between \(c_{1}\) and \(c_{2}\). But \(y/x\) is not constant along the characteristics, so option C is not a general solution.