First Order Partial Differential Equations Questions and Answers

Answer: C

The equation is y + zq = 0.

Revolving a curve about the x-axis gives a surface whose cross-sections perpendicular to the axis are circles, so it has the form

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

Treating z as a function of x and y, differentiate with respect to y: \(2y + 2zq = 0\), where \(q = \partial z/\partial y\). So \(y + zq = 0\), and the arbitrary function \(\phi\) has dropped out.

Differentiating with respect to x instead gives \(2zp = \phi'(x)\), which still contains \(\phi\). The options \(yp - xq = 0\) and \(xp - yq = 0\) describe other families (for example surfaces of revolution about the z-axis satisfy \(yp - xq = 0\)), and \(x + zp = 0\) is the analogue for an axis along y.

Answer: A

The two equations have common solutions.

Solve the pair for p and q. From (2), \(q = xz - x^{2}p\); substituting in (1):

\[ \begin{aligned} p &= \frac{1 + yz}{1 + xy} \\ q &= \frac{x(z - x)}{1 + xy} \end{aligned} \]

The equations are compatible if \(dz = p\,dx + q\,dy\) is integrable, that is, if \(p_{y} + qp_{z} = q_{x} + pq_{z}\). Working it out:

\[ \begin{aligned} p_{y} + qp_{z} &= \frac{(z - x)(1 + xy)}{(1 + xy)^{2}} \\ q_{x} + pq_{z} &= \frac{z - x - x^{2}y + xyz}{(1 + xy)^{2}} \end{aligned} \]

and the second numerator factorises as \((z - x)(1 + xy)\), so the two sides agree. The equations are therefore compatible and share a one-parameter family of solutions. Neither solution set contains the other, since each single equation also has solutions that fail the other.

Answer: A

The characteristics are great circles of the spheres and the envelope is (y − 2x)² + 5z² = 5.

The family is spheres of radius 1 centred at \((a, 2a, 0)\). Differentiate \(F = (x - a)^{2} + (y - 2a)^{2} + z^{2} - 1\) with respect to a:

\[ \begin{aligned} -2(x - a) - 4(y - 2a) &= 0 \\ a &= \frac{x + 2y}{5} \end{aligned} \]

This plane \(x + 2y = 5a\) passes through the centre \((a, 2a, 0)\), so it cuts each sphere in a great circle: that is the characteristic curve. Substituting a back, \(x - a = \frac{2(2x - y)}{5}\) and \(y - 2a = \frac{y - 2x}{5}\), so

\[ \begin{aligned} \frac{4(y - 2x)^{2} + (y - 2x)^{2}}{25} + z^{2} &= 1 \\ (y - 2x)^{2} + 5z^{2} &= 5 \end{aligned} \]

The envelope is this cylinder, not \((y - 2x)^{2} + 2z^{2} = 2\) and not a pair of planes.

Answer: B

The solution \(F(u, v) = 0\) is the general integral.

For a first order equation such as Lagrange's linear equation \(Pp + Qq = R\), if \(u = c_{1}\) and \(v = c_{2}\) are two independent integrals of the auxiliary equations, then \(F(u, v) = 0\) with F an arbitrary function is the general integral (general solution): it contains an arbitrary function, not just arbitrary constants.

A complete integral contains two arbitrary constants, \(\phi(x, y, z, a, b) = 0\). A particular integral is obtained from it by fixing the constants, and a singular integral is the envelope of the complete integral, containing no arbitrary element.

Answer: C

The equations are compatible if ∂(f, g)/∂(x, p) + ∂(f, g)/∂(y, q) = 0.

Two first order equations \(f(x, y, z, p, q) = 0\) and \(g = 0\) are compatible when

\[ \begin{aligned} [f, g] &= \frac{\partial(f, g)}{\partial(x, p)} + p\frac{\partial(f, g)}{\partial(z, p)} \\ &+ \frac{\partial(f, g)}{\partial(y, q)} + q\frac{\partial(f, g)}{\partial(z, q)} = 0 \end{aligned} \]

Here f and g do not involve z, so the Jacobians with respect to z vanish and the condition reduces to the sum in option C.

The options with Jacobians in (x, z) and (y, z) are zero identically for these f and g and test nothing, and the difference of the (x, p) and (y, q) Jacobians has the wrong sign.

Answer: C

The equation has infinitely many surfaces containing the circle.

The Lagrange auxiliary equations are \(\frac{dx}{y} = \frac{dy}{-x} = \frac{dz}{0}\), giving the characteristics \(x^{2} + y^{2} = c_{1}\), \(z = c_{2}\). The general integral is \(z = F(x^{2} + y^{2})\) for an arbitrary function F.

The circle \(x^{2} + y^{2} = 4\), \(z = 2\) is itself a characteristic curve. Any F with \(F(4) = 2\) gives an integral surface through it, for example \(z = 2\), \(z = \frac{x^{2} + y^{2}}{2}\) or \(z = x^{2} + y^{2} - 2\). There are infinitely many such F.

This is the characteristic case of the Cauchy problem, where uniqueness fails; uniqueness holds only when the initial curve is not a characteristic. So 'unique', 'finitely many' and 'none' are all wrong.

Answer: D

(I) and (II) both are true.

For a one-parameter family \(f(x, y, z, a) = 0\), two neighbouring members with parameters a and \(a + \delta a\) meet in a curve. As \(\delta a \to 0\), that curve tends to the set satisfying both \(f = 0\) and \(\partial f / \partial a = 0\). This limiting curve is the characteristic curve of the family, so (I) is the definition.

The envelope is the surface traced out by all the characteristic curves as a varies. It is found by eliminating a between the same two equations, which is statement (II). The two statements describe the same pair of equations, one before and one after the parameter is eliminated.

Answer: C

(II) and (III) are true.

A Pfaffian equation \(\overline{X} \cdot d\overline{r} = 0\) is integrable (has a family of integral surfaces) exactly when \(\overline{X} \cdot \mathrm{curl}\, \overline{X} = 0\). Expanding,

\[ \begin{aligned} \overline{X} \cdot \mathrm{curl}\, \overline{X} = &P(R_{y} - Q_{z}) + Q(P_{z} - R_{x}) \\ &+ R(Q_{x} - P_{y}) \end{aligned} \]

The expression in (III) is the negative of this, so setting it to zero is the same condition; (II) and (III) are equivalent forms of one test.

(I) is false, because many Pfaffian equations fail this test. For instance \(-y\,dx + x\,dy + dz = 0\) has \(\overline{X} = (-y, x, 1)\), \(\mathrm{curl}\, \overline{X} = (0, 0, 2)\) and \(\overline{X} \cdot \mathrm{curl}\, \overline{X} = 2 \neq 0\), so it is not integrable.

Answer: C

The envelope is the cone \(x^{2} + y^{2} = z^{2}\).

A complete integral of \(p^{2} + q^{2} = 1\) is \(z = ax + by + c\) with \(a^{2} + b^{2} = 1\). Planes through the origin have \(c = 0\); write \(a = \cos\alpha\), \(b = \sin\alpha\):

\[ \begin{aligned} z &= x\cos\alpha + y\sin\alpha \\ 0 &= -x\sin\alpha + y\cos\alpha \end{aligned} \]

The second line is the derivative with respect to \(\alpha\). Squaring and adding the two lines eliminates \(\alpha\) and gives \(z^{2} = x^{2} + y^{2}\).

\(x^{2} + y^{2} = 4\) and \(x^{2} - y^{2} = 4\) do not pass through the origin, and \(x^{2} - z^{2} = y^{2}\) has the sign of \(z^{2}\) reversed.