Mth3006 Weekly Problems 5

Question One

The method of characteristics for inhomogeneous equations can also be used when the right-hand side depends on $u$. Use it to solve $y\frac{\partial u}{\partial x}-2xy\frac{\partial u}{\partial y}=2xu$ subject to the boundary condition $u(0,y)=\frac{\sinh y}{y}$.

Solution

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Question Two

Solve $x\frac{\partial u}{\partial x}-y\frac{\partial u}{\partial y}=u$ subject to the boundary condition $u=\frac{1}{\ln x}$ on $xy=1$.

Solution

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Question Three

Use the method of characteristics for homogeneous equations to solve the following:

1. $y\frac{\partial u}{\partial x}+x\frac{\partial u}{\partial y}=0$ given $u=\cos x$ on $x^2+y^2=1$
2. $x^2\frac{\partial u}{\partial x}+y^2\frac{\partial u}{\partial y}=0$ given $u\to e^y$ as $x\to \infty$.

Solution

Step 1: This partial differential equation is homogeneous, since the right-hand side is zero.

Step 2: Write the characteristic equation:

$$\frac{dx}{y}=\frac{dy}{x}$$

Or equivalently,

$$xdx=ydy$$

Step 3: Integrate both sides:

$$\int xdx=\int ydy⟹\frac{1}{2}{x}^2=\frac{1}{2}{y}^2+C$$

So, the characteristic curves are

$$x^2-y^2=P$$

where $P$ is a parameter labeling each characteristic.

Step 4: For a homogeneous partial differential equation, the solution is constant along each characteristic curve, so

$$u=f(P)=f(x^2-y^2)$$

Step 5: Apply the boundary condition $u=\cos x$ on $x^2+y^2=1$.

Let $y^2=1-x^2$ on the boundary, so

$$P=x^2-(1-x^2)=x^2-1+x^2=2x^2-1$$

On the boundary,

$$u=\cos x=f(2x^2-1)$$

So, for every $w=2x^2-1$,

$$f(w)=\cos \left(x(w)\right)$$

where $x(w)$ is the value on the boundary curve solving $w=2x^2-1$.

But to express $u$ for general $(x,y)$, solve for $x$ in terms of the parameter $P=x^2-y^2$:

$$x^2=\frac{P+y^2}{1}$$

Alternatively, express solution in terms of the characteristic curve:

• Each point $(x,y)$ lies on a characteristic with $P=x^2-y^2$.

To match the boundary, use the point on the boundary where $x_b$ satisfies $P=2x_b^2-1$:

$$x_b^2=\frac{P+1}{2}$$

So,

$$u(x,y)=\cos \left(\sqrt{\frac{x^2-y^2+1}{2}}\right)$$

Final Answer:

$$\boxed{u(x,y)=\cos \left(\sqrt{\frac{x^2-y^2+1}{2}}\right)}$$

To solve a homogeneous equation by the method of characteristics, compute the constant of integration along each characteristic curve, relate it to the boundary condition, and substitute to express the solution for any $(x,y)$.

Question Four

Show that transforming to the characteristic coordinates $\xi =x+2y\text{and}\eta =x+y$ reduces the differential equation $2\frac{{\partial }^{2}u}{\partial x^2}-3\frac{{\partial }^{2}u}{\partial x\partial y}+\frac{{\partial }^{2}u}{\partial y^2}=y$ to $\frac{{\partial }^{2}u}{\partial \xi \partial \eta }=\eta -\xi$.

Solution

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Question Five

Confirm that $u(x,t)=\exp (-D{k}^{2}t)\exp [ik(x-\gamma t)]$ is a solution of the advection-diffusion equation $\frac{\partial u}{\partial t}+\gamma \frac{\partial u}{\partial x}=D\frac{{\partial }^{2}u}{\partial x^2}$.

Solution

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