Toronto Math Forum

we have $$u(x,t) = \frac{1}{\sqrt{4\pi t}}\int_{0}^{\infty}[e^\frac{-(x-y)^2}{4t} + e^\frac{-(x+y)^2}{4t}]y dy$$
To solve the first part, we get $$ \frac{1}{\sqrt{4\pi t}}\int_{0}^{\infty}e^\frac{-(x-y)^2}{4t}y dy =  \frac{1}{\sqrt{4\pi t}}\int_{0}^{\infty}e^\frac{-(x-y)^2}{4t}(y-x) dy +  \frac{1}{\sqrt{4\pi t}}x\int_{0}^{\infty}e^\frac{-(x-y)^2}{4t} dy$$
we could integrate the first term, $$\frac{1}{\sqrt{4\pi t}}\int_{0}^{\infty}e^\frac{-(x-y)^2}{4t}(y-x) dy = -\frac{2t}{\sqrt{4\pi t}} e^\frac{-(x-y)^2}{4t}|_{0}^{\infty} = 0 + \frac{2t}{\sqrt{4\pi t}} e^\frac{-x^2}{4t} = \frac{\sqrt{t}}{\sqrt{\pi}} e^\frac{-x^2}{4t}$$
The second term we could substitute $y = x+z\sqrt{4t}$,
$$\frac{x}{\sqrt{\pi}}\int_{\frac{-x}{\sqrt{4t}}}^{\infty} e^{-z^2} dz\\ = \frac{x}{2}[ erf({\infty})-erf({\frac{-x}{\sqrt{4t}}})]$$
For the second part, we get  $$ \frac{1}{\sqrt{4\pi t}}\int_{0}^{\infty}e^\frac{-(x+y)^2}{4t}y dy =  \frac{1}{\sqrt{4\pi t}}\int_{0}^{\infty}e^\frac{-(x+y)^2}{4t}(y+x) dy -  \frac{1}{\sqrt{4\pi t}}x\int_{0}^{\infty}e^\frac{-(x+y)^2}{4t} dy$$
Similarly, we solve first part $$ \frac{1}{\sqrt{4\pi t}}\int_{0}^{\infty}e^\frac{-(x+y)^2}{4t}(y+x) dy=\frac{\sqrt{t}}{\sqrt{\pi}} e^\frac{-x^2}{4t}$$
For the second part, we get $$ -  \frac{1}{\sqrt{4\pi t}}x\int_{0}^{\infty}e^\frac{-(x+y)^2}{4t} dy = \frac{x}{2}[ -erf({\infty})+erf({\frac{-x}{\sqrt{4t}}})]$$
Combine all the solutions, we get $$u(x,t) = 2\frac{\sqrt{t}}{\sqrt{\pi}} e^\frac{-x^2}{4t} + x(erf({\frac{-x}{\sqrt{4t}}}))$$

So, let $$\frac{X''(x)}{X(x)} = -\lambda,\ \frac{T''(t)}{T(t)} = -c^2\lambda$$
Then we have,$$ X''(x) + \lambda X(x) = 0,\ T''(t) + c^2\lambda T(t) = 0$$
And let $-\lambda = \omega^2$, We get $$X(x) = A\cos(\omega x) +B\sin(\omega x),\ T(t) = C\cos(c\omega t) + D\sin(c\omega t)$$
Given conditions that, $u_x(0,t) = 0,\ (u_x + i\alpha u_t)(l,t) = 0$
We have $$X'(0)T(t) = 0\\X'(0) = 0 = \omega B \\B = 0$$
Hence, $$X(x) = A\cos(\omega x)$$
$$X'(l)T(t) + i\alpha X(l)T'(t) = 0\\ -\omega A\sin(\omega l ) T(t) + i\alpha A\cos(\omega l)T'(t) = 0\\T'(t) = \frac{\omega\sin(\omega l)}{i\alpha\cos(\omega l)}T(t)\\ T'(t) = \frac{\omega}{i\alpha}\tan(\omega l)T(t)\\T'(t) = \frac{-i\omega}{\alpha}\tan(\omega l)T(t)$$
Solve this ODE, we get $$ T(t) =Ke^{\frac{-i\omega t}{\alpha}\tan(\omega l)}$$

Oh, $$L= \sqrt{1+u_x^2+u_y^2}\\L_{u_x} = \frac{u_x}{\sqrt{1+u_x^2+u_y^2}},\ L _{u_y} = \frac{u_y}{\sqrt{1+u_x^2+u_y^2}}$$
The Euler-Lagrange PDE of S is
\begin{equation}
-\frac{\partial}{\partial x}(\frac{u_x}{\sqrt{1+u_x^2+u_y^2}}) - \frac{\partial}{\partial y}(\frac{u_y}{\sqrt{1+u_x^2+u_y^2}}) = 0\label{L}
\end{equation}

For part a):
     Given $\Delta^2 u = 0$, we have $u_{xxxx} + 2u_{xxyy} + u_{yyyy} = 0$, we make Fourier transform of $x \mapsto k$, $u(x,y)\mapsto \hat{u}(k,y)$.
     Then we get,$$k^4\hat{u} -2k^2\hat{u}_{yy}+\hat{u}_{yyyy} = 0$$
Solve this ODE, we get$$\hat{u}(k,y) = (A(k)+B(k)y)e^{-|k|y} + (C(k)+D(k)y)e^{|k|y}$$
Because the second term is unbounded, we discard the term $(C(k)+D(k)y)e^{|k|y}$.
Then, $$\hat{u}(k,y) = (A(k)+B(k)y)e^{-|k|y}$$
Given condition that, $$u|_{y=0} = f(x), u_y|_{y=0} = g(x)$$
Then we have $$\hat{u}|_{y=0} = \hat{f}(k),\hat{u}_y|_{y=0} = \hat{g}(k)\\\hat{u}(k,0) = A(k) = \hat{f}(k)\\\hat{u}_y(k,0) = -|k|A(k) + B(k) = \hat{g}(k)$$
Hence, we get $$B(k) = \hat{g}(k)+|k|\hat{f}(k)$$
Then,$$\hat{u}(k,y) = (A(k)+B(k)y)e^{-|k|y}\\= (\hat{f}(k)+ \hat{g}(k)+|k|\hat{f}(k))e^{-|k|y}\\u(x,y)=\int_{-\infty}^{\infty}  (\hat{f}(k)+ \hat{g}(k)+|k|\hat{f}(k))e^{-|k|y}e^{ikx} dk$$

For part b):

     Similarly, we get Fourier transform of x,$$\hat{u}(k,y) = (A(k)+B(k)y)e^{-|k|y} + (C(k)+D(k)y)e^{|k|y}$$
Because the second term is unbounded, we discard the term $(C(k)+D(k)y)e^{|k|y}$.
Then, $$\hat{u}(k,y) = (A(k)+B(k)y)e^{-|k|y}$$
Given condition that, $u_{yy}|_{y=0} = f(x), \Delta u_y|_{y=0} = g(x)$

          we get, $\hat{u}_{yy}|_{y=0} = \hat{f}(k), (\hat{u}_{xxy}+\hat{u}_{yyy})|_{y=0} = \hat{g}(k)$(not sure about this condition)

Then plug in the conditions, we have $$\hat{u}_{yy}(k,0) = |k|^2A(k) -2|k|B(k) = \hat{f}(k) \\(\hat{u}_{xxy}+\hat{u}_{yyy})(k,0)= ?$$
I may figure it out later.

Yes, I found my A(k) is wrong before.

To get Fourier transform,
$$\hat{f}(k) = \frac{1}{2\pi} \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos(x)e^{-ikx} dx$$
Let $g(x) = 1, f(x) = g(x)cos(x) = g(x)\frac{e^{ix}+e^{-ix}}{2} = \frac{1}{2}[g(x)e^{ix} + g(x)e^{-ix}] = \hat{f}(k) = \frac{1}{2}[\hat{g}(k-1) + \hat{g}(k+1)]$
$$\hat{g}(k) = \frac{1}{2\pi}\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} e^{-ikx} dx\\=\frac{1}{2\pi}\frac{e^{\frac{-ik\pi}{2}}-e^{\frac{ik\pi}{2}}}{-ik}\\=\frac{1}{2\pi}\frac{e^{\frac{ik\pi}{2}}-e^{\frac{-ik\pi}{2}}}{ik}\\=\frac{1}{k\pi}\sin(\frac{k\pi}{2})$$
Then, $$\hat{f}(k) = \frac{1}{2}[\hat{g}(k-1) + \hat{g}(k+1)]\\=\frac{1}{2}[\frac{1}{(k-1)\pi}\sin(\frac{(k-1)\pi}{2}) + \frac{1}{(k+1)\pi}\sin(\frac{(k+1)\pi}{2})]\\=\frac{1}{2}[\frac{1}{(k-1)\pi}\sin(\frac{(k\pi-\pi)}{2}) + \frac{1}{(k+1)\pi}\sin(\frac{(k\pi+\pi)}{2}]\\=\frac{1}{2}[\frac{1}{(1-k)\pi}\cos(\frac{k\pi}{2})+\frac{1}{(1+k)\pi}\cos(\frac{k\pi}{2})]\\=\frac{2}{2\pi(1-k^2)}\cos(\frac{k\pi}{2}) \\= \frac{1}{\pi(1-k^2)}\cos(\frac{k\pi}{2})$$
Hence, we write $f(x)$ as a Fourier integral.
$$f(x) = \int_{-\infty}^{\infty} \frac{1}{\pi(1-k^2)}\cos(\frac{k\pi}{2})e^{ikx} dk$$

For part (c),
     Given $r=(x_1^2+...+x_n^2)^\frac{1}{2}$, $$\frac{\partial r}{\partial x_i} = \frac{x_i}{(x_1^2+...+x_n^2)^\frac{1}{2}} = \frac{x_i}{r}\\\frac{\partial u}{\partial x_i} = u_r  \frac{x_i}{r}\\\frac{\partial^2 u}{\partial x_i^2} = \frac{\partial u_r}{\partial x_i}\frac{x_i}{r} + u_r[\frac{1}{r} + \frac{x_i\partial r^-1}{\partial x_i}]\\=u_{rr}(\frac{x_i}{r})^2 + \frac{u_r}{r} - \frac{u_r x_i^2}{r^3}\\\Delta u = \sum_{i=1}^{n} \frac{\partial^2 u}{\partial x_i^2} = \frac{u_{rr}}{r^2}\sum_{i=1}^{n} x_i^2+\frac{nu_r}{r}-\frac{u_r}{r^3}\sum_{i=1}^{n} x_i^2 = 0\\= u_{rr} +\frac{nu_r}{r} - \frac{u_r}{r} = 0\\=u_{rr} +\frac{n-1}{r}u_r = 0 $$

For part (d),
 $\Longleftarrow$ Given $x\neq 0$, suppose $u(r) = Ar^{2-n}+B$, then $$u_r = (2-n)Ar^{1-n}\\u_{rr} = (2-n)(1-n)Ar^{-n}$$
From part (c), we get $$u_{rr} + \frac{n-1}{r}u_r\\=(2-n)(1-n)Ar^{-n}+\frac{n-1}{r}(2-n)Ar^{1-n}\\=(2-n)(1-n)Ar^{-n} - (2-n)(1-n)Ar^{-n} = 0 =\Delta u$$
$\Longrightarrow$ Given $n\neq 2, x\neq 0$, suppose $$\Delta u =  u_{rr} + \frac{n-1}{r}u_r = 0$$
Multiply $r^{n-1}$ on both side,$$r^{n-1}u_{rr} + (n-1)r^{n-2}u_r = 0$$ which is equal to $$(r^{n-1}u_r)' = 0$$ We let $$r^{n-1}u_r = (2-n)A\\u_r = (2-n)Ar^{1-n}\\u=Ar^{2-n}+B$$ which satisfies Laplace equation.

http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter6/S6.3.html
For the fifth line, Cartesian coordinates by $ x=rcos(θ), y=sin(θ)$, there should have $r$ before $sin(\theta)$.

OK